3.5.74 \(\int \frac {\sec ^5(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [474]

Optimal. Leaf size=254 \[ \frac {(2 A-8 B+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {8 (20 A-83 B+216 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A-8 B+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {4 (20 A-83 B+216 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3} \]

[Out]

1/2*(2*A-8*B+21*C)*arctanh(sin(d*x+c))/a^4/d-8/105*(20*A-83*B+216*C)*tan(d*x+c)/a^4/d+1/2*(2*A-8*B+21*C)*sec(d
*x+c)*tan(d*x+c)/a^4/d-1/105*(10*A-52*B+129*C)*sec(d*x+c)^3*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-4/105*(20*A-83*B
+216*C)*sec(d*x+c)^2*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B+C)*sec(d*x+c)^5*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+
1/5*(B-2*C)*sec(d*x+c)^4*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]
time = 0.48, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4169, 4104, 3872, 3852, 8, 3853, 3855} \begin {gather*} -\frac {8 (20 A-83 B+216 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A-8 B+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {(10 A-52 B+129 C) \tan (c+d x) \sec ^3(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {4 (20 A-83 B+216 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)}+\frac {(2 A-8 B+21 C) \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {(A-B+C) \tan (c+d x) \sec ^5(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(B-2 C) \tan (c+d x) \sec ^4(c+d x)}{5 a d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

((2*A - 8*B + 21*C)*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (8*(20*A - 83*B + 216*C)*Tan[c + d*x])/(105*a^4*d) + ((
2*A - 8*B + 21*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d) - ((10*A - 52*B + 129*C)*Sec[c + d*x]^3*Tan[c + d*x])/(
105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(20*A - 83*B + 216*C)*Sec[c + d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c
+ d*x])) - ((A - B + C)*Sec[c + d*x]^5*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((B - 2*C)*Sec[c + d*x]^4*
Tan[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4104

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(
a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rule 4169

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*C
sc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m -
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec ^5(c+d x) (a (2 A+5 B-5 C)+a (2 A-2 B+9 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^4(c+d x) \left (28 a^2 (B-2 C)+a^2 (10 A-24 B+73 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^3(c+d x) \left (-3 a^3 (10 A-52 B+129 C)+a^3 (50 A-176 B+477 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (20 A-83 B+216 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {\int \sec ^2(c+d x) \left (-8 a^4 (20 A-83 B+216 C)+105 a^4 (2 A-8 B+21 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (20 A-83 B+216 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(2 A-8 B+21 C) \int \sec ^3(c+d x) \, dx}{a^4}-\frac {(8 (20 A-83 B+216 C)) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac {(2 A-8 B+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (20 A-83 B+216 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(2 A-8 B+21 C) \int \sec (c+d x) \, dx}{2 a^4}+\frac {(8 (20 A-83 B+216 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac {(2 A-8 B+21 C) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {8 (20 A-83 B+216 C) \tan (c+d x)}{105 a^4 d}+\frac {(2 A-8 B+21 C) \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {(10 A-52 B+129 C) \sec ^3(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^5(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(B-2 C) \sec ^4(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (20 A-83 B+216 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1322\) vs. \(2(254)=508\).
time = 6.51, size = 1322, normalized size = 5.20 \begin {gather*} -\frac {16 (2 A-8 B+21 C) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}+\frac {16 (2 A-8 B+21 C) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}-\frac {4 \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (A \sin \left (\frac {c}{2}\right )-B \sin \left (\frac {c}{2}\right )+C \sin \left (\frac {c}{2}\right )\right )}{7 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}-\frac {8 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (10 A \sin \left (\frac {c}{2}\right )-17 B \sin \left (\frac {c}{2}\right )+24 C \sin \left (\frac {c}{2}\right )\right )}{35 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}-\frac {16 \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (55 A \sin \left (\frac {c}{2}\right )-139 B \sin \left (\frac {c}{2}\right )+258 C \sin \left (\frac {c}{2}\right )\right )}{105 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}-\frac {4 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{7 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}-\frac {8 \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (10 A \sin \left (\frac {d x}{2}\right )-17 B \sin \left (\frac {d x}{2}\right )+24 C \sin \left (\frac {d x}{2}\right )\right )}{35 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}-\frac {16 \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (55 A \sin \left (\frac {d x}{2}\right )-139 B \sin \left (\frac {d x}{2}\right )+258 C \sin \left (\frac {d x}{2}\right )\right )}{105 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}-\frac {32 \cos ^7\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (160 A \sin \left (\frac {d x}{2}\right )-559 B \sin \left (\frac {d x}{2}\right )+1308 C \sin \left (\frac {d x}{2}\right )\right )}{105 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}+\frac {16 C \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (c) \sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sin (d x)}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4}+\frac {16 \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (c) \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) (C \sin (c)+2 B \sin (d x)-8 C \sin (d x))}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^5*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

(-16*(2*A - 8*B + 21*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2*(A +
B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^
4) + (16*(2*A - 8*B + 21*C)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2*(
A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*
x])^4) - (4*Cos[c/2 + (d*x)/2]^2*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*Sin[c/2] -
B*Sin[c/2] + C*Sin[c/2]))/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (8
*Cos[c/2 + (d*x)/2]^4*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(10*A*Sin[c/2] - 17*B*Si
n[c/2] + 24*C*Sin[c/2]))/(35*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (16
*Cos[c/2 + (d*x)/2]^6*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(55*A*Sin[c/2] - 139*B*S
in[c/2] + 258*C*Sin[c/2]))/(105*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) -
(4*Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(A*Sin[(d*x)/2] - B*Sin[
(d*x)/2] + C*Sin[(d*x)/2]))/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (
8*Cos[c/2 + (d*x)/2]^3*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(10*A*Sin[(d*x)/2] - 17
*B*Sin[(d*x)/2] + 24*C*Sin[(d*x)/2]))/(35*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d
*x])^4) - (16*Cos[c/2 + (d*x)/2]^5*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(55*A*Sin[(
d*x)/2] - 139*B*Sin[(d*x)/2] + 258*C*Sin[(d*x)/2]))/(105*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(
a + a*Sec[c + d*x])^4) - (32*Cos[c/2 + (d*x)/2]^7*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]
^2)*(160*A*Sin[(d*x)/2] - 559*B*Sin[(d*x)/2] + 1308*C*Sin[(d*x)/2]))/(105*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Co
s[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) + (16*C*Cos[c/2 + (d*x)/2]^8*Sec[c]*Sec[c + d*x]^4*(A + B*Sec[c + d*x]
+ C*Sec[c + d*x]^2)*Sin[d*x])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) +
(16*Cos[c/2 + (d*x)/2]^8*Sec[c]*Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(C*Sin[c] + 2*B*Sin[d*x
] - 8*C*Sin[d*x]))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4)

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Maple [A]
time = 0.54, size = 294, normalized size = 1.16

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}-A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {9 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {11 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {23 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-13 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-111 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-84 C +32 B -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (84 C -32 B +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{8 d \,a^{4}}\)	\(294\)
default	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}-A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {9 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {11 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {23 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-13 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-111 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-84 C +32 B -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-36 C +8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (84 C -32 B +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{8 d \,a^{4}}\)	\(294\)
risch	\(-\frac {i \left (3456 C +320 A -1328 B -5880 B \,{\mathrm e}^{9 i \left (d x +c \right )}-840 \,{\mathrm e}^{10 i \left (d x +c \right )} B +2205 C \,{\mathrm e}^{10 i \left (d x +c \right )}+15435 C \,{\mathrm e}^{9 i \left (d x +c \right )}+9800 A \,{\mathrm e}^{7 i \left (d x +c \right )}-39200 B \,{\mathrm e}^{7 i \left (d x +c \right )}+102900 C \,{\mathrm e}^{7 i \left (d x +c \right )}-64384 B \,{\mathrm e}^{4 i \left (d x +c \right )}-24664 B \,{\mathrm e}^{2 i \left (d x +c \right )}+14140 A \,{\mathrm e}^{6 i \left (d x +c \right )}+17220 A \,{\mathrm e}^{5 i \left (d x +c \right )}-70896 B \,{\mathrm e}^{5 i \left (d x +c \right )}-8456 B \,{\mathrm e}^{i \left (d x +c \right )}-59248 B \,{\mathrm e}^{6 i \left (d x +c \right )}-46032 B \,{\mathrm e}^{3 i \left (d x +c \right )}+155526 C \,{\mathrm e}^{6 i \left (d x +c \right )}+183162 C \,{\mathrm e}^{5 i \left (d x +c \right )}+119364 C \,{\mathrm e}^{3 i \left (d x +c \right )}+4760 A \,{\mathrm e}^{8 i \left (d x +c \right )}-19040 B \,{\mathrm e}^{8 i \left (d x +c \right )}+49980 C \,{\mathrm e}^{8 i \left (d x +c \right )}+10920 A \,{\mathrm e}^{3 i \left (d x +c \right )}+2030 \,{\mathrm e}^{i \left (d x +c \right )} A +21987 C \,{\mathrm e}^{i \left (d x +c \right )}+15160 A \,{\mathrm e}^{4 i \left (d x +c \right )}+5890 A \,{\mathrm e}^{2 i \left (d x +c \right )}+64053 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1470 A \,{\mathrm e}^{9 i \left (d x +c \right )}+210 A \,{\mathrm e}^{10 i \left (d x +c \right )}+166668 C \,{\mathrm e}^{4 i \left (d x +c \right )}\right )}{105 a^{4} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{4} d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}+\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{4} d}\)	\(538\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/8/d/a^4*(-1/7*tan(1/2*d*x+1/2*c)^7*A+1/7*tan(1/2*d*x+1/2*c)^7*B-1/7*tan(1/2*d*x+1/2*c)^7*C-A*tan(1/2*d*x+1/2
*c)^5+7/5*B*tan(1/2*d*x+1/2*c)^5-9/5*C*tan(1/2*d*x+1/2*c)^5-11/3*A*tan(1/2*d*x+1/2*c)^3+23/3*B*tan(1/2*d*x+1/2
*c)^3-13*C*tan(1/2*d*x+1/2*c)^3-15*A*tan(1/2*d*x+1/2*c)+49*B*tan(1/2*d*x+1/2*c)-111*C*tan(1/2*d*x+1/2*c)-(-36*
C+8*B)/(tan(1/2*d*x+1/2*c)-1)+(-84*C+32*B-8*A)*ln(tan(1/2*d*x+1/2*c)-1)+4*C/(tan(1/2*d*x+1/2*c)-1)^2-(-36*C+8*
B)/(tan(1/2*d*x+1/2*c)+1)+(84*C-32*B+8*A)*ln(tan(1/2*d*x+1/2*c)+1)-4*C/(tan(1/2*d*x+1/2*c)+1)^2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (240) = 480\).
time = 0.32, size = 556, normalized size = 2.19 \begin {gather*} -\frac {3 \, C {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - B {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + 5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/840*(3*C*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c)
+ 1) + 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7)/a^4 - 2940*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x +
c) + 1) - 1)/a^4) - B*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))
+ (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^
4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) + 5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7
)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4))/
d

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Fricas [A]
time = 2.31, size = 400, normalized size = 1.57 \begin {gather*} \frac {105 \, {\left ({\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 8 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (20 \, A - 83 \, B + 216 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (1070 \, A - 4472 \, B + 11619 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (310 \, A - 1318 \, B + 3411 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (130 \, A - 592 \, B + 1509 \, C\right )} \cos \left (d x + c\right )^{2} - 210 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right ) - 105 \, C\right )} \sin \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/420*(105*((2*A - 8*B + 21*C)*cos(d*x + c)^6 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^5 + 6*(2*A - 8*B + 21*C)*cos
(d*x + c)^4 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^3 + (2*A - 8*B + 21*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) -
105*((2*A - 8*B + 21*C)*cos(d*x + c)^6 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^5 + 6*(2*A - 8*B + 21*C)*cos(d*x +
c)^4 + 4*(2*A - 8*B + 21*C)*cos(d*x + c)^3 + (2*A - 8*B + 21*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(1
6*(20*A - 83*B + 216*C)*cos(d*x + c)^5 + (1070*A - 4472*B + 11619*C)*cos(d*x + c)^4 + 4*(310*A - 1318*B + 3411
*C)*cos(d*x + c)^3 + 4*(130*A - 592*B + 1509*C)*cos(d*x + c)^2 - 210*(B - 2*C)*cos(d*x + c) - 105*C)*sin(d*x +
c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*d*
cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{6}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{7}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)

[Out]

(Integral(A*sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(B*sec(c + d*x)**6/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1),
x) + Integral(C*sec(c + d*x)**7/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1)
, x))/a**4

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Giac [A]
time = 0.53, size = 339, normalized size = 1.33 \begin {gather*} \frac {\frac {420 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {420 \, {\left (2 \, A - 8 \, B + 21 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {840 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 189 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1365 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11655 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/840*(420*(2*A - 8*B + 21*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 420*(2*A - 8*B + 21*C)*log(abs(tan(1/2*
d*x + 1/2*c) - 1))/a^4 - 840*(2*B*tan(1/2*d*x + 1/2*c)^3 - 9*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*
c) + 7*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15*B
*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 147*B*a^
24*tan(1/2*d*x + 1/2*c)^5 + 189*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 805*B*a^24
*tan(1/2*d*x + 1/2*c)^3 + 1365*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 1575*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*B*a^24*
tan(1/2*d*x + 1/2*c) + 11655*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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Mupad [B]
time = 3.29, size = 296, normalized size = 1.17 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+5\,B-15\,C\right )}{8\,a^4}-\frac {3\,\left (2\,A-4\,B+6\,C\right )}{4\,a^4}-\frac {5\,\left (A-B+C\right )}{4\,a^4}+\frac {4\,A-20\,C}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-9\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-7\,C\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {2\,A-4\,B+6\,C}{40\,a^4}+\frac {3\,\left (A-B+C\right )}{40\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,A-4\,B+6\,C}{8\,a^4}-\frac {A+5\,B-15\,C}{24\,a^4}+\frac {A-B+C}{4\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-4\,B+\frac {21\,C}{2}\right )}{a^4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^5*(a + a/cos(c + d*x))^4),x)

[Out]

(tan(c/2 + (d*x)/2)*((3*(A + 5*B - 15*C))/(8*a^4) - (3*(2*A - 4*B + 6*C))/(4*a^4) - (5*(A - B + C))/(4*a^4) +
(4*A - 20*C)/(8*a^4)))/d - (tan(c/2 + (d*x)/2)^3*(2*B - 9*C) - tan(c/2 + (d*x)/2)*(2*B - 7*C))/(d*(a^4*tan(c/2
+ (d*x)/2)^4 - 2*a^4*tan(c/2 + (d*x)/2)^2 + a^4)) - (tan(c/2 + (d*x)/2)^5*((2*A - 4*B + 6*C)/(40*a^4) + (3*(A
- B + C))/(40*a^4)))/d - (tan(c/2 + (d*x)/2)^3*((2*A - 4*B + 6*C)/(8*a^4) - (A + 5*B - 15*C)/(24*a^4) + (A -
B + C)/(4*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d) + (2*atanh(tan(c/2 + (d*x)/2))*(A - 4*B + (
21*C)/2))/(a^4*d)

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