3.108 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=169 \[ \frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac{1}{8} a^3 x (15 A+28 C)+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[Out]

(a^3*(15*A + 28*C)*x)/8 + (a^3*C*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(3*A + 4*C)*Sin[c + d*x])/(8*d) + (A*Cos[c
+ d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x
])/(4*a*d) + ((5*A + 4*C)*Cos[c + d*x]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(8*d)

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Rubi [A]  time = 0.410138, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4087, 4017, 3996, 3770} \[ \frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+\frac{A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac{1}{8} a^3 x (15 A+28 C)+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(15*A + 28*C)*x)/8 + (a^3*C*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(3*A + 4*C)*Sin[c + d*x])/(8*d) + (A*Cos[c
+ d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x
])/(4*a*d) + ((5*A + 4*C)*Cos[c + d*x]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(8*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 ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (3 a A+4 a C \sec (c+d x)) \, dx}{4 a}\\ &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (3 a^2 (5 A+4 C)+12 a^2 C \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (3 A+4 C)+24 a^3 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}-\frac{\int \left (-3 a^4 (15 A+28 C)-24 a^4 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac{1}{8} a^3 (15 A+28 C) x+\frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}+\left (a^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^3 (15 A+28 C) x+\frac{a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac{(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}\\ \end{align*}

Mathematica [A]  time = 0.304805, size = 124, normalized size = 0.73 \[ \frac{a^3 \left (8 (13 A+12 C) \sin (c+d x)+8 (4 A+C) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+A \sin (4 (c+d x))+60 A d x-32 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+32 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+112 C d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(60*A*d*x + 112*C*d*x - 32*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*C*Log[Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]] + 8*(13*A + 12*C)*Sin[c + d*x] + 8*(4*A + C)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c + d*x)] + A*Sin[4*(c
+ d*x)]))/(32*d)

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Maple [A]  time = 0.095, size = 175, normalized size = 1. \begin{align*}{\frac{A{a}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,A{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}Ax}{8}}+{\frac{15\,A{a}^{3}c}{8\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{3}Cx}{2}}+{\frac{7\,{a}^{3}Cc}{2\,d}}+{\frac{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{3}}{d}}+3\,{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}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)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)

[Out]

1/4/d*A*a^3*sin(d*x+c)*cos(d*x+c)^3+15/8/d*A*a^3*sin(d*x+c)*cos(d*x+c)+15/8*a^3*A*x+15/8/d*A*a^3*c+1/2/d*a^3*C
*sin(d*x+c)*cos(d*x+c)+7/2*a^3*C*x+7/2/d*C*a^3*c+1/d*A*cos(d*x+c)^2*sin(d*x+c)*a^3+3*a^3*A*sin(d*x+c)/d+3*a^3*
C*sin(d*x+c)/d+1/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.950342, size = 231, normalized size = 1.37 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} -{\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} - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 8 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 96 \,{\left (d x + c\right )} C a^{3} - 16 \, C a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 32 \, A a^{3} \sin \left (d x + c\right ) - 96 \, C a^{3} \sin \left (d x + c\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/32*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*
a^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 8*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 96*(d*x + c)*C*a^
3 - 16*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 32*A*a^3*sin(d*x + c) - 96*C*a^3*sin(d*x + c))/
d

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Fricas [A]  time = 0.532458, size = 284, normalized size = 1.68 \begin{align*} \frac{{\left (15 \, A + 28 \, C\right )} a^{3} d x + 4 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, A a^{3} \cos \left (d x + c\right )^{2} +{\left (15 \, A + 4 \, C\right )} a^{3} \cos \left (d x + c\right ) + 24 \,{\left (A + C\right )} a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/8*((15*A + 28*C)*a^3*d*x + 4*C*a^3*log(sin(d*x + c) + 1) - 4*C*a^3*log(-sin(d*x + c) + 1) + (2*A*a^3*cos(d*x
+ c)^3 + 8*A*a^3*cos(d*x + c)^2 + (15*A + 4*C)*a^3*cos(d*x + c) + 24*(A + C)*a^3)*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)**4*(a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.29159, size = 288, normalized size = 1.7 \begin{align*} \frac{8 \, C a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, C a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (15 \, A a^{3} + 28 \, C a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 20 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 55 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 68 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 73 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 76 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 49 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 28 \, C a^{3} \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 )}^{4}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/8*(8*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (15*A*a^3 + 28*
C*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 20*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 55*A*a^3*tan(1/2*d*x
+ 1/2*c)^5 + 68*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 73*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 76*C*a^3*tan(1/2*d*x + 1/2*c
)^3 + 49*A*a^3*tan(1/2*d*x + 1/2*c) + 28*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d