3.5.66 \(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [466]

Optimal. Leaf size=216 \[ \frac {(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

1/2*(2*A-6*B+13*C)*arctanh(sin(d*x+c))/a^3/d-2/15*(11*A-36*B+76*C)*tan(d*x+c)/a^3/d+1/2*(2*A-6*B+13*C)*sec(d*x
+c)*tan(d*x+c)/a^3/d-1/5*(A-B+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(A-6*B+11*C)*sec(d*x+c)^3*t
an(d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/15*(11*A-36*B+76*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.36, antiderivative size = 216, 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, 3872, 3852, 8, 3853, 3855} \begin {gather*} -\frac {2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {(11 A-36 B+76 C) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(2 A-6 B+13 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {(A-6 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \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])^3,x]

[Out]

((2*A - 6*B + 13*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (2*(11*A - 36*B + 76*C)*Tan[c + d*x])/(15*a^3*d) + ((2*
A - 6*B + 13*C)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*S
ec[c + d*x])^3) - ((A - 6*B + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((11*A - 36
*B + 76*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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 ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^4(c+d x) (a (A+4 B-4 C)+a (2 A-2 B+7 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^3(c+d x) \left (-3 a^2 (A-6 B+11 C)+a^2 (8 A-18 B+43 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sec ^2(c+d x) \left (-2 a^3 (11 A-36 B+76 C)+15 a^3 (2 A-6 B+13 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 A-6 B+13 C) \int \sec ^3(c+d x) \, dx}{a^3}-\frac {(2 (11 A-36 B+76 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=\frac {(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 A-6 B+13 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac {(2 (11 A-36 B+76 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac {(2 A-6 B+13 C) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {2 (11 A-36 B+76 C) \tan (c+d x)}{15 a^3 d}+\frac {(2 A-6 B+13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(A-6 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(11 A-36 B+76 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1081\) vs. \(2(216)=432\).
time = 6.49, size = 1081, normalized size = 5.00 \begin {gather*} -\frac {8 (2 A-6 B+13 C) \cos ^6\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 (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))^3}+\frac {8 (2 A-6 B+13 C) \cos ^6\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 (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))^3}+\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (490 A \sin \left (\frac {d x}{2}\right )-870 B \sin \left (\frac {d x}{2}\right )+1235 C \sin \left (\frac {d x}{2}\right )-530 A \sin \left (\frac {3 d x}{2}\right )+1830 B \sin \left (\frac {3 d x}{2}\right )-3805 C \sin \left (\frac {3 d x}{2}\right )+654 A \sin \left (c-\frac {d x}{2}\right )-2094 B \sin \left (c-\frac {d x}{2}\right )+4329 C \sin \left (c-\frac {d x}{2}\right )-654 A \sin \left (c+\frac {d x}{2}\right )+1314 B \sin \left (c+\frac {d x}{2}\right )-1989 C \sin \left (c+\frac {d x}{2}\right )+490 A \sin \left (2 c+\frac {d x}{2}\right )-1650 B \sin \left (2 c+\frac {d x}{2}\right )+3575 C \sin \left (2 c+\frac {d x}{2}\right )+350 A \sin \left (c+\frac {3 d x}{2}\right )-450 B \sin \left (c+\frac {3 d x}{2}\right )+475 C \sin \left (c+\frac {3 d x}{2}\right )-530 A \sin \left (2 c+\frac {3 d x}{2}\right )+1230 B \sin \left (2 c+\frac {3 d x}{2}\right )-2005 C \sin \left (2 c+\frac {3 d x}{2}\right )+350 A \sin \left (3 c+\frac {3 d x}{2}\right )-1050 B \sin \left (3 c+\frac {3 d x}{2}\right )+2275 C \sin \left (3 c+\frac {3 d x}{2}\right )-378 A \sin \left (c+\frac {5 d x}{2}\right )+1278 B \sin \left (c+\frac {5 d x}{2}\right )-2673 C \sin \left (c+\frac {5 d x}{2}\right )+150 A \sin \left (2 c+\frac {5 d x}{2}\right )-90 B \sin \left (2 c+\frac {5 d x}{2}\right )-105 C \sin \left (2 c+\frac {5 d x}{2}\right )-378 A \sin \left (3 c+\frac {5 d x}{2}\right )+918 B \sin \left (3 c+\frac {5 d x}{2}\right )-1593 C \sin \left (3 c+\frac {5 d x}{2}\right )+150 A \sin \left (4 c+\frac {5 d x}{2}\right )-450 B \sin \left (4 c+\frac {5 d x}{2}\right )+975 C \sin \left (4 c+\frac {5 d x}{2}\right )-190 A \sin \left (2 c+\frac {7 d x}{2}\right )+630 B \sin \left (2 c+\frac {7 d x}{2}\right )-1325 C \sin \left (2 c+\frac {7 d x}{2}\right )+30 A \sin \left (3 c+\frac {7 d x}{2}\right )+60 B \sin \left (3 c+\frac {7 d x}{2}\right )-255 C \sin \left (3 c+\frac {7 d x}{2}\right )-190 A \sin \left (4 c+\frac {7 d x}{2}\right )+480 B \sin \left (4 c+\frac {7 d x}{2}\right )-875 C \sin \left (4 c+\frac {7 d x}{2}\right )+30 A \sin \left (5 c+\frac {7 d x}{2}\right )-90 B \sin \left (5 c+\frac {7 d x}{2}\right )+195 C \sin \left (5 c+\frac {7 d x}{2}\right )-44 A \sin \left (3 c+\frac {9 d x}{2}\right )+144 B \sin \left (3 c+\frac {9 d x}{2}\right )-304 C \sin \left (3 c+\frac {9 d x}{2}\right )+30 B \sin \left (4 c+\frac {9 d x}{2}\right )-90 C \sin \left (4 c+\frac {9 d x}{2}\right )-44 A \sin \left (5 c+\frac {9 d x}{2}\right )+114 B \sin \left (5 c+\frac {9 d x}{2}\right )-214 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{240 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3} \end {gather*}

Warning: Unable to verify antiderivative.

[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])^3,x]

[Out]

(-8*(2*A - 6*B + 13*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]*(A + B*S
ec[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])^3)
+ (8*(2*A - 6*B + 13*C)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]*(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])^3)
+ (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(490*A*Sin[(d*x)
/2] - 870*B*Sin[(d*x)/2] + 1235*C*Sin[(d*x)/2] - 530*A*Sin[(3*d*x)/2] + 1830*B*Sin[(3*d*x)/2] - 3805*C*Sin[(3*
d*x)/2] + 654*A*Sin[c - (d*x)/2] - 2094*B*Sin[c - (d*x)/2] + 4329*C*Sin[c - (d*x)/2] - 654*A*Sin[c + (d*x)/2]
+ 1314*B*Sin[c + (d*x)/2] - 1989*C*Sin[c + (d*x)/2] + 490*A*Sin[2*c + (d*x)/2] - 1650*B*Sin[2*c + (d*x)/2] + 3
575*C*Sin[2*c + (d*x)/2] + 350*A*Sin[c + (3*d*x)/2] - 450*B*Sin[c + (3*d*x)/2] + 475*C*Sin[c + (3*d*x)/2] - 53
0*A*Sin[2*c + (3*d*x)/2] + 1230*B*Sin[2*c + (3*d*x)/2] - 2005*C*Sin[2*c + (3*d*x)/2] + 350*A*Sin[3*c + (3*d*x)
/2] - 1050*B*Sin[3*c + (3*d*x)/2] + 2275*C*Sin[3*c + (3*d*x)/2] - 378*A*Sin[c + (5*d*x)/2] + 1278*B*Sin[c + (5
*d*x)/2] - 2673*C*Sin[c + (5*d*x)/2] + 150*A*Sin[2*c + (5*d*x)/2] - 90*B*Sin[2*c + (5*d*x)/2] - 105*C*Sin[2*c
+ (5*d*x)/2] - 378*A*Sin[3*c + (5*d*x)/2] + 918*B*Sin[3*c + (5*d*x)/2] - 1593*C*Sin[3*c + (5*d*x)/2] + 150*A*S
in[4*c + (5*d*x)/2] - 450*B*Sin[4*c + (5*d*x)/2] + 975*C*Sin[4*c + (5*d*x)/2] - 190*A*Sin[2*c + (7*d*x)/2] + 6
30*B*Sin[2*c + (7*d*x)/2] - 1325*C*Sin[2*c + (7*d*x)/2] + 30*A*Sin[3*c + (7*d*x)/2] + 60*B*Sin[3*c + (7*d*x)/2
] - 255*C*Sin[3*c + (7*d*x)/2] - 190*A*Sin[4*c + (7*d*x)/2] + 480*B*Sin[4*c + (7*d*x)/2] - 875*C*Sin[4*c + (7*
d*x)/2] + 30*A*Sin[5*c + (7*d*x)/2] - 90*B*Sin[5*c + (7*d*x)/2] + 195*C*Sin[5*c + (7*d*x)/2] - 44*A*Sin[3*c +
(9*d*x)/2] + 144*B*Sin[3*c + (9*d*x)/2] - 304*C*Sin[3*c + (9*d*x)/2] + 30*B*Sin[4*c + (9*d*x)/2] - 90*C*Sin[4*
c + (9*d*x)/2] - 44*A*Sin[5*c + (9*d*x)/2] + 114*B*Sin[5*c + (9*d*x)/2] - 214*C*Sin[5*c + (9*d*x)/2]))/(240*d*
(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)

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Maple [A]
time = 0.63, size = 252, normalized size = 1.17 Too large to display

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))^3,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^3*(-1/5*A*tan(1/2*d*x+1/2*c)^5+1/5*B*tan(1/2*d*x+1/2*c)^5-1/5*C*tan(1/2*d*x+1/2*c)^5-4/3*A*tan(1/2*d*x
+1/2*c)^3+2*B*tan(1/2*d*x+1/2*c)^3-8/3*C*tan(1/2*d*x+1/2*c)^3-7*A*tan(1/2*d*x+1/2*c)+17*B*tan(1/2*d*x+1/2*c)-3
1*C*tan(1/2*d*x+1/2*c)-(-14*C+4*B)/(tan(1/2*d*x+1/2*c)-1)+(-26*C+12*B-4*A)*ln(tan(1/2*d*x+1/2*c)-1)+2*C/(tan(1
/2*d*x+1/2*c)-1)^2-(-14*C+4*B)/(tan(1/2*d*x+1/2*c)+1)+(26*C-12*B+4*A)*ln(tan(1/2*d*x+1/2*c)+1)-2*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. 493 vs. \(2 (204) = 408\).
time = 0.29, size = 493, normalized size = 2.28 \begin {gather*} -\frac {C {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \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 {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(40*sin(d*x + c)/((a^3 -
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d
*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

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Fricas [A]
time = 2.81, size = 328, normalized size = 1.52 \begin {gather*} \frac {15 \, {\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, A - 6 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (11 \, A - 36 \, B + 76 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A - 114 \, B + 239 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (64 \, A - 234 \, B + 479 \, C\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} 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)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((2*A - 6*B + 13*C)*cos(d*x + c)^5 + 3*(2*A - 6*B + 13*C)*cos(d*x + c)^4 + 3*(2*A - 6*B + 13*C)*cos(d
*x + c)^3 + (2*A - 6*B + 13*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((2*A - 6*B + 13*C)*cos(d*x + c)^5 +
3*(2*A - 6*B + 13*C)*cos(d*x + c)^4 + 3*(2*A - 6*B + 13*C)*cos(d*x + c)^3 + (2*A - 6*B + 13*C)*cos(d*x + c)^2
)*log(-sin(d*x + c) + 1) - 2*(4*(11*A - 36*B + 76*C)*cos(d*x + c)^4 + 3*(34*A - 114*B + 239*C)*cos(d*x + c)^3
+ (64*A - 234*B + 479*C)*cos(d*x + c)^2 - 15*(2*B - 3*C)*cos(d*x + c) - 15*C)*sin(d*x + c))/(a^3*d*cos(d*x + c
)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*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 ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

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

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Giac [A]
time = 0.53, size = 288, normalized size = 1.33 \begin {gather*} \frac {\frac {30 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (2 \, A - 6 \, B + 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, 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 ) + 5 \, 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^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 465 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, 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))^3,x, algorithm="giac")

[Out]

1/60*(30*(2*A - 6*B + 13*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(2*A - 6*B + 13*C)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1))/a^3 - 60*(2*B*tan(1/2*d*x + 1/2*c)^3 - 7*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) +
5*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*
tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^12*tan(1/
2*d*x + 1/2*c)^3 + 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x
+ 1/2*c) + 465*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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Mupad [B]
time = 3.40, size = 227, normalized size = 1.05 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-3\,B+\frac {13\,C}{2}\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-3\,B+5\,C\right )}{4\,a^3}-\frac {2\,A+2\,B-10\,C}{4\,a^3}+\frac {3\,\left (A-B+C\right )}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-7\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-5\,C\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-3\,B+5\,C}{12\,a^3}+\frac {A-B+C}{4\,a^3}\right )}{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))^3),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(A - 3*B + (13*C)/2))/(a^3*d) - (tan(c/2 + (d*x)/2)*((3*(A - 3*B + 5*C))/(4*a^3)
- (2*A + 2*B - 10*C)/(4*a^3) + (3*(A - B + C))/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^3*(2*B - 7*C) - tan(c/2 + (d*
x)/2)*(2*B - 5*C))/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)^5*(
A - B + C))/(20*a^3*d) - (tan(c/2 + (d*x)/2)^3*((A - 3*B + 5*C)/(12*a^3) + (A - B + C)/(4*a^3)))/d

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