3.172 \(\int (e+f x)^2 \sin (a+b (c+d x)^3) \, dx\)

Optimal. Leaf size=280 \[ \frac{i e^{i a} f (c+d x)^2 (d e-c f) \text{Gamma}\left (\frac{2}{3},-i b (c+d x)^3\right )}{3 d^3 \left (-i b (c+d x)^3\right )^{2/3}}-\frac{i e^{-i a} f (c+d x)^2 (d e-c f) \text{Gamma}\left (\frac{2}{3},i b (c+d x)^3\right )}{3 d^3 \left (i b (c+d x)^3\right )^{2/3}}+\frac{i e^{i a} (c+d x) (d e-c f)^2 \text{Gamma}\left (\frac{1}{3},-i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{-i b (c+d x)^3}}-\frac{i e^{-i a} (c+d x) (d e-c f)^2 \text{Gamma}\left (\frac{1}{3},i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{i b (c+d x)^3}}-\frac{f^2 \cos \left (a+b (c+d x)^3\right )}{3 b d^3} \]

[Out]

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

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Rubi [A]  time = 0.276133, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {3433, 3355, 2208, 3389, 2218, 3379, 2638} \[ \frac{i e^{i a} f (c+d x)^2 (d e-c f) \text{Gamma}\left (\frac{2}{3},-i b (c+d x)^3\right )}{3 d^3 \left (-i b (c+d x)^3\right )^{2/3}}-\frac{i e^{-i a} f (c+d x)^2 (d e-c f) \text{Gamma}\left (\frac{2}{3},i b (c+d x)^3\right )}{3 d^3 \left (i b (c+d x)^3\right )^{2/3}}+\frac{i e^{i a} (c+d x) (d e-c f)^2 \text{Gamma}\left (\frac{1}{3},-i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{-i b (c+d x)^3}}-\frac{i e^{-i a} (c+d x) (d e-c f)^2 \text{Gamma}\left (\frac{1}{3},i b (c+d x)^3\right )}{6 d^3 \sqrt [3]{i b (c+d x)^3}}-\frac{f^2 \cos \left (a+b (c+d x)^3\right )}{3 b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3355

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}, x] && IGtQ[n, 2]

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 3389

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}, x] && IGtQ[n, 0]

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]

Rule 3379

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [F]  time = 40.1766, size = 0, normalized size = 0. \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^3\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.106, size = 0, normalized size = 0. \begin{align*} \int \left ( fx+e \right ) ^{2}\sin \left ( a+ \left ( dx+c \right ) ^{3}b \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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Fricas [A]  time = 1.87395, size = 774, normalized size = 2.76 \begin{align*} -\frac{2 \, d^{2} f^{2} \cos \left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right ) + \left (i \, b d^{3}\right )^{\frac{2}{3}}{\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} e^{\left (-i \, a\right )} \Gamma \left (\frac{1}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + \left (-i \, b d^{3}\right )^{\frac{2}{3}}{\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} e^{\left (i \, a\right )} \Gamma \left (\frac{1}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right ) + 2 \, \left (i \, b d^{3}\right )^{\frac{1}{3}}{\left (d^{2} e f - c d f^{2}\right )} e^{\left (-i \, a\right )} \Gamma \left (\frac{2}{3}, i \, b d^{3} x^{3} + 3 i \, b c d^{2} x^{2} + 3 i \, b c^{2} d x + i \, b c^{3}\right ) + 2 \, \left (-i \, b d^{3}\right )^{\frac{1}{3}}{\left (d^{2} e f - c d f^{2}\right )} e^{\left (i \, a\right )} \Gamma \left (\frac{2}{3}, -i \, b d^{3} x^{3} - 3 i \, b c d^{2} x^{2} - 3 i \, b c^{2} d x - i \, b c^{3}\right )}{6 \, b d^{5}} \end{align*}

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*(2*d^2*f^2*cos(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a) + (I*b*d^3)^(2/3)*(d^2*e^2 - 2*c*d*e*
f + c^2*f^2)*e^(-I*a)*gamma(1/3, I*b*d^3*x^3 + 3*I*b*c*d^2*x^2 + 3*I*b*c^2*d*x + I*b*c^3) + (-I*b*d^3)^(2/3)*(
d^2*e^2 - 2*c*d*e*f + c^2*f^2)*e^(I*a)*gamma(1/3, -I*b*d^3*x^3 - 3*I*b*c*d^2*x^2 - 3*I*b*c^2*d*x - I*b*c^3) +
2*(I*b*d^3)^(1/3)*(d^2*e*f - c*d*f^2)*e^(-I*a)*gamma(2/3, I*b*d^3*x^3 + 3*I*b*c*d^2*x^2 + 3*I*b*c^2*d*x + I*b*
c^3) + 2*(-I*b*d^3)^(1/3)*(d^2*e*f - c*d*f^2)*e^(I*a)*gamma(2/3, -I*b*d^3*x^3 - 3*I*b*c*d^2*x^2 - 3*I*b*c^2*d*
x - I*b*c^3))/(b*d^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e + f x\right )^{2} \sin{\left (a + b c^{3} + 3 b c^{2} d x + 3 b c d^{2} x^{2} + b d^{3} x^{3} \right )}\, dx \end{align*}

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]

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

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

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((d*x + c)^3*b + a), x)