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

Optimal. Leaf size=184 \[ \frac{9 b^2 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^3}-\frac{3 b^2 \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{8 d^3}-\frac{3 b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{8 d^3}+\frac{9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^3}-\frac{3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac{\sin ^3(a+b x)}{2 d (c+d x)^2} \]

[Out]

(9*b^2*CosIntegral[(3*b*c)/d + 3*b*x]*Sin[3*a - (3*b*c)/d])/(8*d^3) - (3*b^2*CosIntegral[(b*c)/d + b*x]*Sin[a
- (b*c)/d])/(8*d^3) - (3*b*Cos[a + b*x]*Sin[a + b*x]^2)/(2*d^2*(c + d*x)) - Sin[a + b*x]^3/(2*d*(c + d*x)^2) -
(3*b^2*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(8*d^3) + (9*b^2*Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)
/d + 3*b*x])/(8*d^3)

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Rubi [A]  time = 0.353643, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3314, 3303, 3299, 3302, 3312} \[ \frac{9 b^2 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^3}-\frac{3 b^2 \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{8 d^3}-\frac{3 b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{8 d^3}+\frac{9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^3}-\frac{3 b \sin ^2(a+b x) \cos (a+b x)}{2 d^2 (c+d x)}-\frac{\sin ^3(a+b x)}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(c + d*x)^3,x]

[Out]

(9*b^2*CosIntegral[(3*b*c)/d + 3*b*x]*Sin[3*a - (3*b*c)/d])/(8*d^3) - (3*b^2*CosIntegral[(b*c)/d + b*x]*Sin[a
- (b*c)/d])/(8*d^3) - (3*b*Cos[a + b*x]*Sin[a + b*x]^2)/(2*d^2*(c + d*x)) - Sin[a + b*x]^3/(2*d*(c + d*x)^2) -
(3*b^2*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(8*d^3) + (9*b^2*Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)
/d + 3*b*x])/(8*d^3)

Rule 3314

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

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]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin{align*} \int \frac{\sin ^3(a+b x)}{(c+d x)^3} \, dx &=-\frac{3 b \cos (a+b x) \sin ^2(a+b x)}{2 d^2 (c+d x)}-\frac{\sin ^3(a+b x)}{2 d (c+d x)^2}+\frac{\left (3 b^2\right ) \int \frac{\sin (a+b x)}{c+d x} \, dx}{d^2}-\frac{\left (9 b^2\right ) \int \frac{\sin ^3(a+b x)}{c+d x} \, dx}{2 d^2}\\ &=-\frac{3 b \cos (a+b x) \sin ^2(a+b x)}{2 d^2 (c+d x)}-\frac{\sin ^3(a+b x)}{2 d (c+d x)^2}-\frac{\left (9 b^2\right ) \int \left (\frac{3 \sin (a+b x)}{4 (c+d x)}-\frac{\sin (3 a+3 b x)}{4 (c+d x)}\right ) \, dx}{2 d^2}+\frac{\left (3 b^2 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{d^2}+\frac{\left (3 b^2 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{d^2}\\ &=\frac{3 b^2 \text{Ci}\left (\frac{b c}{d}+b x\right ) \sin \left (a-\frac{b c}{d}\right )}{d^3}-\frac{3 b \cos (a+b x) \sin ^2(a+b x)}{2 d^2 (c+d x)}-\frac{\sin ^3(a+b x)}{2 d (c+d x)^2}+\frac{3 b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{d^3}+\frac{\left (9 b^2\right ) \int \frac{\sin (3 a+3 b x)}{c+d x} \, dx}{8 d^2}-\frac{\left (27 b^2\right ) \int \frac{\sin (a+b x)}{c+d x} \, dx}{8 d^2}\\ &=\frac{3 b^2 \text{Ci}\left (\frac{b c}{d}+b x\right ) \sin \left (a-\frac{b c}{d}\right )}{d^3}-\frac{3 b \cos (a+b x) \sin ^2(a+b x)}{2 d^2 (c+d x)}-\frac{\sin ^3(a+b x)}{2 d (c+d x)^2}+\frac{3 b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{d^3}+\frac{\left (9 b^2 \cos \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}{8 d^2}-\frac{\left (27 b^2 \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{8 d^2}+\frac{\left (9 b^2 \sin \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}{8 d^2}-\frac{\left (27 b^2 \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{8 d^2}\\ &=\frac{9 b^2 \text{Ci}\left (\frac{3 b c}{d}+3 b x\right ) \sin \left (3 a-\frac{3 b c}{d}\right )}{8 d^3}-\frac{3 b^2 \text{Ci}\left (\frac{b c}{d}+b x\right ) \sin \left (a-\frac{b c}{d}\right )}{8 d^3}-\frac{3 b \cos (a+b x) \sin ^2(a+b x)}{2 d^2 (c+d x)}-\frac{\sin ^3(a+b x)}{2 d (c+d x)^2}-\frac{3 b^2 \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{8 d^3}+\frac{9 b^2 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{8 d^3}\\ \end{align*}

Mathematica [A]  time = 0.79574, size = 221, normalized size = 1.2 \[ \frac{6 b^2 (c+d x)^2 \left (3 \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )-\sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )-\cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )+3 \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )\right )-6 d \cos (b x) (b \cos (a) (c+d x)+d \sin (a))+2 d \cos (3 b x) (3 b \cos (3 a) (c+d x)+d \sin (3 a))+6 d \sin (b x) (b \sin (a) (c+d x)-d \cos (a))+2 d \sin (3 b x) (d \cos (3 a)-3 b \sin (3 a) (c+d x))}{16 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 313, normalized size = 1.7 \begin{align*}{\frac{1}{b} \left ( -{\frac{{b}^{3}}{12} \left ( -{\frac{3\,\sin \left ( 3\,bx+3\,a \right ) }{2\, \left ( \left ( bx+a \right ) d-da+cb \right ) ^{2}d}}+{\frac{3}{2\,d} \left ( -3\,{\frac{\cos \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 ) \cos \left ( 3\,{\frac{-da+cb}{d}} \right ) }-3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-da+cb}{d}} \right ) \sin \left ( 3\,{\frac{-da+cb}{d}} \right ) } \right ) } \right ) } \right ) }+{\frac{3\,{b}^{3}}{4} \left ( -{\frac{\sin \left ( bx+a \right ) }{2\, \left ( \left ( bx+a \right ) d-da+cb \right ) ^{2}d}}+{\frac{1}{2\,d} \left ( -{\frac{\cos \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 ) \cos \left ({\frac{-da+cb}{d}} \right ) }-{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-da+cb}{d}} \right ) \sin \left ({\frac{-da+cb}{d}} \right ) } \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)^3,x)

[Out]

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

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Maxima [C]  time = 1.97265, size = 454, normalized size = 2.47 \begin{align*} \frac{b^{3}{\left (-3 i \, E_{3}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + 3 i \, E_{3}\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^{3}{\left (i \, E_{3}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - i \, E_{3}\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^{3}{\left (E_{3}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\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^{3}{\left (E_{3}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{3}\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^{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(sin(b*x+a)^3/(d*x+c)^3,x, algorithm="maxima")

[Out]

1/8*(b^3*(-3*I*exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 3*I*exp_integral_e(3, -(I*b*c + I*(b*x +
a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^3*(I*exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - I
*exp_integral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) - 3*b^3*(exp_integral_e(3,
(I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(3, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/
d) + b^3*(exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) + exp_integral_e(3, -(3*I*b*c + 3*I*(b*x
+ a)*d - 3*I*a*d)/d))*sin(-3*(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 [B]  time = 1.97958, size = 919, normalized size = 4.99 \begin{align*} \frac{24 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} + 18 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \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^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac{b c - a d}{d}\right ) \operatorname{Si}\left (\frac{b d x + b c}{d}\right ) - 24 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) + 8 \,{\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sin \left (b x + a\right ) - 3 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{b d x + b c}{d}\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (-\frac{b d x + b c}{d}\right )\right )} \sin \left (-\frac{b c - a d}{d}\right ) + 9 \,{\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname{Ci}\left (\frac{3 \,{\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{3 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right )}{16 \,{\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(sin(b*x+a)^3/(d*x+c)^3,x, algorithm="fricas")

[Out]

1/16*(24*(b*d^2*x + b*c*d)*cos(b*x + a)^3 + 18*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-3*(b*c - a*d)/d)*sin
_integral(3*(b*d*x + b*c)/d) - 6*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-(b*c - a*d)/d)*sin_integral((b*d*x
+ b*c)/d) - 24*(b*d^2*x + b*c*d)*cos(b*x + a) + 8*(d^2*cos(b*x + a)^2 - d^2)*sin(b*x + a) - 3*((b^2*d^2*x^2 +
2*b^2*c*d*x + b^2*c^2)*cos_integral((b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-(b
*d*x + b*c)/d))*sin(-(b*c - a*d)/d) + 9*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(3*(b*d*x + b*c)/d)
+ (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-3*(b*d*x + b*c)/d))*sin(-3*(b*c - a*d)/d))/(d^5*x^2 + 2
*c*d^4*x + c^2*d^3)

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

[Out]

Exception raised: NotImplementedError