3.144 \(\int \frac{\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^4} \, dx\)

Optimal. Leaf size=287 \[ -\frac{b^3 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}-\frac{4 b^3 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{3 d^4}+\frac{b^3 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}+\frac{4 b^3 \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{3 d^4}+\frac{b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}+\frac{b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac{b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac{b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac{\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac{\sin (4 a+4 b x)}{24 d (c+d x)^3} \]

[Out]

-(b*Cos[2*a + 2*b*x])/(12*d^2*(c + d*x)^2) - (b*Cos[4*a + 4*b*x])/(12*d^2*(c + d*x)^2) - (b^3*Cos[2*a - (2*b*c
)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/(3*d^4) - (4*b^3*Cos[4*a - (4*b*c)/d]*CosIntegral[(4*b*c)/d + 4*b*x])/(3*
d^4) - Sin[2*a + 2*b*x]/(12*d*(c + d*x)^3) + (b^2*Sin[2*a + 2*b*x])/(6*d^3*(c + d*x)) - Sin[4*a + 4*b*x]/(24*d
*(c + d*x)^3) + (b^2*Sin[4*a + 4*b*x])/(3*d^3*(c + d*x)) + (b^3*Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2
*b*x])/(3*d^4) + (4*b^3*Sin[4*a - (4*b*c)/d]*SinIntegral[(4*b*c)/d + 4*b*x])/(3*d^4)

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Rubi [A]  time = 0.451177, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ -\frac{b^3 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}-\frac{4 b^3 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{3 d^4}+\frac{b^3 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}+\frac{4 b^3 \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{3 d^4}+\frac{b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}+\frac{b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac{b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac{b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac{\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac{\sin (4 a+4 b x)}{24 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]^3*Sin[a + b*x])/(c + d*x)^4,x]

[Out]

-(b*Cos[2*a + 2*b*x])/(12*d^2*(c + d*x)^2) - (b*Cos[4*a + 4*b*x])/(12*d^2*(c + d*x)^2) - (b^3*Cos[2*a - (2*b*c
)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/(3*d^4) - (4*b^3*Cos[4*a - (4*b*c)/d]*CosIntegral[(4*b*c)/d + 4*b*x])/(3*
d^4) - Sin[2*a + 2*b*x]/(12*d*(c + d*x)^3) + (b^2*Sin[2*a + 2*b*x])/(6*d^3*(c + d*x)) - Sin[4*a + 4*b*x]/(24*d
*(c + d*x)^3) + (b^2*Sin[4*a + 4*b*x])/(3*d^3*(c + d*x)) + (b^3*Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2
*b*x])/(3*d^4) + (4*b^3*Sin[4*a - (4*b*c)/d]*SinIntegral[(4*b*c)/d + 4*b*x])/(3*d^4)

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)^4} \, dx &=\int \left (\frac{\sin (2 a+2 b x)}{4 (c+d x)^4}+\frac{\sin (4 a+4 b x)}{8 (c+d x)^4}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\sin (4 a+4 b x)}{(c+d x)^4} \, dx+\frac{1}{4} \int \frac{\sin (2 a+2 b x)}{(c+d x)^4} \, dx\\ &=-\frac{\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac{\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac{b \int \frac{\cos (2 a+2 b x)}{(c+d x)^3} \, dx}{6 d}+\frac{b \int \frac{\cos (4 a+4 b x)}{(c+d x)^3} \, dx}{6 d}\\ &=-\frac{b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac{b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac{\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac{\sin (4 a+4 b x)}{24 d (c+d x)^3}-\frac{b^2 \int \frac{\sin (2 a+2 b x)}{(c+d x)^2} \, dx}{6 d^2}-\frac{b^2 \int \frac{\sin (4 a+4 b x)}{(c+d x)^2} \, dx}{3 d^2}\\ &=-\frac{b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac{b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac{\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac{b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac{\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac{b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac{b^3 \int \frac{\cos (2 a+2 b x)}{c+d x} \, dx}{3 d^3}-\frac{\left (4 b^3\right ) \int \frac{\cos (4 a+4 b x)}{c+d x} \, dx}{3 d^3}\\ &=-\frac{b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac{b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac{\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac{b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac{\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac{b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac{\left (4 b^3 \cos \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}{3 d^3}-\frac{\left (b^3 \cos \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}{3 d^3}+\frac{\left (4 b^3 \sin \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}{3 d^3}+\frac{\left (b^3 \sin \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}{3 d^3}\\ &=-\frac{b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac{b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac{b^3 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Ci}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}-\frac{4 b^3 \cos \left (4 a-\frac{4 b c}{d}\right ) \text{Ci}\left (\frac{4 b c}{d}+4 b x\right )}{3 d^4}-\frac{\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac{b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac{\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac{b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}+\frac{b^3 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}+\frac{4 b^3 \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{3 d^4}\\ \end{align*}

Mathematica [A]  time = 2.54981, size = 316, normalized size = 1.1 \[ -\frac{8 b^3 (c+d x)^3 \left (\cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )-\sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )\right )+32 b^3 (c+d x)^3 \left (\cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b (c+d x)}{d}\right )-\sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b (c+d x)}{d}\right )\right )+2 d \cos (2 b x) \left (\sin (2 a) \left (d^2-2 b^2 (c+d x)^2\right )+b d \cos (2 a) (c+d x)\right )+d \cos (4 b x) \left (\sin (4 a) \left (d^2-8 b^2 (c+d x)^2\right )+2 b d \cos (4 a) (c+d x)\right )-2 d \sin (2 b x) \left (\cos (2 a) \left (2 b^2 (c+d x)^2-d^2\right )+b d \sin (2 a) (c+d x)\right )-d \sin (4 b x) \left (\cos (4 a) \left (8 b^2 (c+d x)^2-d^2\right )+2 b d \sin (4 a) (c+d x)\right )}{24 d^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]^3*Sin[a + b*x])/(c + d*x)^4,x]

[Out]

-(2*d*Cos[2*b*x]*(b*d*(c + d*x)*Cos[2*a] + (d^2 - 2*b^2*(c + d*x)^2)*Sin[2*a]) + d*Cos[4*b*x]*(2*b*d*(c + d*x)
*Cos[4*a] + (d^2 - 8*b^2*(c + d*x)^2)*Sin[4*a]) - 2*d*((-d^2 + 2*b^2*(c + d*x)^2)*Cos[2*a] + b*d*(c + d*x)*Sin
[2*a])*Sin[2*b*x] - d*((-d^2 + 8*b^2*(c + d*x)^2)*Cos[4*a] + 2*b*d*(c + d*x)*Sin[4*a])*Sin[4*b*x] + 8*b^3*(c +
d*x)^3*(Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*(c + d*x))/d] - Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x)
)/d]) + 32*b^3*(c + d*x)^3*(Cos[4*a - (4*b*c)/d]*CosIntegral[(4*b*(c + d*x))/d] - Sin[4*a - (4*b*c)/d]*SinInte
gral[(4*b*(c + d*x))/d]))/(24*d^4*(c + d*x)^3)

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Maple [A]  time = 0.024, size = 404, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{4}}{8} \left ( -{\frac{2\,\sin \left ( 2\,bx+2\,a \right ) }{3\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}+{\frac{2}{3\,d} \left ( -{\frac{\cos \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}-{\frac{1}{d} \left ( -2\,{\frac{\sin \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 ) \sin \left ( 2\,{\frac{-ad+bc}{d}} \right ) }+2\,{\frac{1}{d}{\it Ci} \left ( 2\,bx+2\,a+2\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 2\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \right ) }+{\frac{{b}^{4}}{32} \left ( -{\frac{4\,\sin \left ( 4\,bx+4\,a \right ) }{3\, \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}+{\frac{4}{3\,d} \left ( -2\,{\frac{\cos \left ( 4\,bx+4\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}-2\,{\frac{1}{d} \left ( -4\,{\frac{\sin \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 ) \sin \left ( 4\,{\frac{-ad+bc}{d}} \right ) }+4\,{\frac{1}{d}{\it Ci} \left ( 4\,bx+4\,a+4\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 4\,{\frac{-ad+bc}{d}} \right ) } \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)^4,x)

[Out]

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

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Maxima [C]  time = 2.91, size = 521, normalized size = 1.82 \begin{align*} -\frac{b^{4}{\left (2 i \, E_{4}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 2 i \, E_{4}\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^{4}{\left (i \, E_{4}\left (\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{4}\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^{4}{\left (E_{4}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{4}\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^{4}{\left (E_{4}\left (\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{4}\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^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} +{\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \,{\left (b c d^{3} - a d^{4}\right )}{\left (b x + a\right )}^{2} + 3 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\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)^4,x, algorithm="maxima")

[Out]

-1/16*(b^4*(2*I*exp_integral_e(4, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) - 2*I*exp_integral_e(4, -(2*I*b*c +
2*I*(b*x + a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) + b^4*(I*exp_integral_e(4, (4*I*b*c + 4*I*(b*x + a)*d -
4*I*a*d)/d) - I*exp_integral_e(4, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*cos(-4*(b*c - a*d)/d) + 2*b^4*(ex
p_integral_e(4, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + exp_integral_e(4, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I
*a*d)/d))*sin(-2*(b*c - a*d)/d) + b^4*(exp_integral_e(4, (4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d) + exp_integr
al_e(4, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*sin(-4*(b*c - a*d)/d))/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^
2*b*c*d^3 + (b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3 - a*d^4)*(b*x + a)^2 + 3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d
^4)*(b*x + a))*b)

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Fricas [B]  time = 0.682596, size = 1256, normalized size = 4.38 \begin{align*} -\frac{4 \,{\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{4} - 3 \,{\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{2} - 8 \,{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \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^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \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^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + 4 \,{\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (\frac{4 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (-\frac{4 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) - 2 \,{\left ({\left (8 \, b^{2} d^{3} x^{2} + 16 \, b^{2} c d^{2} x + 8 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{3} - 3 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{6 \,{\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\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)^4,x, algorithm="fricas")

[Out]

-1/6*(4*(b*d^3*x + b*c*d^2)*cos(b*x + a)^4 - 3*(b*d^3*x + b*c*d^2)*cos(b*x + a)^2 - 8*(b^3*d^3*x^3 + 3*b^3*c*d
^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*sin(-4*(b*c - a*d)/d)*sin_integral(4*(b*d*x + b*c)/d) - 2*(b^3*d^3*x^3 + 3*b
^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*sin(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) + ((b^3*d^3*x^3
+ 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(2*(b*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2
+ 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-2*(b*d*x + b*c)/d))*cos(-2*(b*c - a*d)/d) + 4*((b^3*d^3*x^3 + 3*b^3*c
*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(4*(b*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c
^2*d*x + b^3*c^3)*cos_integral(-4*(b*d*x + b*c)/d))*cos(-4*(b*c - a*d)/d) - 2*((8*b^2*d^3*x^2 + 16*b^2*c*d^2*x
+ 8*b^2*c^2*d - d^3)*cos(b*x + a)^3 - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cos(b*x + a))*sin(b*x + a))
/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^3*d^4)

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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)/(d*x+c)**4,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)/(d*x+c)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError