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

Optimal. Leaf size=231 \[ -\frac{b^2 \sin \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{d^3}-\frac{b^2 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^3}-\frac{b^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^3}-\frac{b^2 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{d^3}-\frac{b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac{b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac{\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac{\sin (4 a+4 b x)}{16 d (c+d x)^2} \]

[Out]

-(b*Cos[2*a + 2*b*x])/(4*d^2*(c + d*x)) - (b*Cos[4*a + 4*b*x])/(4*d^2*(c + d*x)) - (b^2*CosIntegral[(4*b*c)/d
+ 4*b*x]*Sin[4*a - (4*b*c)/d])/d^3 - (b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/(2*d^3) - Sin[2
*a + 2*b*x]/(8*d*(c + d*x)^2) - Sin[4*a + 4*b*x]/(16*d*(c + d*x)^2) - (b^2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2
*b*c)/d + 2*b*x])/(2*d^3) - (b^2*Cos[4*a - (4*b*c)/d]*SinIntegral[(4*b*c)/d + 4*b*x])/d^3

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Rubi [A]  time = 0.32919, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4406, 3297, 3303, 3299, 3302} \[ -\frac{b^2 \sin \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{d^3}-\frac{b^2 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^3}-\frac{b^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^3}-\frac{b^2 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{d^3}-\frac{b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac{b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac{\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac{\sin (4 a+4 b x)}{16 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[2*a + 2*b*x])/(4*d^2*(c + d*x)) - (b*Cos[4*a + 4*b*x])/(4*d^2*(c + d*x)) - (b^2*CosIntegral[(4*b*c)/d
+ 4*b*x]*Sin[4*a - (4*b*c)/d])/d^3 - (b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/(2*d^3) - Sin[2
*a + 2*b*x]/(8*d*(c + d*x)^2) - Sin[4*a + 4*b*x]/(16*d*(c + d*x)^2) - (b^2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2
*b*c)/d + 2*b*x])/(2*d^3) - (b^2*Cos[4*a - (4*b*c)/d]*SinIntegral[(4*b*c)/d + 4*b*x])/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 (a+b x)}{(c+d x)^3} \, dx &=\int \left (\frac{\sin (2 a+2 b x)}{4 (c+d x)^3}+\frac{\sin (4 a+4 b x)}{8 (c+d x)^3}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\sin (4 a+4 b x)}{(c+d x)^3} \, dx+\frac{1}{4} \int \frac{\sin (2 a+2 b x)}{(c+d x)^3} \, dx\\ &=-\frac{\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac{\sin (4 a+4 b x)}{16 d (c+d x)^2}+\frac{b \int \frac{\cos (2 a+2 b x)}{(c+d x)^2} \, dx}{4 d}+\frac{b \int \frac{\cos (4 a+4 b x)}{(c+d x)^2} \, dx}{4 d}\\ &=-\frac{b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac{b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac{\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac{\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac{b^2 \int \frac{\sin (2 a+2 b x)}{c+d x} \, dx}{2 d^2}-\frac{b^2 \int \frac{\sin (4 a+4 b x)}{c+d x} \, dx}{d^2}\\ &=-\frac{b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac{b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac{\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac{\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac{\left (b^2 \cos \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{4 b c}{d}+4 b x\right )}{c+d x} \, dx}{d^2}-\frac{\left (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}{2 d^2}-\frac{\left (b^2 \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{4 b c}{d}+4 b x\right )}{c+d x} \, dx}{d^2}-\frac{\left (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}{2 d^2}\\ &=-\frac{b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac{b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac{b^2 \text{Ci}\left (\frac{4 b c}{d}+4 b x\right ) \sin \left (4 a-\frac{4 b c}{d}\right )}{d^3}-\frac{b^2 \text{Ci}\left (\frac{2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{2 d^3}-\frac{\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac{\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac{b^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^3}-\frac{b^2 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 3.79172, size = 197, normalized size = 0.85 \[ -\frac{16 b^2 \sin \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b (c+d x)}{d}\right )+8 b^2 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )+8 b^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )+16 b^2 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b (c+d x)}{d}\right )+\frac{2 d (2 b (c+d x) \cos (2 (a+b x))+d \sin (2 (a+b x)))}{(c+d x)^2}+\frac{d (4 b (c+d x) \cos (4 (a+b x))+d \sin (4 (a+b x)))}{(c+d x)^2}}{16 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(16*b^2*CosIntegral[(4*b*(c + d*x))/d]*Sin[4*a - (4*b*c)/d] + 8*b^2*CosIntegral[(2*b*(c + d*x))/d]*Sin[2*a -
(2*b*c)/d] + (2*d*(2*b*(c + d*x)*Cos[2*(a + b*x)] + d*Sin[2*(a + b*x)]))/(c + d*x)^2 + (d*(4*b*(c + d*x)*Cos[4
*(a + b*x)] + d*Sin[4*(a + b*x)]))/(c + d*x)^2 + 8*b^2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d] + 1
6*b^2*Cos[4*a - (4*b*c)/d]*SinIntegral[(4*b*(c + d*x))/d])/(16*d^3)

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Maple [A]  time = 0.023, size = 329, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{3}}{8} \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 ) }+{\frac{{b}^{3}}{32} \left ( -2\,{\frac{\sin \left ( 4\,bx+4\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}+2\,{\frac{1}{d} \left ( -4\,{\frac{\cos \left ( 4\,bx+4\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-4\,{\frac{1}{d} \left ( 4\,{\frac{1}{d}{\it Si} \left ( 4\,bx+4\,a+4\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 4\,{\frac{-ad+bc}{d}} \right ) }-4\,{\frac{1}{d}{\it Ci} \left ( 4\,bx+4\,a+4\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 4\,{\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)/(d*x+c)^3,x)

[Out]

1/b*(1/8*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)+1/32*b^3*
(-2*sin(4*b*x+4*a)/((b*x+a)*d-a*d+b*c)^2/d+2*(-4*cos(4*b*x+4*a)/((b*x+a)*d-a*d+b*c)/d-4*(4*Si(4*b*x+4*a+4*(-a*
d+b*c)/d)*cos(4*(-a*d+b*c)/d)/d-4*Ci(4*b*x+4*a+4*(-a*d+b*c)/d)*sin(4*(-a*d+b*c)/d)/d)/d)/d))

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Maxima [C]  time = 2.2536, size = 454, normalized size = 1.97 \begin{align*} -\frac{b^{3}{\left (2 i \, E_{3}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 2 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{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{3}\left (-\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) + 2 \, 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{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{3}\left (-\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \sin \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right )}{16 \,{\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)/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/16*(b^3*(2*I*exp_integral_e(3, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) - 2*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, (4*I*b*c + 4*I*(b*x + a)*d -
4*I*a*d)/d) - I*exp_integral_e(3, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*cos(-4*(b*c - a*d)/d) + 2*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, (4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d) + exp_integr
al_e(3, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*sin(-4*(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 = 0.624906, size = 918, normalized size = 3.97 \begin{align*} -\frac{2 \, d^{2} \cos \left (b x + a\right )^{3} \sin \left (b x + a\right ) + 8 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} - 6 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 4 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{4 \,{\left (b d x + b c\right )}}{d}\right ) + 2 \,{\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 ) +{\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 ) + 2 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{4 \,{\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{4 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right )}{4 \,{\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)/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/4*(2*d^2*cos(b*x + a)^3*sin(b*x + a) + 8*(b*d^2*x + b*c*d)*cos(b*x + a)^4 - 6*(b*d^2*x + b*c*d)*cos(b*x + a
)^2 + 4*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-4*(b*c - a*d)/d)*sin_integral(4*(b*d*x + b*c)/d) + 2*(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) + ((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) + 2*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(4*(b*d*x + b*c)/d)
+ (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-4*(b*d*x + b*c)/d))*sin(-4*(b*c - a*d)/d))/(d^5*x^2 + 2*
c*d^4*x + c^2*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(d*x+c)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError