3.5.75 \(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [475]

Optimal. Leaf size=204 \[ \frac {(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {(6 A-55 B+244 C) \tan (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(B-4 C) \tan (c+d x)}{a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]

[Out]

(B-4*C)*arctanh(sin(d*x+c))/a^4/d+1/105*(6*A-55*B+244*C)*tan(d*x+c)/a^4/d+1/105*(3*A+25*B-88*C)*sec(d*x+c)^2*t
an(d*x+c)/a^4/d/(1+sec(d*x+c))^2-(B-4*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B+C)*sec(d*x+c)^4*tan(d*x+c)/d
/(a+a*sec(d*x+c))^4+1/35*(2*A+5*B-12*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]
time = 0.43, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4169, 4104, 4093, 3872, 3855, 3852, 8} \begin {gather*} \frac {(6 A-55 B+244 C) \tan (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(B-4 C) \tan (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((B - 4*C)*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((6*A - 55*B + 244*C)*Tan[c + d*x])/(105*a^4*d) + ((3*A + 25*B - 8
8*C)*Sec[c + d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - ((B - 4*C)*Tan[c + d*x])/(a^4*d*(1 + Sec[
c + d*x])) - ((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((2*A + 5*B - 12*C)*Sec[
c + d*x]^3*Tan[c + d*x])/(35*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 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 4093

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

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 ^4(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 ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec ^4(c+d x) (a (3 A+4 B-4 C)+a (A-B+8 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (2 A+5 B-12 C)+a^2 (3 A-10 B+52 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a^3 (3 A+25 B-88 C)+a^3 (6 A-55 B+244 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac {(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(B-4 C) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {\int \sec (c+d x) \left (-105 a^4 (B-4 C)-a^4 (6 A-55 B+244 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=\frac {(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(B-4 C) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(B-4 C) \int \sec (c+d x) \, dx}{a^4}+\frac {(6 A-55 B+244 C) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac {(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(B-4 C) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {(6 A-55 B+244 C) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac {(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {(6 A-55 B+244 C) \tan (c+d x)}{105 a^4 d}+\frac {(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(B-4 C) \tan (c+d x)}{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. \(1208\) vs. \(2(204)=408\).
time = 6.44, size = 1208, normalized size = 5.92 \begin {gather*} \frac {32 (-B+4 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 {32 (-B+4 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 (3 A \sin \left (\frac {c}{2}\right )-10 B \sin \left (\frac {c}{2}\right )+17 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 (6 A \sin \left (\frac {c}{2}\right )-55 B \sin \left (\frac {c}{2}\right )+139 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 (3 A \sin \left (\frac {d x}{2}\right )-10 B \sin \left (\frac {d x}{2}\right )+17 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 (6 A \sin \left (\frac {d x}{2}\right )-55 B \sin \left (\frac {d x}{2}\right )+139 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 (6 A \sin \left (\frac {d x}{2}\right )-160 B \sin \left (\frac {d x}{2}\right )+559 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 C \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 ) \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]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

(32*(-B + 4*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) - (32*
(-B + 4*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*S
in[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)*(3*A*Sin[c/2] - 10*B*Sin[c/2] + 17*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)*(6*A*Sin[c/2] - 55*B*Sin[c/2] + 139*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)*(3*A*Sin[(d*x)/2] - 10*B*Sin[(d*x)/2] + 17*
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)*(6*A*Sin[(d*x)/2] - 55*B*Sin[(d*
x)/2] + 139*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)*(6*A*Sin[(d*x)/2] -
160*B*Sin[(d*x)/2] + 559*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*C*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)*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 = 0.59, size = 242, normalized size = 1.19

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}+\frac {3 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B +\frac {7 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {11 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {23 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (32 C -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-32 C +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\)	\(242\)
default	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{7}+\frac {3 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B +\frac {7 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {11 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {23 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (32 C -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-32 C +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\)	\(242\)
risch	\(-\frac {2 i \left (105 B \,{\mathrm e}^{8 i \left (d x +c \right )}-420 C \,{\mathrm e}^{8 i \left (d x +c \right )}+735 B \,{\mathrm e}^{7 i \left (d x +c \right )}-2940 C \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}-9100 C \,{\mathrm e}^{6 i \left (d x +c \right )}-210 A \,{\mathrm e}^{5 i \left (d x +c \right )}+4165 B \,{\mathrm e}^{5 i \left (d x +c \right )}-16660 C \,{\mathrm e}^{5 i \left (d x +c \right )}-126 A \,{\mathrm e}^{4 i \left (d x +c \right )}+4795 B \,{\mathrm e}^{4 i \left (d x +c \right )}-20524 C \,{\mathrm e}^{4 i \left (d x +c \right )}-252 A \,{\mathrm e}^{3 i \left (d x +c \right )}+4445 B \,{\mathrm e}^{3 i \left (d x +c \right )}-18788 C \,{\mathrm e}^{3 i \left (d x +c \right )}-132 A \,{\mathrm e}^{2 i \left (d x +c \right )}+2785 B \,{\mathrm e}^{2 i \left (d x +c \right )}-11668 C \,{\mathrm e}^{2 i \left (d x +c \right )}-42 \,{\mathrm e}^{i \left (d x +c \right )} A +1015 B \,{\mathrm e}^{i \left (d x +c \right )}-4228 C \,{\mathrm e}^{i \left (d x +c \right )}-6 A +160 B -664 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{4} d}\)	\(386\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(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+3/5*A*tan(1/2*d*x+
1/2*c)^5-B*tan(1/2*d*x+1/2*c)^5+7/5*C*tan(1/2*d*x+1/2*c)^5+A*tan(1/2*d*x+1/2*c)^3-11/3*B*tan(1/2*d*x+1/2*c)^3+
23/3*C*tan(1/2*d*x+1/2*c)^3+A*tan(1/2*d*x+1/2*c)-15*B*tan(1/2*d*x+1/2*c)+49*C*tan(1/2*d*x+1/2*c)+(32*C-8*B)*ln
(tan(1/2*d*x+1/2*c)-1)-8*C/(tan(1/2*d*x+1/2*c)-1)+(-32*C+8*B)*ln(tan(1/2*d*x+1/2*c)+1)-8*C/(tan(1/2*d*x+1/2*c)
+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (196) = 392\).
time = 0.30, size = 411, normalized size = 2.01 \begin {gather*} \frac {C {\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 \, B {\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 )} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \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 {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(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*(C*(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*B*((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*l
og(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) + 3*A*(35*sin(
d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d

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Fricas [A]
time = 2.91, size = 328, normalized size = 1.61 \begin {gather*} \frac {105 \, {\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A - 80 \, B + 332 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (24 \, A - 535 \, B + 2236 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 620 \, B + 2636 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (9 \, A - 65 \, B + 296 \, C\right )} \cos \left (d x + c\right ) + 105 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*((B - 4*C)*cos(d*x + c)^5 + 4*(B - 4*C)*cos(d*x + c)^4 + 6*(B - 4*C)*cos(d*x + c)^3 + 4*(B - 4*C)*c
os(d*x + c)^2 + (B - 4*C)*cos(d*x + c))*log(sin(d*x + c) + 1) - 105*((B - 4*C)*cos(d*x + c)^5 + 4*(B - 4*C)*co
s(d*x + c)^4 + 6*(B - 4*C)*cos(d*x + c)^3 + 4*(B - 4*C)*cos(d*x + c)^2 + (B - 4*C)*cos(d*x + c))*log(-sin(d*x
+ c) + 1) + 2*(2*(3*A - 80*B + 332*C)*cos(d*x + c)^4 + (24*A - 535*B + 2236*C)*cos(d*x + c)^3 + (39*A - 620*B
+ 2636*C)*cos(d*x + c)^2 + 4*(9*A - 65*B + 296*C)*cos(d*x + c) + 105*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 +
4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A \sec ^{4}{\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 ^{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 {C \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}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(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)**4/(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)**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(C*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))/a**4

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Giac [A]
time = 0.51, size = 285, normalized size = 1.40 \begin {gather*} \frac {\frac {840 \, {\left (B - 4 \, 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 (B - 4 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {1680 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} 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} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 805 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, 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)^4*(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*(840*(B - 4*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*(B - 4*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))
/a^4 - 1680*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*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 + 63*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 105*B*a
^24*tan(1/2*d*x + 1/2*c)^5 + 147*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^2
4*tan(1/2*d*x + 1/2*c)^3 + 805*C*a^24*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^24*tan(1/2*d*x + 1/2*c) - 1575*B*a^24*t
an(1/2*d*x + 1/2*c) + 5145*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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Mupad [B]
time = 3.25, size = 250, normalized size = 1.23 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A-3\,B+5\,C}{40\,a^4}+\frac {A-B+C}{20\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-3\,B+5\,C}{12\,a^4}-\frac {2\,A+2\,B-10\,C}{24\,a^4}+\frac {A-B+C}{8\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-3\,B+5\,C\right )}{8\,a^4}+\frac {2\,B-2\,A+10\,C}{8\,a^4}-\frac {2\,A+2\,B-10\,C}{4\,a^4}+\frac {A-B+C}{2\,a^4}\right )}{d}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B-4\,C\right )}{a^4\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (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 )}^7\,\left (A-B+C\right )}{56\,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)^4*(a + a/cos(c + d*x))^4),x)

[Out]

(tan(c/2 + (d*x)/2)^5*((A - 3*B + 5*C)/(40*a^4) + (A - B + C)/(20*a^4)))/d + (tan(c/2 + (d*x)/2)^3*((A - 3*B +
5*C)/(12*a^4) - (2*A + 2*B - 10*C)/(24*a^4) + (A - B + C)/(8*a^4)))/d + (tan(c/2 + (d*x)/2)*((3*(A - 3*B + 5*
C))/(8*a^4) + (2*B - 2*A + 10*C)/(8*a^4) - (2*A + 2*B - 10*C)/(4*a^4) + (A - B + C)/(2*a^4)))/d + (2*atanh(tan
(c/2 + (d*x)/2))*(B - 4*C))/(a^4*d) - (2*C*tan(c/2 + (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^2 - a^4)) + (tan(c/2
+ (d*x)/2)^7*(A - B + C))/(56*a^4*d)

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