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

Optimal. Leaf size=225 \[ \frac{a^4 (28 A+35 B+40 C) \sin (c+d x)}{8 d}+\frac{(28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{60 d}+\frac{(28 A+35 B+32 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac{1}{8} a^4 x (28 A+35 B+48 C)+\frac{a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

[Out]

(a^4*(28*A + 35*B + 48*C)*x)/8 + (a^4*C*ArcTanh[Sin[c + d*x]])/d + (a^4*(28*A + 35*B + 40*C)*Sin[c + d*x])/(8*
d) + (a*(4*A + 5*B)*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + a*Sec[
c + d*x])^4*Sin[c + d*x])/(5*d) + ((28*A + 35*B + 20*C)*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x]
)/(60*d) + ((28*A + 35*B + 32*C)*Cos[c + d*x]*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(24*d)

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Rubi [A]  time = 0.582938, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {4086, 4017, 3996, 3770} \[ \frac{a^4 (28 A+35 B+40 C) \sin (c+d x)}{8 d}+\frac{(28 A+35 B+20 C) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{60 d}+\frac{(28 A+35 B+32 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+\frac{1}{8} a^4 x (28 A+35 B+48 C)+\frac{a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (4 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(28*A + 35*B + 48*C)*x)/8 + (a^4*C*ArcTanh[Sin[c + d*x]])/d + (a^4*(28*A + 35*B + 40*C)*Sin[c + d*x])/(8*
d) + (a*(4*A + 5*B)*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + a*Sec[
c + d*x])^4*Sin[c + d*x])/(5*d) + ((28*A + 35*B + 20*C)*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x]
)/(60*d) + ((28*A + 35*B + 32*C)*Cos[c + d*x]*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(24*d)

Rule 4086

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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 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 ^5(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{\int \cos ^4(c+d x) (a+a \sec (c+d x))^4 (a (4 A+5 B)+5 a C \sec (c+d x)) \, dx}{5 a}\\ &=\frac{a (4 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (a^2 (28 A+35 B+20 C)+20 a^2 C \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac{a (4 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(28 A+35 B+20 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (5 a^3 (28 A+35 B+32 C)+60 a^3 C \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac{a (4 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(28 A+35 B+20 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(28 A+35 B+32 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^4 (28 A+35 B+40 C)+120 a^4 C \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac{a^4 (28 A+35 B+40 C) \sin (c+d x)}{8 d}+\frac{a (4 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(28 A+35 B+20 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(28 A+35 B+32 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}-\frac{\int \left (-15 a^5 (28 A+35 B+48 C)-120 a^5 C \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac{1}{8} a^4 (28 A+35 B+48 C) x+\frac{a^4 (28 A+35 B+40 C) \sin (c+d x)}{8 d}+\frac{a (4 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(28 A+35 B+20 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(28 A+35 B+32 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^4 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^4 (28 A+35 B+48 C) x+\frac{a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^4 (28 A+35 B+40 C) \sin (c+d x)}{8 d}+\frac{a (4 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(28 A+35 B+20 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{60 d}+\frac{(28 A+35 B+32 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end{align*}

Mathematica [A]  time = 0.617114, size = 182, normalized size = 0.81 \[ \frac{a^4 \left (60 (49 A+56 B+54 C) \sin (c+d x)+120 (8 A+7 B+4 C) \sin (2 (c+d x))+290 A \sin (3 (c+d x))+60 A \sin (4 (c+d x))+6 A \sin (5 (c+d x))+1680 A d x+160 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+2100 B d x+40 C \sin (3 (c+d x))-480 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+480 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2880 C d x\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(1680*A*d*x + 2100*B*d*x + 2880*C*d*x - 480*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 480*C*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]] + 60*(49*A + 56*B + 54*C)*Sin[c + d*x] + 120*(8*A + 7*B + 4*C)*Sin[2*(c + d*x)]
+ 290*A*Sin[3*(c + d*x)] + 160*B*Sin[3*(c + d*x)] + 40*C*Sin[3*(c + d*x)] + 60*A*Sin[4*(c + d*x)] + 15*B*Sin[
4*(c + d*x)] + 6*A*Sin[5*(c + d*x)]))/(480*d)

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Maple [A]  time = 0.125, size = 320, normalized size = 1.4 \begin{align*}{\frac{4\,B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{20\,B{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{35\,B{a}^{4}c}{8\,d}}+{\frac{7\,A{a}^{4}c}{2\,d}}+{\frac{35\,B{a}^{4}x}{8}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{27\,B{a}^{4}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}+{\frac{7\,{a}^{4}Ax}{2}}+{\frac{A{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{7\,A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+6\,{a}^{4}Cx+2\,{\frac{{a}^{4}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+{\frac{20\,{a}^{4}C\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{83\,A{a}^{4}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{34\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{15\,d}}+6\,{\frac{{a}^{4}Cc}{d}}+{\frac{A{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3/d*B*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*B*a^4*sin(d*x+c)+35/8/d*B*a^4*c+7/2/d*A*a^4*c+35/8*B*a^4*x+1/4/d*B*
a^4*sin(d*x+c)*cos(d*x+c)^3+27/8/d*B*a^4*sin(d*x+c)*cos(d*x+c)+7/2*a^4*A*x+1/d*A*a^4*sin(d*x+c)*cos(d*x+c)^3+7
/2/d*A*a^4*cos(d*x+c)*sin(d*x+c)+6*a^4*C*x+2/d*a^4*C*sin(d*x+c)*cos(d*x+c)+20/3/d*a^4*C*sin(d*x+c)+1/d*a^4*C*l
n(sec(d*x+c)+tan(d*x+c))+83/15/d*A*a^4*sin(d*x+c)+34/15/d*A*sin(d*x+c)*cos(d*x+c)^2*a^4+6/d*C*a^4*c+1/5/d*A*a^
4*sin(d*x+c)*cos(d*x+c)^4+1/3/d*C*sin(d*x+c)*cos(d*x+c)^2*a^4

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Maxima [A]  time = 0.967566, size = 448, normalized size = 1.99 \begin{align*} \frac{32 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} - 960 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} + 60 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 480 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 640 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 720 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 480 \,{\left (d x + c\right )} B a^{4} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 480 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 1920 \,{\left (d x + c\right )} C a^{4} + 240 \, 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 )} + 480 \, A a^{4} \sin \left (d x + c\right ) + 1920 \, B a^{4} \sin \left (d x + c\right ) + 2880 \, C a^{4} \sin \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c
))*A*a^4 + 60*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 480*(2*d*x + 2*c + sin(2*d*x + 2
*c))*A*a^4 - 640*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x
+ 2*c))*B*a^4 + 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 480*(d*x + c)*B*a^4 - 160*(sin(d*x + c)^3 - 3*sin
(d*x + c))*C*a^4 + 480*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 1920*(d*x + c)*C*a^4 + 240*C*a^4*(log(sin(d*x
+ c) + 1) - log(sin(d*x + c) - 1)) + 480*A*a^4*sin(d*x + c) + 1920*B*a^4*sin(d*x + c) + 2880*C*a^4*sin(d*x + c
))/d

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Fricas [A]  time = 0.561305, size = 408, normalized size = 1.81 \begin{align*} \frac{15 \,{\left (28 \, A + 35 \, B + 48 \, C\right )} a^{4} d x + 60 \, C a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 60 \, C a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 8 \,{\left (34 \, A + 20 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \,{\left (28 \, A + 27 \, B + 16 \, C\right )} a^{4} \cos \left (d x + c\right ) + 8 \,{\left (83 \, A + 100 \, B + 100 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(28*A + 35*B + 48*C)*a^4*d*x + 60*C*a^4*log(sin(d*x + c) + 1) - 60*C*a^4*log(-sin(d*x + c) + 1) + (2
4*A*a^4*cos(d*x + c)^4 + 30*(4*A + B)*a^4*cos(d*x + c)^3 + 8*(34*A + 20*B + 5*C)*a^4*cos(d*x + c)^2 + 15*(28*A
+ 27*B + 16*C)*a^4*cos(d*x + c) + 8*(83*A + 100*B + 100*C)*a^4)*sin(d*x + c))/d

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

[Out]

Timed out

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Giac [A]  time = 1.34333, size = 455, normalized size = 2.02 \begin{align*} \frac{120 \, C a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 120 \, C a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 15 \,{\left (28 \, A a^{4} + 35 \, B a^{4} + 48 \, C a^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (420 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 600 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 1960 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 2450 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 2720 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4720 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3160 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3950 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3680 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1080 \, 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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(120*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 120*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 15*(28*A*
a^4 + 35*B*a^4 + 48*C*a^4)*(d*x + c) + 2*(420*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a^4*tan(1/2*d*x + 1/2*c)^9
+ 600*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 1960*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 2450*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 2
720*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 3584*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 472
0*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 3160*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 3950*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 3680*
C*a^4*tan(1/2*d*x + 1/2*c)^3 + 1500*A*a^4*tan(1/2*d*x + 1/2*c) + 1395*B*a^4*tan(1/2*d*x + 1/2*c) + 1080*C*a^4*
tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d