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

Optimal. Leaf size=250 \[ \frac{\left (a^2 b^2 (56 A+85 C)+2 a^4 (4 A+5 C)+6 A b^4\right ) \sin (c+d x)}{15 d}+\frac{a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{30 d}+\frac{\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{15 d}+\frac{1}{2} a b x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac{A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}+\frac{b^4 C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*x)/2 + (b^4*C*ArcTanh[Sin[c + d*x]])/d + ((6*A*b^4 + 2*a^4*(4*A + 5*C
) + a^2*b^2*(56*A + 85*C))*Sin[c + d*x])/(15*d) + (a*b*(6*A*b^2 + a^2*(29*A + 40*C))*Cos[c + d*x]*Sin[c + d*x]
)/(30*d) + ((3*A*b^2 + a^2*(4*A + 5*C))*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + (A*b*Cos[
c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) + (A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*Sin[c + d*x])
/(5*d)

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Rubi [A]  time = 0.884453, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4094, 4074, 4047, 8, 4045, 3770} \[ \frac{\left (a^2 b^2 (56 A+85 C)+2 a^4 (4 A+5 C)+6 A b^4\right ) \sin (c+d x)}{15 d}+\frac{a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \sin (c+d x) \cos (c+d x)}{30 d}+\frac{\left (a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{15 d}+\frac{1}{2} a b x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac{A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^4}{5 d}+\frac{b^4 C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*x)/2 + (b^4*C*ArcTanh[Sin[c + d*x]])/d + ((6*A*b^4 + 2*a^4*(4*A + 5*C
) + a^2*b^2*(56*A + 85*C))*Sin[c + d*x])/(15*d) + (a*b*(6*A*b^2 + a^2*(29*A + 40*C))*Cos[c + d*x]*Sin[c + d*x]
)/(30*d) + ((3*A*b^2 + a^2*(4*A + 5*C))*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + (A*b*Cos[
c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) + (A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^4*Sin[c + d*x])
/(5*d)

Rule 4095

Int[((A_.) + 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] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - 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, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && 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 4074

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[(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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, 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 ^5(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \left (4 A b+a (4 A+5 C) \sec (c+d x)+5 b C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (4 \left (3 A b^2+a^2 (4 A+5 C)\right )+4 a b (7 A+10 C) \sec (c+d x)+20 b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{\left (3 A b^2+a^2 (4 A+5 C)\right ) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{A b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{60} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (4 b \left (6 A b^2+a^2 (29 A+40 C)\right )+4 a \left (9 b^2 (3 A+5 C)+2 a^2 (4 A+5 C)\right ) \sec (c+d x)+60 b^3 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 A b^2+a^2 (4 A+5 C)\right ) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{A b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d}-\frac{1}{120} \int \cos (c+d x) \left (-8 \left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right )-60 a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \sec (c+d x)-120 b^4 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 A b^2+a^2 (4 A+5 C)\right ) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{A b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d}-\frac{1}{120} \int \cos (c+d x) \left (-8 \left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right )-120 b^4 C \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right )\right ) \int 1 \, dx\\ &=\frac{1}{2} a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac{\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 A b^2+a^2 (4 A+5 C)\right ) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{A b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d}+\left (b^4 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac{b^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 A b^2+a^2 (4 A+5 C)\right ) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{15 d}+\frac{A b \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.83344, size = 223, normalized size = 0.89 \[ \frac{120 a b (c+d x) \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+5 a^2 \left (a^2 (5 A+4 C)+24 A b^2\right ) \sin (3 (c+d x))+240 a b \left (a^2 (A+C)+A b^2\right ) \sin (2 (c+d x))+30 \left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 A b^4\right ) \sin (c+d x)+30 a^3 A b \sin (4 (c+d x))+3 a^4 A \sin (5 (c+d x))-240 b^4 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+240 b^4 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*(c + d*x) - 240*b^4*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
240*b^4*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 30*(8*A*b^4 + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*S
in[c + d*x] + 240*a*b*(A*b^2 + a^2*(A + C))*Sin[2*(c + d*x)] + 5*a^2*(24*A*b^2 + a^2*(5*A + 4*C))*Sin[3*(c + d
*x)] + 30*a^3*A*b*Sin[4*(c + d*x)] + 3*a^4*A*Sin[5*(c + d*x)])/(240*d)

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Maple [A]  time = 0.085, size = 364, normalized size = 1.5 \begin{align*}{\frac{8\,A{a}^{4}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{15\,d}}+{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{2\,{a}^{4}C\sin \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{3}b\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,A{a}^{3}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{3}Abx}{2}}+{\frac{3\,A{a}^{3}bc}{2\,d}}+2\,{\frac{{a}^{3}bC\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,{a}^{3}bCx+2\,{\frac{{a}^{3}bCc}{d}}+2\,{\frac{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{2}{b}^{2}}{d}}+4\,{\frac{A{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+6\,{\frac{C{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{A\cos \left ( dx+c \right ) a{b}^{3}\sin \left ( dx+c \right ) }{d}}+2\,Aa{b}^{3}x+2\,{\frac{Aa{b}^{3}c}{d}}+4\,Ca{b}^{3}x+4\,{\frac{Ca{b}^{3}c}{d}}+{\frac{A{b}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{C{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/15/d*A*a^4*sin(d*x+c)+1/5/d*A*a^4*sin(d*x+c)*cos(d*x+c)^4+4/15/d*A*cos(d*x+c)^2*sin(d*x+c)*a^4+1/3/d*C*sin(d
*x+c)*cos(d*x+c)^2*a^4+2/3/d*a^4*C*sin(d*x+c)+1/d*A*a^3*b*sin(d*x+c)*cos(d*x+c)^3+3/2/d*A*a^3*b*sin(d*x+c)*cos
(d*x+c)+3/2*a^3*A*b*x+3/2/d*A*a^3*b*c+2/d*a^3*b*C*cos(d*x+c)*sin(d*x+c)+2*a^3*b*C*x+2/d*C*a^3*b*c+2/d*A*sin(d*
x+c)*cos(d*x+c)^2*a^2*b^2+4/d*A*a^2*b^2*sin(d*x+c)+6/d*C*a^2*b^2*sin(d*x+c)+2/d*A*a*b^3*cos(d*x+c)*sin(d*x+c)+
2*A*a*b^3*x+2/d*A*a*b^3*c+4*C*a*b^3*x+4/d*C*a*b^3*c+1/d*A*b^4*sin(d*x+c)+1/d*C*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.987768, size = 323, normalized size = 1.29 \begin{align*} \frac{8 \,{\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} - 40 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C 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 )} A a^{3} b + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} b - 240 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{3} + 480 \,{\left (d x + c\right )} C a b^{3} + 60 \, C b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a^{2} b^{2} \sin \left (d x + c\right ) + 120 \, A b^{4} \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+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/120*(8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 - 40*(sin(d*x + c)^3 - 3*sin(d*x + c))
*C*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3*b + 120*(2*d*x + 2*c + sin(2*d*x + 2
*c))*C*a^3*b - 240*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b^3
+ 480*(d*x + c)*C*a*b^3 + 60*C*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 720*C*a^2*b^2*sin(d*x + c
) + 120*A*b^4*sin(d*x + c))/d

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Fricas [A]  time = 0.57952, size = 473, normalized size = 1.89 \begin{align*} \frac{15 \, C b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, C b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 15 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \,{\left (A + 2 \, C\right )} a b^{3}\right )} d x +{\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \, A a^{3} b \cos \left (d x + c\right )^{3} + 4 \,{\left (4 \, A + 5 \, C\right )} a^{4} + 60 \,{\left (2 \, A + 3 \, C\right )} a^{2} b^{2} + 30 \, A b^{4} + 2 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{4} + 30 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*C*b^4*log(sin(d*x + c) + 1) - 15*C*b^4*log(-sin(d*x + c) + 1) + 15*((3*A + 4*C)*a^3*b + 4*(A + 2*C)*a
*b^3)*d*x + (6*A*a^4*cos(d*x + c)^4 + 30*A*a^3*b*cos(d*x + c)^3 + 4*(4*A + 5*C)*a^4 + 60*(2*A + 3*C)*a^2*b^2 +
30*A*b^4 + 2*((4*A + 5*C)*a^4 + 30*A*a^2*b^2)*cos(d*x + c)^2 + 15*((3*A + 4*C)*a^3*b + 4*A*a*b^3)*cos(d*x + c
))*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+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.28982, size = 1017, normalized size = 4.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(30*C*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*C*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 15*(3*A*a^3*
b + 4*C*a^3*b + 4*A*a*b^3 + 8*C*a*b^3)*(d*x + c) + 2*(30*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 30*C*a^4*tan(1/2*d*x +
1/2*c)^9 - 75*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 180*A*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^9 + 180*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*A*b^4*tan(1/2*d*x +
1/2*c)^9 + 40*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 80*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 30*A*a^3*b*tan(1/2*d*x + 1/2*c
)^7 - 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 480*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*C*a^2*b^2*tan(1/2*d*x +
1/2*c)^7 - 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 120*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 116*A*a^4*tan(1/2*d*x + 1/2
*c)^5 + 100*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 1080*C*a^2*b^2*tan(1/2*d*x +
1/2*c)^5 + 180*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 80*C*a^4*tan(1/2*d*x + 1/2*c)
^3 + 30*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 480*A*a^2*b^2*tan(1/2*d*x + 1/2*
c)^3 + 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*b^4*tan(1/2*d*x + 1/2
*c)^3 + 30*A*a^4*tan(1/2*d*x + 1/2*c) + 30*C*a^4*tan(1/2*d*x + 1/2*c) + 75*A*a^3*b*tan(1/2*d*x + 1/2*c) + 60*C
*a^3*b*tan(1/2*d*x + 1/2*c) + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*A*a
*b^3*tan(1/2*d*x + 1/2*c) + 30*A*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d