3.28 \(\int \frac{\cos (a+b x) \sin ^3(a+b x)}{(c+d x)^2} \, dx\)

Optimal. Leaf size=179 \[ \frac{b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}-\frac{b \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}+\frac{b \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{\sin (2 a+2 b x)}{4 d (c+d x)}+\frac{\sin (4 a+4 b x)}{8 d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.280256, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}-\frac{b \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}+\frac{b \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{\sin (2 a+2 b x)}{4 d (c+d x)}+\frac{\sin (4 a+4 b x)}{8 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]*Sin[a + b*x]^3)/(c + d*x)^2,x]

[Out]

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

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

Mathematica [A]  time = 1.29574, size = 151, normalized size = 0.84 \[ \frac{4 b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )-4 b \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b (c+d x)}{d}\right )-4 b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )+4 b \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b (c+d x)}{d}\right )-\frac{2 d \sin (2 (a+b x))}{c+d x}+\frac{d \sin (4 (a+b x))}{c+d x}}{8 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]*Sin[a + b*x]^3)/(c + d*x)^2,x]

[Out]

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

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Maple [A]  time = 0.023, size = 256, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{2}}{8} \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 ) }-{\frac{{b}^{2}}{32} \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*sin(b*x+a)^3/(d*x+c)^2,x)

[Out]

1/b*(1/8*b^2*(-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)-1/32*b^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))

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Maxima [C]  time = 1.77806, size = 406, normalized size = 2.27 \begin{align*} \frac{b^{2}{\left (-2 i \, E_{2}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + 2 i \, E_{2}\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^{2}{\left (i \, E_{2}\left (\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{2}\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^{2}{\left (E_{2}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{2}\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^{2}{\left (E_{2}\left (\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{2}\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 c d +{\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^3/(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.553995, size = 621, normalized size = 3.47 \begin{align*} \frac{2 \,{\left (b d x + b c\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 d x + b c\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 d x + b c\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\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 ) -{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{4 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\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 ) + 4 \,{\left (d \cos \left (b x + a\right )^{3} - d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{4 \,{\left (d^{3} x + c d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^3/(d*x+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)**3/(d*x+c)**2,x)

[Out]

Integral(sin(a + b*x)**3*cos(a + b*x)/(c + d*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right ) \sin \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^3/(d*x+c)^2,x, algorithm="giac")

[Out]

integrate(cos(b*x + a)*sin(b*x + a)^3/(d*x + c)^2, x)