3.35 \(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=200 \[ -\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{a^4 (35 A+52 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(35 A+36 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{12 d}+\frac{(7 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{8 d}+4 a^4 C x+\frac{a A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^4}{4 d} \]

[Out]

4*a^4*C*x + (a^4*(35*A + 52*C)*ArcTanh[Sin[c + d*x]])/(8*d) - (5*a^4*(7*A + 4*C)*Sin[c + d*x])/(8*d) + ((35*A
+ 36*C)*(a^4 + a^4*Cos[c + d*x])*Tan[c + d*x])/(12*d) + ((7*A + 4*C)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]*T
an[c + d*x])/(8*d) + (a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (A*(a + a*Cos[c + d*x])^
4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.668955, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3044, 2975, 2968, 3023, 2735, 3770} \[ -\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{a^4 (35 A+52 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(35 A+36 C) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{12 d}+\frac{(7 A+4 C) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{8 d}+4 a^4 C x+\frac{a A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4*a^4*C*x + (a^4*(35*A + 52*C)*ArcTanh[Sin[c + d*x]])/(8*d) - (5*a^4*(7*A + 4*C)*Sin[c + d*x])/(8*d) + ((35*A
+ 36*C)*(a^4 + a^4*Cos[c + d*x])*Tan[c + d*x])/(12*d) + ((7*A + 4*C)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]*T
an[c + d*x])/(8*d) + (a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (A*(a + a*Cos[c + d*x])^
4*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rule 2975

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

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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 (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^4 (4 a A-a (A-4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^3 \left (3 a^2 (7 A+4 C)-a^2 (7 A-12 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac{(7 A+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^2 \left (2 a^3 (35 A+36 C)-a^3 (35 A-12 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac{(35 A+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac{(7 A+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x)) \left (3 a^4 (35 A+52 C)-15 a^4 (7 A+4 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{(35 A+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac{(7 A+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (3 a^5 (35 A+52 C)+\left (-15 a^5 (7 A+4 C)+3 a^5 (35 A+52 C)\right ) \cos (c+d x)-15 a^5 (7 A+4 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=-\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{(35 A+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac{(7 A+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (3 a^5 (35 A+52 C)+96 a^5 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=4 a^4 C x-\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{(35 A+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac{(7 A+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} \left (a^4 (35 A+52 C)\right ) \int \sec (c+d x) \, dx\\ &=4 a^4 C x+\frac{a^4 (35 A+52 C) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{5 a^4 (7 A+4 C) \sin (c+d x)}{8 d}+\frac{(35 A+36 C) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{12 d}+\frac{(7 A+4 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 2.11598, size = 350, normalized size = 1.75 \[ \frac{a^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^4 \left (\sec (c) (105 A \sin (2 c+d x)+544 A \sin (c+2 d x)-96 A \sin (3 c+2 d x)+81 A \sin (2 c+3 d x)+81 A \sin (4 c+3 d x)+160 A \sin (3 c+4 d x)-480 A \sin (c)+105 A \sin (d x)+24 C \sin (2 c+d x)+288 C \sin (c+2 d x)-96 C \sin (3 c+2 d x)+30 C \sin (2 c+3 d x)+30 C \sin (4 c+3 d x)+96 C \sin (3 c+4 d x)+6 C \sin (4 c+5 d x)+6 C \sin (6 c+5 d x)+288 C d x \cos (c)+192 C d x \cos (c+2 d x)+192 C d x \cos (3 c+2 d x)+48 C d x \cos (3 c+4 d x)+48 C d x \cos (5 c+4 d x)-288 C \sin (c)+24 C \sin (d x))-24 (35 A+52 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{3072 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(-24*(35*A + 52*C)*Cos[c + d*x]^4*(Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]*(288*C*d*x*Cos[c] + 192*C*d*x*Cos[c + 2*d*x] +
192*C*d*x*Cos[3*c + 2*d*x] + 48*C*d*x*Cos[3*c + 4*d*x] + 48*C*d*x*Cos[5*c + 4*d*x] - 480*A*Sin[c] - 288*C*Sin
[c] + 105*A*Sin[d*x] + 24*C*Sin[d*x] + 105*A*Sin[2*c + d*x] + 24*C*Sin[2*c + d*x] + 544*A*Sin[c + 2*d*x] + 288
*C*Sin[c + 2*d*x] - 96*A*Sin[3*c + 2*d*x] - 96*C*Sin[3*c + 2*d*x] + 81*A*Sin[2*c + 3*d*x] + 30*C*Sin[2*c + 3*d
*x] + 81*A*Sin[4*c + 3*d*x] + 30*C*Sin[4*c + 3*d*x] + 160*A*Sin[3*c + 4*d*x] + 96*C*Sin[3*c + 4*d*x] + 6*C*Sin
[4*c + 5*d*x] + 6*C*Sin[6*c + 5*d*x])))/(3072*d)

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Maple [A]  time = 0.086, size = 197, normalized size = 1. \begin{align*}{\frac{35\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) }{d}}+{\frac{20\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+4\,{a}^{4}Cx+4\,{\frac{C{a}^{4}c}{d}}+{\frac{27\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{13\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{4\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+4\,{\frac{{a}^{4}C\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,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((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

35/8/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^4*C*sin(d*x+c)+20/3/d*A*a^4*tan(d*x+c)+4*a^4*C*x+4/d*a^4*C*c+27/8
/d*A*a^4*sec(d*x+c)*tan(d*x+c)+13/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+4/d*
a^4*C*tan(d*x+c)+1/4/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+1/2/d*a^4*C*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 1.07609, size = 400, normalized size = 2. \begin{align*} \frac{64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 192 \,{\left (d x + c\right )} C a^{4} - 3 \, A a^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A 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 )} - 12 \, 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 )} + 24 \, 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 )} + 144 \, 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 )} + 48 \, C a^{4} \sin \left (d x + c\right ) + 192 \, A a^{4} \tan \left (d x + c\right ) + 192 \, C a^{4} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 192*(d*x + c)*C*a^4 - 3*A*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(
d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 72*A*
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)) - 12*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)) + 24*A*a^4*(log(sin(d*x + c) + 1)
- log(sin(d*x + c) - 1)) + 144*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*C*a^4*sin(d*x + c)
+ 192*A*a^4*tan(d*x + c) + 192*C*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.62247, size = 437, normalized size = 2.18 \begin{align*} \frac{192 \, C a^{4} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (35 \, A + 52 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (35 \, A + 52 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, C a^{4} \cos \left (d x + c\right )^{4} + 32 \,{\left (5 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \,{\left (27 \, A + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, A a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(192*C*a^4*d*x*cos(d*x + c)^4 + 3*(35*A + 52*C)*a^4*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(35*A + 52*C
)*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*C*a^4*cos(d*x + c)^4 + 32*(5*A + 3*C)*a^4*cos(d*x + c)^3 +
3*(27*A + 4*C)*a^4*cos(d*x + c)^2 + 32*A*a^4*cos(d*x + c) + 6*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^4)

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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((a+a*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.40601, size = 342, normalized size = 1.71 \begin{align*} \frac{96 \,{\left (d x + c\right )} C a^{4} + \frac{48 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 3 \,{\left (35 \, A a^{4} + 52 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (35 \, A a^{4} + 52 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (105 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 84 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 385 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 276 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 511 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 300 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 279 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 108 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(96*(d*x + c)*C*a^4 + 48*C*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 3*(35*A*a^4 + 52*C*a^4
)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(35*A*a^4 + 52*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(105*A*a
^4*tan(1/2*d*x + 1/2*c)^7 + 84*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 385*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 276*C*a^4*tan
(1/2*d*x + 1/2*c)^5 + 511*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 300*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 279*A*a^4*tan(1/2*
d*x + 1/2*c) - 108*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d