3.244 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=195 \[ -\frac{b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{3 d}-\frac{b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} b x \left (12 a^2 A b+8 a^3 B+4 a b^2 B+A b^3\right )+\frac{a^3 (a B+4 A b) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{a A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]

[Out]

(b*(12*a^2*A*b + A*b^3 + 8*a^3*B + 4*a*b^2*B)*x)/2 + (a^3*(4*A*b + a*B)*ArcTanh[Sin[c + d*x]])/d - (b*(6*a^3*A
- 12*a*A*b^2 - 17*a^2*b*B - 2*b^3*B)*Sin[c + d*x])/(3*d) - (b^2*(6*a^2*A - 3*A*b^2 - 8*a*b*B)*Cos[c + d*x]*Si
n[c + d*x])/(6*d) - (b*(3*a*A - b*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + (a*A*(a + b*Cos[c + d*x])^3*
Tan[c + d*x])/d

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Rubi [A]  time = 0.569693, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2989, 3049, 3033, 3023, 2735, 3770} \[ -\frac{b \left (6 a^3 A-17 a^2 b B-12 a A b^2-2 b^3 B\right ) \sin (c+d x)}{3 d}-\frac{b^2 \left (6 a^2 A-8 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} b x \left (12 a^2 A b+8 a^3 B+4 a b^2 B+A b^3\right )+\frac{a^3 (a B+4 A b) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b (3 a A-b B) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{a A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(12*a^2*A*b + A*b^3 + 8*a^3*B + 4*a*b^2*B)*x)/2 + (a^3*(4*A*b + a*B)*ArcTanh[Sin[c + d*x]])/d - (b*(6*a^3*A
- 12*a*A*b^2 - 17*a^2*b*B - 2*b^3*B)*Sin[c + d*x])/(3*d) - (b^2*(6*a^2*A - 3*A*b^2 - 8*a*b*B)*Cos[c + d*x]*Si
n[c + d*x])/(6*d) - (b*(3*a*A - b*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + (a*A*(a + b*Cos[c + d*x])^3*
Tan[c + d*x])/d

Rule 2989

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

Rule 3049

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

Rule 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[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))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx &=\frac{a A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^2 \left (a (4 A b+a B)+b (A b+2 a B) \cos (c+d x)-b (3 a A-b B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b (3 a A-b B) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{a A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{1}{3} \int (a+b \cos (c+d x)) \left (3 a^2 (4 A b+a B)+b \left (9 a A b+9 a^2 B+2 b^2 B\right ) \cos (c+d x)-b \left (6 a^2 A-3 A b^2-8 a b B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{a A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{1}{6} \int \left (6 a^3 (4 A b+a B)+3 b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) \cos (c+d x)-2 b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{3 d}-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{a A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{1}{6} \int \left (6 a^3 (4 A b+a B)+3 b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) x-\frac{b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{3 d}-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{a A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\left (a^3 (4 A b+a B)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} b \left (12 a^2 A b+A b^3+8 a^3 B+4 a b^2 B\right ) x+\frac{a^3 (4 A b+a B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b \left (6 a^3 A-12 a A b^2-17 a^2 b B-2 b^3 B\right ) \sin (c+d x)}{3 d}-\frac{b^2 \left (6 a^2 A-3 A b^2-8 a b B\right ) \cos (c+d x) \sin (c+d x)}{6 d}-\frac{b (3 a A-b B) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{a A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 1.01931, size = 257, normalized size = 1.32 \[ \frac{6 b (c+d x) \left (12 a^2 A b+8 a^3 B+4 a b^2 B+A b^3\right )+3 b^2 \left (24 a^2 B+16 a A b+3 b^2 B\right ) \sin (c+d x)-12 a^3 (a B+4 A b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^3 (a B+4 A b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{12 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{12 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+3 b^3 (4 a B+A b) \sin (2 (c+d x))+b^4 B \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*(12*a^2*A*b + A*b^3 + 8*a^3*B + 4*a*b^2*B)*(c + d*x) - 12*a^3*(4*A*b + a*B)*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] + 12*a^3*(4*A*b + a*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*a^4*A*Sin[(c + d*x)/2])/(Cos
[(c + d*x)/2] - Sin[(c + d*x)/2]) + (12*a^4*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*b^2*
(16*a*A*b + 24*a^2*B + 3*b^2*B)*Sin[c + d*x] + 3*b^3*(A*b + 4*a*B)*Sin[2*(c + d*x)] + b^4*B*Sin[3*(c + d*x)])/
(12*d)

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Maple [A]  time = 0.084, size = 255, normalized size = 1.3 \begin{align*}{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,B{a}^{3}bx+4\,{\frac{B{a}^{3}bc}{d}}+6\,A{a}^{2}{b}^{2}x+6\,{\frac{A{a}^{2}{b}^{2}c}{d}}+6\,{\frac{B{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{Aa{b}^{3}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{Ba{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,Ba{b}^{3}x+2\,{\frac{Ba{b}^{3}c}{d}}+{\frac{A{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{4}x}{2}}+{\frac{A{b}^{4}c}{2\,d}}+{\frac{B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{4}}{3\,d}}+{\frac{2\,B{b}^{4}\sin \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12913, size = 266, normalized size = 1.36 \begin{align*} \frac{48 \,{\left (d x + c\right )} B a^{3} b + 72 \,{\left (d x + c\right )} A a^{2} b^{2} + 12 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{4} + 6 \, B a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 72 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 48 \, A a b^{3} \sin \left (d x + c\right ) + 12 \, A a^{4} \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))^4*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

1/12*(48*(d*x + c)*B*a^3*b + 72*(d*x + c)*A*a^2*b^2 + 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b^3 + 3*(2*d*x +
2*c + sin(2*d*x + 2*c))*A*b^4 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b^4 + 6*B*a^4*(log(sin(d*x + c) + 1) -
log(sin(d*x + c) - 1)) + 24*A*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 72*B*a^2*b^2*sin(d*x + c
) + 48*A*a*b^3*sin(d*x + c) + 12*A*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.60192, size = 471, normalized size = 2.42 \begin{align*} \frac{3 \,{\left (8 \, B a^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, B b^{4} \cos \left (d x + c\right )^{3} + 6 \, A a^{4} + 3 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (9 \, B a^{2} b^{2} + 6 \, A a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(8*B*a^3*b + 12*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*d*x*cos(d*x + c) + 3*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)*lo
g(sin(d*x + c) + 1) - 3*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*B*b^4*cos(d*x + c)^3 + 6*
A*a^4 + 3*(4*B*a*b^3 + A*b^4)*cos(d*x + c)^2 + 4*(9*B*a^2*b^2 + 6*A*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x + c))
/(d*cos(d*x + c))

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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))**4*(A+B*cos(d*x+c))*sec(d*x+c)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.52787, size = 501, normalized size = 2.57 \begin{align*} -\frac{\frac{12 \, A 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 (8 \, B a^{3} b + 12 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )}{\left (d x + c\right )} - 6 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 6 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (36 \, B a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, A a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, B b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 72 \, B a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 48 \, A a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, B b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, B a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, A a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, B a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B b^{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((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="giac")

[Out]

-1/6*(12*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(8*B*a^3*b + 12*A*a^2*b^2 + 4*B*a*b^3 + A
*b^4)*(d*x + c) - 6*(B*a^4 + 4*A*a^3*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(B*a^4 + 4*A*a^3*b)*log(abs(tan
(1/2*d*x + 1/2*c) - 1)) - 2*(36*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 24*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*
b^3*tan(1/2*d*x + 1/2*c)^5 - 3*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 72*B*a^2*b^2*ta
n(1/2*d*x + 1/2*c)^3 + 48*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*B*b^4*tan(1/2*d*x + 1/2*c)^3 + 36*B*a^2*b^2*tan(1
/2*d*x + 1/2*c) + 24*A*a*b^3*tan(1/2*d*x + 1/2*c) + 12*B*a*b^3*tan(1/2*d*x + 1/2*c) + 3*A*b^4*tan(1/2*d*x + 1/
2*c) + 6*B*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d