3.2.83 \(\int (e+f x) \sin (a+\frac {b}{(c+d x)^3}) \, dx\) [183]

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

[Out]

-1/6*I*exp(I*a)*f*(-I*b/(d*x+c)^3)^(2/3)*(d*x+c)^2*GAMMA(-2/3,-I*b/(d*x+c)^3)/d^2+1/6*I*f*(I*b/(d*x+c)^3)^(2/3
)*(d*x+c)^2*GAMMA(-2/3,I*b/(d*x+c)^3)/d^2/exp(I*a)-1/6*I*exp(I*a)*(-c*f+d*e)*(-I*b/(d*x+c)^3)^(1/3)*(d*x+c)*GA
MMA(-1/3,-I*b/(d*x+c)^3)/d^2+1/6*I*(-c*f+d*e)*(I*b/(d*x+c)^3)^(1/3)*(d*x+c)*GAMMA(-1/3,I*b/(d*x+c)^3)/d^2/exp(
I*a)

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Rubi [A]
time = 0.10, antiderivative size = 235, normalized size of antiderivative = 1.00, 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 = {3514, 3446, 2239, 3504, 2250} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/6*I)*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 2239

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 2250

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

Rule 3446

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 3504

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 3514

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]

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(700\) vs. \(2(235)=470\).
time = 1.49, size = 700, normalized size = 2.98 \begin {gather*} \frac {e (c+d x) \cos \left (\frac {b}{(c+d x)^3}\right ) \sin (a)}{d}+\frac {f (-c+d x) (c+d x) \cos \left (\frac {b}{(c+d x)^3}\right ) \sin (a)}{2 d^2}+\frac {3 b f \left (\frac {1}{2} \cos (a) \left (\frac {\Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{3 \sqrt [3]{-\frac {i b}{(c+d x)^3}} (c+d x)}+\frac {\Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{3 \sqrt [3]{\frac {i b}{(c+d x)^3}} (c+d x)}\right )+\frac {1}{2} i \left (\frac {\Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{3 \sqrt [3]{-\frac {i b}{(c+d x)^3}} (c+d x)}-\frac {\Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{3 \sqrt [3]{\frac {i b}{(c+d x)^3}} (c+d x)}\right ) \sin (a)\right )}{2 d^2}+\frac {3 b e \left (\frac {1}{2} \cos (a) \left (\frac {\Gamma \left (\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )}{3 \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}+\frac {\Gamma \left (\frac {2}{3},\frac {i b}{(c+d x)^3}\right )}{3 \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}\right )+\frac {1}{2} i \left (\frac {\Gamma \left (\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )}{3 \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}-\frac {\Gamma \left (\frac {2}{3},\frac {i b}{(c+d x)^3}\right )}{3 \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}\right ) \sin (a)\right )}{d}-\frac {3 b c f \left (\frac {1}{2} \cos (a) \left (\frac {\Gamma \left (\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )}{3 \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}+\frac {\Gamma \left (\frac {2}{3},\frac {i b}{(c+d x)^3}\right )}{3 \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}\right )+\frac {1}{2} i \left (\frac {\Gamma \left (\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )}{3 \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}-\frac {\Gamma \left (\frac {2}{3},\frac {i b}{(c+d x)^3}\right )}{3 \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2}\right ) \sin (a)\right )}{d^2}+\frac {e (c+d x) \cos (a) \sin \left (\frac {b}{(c+d x)^3}\right )}{d}+\frac {f (-c+d x) (c+d x) \cos (a) \sin \left (\frac {b}{(c+d x)^3}\right )}{2 d^2} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \left (f x +e \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{3}}\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*x*e)*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*x*e)*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*x*e)*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 = 0.12, size = 326, normalized size = 1.39 \begin {gather*} \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 ) - 2 \, {\left (-i \, c d f + i \, d^{2} e\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 (i \, c d f - i \, d^{2} e\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} - c^{2} f + 2 \, {\left (d^{2} x + c d\right )} e\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 {gather*}

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*c*d*f + I*d^2*e)*(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*c*d*f - I*d^2*e)*(-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 - c^2*f + 2*(d^2*x
+ c*d)*e)*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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e + f x\right ) \sin {\left (a + \frac {b}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^3}\right )\,\left (e+f\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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