3.2.82 \(\int (e+f x)^2 \sin (a+\frac {b}{(c+d x)^3}) \, dx\) [182]

Optimal. Leaf size=330 \[ -\frac {b f^2 \cos (a) \text {Ci}\left (\frac {b}{(c+d x)^3}\right )}{3 d^3}-\frac {i e^{i a} f (d e-c f) \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2 \Gamma \left (-\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )}{3 d^3}+\frac {i e^{-i a} f (d e-c f) \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2 \Gamma \left (-\frac {2}{3},\frac {i b}{(c+d x)^3}\right )}{3 d^3}-\frac {i e^{i a} (d e-c f)^2 \sqrt [3]{-\frac {i b}{(c+d x)^3}} (c+d x) \Gamma \left (-\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{6 d^3}+\frac {i e^{-i a} (d e-c f)^2 \sqrt [3]{\frac {i b}{(c+d x)^3}} (c+d x) \Gamma \left (-\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{6 d^3}+\frac {f^2 (c+d x)^3 \sin \left (a+\frac {b}{(c+d x)^3}\right )}{3 d^3}+\frac {b f^2 \sin (a) \text {Si}\left (\frac {b}{(c+d x)^3}\right )}{3 d^3} \]

[Out]

-1/3*b*f^2*Ci(b/(d*x+c)^3)*cos(a)/d^3-1/3*I*exp(I*a)*f*(-c*f+d*e)*(-I*b/(d*x+c)^3)^(2/3)*(d*x+c)^2*GAMMA(-2/3,
-I*b/(d*x+c)^3)/d^3+1/3*I*f*(-c*f+d*e)*(I*b/(d*x+c)^3)^(2/3)*(d*x+c)^2*GAMMA(-2/3,I*b/(d*x+c)^3)/d^3/exp(I*a)-
1/6*I*exp(I*a)*(-c*f+d*e)^2*(-I*b/(d*x+c)^3)^(1/3)*(d*x+c)*GAMMA(-1/3,-I*b/(d*x+c)^3)/d^3+1/6*I*(-c*f+d*e)^2*(
I*b/(d*x+c)^3)^(1/3)*(d*x+c)*GAMMA(-1/3,I*b/(d*x+c)^3)/d^3/exp(I*a)+1/3*b*f^2*Si(b/(d*x+c)^3)*sin(a)/d^3+1/3*f
^2*(d*x+c)^3*sin(a+b/(d*x+c)^3)/d^3

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Rubi [A]
time = 0.20, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3514, 3446, 2239, 3504, 2250, 3460, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {i e^{i a} f (c+d x)^2 \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (d e-c f) \text {Gamma}\left (-\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )}{3 d^3}+\frac {i e^{-i a} f (c+d x)^2 \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (d e-c f) \text {Gamma}\left (-\frac {2}{3},\frac {i b}{(c+d x)^3}\right )}{3 d^3}-\frac {i e^{i a} (c+d x) \sqrt [3]{-\frac {i b}{(c+d x)^3}} (d e-c f)^2 \text {Gamma}\left (-\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{6 d^3}+\frac {i e^{-i a} (c+d x) \sqrt [3]{\frac {i b}{(c+d x)^3}} (d e-c f)^2 \text {Gamma}\left (-\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{6 d^3}-\frac {b f^2 \cos (a) \text {CosIntegral}\left (\frac {b}{(c+d x)^3}\right )}{3 d^3}+\frac {b f^2 \sin (a) \text {Si}\left (\frac {b}{(c+d x)^3}\right )}{3 d^3}+\frac {f^2 (c+d x)^3 \sin \left (a+\frac {b}{(c+d x)^3}\right )}{3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*Sin[a + b/(c + d*x)^3],x]

[Out]

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

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 3378

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m
+ 1))), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3380

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 3383

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 3384

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 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 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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

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Mathematica [A]
time = 1.59, size = 405, normalized size = 1.23 \begin {gather*} \frac {\frac {3 b f (d e-c f) \left (\sqrt [3]{-\frac {i b}{(c+d x)^3}} \Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^3}\right ) (\cos (a)-i \sin (a))+\sqrt [3]{\frac {i b}{(c+d x)^3}} \Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^3}\right ) (\cos (a)+i \sin (a))\right )}{2 \sqrt [3]{\frac {b^2}{(c+d x)^6}} (c+d x)}+\frac {3 b (d e-c f)^2 \left (\left (-\frac {i b}{(c+d x)^3}\right )^{2/3} \Gamma \left (\frac {2}{3},\frac {i b}{(c+d x)^3}\right ) (\cos (a)-i \sin (a))+\left (\frac {i b}{(c+d x)^3}\right )^{2/3} \Gamma \left (\frac {2}{3},-\frac {i b}{(c+d x)^3}\right ) (\cos (a)+i \sin (a))\right )}{2 \left (\frac {b^2}{(c+d x)^6}\right )^{2/3} (c+d x)^2}+(c+d x) \left (c^2 f^2-c d f (3 e+f x)+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right ) \cos \left (\frac {b}{(c+d x)^3}\right ) \sin (a)+(c+d x) \left (c^2 f^2-c d f (3 e+f x)+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right ) \cos (a) \sin \left (\frac {b}{(c+d x)^3}\right )-b f^2 \left (\cos (a) \text {Ci}\left (\frac {b}{(c+d x)^3}\right )-\sin (a) \text {Si}\left (\frac {b}{(c+d x)^3}\right )\right )}{3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^2*Sin[a + b/(c + d*x)^3],x]

[Out]

((3*b*f*(d*e - c*f)*((((-I)*b)/(c + d*x)^3)^(1/3)*Gamma[1/3, (I*b)/(c + d*x)^3]*(Cos[a] - I*Sin[a]) + ((I*b)/(
c + d*x)^3)^(1/3)*Gamma[1/3, ((-I)*b)/(c + d*x)^3]*(Cos[a] + I*Sin[a])))/(2*(b^2/(c + d*x)^6)^(1/3)*(c + d*x))
+ (3*b*(d*e - c*f)^2*((((-I)*b)/(c + d*x)^3)^(2/3)*Gamma[2/3, (I*b)/(c + d*x)^3]*(Cos[a] - I*Sin[a]) + ((I*b)
/(c + d*x)^3)^(2/3)*Gamma[2/3, ((-I)*b)/(c + d*x)^3]*(Cos[a] + I*Sin[a])))/(2*(b^2/(c + d*x)^6)^(2/3)*(c + d*x
)^2) + (c + d*x)*(c^2*f^2 - c*d*f*(3*e + f*x) + d^2*(3*e^2 + 3*e*f*x + f^2*x^2))*Cos[b/(c + d*x)^3]*Sin[a] + (
c + d*x)*(c^2*f^2 - c*d*f*(3*e + f*x) + d^2*(3*e^2 + 3*e*f*x + f^2*x^2))*Cos[a]*Sin[b/(c + d*x)^3] - b*f^2*(Co
s[a]*CosIntegral[b/(c + d*x)^3] - Sin[a]*SinIntegral[b/(c + d*x)^3]))/(3*d^3)

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \left (f x +e \right )^{2} \sin \left (a +\frac {b}{\left (d x +c \right )^{3}}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x+e)^2*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)^2*sin(a+b/(d*x+c)^3),x, algorithm="maxima")

[Out]

1/3*(f^2*x^3 + 3*f*x^2*e + 3*x*e^2)*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(1/2*(b*d*f^2*x^3 + 3*b*d*f*x^2*e + 3*b*d*x*e^2)*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) + integrate(1/2*(b*d*f^2*x^3 + 3*b*d*f*x^2*e + 3*b*d*x*e^2)*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 = 0.13, size = 489, normalized size = 1.48 \begin {gather*} -\frac {b f^{2} {\rm Ei}\left (\frac {i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) e^{\left (i \, a\right )} + b f^{2} {\rm Ei}\left (-\frac {i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) e^{\left (-i \, a\right )} + 3 \, {\left (-i \, c d^{2} f^{2} + i \, d^{3} f e\right )} \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 ) + 3 \, {\left (i \, c d^{2} f^{2} - i \, d^{3} f e\right )} \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 ) + 3 \, {\left (i \, c^{2} d f^{2} - 2 i \, c d^{2} f e + i \, d^{3} e^{2}\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 ) + 3 \, {\left (-i \, c^{2} d f^{2} + 2 i \, c d^{2} f e - i \, d^{3} e^{2}\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^{3} f^{2} x^{3} + c^{3} f^{2} + 3 \, {\left (d^{3} x + c d^{2}\right )} e^{2} + 3 \, {\left (d^{3} f x^{2} - c^{2} d f\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 )}{6 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sin(a+b/(d*x+c)^3),x, algorithm="fricas")

[Out]

-1/6*(b*f^2*Ei(I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))*e^(I*a) + b*f^2*Ei(-I*b/(d^3*x^3 + 3*c*d^2*x^2 +
3*c^2*d*x + c^3))*e^(-I*a) + 3*(-I*c*d^2*f^2 + I*d^3*f*e)*(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)) + 3*(I*c*d^2*f^2 - I*d^3*f*e)*(-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)) + 3*(I*c^2*d*f^2 - 2*I*c*d^2*f*e + I*d^3*e^2)*(I*b/d^3)^(1/3)*e^(-I*a)*gam
ma(2/3, I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + 3*(-I*c^2*d*f^2 + 2*I*c*d^2*f*e - I*d^3*e^2)*(-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^3*f^2*x^3 + c^3*f^2 + 3*(d^
3*x + c*d^2)*e^2 + 3*(d^3*f*x^2 - c^2*d*f)*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^3

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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)^2*sin(a+b/(d*x+c)^3),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*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 )}^2 \,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)^2,x)

[Out]

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

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