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

Optimal. Leaf size=107 \[ \frac{i e^{-i a} (c+d x) \sqrt [3]{\frac{i b}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},\frac{i b}{(c+d x)^3}\right )}{6 d}-\frac{i e^{i a} (c+d x) \sqrt [3]{-\frac{i b}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{i b}{(c+d x)^3}\right )}{6 d} \]

[Out]

((-I/6)*E^(I*a)*(((-I)*b)/(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, ((-I)*b)/(c + d*x)^3])/d + ((I/6)*((I*b)/(c
+ d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, (I*b)/(c + d*x)^3])/(d*E^(I*a))

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Rubi [A]  time = 0.0267101, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3365, 2208} \[ \frac{i e^{-i a} (c+d x) \sqrt [3]{\frac{i b}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},\frac{i b}{(c+d x)^3}\right )}{6 d}-\frac{i e^{i a} (c+d x) \sqrt [3]{-\frac{i b}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{i b}{(c+d x)^3}\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^3],x]

[Out]

((-I/6)*E^(I*a)*(((-I)*b)/(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, ((-I)*b)/(c + d*x)^3])/d + ((I/6)*((I*b)/(c
+ d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, (I*b)/(c + d*x)^3])/(d*E^(I*a))

Rule 3365

Int[Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[I/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] - Dist[I/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f, n}, x]

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rubi steps

\begin{align*} \int \sin \left (a+\frac{b}{(c+d x)^3}\right ) \, dx &=\frac{1}{2} i \int e^{-i a-\frac{i b}{(c+d x)^3}} \, dx-\frac{1}{2} i \int e^{i a+\frac{i b}{(c+d x)^3}} \, dx\\ &=-\frac{i e^{i a} \sqrt [3]{-\frac{i b}{(c+d x)^3}} (c+d x) \Gamma \left (-\frac{1}{3},-\frac{i b}{(c+d x)^3}\right )}{6 d}+\frac{i e^{-i a} \sqrt [3]{\frac{i b}{(c+d x)^3}} (c+d x) \Gamma \left (-\frac{1}{3},\frac{i b}{(c+d x)^3}\right )}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.487147, size = 203, normalized size = 1.9 \[ \frac{b \cos (a) \left (\frac{\text{Gamma}\left (\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{\left (-\frac{i b}{(c+d x)^3}\right )^{2/3}}+\frac{\text{Gamma}\left (\frac{2}{3},\frac{i b}{(c+d x)^3}\right )}{\left (\frac{i b}{(c+d x)^3}\right )^{2/3}}\right )+i b \sin (a) \left (\frac{\text{Gamma}\left (\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{\left (-\frac{i b}{(c+d x)^3}\right )^{2/3}}-\frac{\text{Gamma}\left (\frac{2}{3},\frac{i b}{(c+d x)^3}\right )}{\left (\frac{i b}{(c+d x)^3}\right )^{2/3}}\right )+2 \sin (a) (c+d x)^3 \cos \left (\frac{b}{(c+d x)^3}\right )+2 \cos (a) (c+d x)^3 \sin \left (\frac{b}{(c+d x)^3}\right )}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^3],x]

[Out]

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

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Maple [F]  time = 0.101, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{3}}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 3 \, b d \int \frac{x \cos \left (\frac{a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}{2 \,{\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}\right )}}\,{d x} + 3 \, b d \int \frac{x \cos \left (\frac{a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}{2 \,{\left ({\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}\right )} \cos \left (\frac{a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )^{2} +{\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}\right )} \sin \left (\frac{a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )^{2}\right )}}\,{d x} + x \sin \left (\frac{a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b*d*integrate(1/2*x*cos((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2
*d*x + c^3))/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x) + 3*b*d*integrate(1/2*x*cos((a*d^3*
x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/((d^4*x^4 + 4*c*d^3*
x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4)*cos((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*
c*d^2*x^2 + 3*c^2*d*x + c^3))^2 + (d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4)*sin((a*d^3*x^3 + 3
*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))^2), x) + x*sin((a*d^3*x^3 +
3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))

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Fricas [B]  time = 1.86678, size = 408, normalized size = 3.81 \begin{align*} \frac{-i \, d \left (\frac{i \, b}{d^{3}}\right )^{\frac{1}{3}} e^{\left (-i \, a\right )} \Gamma \left (\frac{2}{3}, \frac{i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + i \, d \left (-\frac{i \, b}{d^{3}}\right )^{\frac{1}{3}} e^{\left (i \, a\right )} \Gamma \left (\frac{2}{3}, -\frac{i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + 2 \,{\left (d x + c\right )} \sin \left (\frac{a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-I*d*(I*b/d^3)^(1/3)*e^(-I*a)*gamma(2/3, I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + I*d*(-I*b/d^3)^
(1/3)*e^(I*a)*gamma(2/3, -I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + 2*(d*x + c)*sin((a*d^3*x^3 + 3*a*c*
d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b/(c + d*x)**3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a + b/(d*x + c)^3), x)