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

Optimal. Leaf size=163 \[ \frac{a \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{3 d}+\frac{b \left (3 a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+\frac{A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+3 a b^2 C x-\frac{b^3 (5 A-6 C) \sin (c+d x)}{6 d} \]

[Out]

3*a*b^2*C*x + (b*(2*A*b^2 + 3*a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) - (b^3*(5*A - 6*C)*Sin[c + d*x])/(6*
d) + (a*(3*A*b^2 + a^2*(2*A + 3*C))*Tan[c + d*x])/(3*d) + (A*b*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x
])/(2*d) + (A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.535914, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3047, 3031, 3023, 2735, 3770} \[ \frac{a \left (a^2 (2 A+3 C)+3 A b^2\right ) \tan (c+d x)}{3 d}+\frac{b \left (3 a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}+\frac{A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+3 a b^2 C x-\frac{b^3 (5 A-6 C) \sin (c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*a*b^2*C*x + (b*(2*A*b^2 + 3*a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) - (b^3*(5*A - 6*C)*Sin[c + d*x])/(6*
d) + (a*(3*A*b^2 + a^2*(2*A + 3*C))*Tan[c + d*x])/(3*d) + (A*b*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x
])/(2*d) + (A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 3048

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/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3047

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

Rule 3031

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

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+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x))^2 \left (3 A b+a (2 A+3 C) \cos (c+d x)-b (A-3 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{A b (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (a+b \cos (c+d x)) \left (2 \left (3 A b^2+\frac{1}{2} a^2 (4 A+6 C)\right )+a b (5 A+12 C) \cos (c+d x)-b^2 (5 A-6 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a \left (3 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{A b (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 b \left (2 A b^2+3 a^2 (A+2 C)\right )-18 a b^2 C \cos (c+d x)+b^3 (5 A-6 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^3 (5 A-6 C) \sin (c+d x)}{6 d}+\frac{a \left (3 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{A b (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 b \left (2 A b^2+3 a^2 (A+2 C)\right )-18 a b^2 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=3 a b^2 C x-\frac{b^3 (5 A-6 C) \sin (c+d x)}{6 d}+\frac{a \left (3 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{A b (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} \left (b \left (2 A b^2+3 a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=3 a b^2 C x+\frac{b \left (2 A b^2+3 a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^3 (5 A-6 C) \sin (c+d x)}{6 d}+\frac{a \left (3 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac{A b (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 4.29332, size = 377, normalized size = 2.31 \[ \frac{\frac{4 a \left (a^2 (2 A+3 C)+9 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a \left (a^2 (2 A+3 C)+9 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-6 b \left (3 a^2 (A+2 C)+2 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 b \left (3 a^2 (A+2 C)+2 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^2 A (a+9 b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2 A (a+9 b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+36 a b^2 C (c+d x)+12 b^3 C \sin (c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a*b^2*C*(c + d*x) - 6*b*(2*A*b^2 + 3*a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*b*(2*A*b^
2 + 3*a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*A*(a + 9*b))/(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^2 + (2*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*a*(9*A*b^2 + a^2*(2*A +
3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^3 - (a^2*A*(a + 9*b))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a*(9*A*b^2 + a^2*(2*A +
3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 12*b^3*C*Sin[c + d*x])/(12*d)

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Maple [A]  time = 0.062, size = 195, normalized size = 1.2 \begin{align*}{\frac{A{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{C{b}^{3}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{aA{b}^{2}\tan \left ( dx+c \right ) }{d}}+3\,a{b}^{2}Cx+3\,{\frac{Ca{b}^{2}c}{d}}+{\frac{3\,A{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,A{a}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*A*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*C*b^3*sin(d*x+c)+3/d*a*A*b^2*tan(d*x+c)+3*a*b^2*C*x+3/d*C*a*b^2*c+3/2/
d*A*a^2*b*sec(d*x+c)*tan(d*x+c)+3/2/d*A*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^2*b*C*ln(sec(d*x+c)+tan(d*x+c))+
2/3/d*A*a^3*tan(d*x+c)+1/3/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+1/d*a^3*C*tan(d*x+c)

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Maxima [A]  time = 1.03364, size = 244, normalized size = 1.5 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 36 \,{\left (d x + c\right )} C a b^{2} - 9 \, A a^{2} b{\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 )} + 18 \, C a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C b^{3} \sin \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right ) + 36 \, A a b^{2} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 36*(d*x + c)*C*a*b^2 - 9*A*a^2*b*(2*sin(d*x + c)/(sin(d*x +
c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 18*C*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x +
c) - 1)) + 6*A*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*b^3*sin(d*x + c) + 12*C*a^3*tan(d*x
+ c) + 36*A*a*b^2*tan(d*x + c))/d

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Fricas [A]  time = 1.52551, size = 440, normalized size = 2.7 \begin{align*} \frac{36 \, C a b^{2} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \,{\left (A + 2 \, C\right )} a^{2} b + 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \,{\left (A + 2 \, C\right )} a^{2} b + 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, C b^{3} \cos \left (d x + c\right )^{3} + 9 \, A a^{2} b \cos \left (d x + c\right ) + 2 \, A a^{3} + 2 \,{\left ({\left (2 \, A + 3 \, C\right )} a^{3} + 9 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(36*C*a*b^2*d*x*cos(d*x + c)^3 + 3*(3*(A + 2*C)*a^2*b + 2*A*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3
*(3*(A + 2*C)*a^2*b + 2*A*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*C*b^3*cos(d*x + c)^3 + 9*A*a^2*b*c
os(d*x + c) + 2*A*a^3 + 2*((2*A + 3*C)*a^3 + 9*A*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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

[Out]

Timed out

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Giac [B]  time = 1.32864, size = 435, normalized size = 2.67 \begin{align*} \frac{18 \,{\left (d x + c\right )} C a b^{2} + \frac{12 \, C b^{3} \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 (3 \, A a^{2} b + 6 \, C a^{2} b + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (3 \, A a^{2} b + 6 \, C a^{2} b + 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, A a b^{2} \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((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(18*(d*x + c)*C*a*b^2 + 12*C*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 3*(3*A*a^2*b + 6*C*a^
2*b + 2*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a^2*b + 6*C*a^2*b + 2*A*b^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^2*b*tan(1/2*d*x + 1/
2*c)^5 + 18*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^3
- 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1/2*d*x + 1/2*c) + 6*C*a^3*tan(1/2*d*x + 1/2*c) + 9*A*a^2*b*
tan(1/2*d*x + 1/2*c) + 18*A*a*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d