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

Optimal. Leaf size=145 \[ \frac{3 b \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{4 d^2}-\frac{3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{3 b \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{4 d^2}+\frac{3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{\sin ^3(a+b x)}{d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.242383, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3313, 3303, 3299, 3302} \[ \frac{3 b \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{4 d^2}-\frac{3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{3 b \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{4 d^2}+\frac{3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{\sin ^3(a+b x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(c + d*x)^2,x]

[Out]

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

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && 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{\sin ^3(a+b x)}{(c+d x)^2} \, dx &=-\frac{\sin ^3(a+b x)}{d (c+d x)}+\frac{(3 b) \int \left (\frac{\cos (a+b x)}{4 (c+d x)}-\frac{\cos (3 a+3 b x)}{4 (c+d x)}\right ) \, dx}{d}\\ &=-\frac{\sin ^3(a+b x)}{d (c+d x)}+\frac{(3 b) \int \frac{\cos (a+b x)}{c+d x} \, dx}{4 d}-\frac{(3 b) \int \frac{\cos (3 a+3 b x)}{c+d x} \, dx}{4 d}\\ &=-\frac{\sin ^3(a+b x)}{d (c+d x)}-\frac{\left (3 b \cos \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{4 d}+\frac{\left (3 b \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{4 d}+\frac{\left (3 b \sin \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{4 d}-\frac{\left (3 b \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{4 d}\\ &=\frac{3 b \cos \left (a-\frac{b c}{d}\right ) \text{Ci}\left (\frac{b c}{d}+b x\right )}{4 d^2}-\frac{3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Ci}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{\sin ^3(a+b x)}{d (c+d x)}-\frac{3 b \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{4 d^2}+\frac{3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}\\ \end{align*}

Mathematica [A]  time = 1.05066, size = 175, normalized size = 1.21 \[ \frac{3 b (c+d x) \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )-3 b (c+d x) \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )-3 b (c+d x) \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )+3 b (c+d x) \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )-3 d \sin (a) \cos (b x)+d \sin (3 a) \cos (3 b x)-3 d \cos (a) \sin (b x)+d \cos (3 a) \sin (3 b x)}{4 d^2 (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 240, normalized size = 1.7 \begin{align*}{\frac{1}{b} \left ( -{\frac{{b}^{2}}{12} \left ( -3\,{\frac{\sin \left ( 3\,bx+3\,a \right ) }{ \left ( \left ( bx+a \right ) d-da+cb \right ) d}}+3\,{\frac{1}{d} \left ( 3\,{\frac{1}{d}{\it Si} \left ( 3\,bx+3\,a+3\,{\frac{-da+cb}{d}} \right ) \sin \left ( 3\,{\frac{-da+cb}{d}} \right ) }+3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-da+cb}{d}} \right ) \cos \left ( 3\,{\frac{-da+cb}{d}} \right ) } \right ) } \right ) }+{\frac{3\,{b}^{2}}{4} \left ( -{\frac{\sin \left ( bx+a \right ) }{ \left ( \left ( bx+a \right ) d-da+cb \right ) d}}+{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ( bx+a+{\frac{-da+cb}{d}} \right ) \sin \left ({\frac{-da+cb}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-da+cb}{d}} \right ) \cos \left ({\frac{-da+cb}{d}} \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/12*b^2*(-3*sin(3*b*x+3*a)/((b*x+a)*d-d*a+c*b)/d+3*(3*Si(3*b*x+3*a+3*(-a*d+b*c)/d)*sin(3*(-a*d+b*c)/d)/
d+3*Ci(3*b*x+3*a+3*(-a*d+b*c)/d)*cos(3*(-a*d+b*c)/d)/d)/d)+3/4*b^2*(-sin(b*x+a)/((b*x+a)*d-d*a+c*b)/d+(Si(b*x+
a+(-a*d+b*c)/d)*sin((-a*d+b*c)/d)/d+Ci(b*x+a+(-a*d+b*c)/d)*cos((-a*d+b*c)/d)/d)/d))

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Maxima [C]  time = 1.81778, size = 406, normalized size = 2.8 \begin{align*} \frac{b^{2}{\left (-3 i \, E_{2}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + 3 i \, E_{2}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac{b c - a d}{d}\right ) + b^{2}{\left (i \, E_{2}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - i \, E_{2}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) - 3 \, b^{2}{\left (E_{2}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac{b c - a d}{d}\right ) + b^{2}{\left (E_{2}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{2}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right )}{8 \,{\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(sin(b*x+a)^3/(d*x+c)^2,x, algorithm="maxima")

[Out]

1/8*(b^2*(-3*I*exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 3*I*exp_integral_e(2, -(I*b*c + I*(b*x +
a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^2*(I*exp_integral_e(2, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - I
*exp_integral_e(2, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) - 3*b^2*(exp_integral_e(2,
(I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(2, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/
d) + b^2*(exp_integral_e(2, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) + exp_integral_e(2, -(3*I*b*c + 3*I*(b*x
+ a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d))/((b*c*d + (b*x + a)*d^2 - a*d^2)*b)

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Fricas [A]  time = 1.99735, size = 595, normalized size = 4.1 \begin{align*} \frac{6 \,{\left (b d x + b c\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) - 6 \,{\left (b d x + b c\right )} \sin \left (-\frac{b c - a d}{d}\right ) \operatorname{Si}\left (\frac{b d x + b c}{d}\right ) + 3 \,{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{b d x + b c}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{b d x + b c}{d}\right )\right )} \cos \left (-\frac{b c - a d}{d}\right ) - 3 \,{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{3 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + 8 \,{\left (d \cos \left (b x + a\right )^{2} - d\right )} \sin \left (b x + a\right )}{8 \,{\left (d^{3} x + c d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(6*(b*d*x + b*c)*sin(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d) - 6*(b*d*x + b*c)*sin(-(b*c - a*d)/
d)*sin_integral((b*d*x + b*c)/d) + 3*((b*d*x + b*c)*cos_integral((b*d*x + b*c)/d) + (b*d*x + b*c)*cos_integral
(-(b*d*x + b*c)/d))*cos(-(b*c - a*d)/d) - 3*((b*d*x + b*c)*cos_integral(3*(b*d*x + b*c)/d) + (b*d*x + b*c)*cos
_integral(-3*(b*d*x + b*c)/d))*cos(-3*(b*c - a*d)/d) + 8*(d*cos(b*x + a)^2 - d)*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 )}}{\left (c + d x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out