3.116 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=198 \[ \frac{5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac{a^4 (2 A+13 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(2 A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}-\frac{(4 A-9 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+2 a^4 x (3 A+2 C)+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}+\frac{2 a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

[Out]

2*a^4*(3*A + 2*C)*x + (a^4*(2*A + 13*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^4*(2*A - C)*Sin[c + d*x])/(2*d) +
(2*a*A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*Sin[
c + d*x])/(3*d) - ((2*A - C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - ((4*A - 9*C)*(a^4 + a^4*Sec[c +
d*x])*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.546137, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4087, 4017, 4018, 3996, 3770} \[ \frac{5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac{a^4 (2 A+13 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(2 A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}-\frac{(4 A-9 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+2 a^4 x (3 A+2 C)+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}+\frac{2 a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

2*a^4*(3*A + 2*C)*x + (a^4*(2*A + 13*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^4*(2*A - C)*Sin[c + d*x])/(2*d) +
(2*a*A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*Sin[
c + d*x])/(3*d) - ((2*A - C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - ((4*A - 9*C)*(a^4 + a^4*Sec[c +
d*x])*Sin[c + d*x])/(3*d)

Rule 4087

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

Rule 4017

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

Rule 4018

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

Rule 3996

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

Rule 3770

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

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (4 a A-a (2 A-3 C) \sec (c+d x)) \, dx}{3 a}\\ &=\frac{2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^3 \left (2 a^2 (8 A+3 C)-6 a^2 (2 A-C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac{2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^3 (22 A+3 C)-4 a^3 (4 A-9 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (30 a^4 (2 A-C)+6 a^4 (2 A+13 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac{2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}-\frac{\int \left (-24 a^5 (3 A+2 C)-6 a^5 (2 A+13 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=2 a^4 (3 A+2 C) x+\frac{5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac{2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac{1}{2} \left (a^4 (2 A+13 C)\right ) \int \sec (c+d x) \, dx\\ &=2 a^4 (3 A+2 C) x+\frac{a^4 (2 A+13 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac{2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac{(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac{(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 6.21813, size = 1250, normalized size = 6.31 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

((3*A + 2*C)*x*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(4*(A + 2*C
+ A*Cos[2*c + 2*d*x])) + ((-2*A - 13*C)*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 +
(d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((2*A + 13*C
)*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A +
C*Sec[c + d*x]^2))/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((27*A + 4*C)*Cos[d*x]*Cos[c + d*x]^6*Sec[c/2 + (d
*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[c])/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[
2*d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2*c])/(8*d*(A + 2
*C + A*Cos[2*c + 2*d*x])) + (A*Cos[3*d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Se
c[c + d*x]^2)*Sin[3*c])/(96*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((27*A + 4*C)*Cos[c]*Cos[c + d*x]^6*Sec[c/2 +
(d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[d*x])/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*
Cos[2*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2*d*x])/(8*d*(A
+ 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[3*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*
Sec[c + d*x]^2)*Sin[3*d*x])/(96*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a
+ a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2 + (d*x)/2] - Sin[c/2
+ (d*x)/2])^2) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(d*
x)/2])/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) -
(C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(32*d*(A + 2*C + A*Cos[2
*c + 2*d*x])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[
c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] + Sin[c/2])*(Co
s[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.108, size = 190, normalized size = 1. \begin{align*}{\frac{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{3\,d}}+{\frac{20\,A{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+6\,{a}^{4}Ax+6\,{\frac{A{a}^{4}c}{d}}+4\,{a}^{4}Cx+4\,{\frac{C{a}^{4}c}{d}}+{\frac{13\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+4\,{\frac{{a}^{4}C\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

1/3/d*A*cos(d*x+c)^2*sin(d*x+c)*a^4+20/3/d*A*a^4*sin(d*x+c)+1/d*a^4*C*sin(d*x+c)+2/d*A*a^4*cos(d*x+c)*sin(d*x+
c)+6*a^4*A*x+6/d*A*a^4*c+4*a^4*C*x+4/d*C*a^4*c+13/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4/d*a^4*C*tan(d*x+c)+1/d
*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^4*C*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 0.952799, size = 285, normalized size = 1.44 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 48 \,{\left (d x + c\right )} A a^{4} - 48 \,{\left (d x + c\right )} C a^{4} + 3 \, C a^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} \sin \left (d x + c\right ) - 12 \, C a^{4} \sin \left (d x + c\right ) - 48 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 48*(d*x + c)*A*
a^4 - 48*(d*x + c)*C*a^4 + 3*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x
+ c) - 1)) - 6*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*C*a^4*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) - 72*A*a^4*sin(d*x + c) - 12*C*a^4*sin(d*x + c) - 48*C*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.547472, size = 432, normalized size = 2.18 \begin{align*} \frac{24 \,{\left (3 \, A + 2 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, A + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 12 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (20 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 24 \, C a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(24*(3*A + 2*C)*a^4*d*x*cos(d*x + c)^2 + 3*(2*A + 13*C)*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(2*A
+ 13*C)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^4*cos(d*x + c)^4 + 12*A*a^4*cos(d*x + c)^3 + 2*(
20*A + 3*C)*a^4*cos(d*x + c)^2 + 24*C*a^4*cos(d*x + c) + 3*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.24505, size = 335, normalized size = 1.69 \begin{align*} \frac{12 \,{\left (3 \, A a^{4} + 2 \, C a^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (2 \, A a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, A a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{6 \,{\left (7 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, C a^{4} \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}} + \frac{4 \,{\left (15 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 38 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{4} \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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(12*(3*A*a^4 + 2*C*a^4)*(d*x + c) + 3*(2*A*a^4 + 13*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a^4
+ 13*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(7*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*C*a^4*tan(1/2*d*x + 1/
2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 4*(15*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 3
8*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) + 3*C*a^4*tan(
1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d