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

Optimal. Leaf size=207 \[ \frac{a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac{(7 A+5 C) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 d}+\frac{(7 A+8 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (7 A+12 C)+\frac{a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

[Out]

(a^4*(7*A + 12*C)*x)/2 + (a^4*C*ArcTanh[Sin[c + d*x]])/d + (a^4*(7*A + 10*C)*Sin[c + d*x])/(2*d) + (a*A*Cos[c
+ d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(
5*d) + ((7*A + 5*C)*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + ((7*A + 8*C)*Cos[c + d*x]
*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(6*d)

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Rubi [A]  time = 0.551686, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4087, 4017, 3996, 3770} \[ \frac{a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac{(7 A+5 C) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{15 d}+\frac{(7 A+8 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (7 A+12 C)+\frac{a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(7*A + 12*C)*x)/2 + (a^4*C*ArcTanh[Sin[c + d*x]])/d + (a^4*(7*A + 10*C)*Sin[c + d*x])/(2*d) + (a*A*Cos[c
+ d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(
5*d) + ((7*A + 5*C)*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(15*d) + ((7*A + 8*C)*Cos[c + d*x]
*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(6*d)

Rule 4087

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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+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 (4 a A+5 a C \sec (c+d x)) \, dx}{5 a}\\ &=\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 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 (4 a^2 (7 A+5 C)+20 a^2 C \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (20 a^3 (7 A+8 C)+60 a^3 C \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac{(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (60 a^4 (7 A+10 C)+120 a^4 C \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac{a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac{(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac{\int \left (-60 a^5 (7 A+12 C)-120 a^5 C \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac{1}{2} a^4 (7 A+12 C) x+\frac{a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac{(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^4 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^4 (7 A+12 C) x+\frac{a^4 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^4 (7 A+10 C) \sin (c+d x)}{2 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{(7 A+5 C) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac{(7 A+8 C) \cos (c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.422612, size = 147, normalized size = 0.71 \[ \frac{a^4 \left (30 (49 A+54 C) \sin (c+d x)+240 (2 A+C) \sin (2 (c+d x))+145 A \sin (3 (c+d x))+30 A \sin (4 (c+d x))+3 A \sin (5 (c+d x))+840 A d x+20 C \sin (3 (c+d x))-240 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+240 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+1440 C d x\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(840*A*d*x + 1440*C*d*x - 240*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 240*C*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] + 30*(49*A + 54*C)*Sin[c + d*x] + 240*(2*A + C)*Sin[2*(c + d*x)] + 145*A*Sin[3*(c + d*x)] + 2
0*C*Sin[3*(c + d*x)] + 30*A*Sin[4*(c + d*x)] + 3*A*Sin[5*(c + d*x)]))/(240*d)

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Maple [A]  time = 0.132, size = 221, normalized size = 1.1 \begin{align*}{\frac{83\,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{34\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{15\,d}}+{\frac{C \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{3\,d}}+{\frac{20\,{a}^{4}C\sin \left ( dx+c \right ) }{3\,d}}+{\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}}+{\frac{7\,{a}^{4}Ax}{2}}+{\frac{7\,A{a}^{4}c}{2\,d}}+2\,{\frac{{a}^{4}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{d}}+6\,{a}^{4}Cx+6\,{\frac{{a}^{4}Cc}{d}}+{\frac{{a}^{4}C\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+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

83/15/d*A*a^4*sin(d*x+c)+1/5/d*A*a^4*sin(d*x+c)*cos(d*x+c)^4+34/15/d*A*cos(d*x+c)^2*sin(d*x+c)*a^4+1/3/d*C*cos
(d*x+c)^2*sin(d*x+c)*a^4+20/3/d*a^4*C*sin(d*x+c)+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)+7/2*a^4*A*x+7/2/d*A*a^4*c+2/d*a^4*C*sin(d*x+c)*cos(d*x+c)+6*a^4*C*x+6/d*C*a^4*c+1/d*a^4*C*ln(sec(d*x+c)
+tan(d*x+c))

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Maxima [A]  time = 0.954818, size = 309, normalized size = 1.49 \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} - 240 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A 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^{4} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, 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} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 480 \,{\left (d x + c\right )} C a^{4} + 60 \, 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 )} + 120 \, A a^{4} \sin \left (d x + c\right ) + 720 \, C a^{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+a*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 - 240*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*A*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 120*(2*d*x + 2*c + sin(2*d*x + 2*
c))*A*a^4 - 40*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 480*(d*x
+ c)*C*a^4 + 60*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 120*A*a^4*sin(d*x + c) + 720*C*a^4*si
n(d*x + c))/d

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Fricas [A]  time = 0.551146, size = 351, normalized size = 1.7 \begin{align*} \frac{15 \,{\left (7 \, A + 12 \, C\right )} a^{4} d x + 15 \, C a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, C a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (34 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \,{\left (7 \, A + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 2 \,{\left (83 \, A + 100 \, C\right )} a^{4}\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+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/30*(15*(7*A + 12*C)*a^4*d*x + 15*C*a^4*log(sin(d*x + c) + 1) - 15*C*a^4*log(-sin(d*x + c) + 1) + (6*A*a^4*co
s(d*x + c)^4 + 30*A*a^4*cos(d*x + c)^3 + 2*(34*A + 5*C)*a^4*cos(d*x + c)^2 + 15*(7*A + 4*C)*a^4*cos(d*x + c) +
2*(83*A + 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+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.32693, size = 335, normalized size = 1.62 \begin{align*} \frac{30 \, C a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 30 \, C a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 15 \,{\left (7 \, A a^{4} + 12 \, C a^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (105 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 150 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 490 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 680 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 896 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1180 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 790 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 920 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 375 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 270 \, 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}}}{30 \, 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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/30*(30*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*C*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 15*(7*A*a^4
+ 12*C*a^4)*(d*x + c) + 2*(105*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 150*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 490*A*a^4*tan
(1/2*d*x + 1/2*c)^7 + 680*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 896*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 1180*C*a^4*tan(1/2
*d*x + 1/2*c)^5 + 790*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 920*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 375*A*a^4*tan(1/2*d*x
+ 1/2*c) + 270*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d