3.161 \(\int \frac{\cos ^3(a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=235 \[ \frac{9 b^2 \sin \left (6 a-\frac{6 b c}{d}\right ) \text{CosIntegral}\left (\frac{6 b c}{d}+6 b x\right )}{16 d^3}-\frac{3 b^2 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{16 d^3}-\frac{3 b^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{16 d^3}+\frac{9 b^2 \cos \left (6 a-\frac{6 b c}{d}\right ) \text{Si}\left (\frac{6 b c}{d}+6 b x\right )}{16 d^3}-\frac{3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac{3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac{3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac{\sin (6 a+6 b x)}{64 d (c+d x)^2} \]

[Out]

(-3*b*Cos[2*a + 2*b*x])/(32*d^2*(c + d*x)) + (3*b*Cos[6*a + 6*b*x])/(32*d^2*(c + d*x)) + (9*b^2*CosIntegral[(6
*b*c)/d + 6*b*x]*Sin[6*a - (6*b*c)/d])/(16*d^3) - (3*b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/
(16*d^3) - (3*Sin[2*a + 2*b*x])/(64*d*(c + d*x)^2) + Sin[6*a + 6*b*x]/(64*d*(c + d*x)^2) - (3*b^2*Cos[2*a - (2
*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/(16*d^3) + (9*b^2*Cos[6*a - (6*b*c)/d]*SinIntegral[(6*b*c)/d + 6*b*x]
)/(16*d^3)

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Rubi [A]  time = 0.353223, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4406, 3297, 3303, 3299, 3302} \[ \frac{9 b^2 \sin \left (6 a-\frac{6 b c}{d}\right ) \text{CosIntegral}\left (\frac{6 b c}{d}+6 b x\right )}{16 d^3}-\frac{3 b^2 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{16 d^3}-\frac{3 b^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{16 d^3}+\frac{9 b^2 \cos \left (6 a-\frac{6 b c}{d}\right ) \text{Si}\left (\frac{6 b c}{d}+6 b x\right )}{16 d^3}-\frac{3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac{3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac{3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac{\sin (6 a+6 b x)}{64 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]^3*Sin[a + b*x]^3)/(c + d*x)^3,x]

[Out]

(-3*b*Cos[2*a + 2*b*x])/(32*d^2*(c + d*x)) + (3*b*Cos[6*a + 6*b*x])/(32*d^2*(c + d*x)) + (9*b^2*CosIntegral[(6
*b*c)/d + 6*b*x]*Sin[6*a - (6*b*c)/d])/(16*d^3) - (3*b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/
(16*d^3) - (3*Sin[2*a + 2*b*x])/(64*d*(c + d*x)^2) + Sin[6*a + 6*b*x]/(64*d*(c + d*x)^2) - (3*b^2*Cos[2*a - (2
*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/(16*d^3) + (9*b^2*Cos[6*a - (6*b*c)/d]*SinIntegral[(6*b*c)/d + 6*b*x]
)/(16*d^3)

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^3(a+b x) \sin ^3(a+b x)}{(c+d x)^3} \, dx &=\int \left (\frac{3 \sin (2 a+2 b x)}{32 (c+d x)^3}-\frac{\sin (6 a+6 b x)}{32 (c+d x)^3}\right ) \, dx\\ &=-\left (\frac{1}{32} \int \frac{\sin (6 a+6 b x)}{(c+d x)^3} \, dx\right )+\frac{3}{32} \int \frac{\sin (2 a+2 b x)}{(c+d x)^3} \, dx\\ &=-\frac{3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac{\sin (6 a+6 b x)}{64 d (c+d x)^2}+\frac{(3 b) \int \frac{\cos (2 a+2 b x)}{(c+d x)^2} \, dx}{32 d}-\frac{(3 b) \int \frac{\cos (6 a+6 b x)}{(c+d x)^2} \, dx}{32 d}\\ &=-\frac{3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac{3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac{3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac{\sin (6 a+6 b x)}{64 d (c+d x)^2}-\frac{\left (3 b^2\right ) \int \frac{\sin (2 a+2 b x)}{c+d x} \, dx}{16 d^2}+\frac{\left (9 b^2\right ) \int \frac{\sin (6 a+6 b x)}{c+d x} \, dx}{16 d^2}\\ &=-\frac{3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac{3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac{3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac{\sin (6 a+6 b x)}{64 d (c+d x)^2}+\frac{\left (9 b^2 \cos \left (6 a-\frac{6 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{6 b c}{d}+6 b x\right )}{c+d x} \, dx}{16 d^2}-\frac{\left (3 b^2 \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{2 b c}{d}+2 b x\right )}{c+d x} \, dx}{16 d^2}+\frac{\left (9 b^2 \sin \left (6 a-\frac{6 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{6 b c}{d}+6 b x\right )}{c+d x} \, dx}{16 d^2}-\frac{\left (3 b^2 \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{2 b c}{d}+2 b x\right )}{c+d x} \, dx}{16 d^2}\\ &=-\frac{3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac{3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}+\frac{9 b^2 \text{Ci}\left (\frac{6 b c}{d}+6 b x\right ) \sin \left (6 a-\frac{6 b c}{d}\right )}{16 d^3}-\frac{3 b^2 \text{Ci}\left (\frac{2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{16 d^3}-\frac{3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac{\sin (6 a+6 b x)}{64 d (c+d x)^2}-\frac{3 b^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{16 d^3}+\frac{9 b^2 \cos \left (6 a-\frac{6 b c}{d}\right ) \text{Si}\left (\frac{6 b c}{d}+6 b x\right )}{16 d^3}\\ \end{align*}

Mathematica [A]  time = 1.09066, size = 239, normalized size = 1.02 \[ \frac{6 b^2 (c+d x)^2 \left (6 \sin \left (6 a-\frac{6 b c}{d}\right ) \text{CosIntegral}\left (\frac{6 b (c+d x)}{d}\right )-2 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )-2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )+6 \cos \left (6 a-\frac{6 b c}{d}\right ) \text{Si}\left (\frac{6 b (c+d x)}{d}\right )\right )-3 d \cos (2 b x) (2 b \cos (2 a) (c+d x)+d \sin (2 a))+d \cos (6 b x) (6 b \cos (6 a) (c+d x)+d \sin (6 a))+3 d \sin (2 b x) (2 b \sin (2 a) (c+d x)-d \cos (2 a))+d \sin (6 b x) (d \cos (6 a)-6 b \sin (6 a) (c+d x))}{64 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]^3*Sin[a + b*x]^3)/(c + d*x)^3,x]

[Out]

(-3*d*Cos[2*b*x]*(2*b*(c + d*x)*Cos[2*a] + d*Sin[2*a]) + d*Cos[6*b*x]*(6*b*(c + d*x)*Cos[6*a] + d*Sin[6*a]) +
3*d*(-(d*Cos[2*a]) + 2*b*(c + d*x)*Sin[2*a])*Sin[2*b*x] + d*(d*Cos[6*a] - 6*b*(c + d*x)*Sin[6*a])*Sin[6*b*x] +
6*b^2*(c + d*x)^2*(6*CosIntegral[(6*b*(c + d*x))/d]*Sin[6*a - (6*b*c)/d] - 2*CosIntegral[(2*b*(c + d*x))/d]*S
in[2*a - (2*b*c)/d] - 2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d] + 6*Cos[6*a - (6*b*c)/d]*SinIntegr
al[(6*b*(c + d*x))/d]))/(64*d^3*(c + d*x)^2)

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Maple [A]  time = 0.028, size = 329, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ( -{\frac{{b}^{3}}{192} \left ( -3\,{\frac{\sin \left ( 6\,bx+6\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}+3\,{\frac{1}{d} \left ( -6\,{\frac{\cos \left ( 6\,bx+6\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-6\,{\frac{1}{d} \left ( 6\,{\frac{1}{d}{\it Si} \left ( 6\,bx+6\,a+6\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 6\,{\frac{-ad+bc}{d}} \right ) }-6\,{\frac{1}{d}{\it Ci} \left ( 6\,bx+6\,a+6\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 6\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) }+{\frac{3\,{b}^{3}}{64} \left ( -{\frac{\sin \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}+{\frac{1}{d} \left ( -2\,{\frac{\cos \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,bx+2\,a+2\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 2\,{\frac{-ad+bc}{d}} \right ) }-2\,{\frac{1}{d}{\it Ci} \left ( 2\,bx+2\,a+2\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 2\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*sin(b*x+a)^3/(d*x+c)^3,x)

[Out]

1/b*(-1/192*b^3*(-3*sin(6*b*x+6*a)/((b*x+a)*d-a*d+b*c)^2/d+3*(-6*cos(6*b*x+6*a)/((b*x+a)*d-a*d+b*c)/d-6*(6*Si(
6*b*x+6*a+6*(-a*d+b*c)/d)*cos(6*(-a*d+b*c)/d)/d-6*Ci(6*b*x+6*a+6*(-a*d+b*c)/d)*sin(6*(-a*d+b*c)/d)/d)/d)/d)+3/
64*b^3*(-sin(2*b*x+2*a)/((b*x+a)*d-a*d+b*c)^2/d+(-2*cos(2*b*x+2*a)/((b*x+a)*d-a*d+b*c)/d-2*(2*Si(2*b*x+2*a+2*(
-a*d+b*c)/d)*cos(2*(-a*d+b*c)/d)/d-2*Ci(2*b*x+2*a+2*(-a*d+b*c)/d)*sin(2*(-a*d+b*c)/d)/d)/d)/d))

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Maxima [C]  time = 2.3008, size = 454, normalized size = 1.93 \begin{align*} \frac{b^{3}{\left (-3 i \, E_{3}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + 3 i \, E_{3}\left (-\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + b^{3}{\left (i \, E_{3}\left (\frac{6 i \, b c + 6 i \,{\left (b x + a\right )} d - 6 i \, a d}{d}\right ) - i \, E_{3}\left (-\frac{6 i \, b c + 6 i \,{\left (b x + a\right )} d - 6 i \, a d}{d}\right )\right )} \cos \left (-\frac{6 \,{\left (b c - a d\right )}}{d}\right ) - 3 \, b^{3}{\left (E_{3}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{3}\left (-\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + b^{3}{\left (E_{3}\left (\frac{6 i \, b c + 6 i \,{\left (b x + a\right )} d - 6 i \, a d}{d}\right ) + E_{3}\left (-\frac{6 i \, b c + 6 i \,{\left (b x + a\right )} d - 6 i \, a d}{d}\right )\right )} \sin \left (-\frac{6 \,{\left (b c - a d\right )}}{d}\right )}{64 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} +{\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \,{\left (b c d^{2} - a d^{3}\right )}{\left (b x + a\right )}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^3/(d*x+c)^3,x, algorithm="maxima")

[Out]

1/64*(b^3*(-3*I*exp_integral_e(3, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + 3*I*exp_integral_e(3, -(2*I*b*c +
2*I*(b*x + a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) + b^3*(I*exp_integral_e(3, (6*I*b*c + 6*I*(b*x + a)*d -
6*I*a*d)/d) - I*exp_integral_e(3, -(6*I*b*c + 6*I*(b*x + a)*d - 6*I*a*d)/d))*cos(-6*(b*c - a*d)/d) - 3*b^3*(ex
p_integral_e(3, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + exp_integral_e(3, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I
*a*d)/d))*sin(-2*(b*c - a*d)/d) + b^3*(exp_integral_e(3, (6*I*b*c + 6*I*(b*x + a)*d - 6*I*a*d)/d) + exp_integr
al_e(3, -(6*I*b*c + 6*I*(b*x + a)*d - 6*I*a*d)/d))*sin(-6*(b*c - a*d)/d))/((b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a
)^2*d^3 + a^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*b)

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Fricas [A]  time = 0.705712, size = 1010, normalized size = 4.3 \begin{align*} \frac{96 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{6} - 144 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} + 48 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 18 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac{6 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{6 \,{\left (b d x + b c\right )}}{d}\right ) - 6 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) + 16 \,{\left (d^{2} \cos \left (b x + a\right )^{5} - d^{2} \cos \left (b x + a\right )^{3}\right )} \sin \left (b x + a\right ) - 3 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + 9 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{6 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (-\frac{6 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{6 \,{\left (b c - a d\right )}}{d}\right )}{32 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^3/(d*x+c)^3,x, algorithm="fricas")

[Out]

1/32*(96*(b*d^2*x + b*c*d)*cos(b*x + a)^6 - 144*(b*d^2*x + b*c*d)*cos(b*x + a)^4 + 48*(b*d^2*x + b*c*d)*cos(b*
x + a)^2 + 18*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-6*(b*c - a*d)/d)*sin_integral(6*(b*d*x + b*c)/d) - 6*
(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) + 16*(d^2*cos(b*x
+ a)^5 - d^2*cos(b*x + a)^3)*sin(b*x + a) - 3*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(2*(b*d*x + b
*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-2*(b*d*x + b*c)/d))*sin(-2*(b*c - a*d)/d) + 9*((b
^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(6*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*co
s_integral(-6*(b*d*x + b*c)/d))*sin(-6*(b*c - a*d)/d))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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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(b*x+a)**3*sin(b*x+a)**3/(d*x+c)**3,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^3/(d*x+c)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError