3.307 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=198 \[ \frac{a^2 \left (2 a^2 A+9 a b B+9 A b^2\right ) \sin (c+d x)}{3 d}-\frac{b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{6 d}+\frac{1}{2} a x \left (4 a^2 A b+a^3 B+12 a b^2 B+8 A b^3\right )+\frac{b^3 (4 a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d} \]

[Out]

(a*(4*a^2*A*b + 8*A*b^3 + a^3*B + 12*a*b^2*B)*x)/2 + (b^3*(A*b + 4*a*B)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2
*A + 9*A*b^2 + 9*a*b*B)*Sin[c + d*x])/(3*d) + (a*(2*A*b + a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x
])/(2*d) + (a*A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) - (b^2*(8*a*A*b + 3*a^2*B - 6*b^2*B)
*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.590815, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4025, 4094, 4076, 4047, 8, 4045, 3770} \[ \frac{a^2 \left (2 a^2 A+9 a b B+9 A b^2\right ) \sin (c+d x)}{3 d}-\frac{b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \tan (c+d x)}{6 d}+\frac{1}{2} a x \left (4 a^2 A b+a^3 B+12 a b^2 B+8 A b^3\right )+\frac{b^3 (4 a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (a B+2 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{a A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*a^2*A*b + 8*A*b^3 + a^3*B + 12*a*b^2*B)*x)/2 + (b^3*(A*b + 4*a*B)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2
*A + 9*A*b^2 + 9*a*b*B)*Sin[c + d*x])/(3*d) + (a*(2*A*b + a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x
])/(2*d) + (a*A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) - (b^2*(8*a*A*b + 3*a^2*B - 6*b^2*B)
*Tan[c + d*x])/(6*d)

Rule 4025

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[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

Rule 4094

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

Rule 4076

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_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -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+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^2 \left (-3 a (2 A b+a B)-\left (2 a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x)+b (a A-3 b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (2 A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{1}{6} \int \cos (c+d x) (a+b \sec (c+d x)) \left (-2 a \left (2 a^2 A+9 A b^2+9 a b B\right )-\left (8 a^2 A b+6 A b^3+3 a^3 B+18 a b^2 B\right ) \sec (c+d x)+b \left (8 a A b+3 a^2 B-6 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (2 A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \tan (c+d x)}{6 d}-\frac{1}{6} \int \cos (c+d x) \left (-2 a^2 \left (2 a^2 A+9 A b^2+9 a b B\right )-3 a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right ) \sec (c+d x)-6 b^3 (A b+4 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a (2 A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \tan (c+d x)}{6 d}-\frac{1}{6} \int \cos (c+d x) \left (-2 a^2 \left (2 a^2 A+9 A b^2+9 a b B\right )-6 b^3 (A b+4 a B) \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right )\right ) \int 1 \, dx\\ &=\frac{1}{2} a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right ) x+\frac{a^2 \left (2 a^2 A+9 A b^2+9 a b B\right ) \sin (c+d x)}{3 d}+\frac{a (2 A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \tan (c+d x)}{6 d}+\left (b^3 (A b+4 a B)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right ) x+\frac{b^3 (A b+4 a B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \left (2 a^2 A+9 A b^2+9 a b B\right ) \sin (c+d x)}{3 d}+\frac{a (2 A b+a B) \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{a A \cos ^2(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}-\frac{b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 1.07503, size = 257, normalized size = 1.3 \[ \frac{6 a (c+d x) \left (4 a^2 A b+a^3 B+12 a b^2 B+8 A b^3\right )+3 a^2 \left (3 a^2 A+16 a b B+24 A b^2\right ) \sin (c+d x)+3 a^3 (a B+4 A b) \sin (2 (c+d x))+a^4 A \sin (3 (c+d x))-12 b^3 (4 a B+A b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^3 (4 a B+A b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{12 b^4 B \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{12 b^4 B \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 255, normalized size = 1.3 \begin{align*}{\frac{A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{2\,A{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{4}x}{2}}+{\frac{B{a}^{4}c}{2\,d}}+2\,{\frac{A{a}^{3}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+2\,A{a}^{3}bx+2\,{\frac{A{a}^{3}bc}{d}}+4\,{\frac{B{a}^{3}b\sin \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+6\,B{a}^{2}{b}^{2}x+6\,{\frac{B{a}^{2}{b}^{2}c}{d}}+4\,Aa{b}^{3}x+4\,{\frac{Aa{b}^{3}c}{d}}+4\,{\frac{Ba{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{b}^{4}\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.989478, size = 266, normalized size = 1.34 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 12 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 72 \,{\left (d x + c\right )} B a^{2} b^{2} - 48 \,{\left (d x + c\right )} A a b^{3} - 24 \, B 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 )} - 6 \, A b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{3} b \sin \left (d x + c\right ) - 72 \, A a^{2} b^{2} \sin \left (d x + c\right ) - 12 \, B b^{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+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.593142, size = 471, normalized size = 2.38 \begin{align*} \frac{3 \,{\left (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, A a^{4} \cos \left (d x + c\right )^{3} + 6 \, B b^{4} + 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (A a^{4} + 6 \, B a^{3} b + 9 \, A a^{2} b^{2}\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(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.28818, size = 501, normalized size = 2.53 \begin{align*} -\frac{\frac{12 \, B b^{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 (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )}{\left (d x + c\right )} - 6 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 6 \,{\left (4 \, B a b^{3} + A b^{4}\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^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, A a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 48 \, B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 72 \, A a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \, A a^{2} 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(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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