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

Optimal. Leaf size=235 \[ -\frac{i e^{i a} (c+d x) \sqrt [3]{-\frac{i b}{(c+d x)^3}} (d e-c f) \text{Gamma}\left (-\frac{1}{3},-\frac{i b}{(c+d x)^3}\right )}{6 d^2}+\frac{i e^{-i a} (c+d x) \sqrt [3]{\frac{i b}{(c+d x)^3}} (d e-c f) \text{Gamma}\left (-\frac{1}{3},\frac{i b}{(c+d x)^3}\right )}{6 d^2}-\frac{i e^{i a} f (c+d x)^2 \left (-\frac{i b}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{6 d^2}+\frac{i e^{-i a} f (c+d x)^2 \left (\frac{i b}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},\frac{i b}{(c+d x)^3}\right )}{6 d^2} \]

[Out]

((-I/6)*E^(I*a)*f*(((-I)*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2*Gamma[-2/3, ((-I)*b)/(c + d*x)^3])/d^2 + ((I/6)*f*(
(I*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2*Gamma[-2/3, (I*b)/(c + d*x)^3])/(d^2*E^(I*a)) - ((I/6)*E^(I*a)*(d*e - c*f
)*(((-I)*b)/(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, ((-I)*b)/(c + d*x)^3])/d^2 + ((I/6)*(d*e - c*f)*((I*b)/(c
+ d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, (I*b)/(c + d*x)^3])/(d^2*E^(I*a))

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Rubi [A]  time = 0.14581, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {3433, 3365, 2208, 3423, 2218} \[ -\frac{i e^{i a} (c+d x) \sqrt [3]{-\frac{i b}{(c+d x)^3}} (d e-c f) \text{Gamma}\left (-\frac{1}{3},-\frac{i b}{(c+d x)^3}\right )}{6 d^2}+\frac{i e^{-i a} (c+d x) \sqrt [3]{\frac{i b}{(c+d x)^3}} (d e-c f) \text{Gamma}\left (-\frac{1}{3},\frac{i b}{(c+d x)^3}\right )}{6 d^2}-\frac{i e^{i a} f (c+d x)^2 \left (-\frac{i b}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{6 d^2}+\frac{i e^{-i a} f (c+d x)^2 \left (\frac{i b}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},\frac{i b}{(c+d x)^3}\right )}{6 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/6)*E^(I*a)*f*(((-I)*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2*Gamma[-2/3, ((-I)*b)/(c + d*x)^3])/d^2 + ((I/6)*f*(
(I*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2*Gamma[-2/3, (I*b)/(c + d*x)^3])/(d^2*E^(I*a)) - ((I/6)*E^(I*a)*(d*e - c*f
)*(((-I)*b)/(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, ((-I)*b)/(c + d*x)^3])/d^2 + ((I/6)*(d*e - c*f)*((I*b)/(c
+ d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, (I*b)/(c + d*x)^3])/(d^2*E^(I*a))

Rule 3433

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Sin[c +
d*x^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]

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]

Rule 3423

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

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

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

Mathematica [B]  time = 2.27234, size = 700, normalized size = 2.98 \[ \frac{3 b f \left (\frac{1}{2} \cos (a) \left (\frac{\text{Gamma}\left (\frac{1}{3},-\frac{i b}{(c+d x)^3}\right )}{3 (c+d x) \sqrt [3]{-\frac{i b}{(c+d x)^3}}}+\frac{\text{Gamma}\left (\frac{1}{3},\frac{i b}{(c+d x)^3}\right )}{3 (c+d x) \sqrt [3]{\frac{i b}{(c+d x)^3}}}\right )+\frac{1}{2} i \sin (a) \left (\frac{\text{Gamma}\left (\frac{1}{3},-\frac{i b}{(c+d x)^3}\right )}{3 (c+d x) \sqrt [3]{-\frac{i b}{(c+d x)^3}}}-\frac{\text{Gamma}\left (\frac{1}{3},\frac{i b}{(c+d x)^3}\right )}{3 (c+d x) \sqrt [3]{\frac{i b}{(c+d x)^3}}}\right )\right )}{2 d^2}-\frac{3 b c f \left (\frac{1}{2} \cos (a) \left (\frac{\text{Gamma}\left (\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{3 (c+d x)^2 \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 )}{3 (c+d x)^2 \left (\frac{i b}{(c+d x)^3}\right )^{2/3}}\right )+\frac{1}{2} i \sin (a) \left (\frac{\text{Gamma}\left (\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{3 (c+d x)^2 \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 )}{3 (c+d x)^2 \left (\frac{i b}{(c+d x)^3}\right )^{2/3}}\right )\right )}{d^2}+\frac{3 b e \left (\frac{1}{2} \cos (a) \left (\frac{\text{Gamma}\left (\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{3 (c+d x)^2 \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 )}{3 (c+d x)^2 \left (\frac{i b}{(c+d x)^3}\right )^{2/3}}\right )+\frac{1}{2} i \sin (a) \left (\frac{\text{Gamma}\left (\frac{2}{3},-\frac{i b}{(c+d x)^3}\right )}{3 (c+d x)^2 \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 )}{3 (c+d x)^2 \left (\frac{i b}{(c+d x)^3}\right )^{2/3}}\right )\right )}{d}+\frac{f \sin (a) (d x-c) (c+d x) \cos \left (\frac{b}{(c+d x)^3}\right )}{2 d^2}+\frac{f \cos (a) (d x-c) (c+d x) \sin \left (\frac{b}{(c+d x)^3}\right )}{2 d^2}+\frac{e \sin (a) (c+d x) \cos \left (\frac{b}{(c+d x)^3}\right )}{d}+\frac{e \cos (a) (c+d x) \sin \left (\frac{b}{(c+d x)^3}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(c + d*x)*Cos[b/(c + d*x)^3]*Sin[a])/d + (f*(-c + d*x)*(c + d*x)*Cos[b/(c + d*x)^3]*Sin[a])/(2*d^2) + (3*b*
f*((Cos[a]*(Gamma[1/3, ((-I)*b)/(c + d*x)^3]/(3*(((-I)*b)/(c + d*x)^3)^(1/3)*(c + d*x)) + Gamma[1/3, (I*b)/(c
+ d*x)^3]/(3*((I*b)/(c + d*x)^3)^(1/3)*(c + d*x))))/2 + (I/2)*(Gamma[1/3, ((-I)*b)/(c + d*x)^3]/(3*(((-I)*b)/(
c + d*x)^3)^(1/3)*(c + d*x)) - Gamma[1/3, (I*b)/(c + d*x)^3]/(3*((I*b)/(c + d*x)^3)^(1/3)*(c + d*x)))*Sin[a]))
/(2*d^2) + (3*b*e*((Cos[a]*(Gamma[2/3, ((-I)*b)/(c + d*x)^3]/(3*(((-I)*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2) + Ga
mma[2/3, (I*b)/(c + d*x)^3]/(3*((I*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2)))/2 + (I/2)*(Gamma[2/3, ((-I)*b)/(c + d*
x)^3]/(3*(((-I)*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2) - Gamma[2/3, (I*b)/(c + d*x)^3]/(3*((I*b)/(c + d*x)^3)^(2/3
)*(c + d*x)^2))*Sin[a]))/d - (3*b*c*f*((Cos[a]*(Gamma[2/3, ((-I)*b)/(c + d*x)^3]/(3*(((-I)*b)/(c + d*x)^3)^(2/
3)*(c + d*x)^2) + Gamma[2/3, (I*b)/(c + d*x)^3]/(3*((I*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2)))/2 + (I/2)*(Gamma[2
/3, ((-I)*b)/(c + d*x)^3]/(3*(((-I)*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2) - Gamma[2/3, (I*b)/(c + d*x)^3]/(3*((I*
b)/(c + d*x)^3)^(2/3)*(c + d*x)^2))*Sin[a]))/d^2 + (e*(c + d*x)*Cos[a]*Sin[b/(c + d*x)^3])/d + (f*(-c + d*x)*(
c + d*x)*Cos[a]*Sin[b/(c + d*x)^3])/(2*d^2)

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Maple [F]  time = 0.21, size = 0, normalized size = 0. \begin{align*} \int \left ( fx+e \right ) \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((f*x+e)*sin(a+b/(d*x+c)^3),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (f x^{2} + 2 \, e x\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 ) + \int \frac{3 \,{\left (b d f x^{2} + 2 \, b d e x\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 )}{4 \,{\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} + \int \frac{3 \,{\left (b d f x^{2} + 2 \, b d e x\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 )}{4 \,{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(f*x^2 + 2*e*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)) + integrate(3/4*(b*d*f*x^2 + 2*b*d*e*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) + inte
grate(3/4*(b*d*f*x^2 + 2*b*d*e*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)

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Fricas [A]  time = 2.0135, size = 761, normalized size = 3.24 \begin{align*} \frac{-i \, d^{2} f \left (\frac{i \, b}{d^{3}}\right )^{\frac{2}{3}} e^{\left (-i \, a\right )} \Gamma \left (\frac{1}{3}, \frac{i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + i \, d^{2} f \left (-\frac{i \, b}{d^{3}}\right )^{\frac{2}{3}} e^{\left (i \, a\right )} \Gamma \left (\frac{1}{3}, -\frac{i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) +{\left (-2 i \, d^{2} e + 2 i \, c d f\right )} \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 ) +{\left (2 i \, d^{2} e - 2 i \, c d f\right )} \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^{2} f x^{2} + 2 \, d^{2} e x + 2 \, c d e - c^{2} f\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 )}{4 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-I*d^2*f*(I*b/d^3)^(2/3)*e^(-I*a)*gamma(1/3, I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + I*d^2*f*(-I
*b/d^3)^(2/3)*e^(I*a)*gamma(1/3, -I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + (-2*I*d^2*e + 2*I*c*d*f)*(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*I*d^2*e - 2*I*c*d*f)*(-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^2*f*x^2 + 2*d^2*e*x +
2*c*d*e - c^2*f)*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^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((f*x+e)*sin(a+b/(d*x+c)**3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )} \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((f*x+e)*sin(a+b/(d*x+c)^3),x, algorithm="giac")

[Out]

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