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

Optimal. Leaf size=194 \[ \frac{3 \cos \left (\frac{b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac{3 \cos \left (\frac{3 b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac{3 \sin \left (\frac{b}{d}\right ) (b c-a d) \text{Si}\left (\frac{b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac{3 \sin \left (\frac{3 b}{d}\right ) (b c-a d) \text{Si}\left (\frac{3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac{(c+d x) \sin ^3\left (\frac{a+b x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.322127, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4563, 3313, 3303, 3299, 3302} \[ \frac{3 \cos \left (\frac{b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac{3 \cos \left (\frac{3 b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac{3 \sin \left (\frac{b}{d}\right ) (b c-a d) \text{Si}\left (\frac{b c-a d}{d (c+d x)}\right )}{4 d^2}-\frac{3 \sin \left (\frac{3 b}{d}\right ) (b c-a d) \text{Si}\left (\frac{3 (b c-a d)}{d (c+d x)}\right )}{4 d^2}+\frac{(c+d x) \sin ^3\left (\frac{a+b x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[(a + b*x)/(c + d*x)]^3,x]

[Out]

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

Rule 4563

Int[Sin[((e_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_))]^(n_.), x_Symbol] :> -Dist[d^(-1), Subst[Int[Sin[(b*
e)/d - (e*(b*c - a*d)*x)/d]^n/x^2, x], x, 1/(c + d*x)], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && NeQ[b*c
- a*d, 0]

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

Mathematica [C]  time = 7.72116, size = 657, normalized size = 3.39 \[ -\frac{3 \left (a c d-b c^2\right ) \left (\frac{i \left (1+e^{\frac{2 i b}{d}}\right ) \left (e^{\frac{2 i b c}{d (c+d x)}}-e^{\frac{2 i a}{c+d x}}\right ) \exp \left (-\frac{i (a d+2 b c+b d x)}{d (c+d x)}\right )}{4 (b c-a d)}-\frac{i \left (-1+e^{\frac{2 i b}{d}}\right ) \left (e^{\frac{2 i a}{c+d x}}+e^{\frac{2 i b c}{d (c+d x)}}\right ) \exp \left (-\frac{i (a d+2 b c+b d x)}{d (c+d x)}\right )}{4 (b c-a d)}\right )}{4 d}+\frac{3 \left (a c d-b c^2\right ) \left (\frac{i \left (1+e^{\frac{6 i b}{d}}\right ) \left (e^{\frac{6 i b c}{d (c+d x)}}-e^{\frac{6 i a}{c+d x}}\right ) \exp \left (-\frac{3 i (a d+2 b c+b d x)}{d (c+d x)}\right )}{12 (b c-a d)}-\frac{i \left (-1+e^{\frac{6 i b}{d}}\right ) \left (e^{\frac{6 i a}{c+d x}}+e^{\frac{6 i b c}{d (c+d x)}}\right ) \exp \left (-\frac{3 i (a d+2 b c+b d x)}{d (c+d x)}\right )}{12 (b c-a d)}\right )}{4 d}+\frac{3 (a d-b c) \left (-\cos \left (\frac{b}{d}\right ) \text{CosIntegral}\left (\frac{a d-b c}{d (c+d x)}\right )+\cos \left (\frac{3 b}{d}\right ) \text{CosIntegral}\left (\frac{3 (a d-b c)}{d (c+d x)}\right )+\sin \left (\frac{b}{d}\right ) \text{Si}\left (\frac{a d-b c}{d (c+d x)}\right )-\sin \left (\frac{3 b}{d}\right ) \text{Si}\left (\frac{3 (a d-b c)}{d (c+d x)}\right )\right )}{4 d^2}+\frac{3}{4} x \sin \left (\frac{b}{d}\right ) \cos \left (\frac{a d-b c}{d (c+d x)}\right )-\frac{1}{4} x \sin \left (\frac{3 b}{d}\right ) \cos \left (\frac{3 (a d-b c)}{d (c+d x)}\right )+\frac{3}{4} x \cos \left (\frac{b}{d}\right ) \sin \left (\frac{a d-b c}{d (c+d x)}\right )-\frac{1}{4} x \cos \left (\frac{3 b}{d}\right ) \sin \left (\frac{3 (a d-b c)}{d (c+d x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(-(b*c^2) + a*c*d)*(((I/4)*(1 + E^(((2*I)*b)/d))*(-E^(((2*I)*a)/(c + d*x)) + E^(((2*I)*b*c)/(d*(c + d*x)))
))/((b*c - a*d)*E^((I*(2*b*c + a*d + b*d*x))/(d*(c + d*x)))) - ((I/4)*(-1 + E^(((2*I)*b)/d))*(E^(((2*I)*a)/(c
+ d*x)) + E^(((2*I)*b*c)/(d*(c + d*x)))))/((b*c - a*d)*E^((I*(2*b*c + a*d + b*d*x))/(d*(c + d*x))))))/(4*d) +
(3*(-(b*c^2) + a*c*d)*(((I/12)*(1 + E^(((6*I)*b)/d))*(-E^(((6*I)*a)/(c + d*x)) + E^(((6*I)*b*c)/(d*(c + d*x)))
))/((b*c - a*d)*E^(((3*I)*(2*b*c + a*d + b*d*x))/(d*(c + d*x)))) - ((I/12)*(-1 + E^(((6*I)*b)/d))*(E^(((6*I)*a
)/(c + d*x)) + E^(((6*I)*b*c)/(d*(c + d*x)))))/((b*c - a*d)*E^(((3*I)*(2*b*c + a*d + b*d*x))/(d*(c + d*x))))))
/(4*d) + (3*x*Cos[(-(b*c) + a*d)/(d*(c + d*x))]*Sin[b/d])/4 - (x*Cos[(3*(-(b*c) + a*d))/(d*(c + d*x))]*Sin[(3*
b)/d])/4 + (3*x*Cos[b/d]*Sin[(-(b*c) + a*d)/(d*(c + d*x))])/4 - (x*Cos[(3*b)/d]*Sin[(3*(-(b*c) + a*d))/(d*(c +
d*x))])/4 + (3*(-(b*c) + a*d)*(-(Cos[b/d]*CosIntegral[(-(b*c) + a*d)/(d*(c + d*x))]) + Cos[(3*b)/d]*CosIntegr
al[(3*(-(b*c) + a*d))/(d*(c + d*x))] + Sin[b/d]*SinIntegral[(-(b*c) + a*d)/(d*(c + d*x))] - Sin[(3*b)/d]*SinIn
tegral[(3*(-(b*c) + a*d))/(d*(c + d*x))]))/(4*d^2)

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Maple [A]  time = 0.015, size = 295, normalized size = 1.5 \begin{align*} -{\frac{ad-cb}{{d}^{2}} \left ( -{\frac{{d}^{2}}{12} \left ( -3\,{\frac{1}{d}\sin \left ( 3\,{\frac{ad-cb}{d \left ( dx+c \right ) }}+3\,{\frac{b}{d}} \right ) \left ( d \left ({\frac{b}{d}}+{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) -b \right ) ^{-1}}+3\,{\frac{1}{d} \left ( -3\,{\frac{1}{d}{\it Si} \left ( 3\,{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \sin \left ( 3\,{\frac{b}{d}} \right ) }+3\,{\frac{1}{d}{\it Ci} \left ( 3\,{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \cos \left ( 3\,{\frac{b}{d}} \right ) } \right ) } \right ) }+{\frac{3\,{d}^{2}}{4} \left ( -{\frac{1}{d}\sin \left ({\frac{b}{d}}+{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \left ( d \left ({\frac{b}{d}}+{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) -b \right ) ^{-1}}+{\frac{1}{d} \left ( -{\frac{1}{d}{\it Si} \left ({\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \sin \left ({\frac{b}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ({\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \cos \left ({\frac{b}{d}} \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.47593, size = 651, normalized size = 3.36 \begin{align*} -\frac{6 \,{\left (b c - a d\right )} \sin \left (\frac{b}{d}\right ) \operatorname{Si}\left (-\frac{b c - a d}{d^{2} x + c d}\right ) - 6 \,{\left (b c - a d\right )} \sin \left (\frac{3 \, b}{d}\right ) \operatorname{Si}\left (-\frac{3 \,{\left (b c - a d\right )}}{d^{2} x + c d}\right ) + 3 \,{\left ({\left (b c - a d\right )} \operatorname{Ci}\left (\frac{3 \,{\left (b c - a d\right )}}{d^{2} x + c d}\right ) +{\left (b c - a d\right )} \operatorname{Ci}\left (-\frac{3 \,{\left (b c - a d\right )}}{d^{2} x + c d}\right )\right )} \cos \left (\frac{3 \, b}{d}\right ) - 3 \,{\left ({\left (b c - a d\right )} \operatorname{Ci}\left (\frac{b c - a d}{d^{2} x + c d}\right ) +{\left (b c - a d\right )} \operatorname{Ci}\left (-\frac{b c - a d}{d^{2} x + c d}\right )\right )} \cos \left (\frac{b}{d}\right ) - 8 \,{\left (d^{2} x -{\left (d^{2} x + c d\right )} \cos \left (\frac{b x + a}{d x + c}\right )^{2} + c d\right )} \sin \left (\frac{b x + a}{d x + c}\right )}{8 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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