3.260 \(\int x^3 \sin (a+b (c+d x)^n) \, dx\)

Optimal. Leaf size=503 \[ \frac{3 i e^{i a} c^2 (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-i b (c+d x)^n\right )}{2 d^4 n}-\frac{i e^{i a} c^3 (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-i b (c+d x)^n\right )}{2 d^4 n}+\frac{i e^{-i a} c^3 (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},i b (c+d x)^n\right )}{2 d^4 n}-\frac{3 i e^{-i a} c^2 (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},i b (c+d x)^n\right )}{2 d^4 n}+\frac{i e^{i a} (c+d x)^4 \left (-i b (c+d x)^n\right )^{-4/n} \text{Gamma}\left (\frac{4}{n},-i b (c+d x)^n\right )}{2 d^4 n}-\frac{3 i e^{i a} c (c+d x)^3 \left (-i b (c+d x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-i b (c+d x)^n\right )}{2 d^4 n}+\frac{3 i e^{-i a} c (c+d x)^3 \left (i b (c+d x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},i b (c+d x)^n\right )}{2 d^4 n}-\frac{i e^{-i a} (c+d x)^4 \left (i b (c+d x)^n\right )^{-4/n} \text{Gamma}\left (\frac{4}{n},i b (c+d x)^n\right )}{2 d^4 n} \]

[Out]

((-I/2)*c^3*E^(I*a)*(c + d*x)*Gamma[n^(-1), (-I)*b*(c + d*x)^n])/(d^4*n*((-I)*b*(c + d*x)^n)^n^(-1)) + ((I/2)*
c^3*(c + d*x)*Gamma[n^(-1), I*b*(c + d*x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*x)^n)^n^(-1)) + (((3*I)/2)*c^2*E^(I*a
)*(c + d*x)^2*Gamma[2/n, (-I)*b*(c + d*x)^n])/(d^4*n*((-I)*b*(c + d*x)^n)^(2/n)) - (((3*I)/2)*c^2*(c + d*x)^2*
Gamma[2/n, I*b*(c + d*x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*x)^n)^(2/n)) - (((3*I)/2)*c*E^(I*a)*(c + d*x)^3*Gamma[
3/n, (-I)*b*(c + d*x)^n])/(d^4*n*((-I)*b*(c + d*x)^n)^(3/n)) + (((3*I)/2)*c*(c + d*x)^3*Gamma[3/n, I*b*(c + d*
x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*x)^n)^(3/n)) + ((I/2)*E^(I*a)*(c + d*x)^4*Gamma[4/n, (-I)*b*(c + d*x)^n])/(d
^4*n*((-I)*b*(c + d*x)^n)^(4/n)) - ((I/2)*(c + d*x)^4*Gamma[4/n, I*b*(c + d*x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*
x)^n)^(4/n))

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Rubi [A]  time = 0.391409, antiderivative size = 503, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3433, 3365, 2208, 3423, 2218} \[ \frac{3 i e^{i a} c^2 (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-i b (c+d x)^n\right )}{2 d^4 n}-\frac{i e^{i a} c^3 (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-i b (c+d x)^n\right )}{2 d^4 n}+\frac{i e^{-i a} c^3 (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},i b (c+d x)^n\right )}{2 d^4 n}-\frac{3 i e^{-i a} c^2 (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},i b (c+d x)^n\right )}{2 d^4 n}+\frac{i e^{i a} (c+d x)^4 \left (-i b (c+d x)^n\right )^{-4/n} \text{Gamma}\left (\frac{4}{n},-i b (c+d x)^n\right )}{2 d^4 n}-\frac{3 i e^{i a} c (c+d x)^3 \left (-i b (c+d x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-i b (c+d x)^n\right )}{2 d^4 n}+\frac{3 i e^{-i a} c (c+d x)^3 \left (i b (c+d x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},i b (c+d x)^n\right )}{2 d^4 n}-\frac{i e^{-i a} (c+d x)^4 \left (i b (c+d x)^n\right )^{-4/n} \text{Gamma}\left (\frac{4}{n},i b (c+d x)^n\right )}{2 d^4 n} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sin[a + b*(c + d*x)^n],x]

[Out]

((-I/2)*c^3*E^(I*a)*(c + d*x)*Gamma[n^(-1), (-I)*b*(c + d*x)^n])/(d^4*n*((-I)*b*(c + d*x)^n)^n^(-1)) + ((I/2)*
c^3*(c + d*x)*Gamma[n^(-1), I*b*(c + d*x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*x)^n)^n^(-1)) + (((3*I)/2)*c^2*E^(I*a
)*(c + d*x)^2*Gamma[2/n, (-I)*b*(c + d*x)^n])/(d^4*n*((-I)*b*(c + d*x)^n)^(2/n)) - (((3*I)/2)*c^2*(c + d*x)^2*
Gamma[2/n, I*b*(c + d*x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*x)^n)^(2/n)) - (((3*I)/2)*c*E^(I*a)*(c + d*x)^3*Gamma[
3/n, (-I)*b*(c + d*x)^n])/(d^4*n*((-I)*b*(c + d*x)^n)^(3/n)) + (((3*I)/2)*c*(c + d*x)^3*Gamma[3/n, I*b*(c + d*
x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*x)^n)^(3/n)) + ((I/2)*E^(I*a)*(c + d*x)^4*Gamma[4/n, (-I)*b*(c + d*x)^n])/(d
^4*n*((-I)*b*(c + d*x)^n)^(4/n)) - ((I/2)*(c + d*x)^4*Gamma[4/n, I*b*(c + d*x)^n])/(d^4*E^(I*a)*n*(I*b*(c + d*
x)^n)^(4/n))

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 x^3 \sin \left (a+b (c+d x)^n\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-c^3 \sin \left (a+b x^n\right )+3 c^2 x \sin \left (a+b x^n\right )-3 c x^2 \sin \left (a+b x^n\right )+x^3 \sin \left (a+b x^n\right )\right ) \, dx,x,c+d x\right )}{d^4}\\ &=\frac{\operatorname{Subst}\left (\int x^3 \sin \left (a+b x^n\right ) \, dx,x,c+d x\right )}{d^4}-\frac{(3 c) \operatorname{Subst}\left (\int x^2 \sin \left (a+b x^n\right ) \, dx,x,c+d x\right )}{d^4}+\frac{\left (3 c^2\right ) \operatorname{Subst}\left (\int x \sin \left (a+b x^n\right ) \, dx,x,c+d x\right )}{d^4}-\frac{c^3 \operatorname{Subst}\left (\int \sin \left (a+b x^n\right ) \, dx,x,c+d x\right )}{d^4}\\ &=\frac{i \operatorname{Subst}\left (\int e^{-i a-i b x^n} x^3 \, dx,x,c+d x\right )}{2 d^4}-\frac{i \operatorname{Subst}\left (\int e^{i a+i b x^n} x^3 \, dx,x,c+d x\right )}{2 d^4}-\frac{(3 i c) \operatorname{Subst}\left (\int e^{-i a-i b x^n} x^2 \, dx,x,c+d x\right )}{2 d^4}+\frac{(3 i c) \operatorname{Subst}\left (\int e^{i a+i b x^n} x^2 \, dx,x,c+d x\right )}{2 d^4}+\frac{\left (3 i c^2\right ) \operatorname{Subst}\left (\int e^{-i a-i b x^n} x \, dx,x,c+d x\right )}{2 d^4}-\frac{\left (3 i c^2\right ) \operatorname{Subst}\left (\int e^{i a+i b x^n} x \, dx,x,c+d x\right )}{2 d^4}-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int e^{-i a-i b x^n} \, dx,x,c+d x\right )}{2 d^4}+\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int e^{i a+i b x^n} \, dx,x,c+d x\right )}{2 d^4}\\ &=-\frac{i c^3 e^{i a} (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac{1}{n},-i b (c+d x)^n\right )}{2 d^4 n}+\frac{i c^3 e^{-i a} (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac{1}{n},i b (c+d x)^n\right )}{2 d^4 n}+\frac{3 i c^2 e^{i a} (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac{2}{n},-i b (c+d x)^n\right )}{2 d^4 n}-\frac{3 i c^2 e^{-i a} (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac{2}{n},i b (c+d x)^n\right )}{2 d^4 n}-\frac{3 i c e^{i a} (c+d x)^3 \left (-i b (c+d x)^n\right )^{-3/n} \Gamma \left (\frac{3}{n},-i b (c+d x)^n\right )}{2 d^4 n}+\frac{3 i c e^{-i a} (c+d x)^3 \left (i b (c+d x)^n\right )^{-3/n} \Gamma \left (\frac{3}{n},i b (c+d x)^n\right )}{2 d^4 n}+\frac{i e^{i a} (c+d x)^4 \left (-i b (c+d x)^n\right )^{-4/n} \Gamma \left (\frac{4}{n},-i b (c+d x)^n\right )}{2 d^4 n}-\frac{i e^{-i a} (c+d x)^4 \left (i b (c+d x)^n\right )^{-4/n} \Gamma \left (\frac{4}{n},i b (c+d x)^n\right )}{2 d^4 n}\\ \end{align*}

Mathematica [F]  time = 11.5296, size = 0, normalized size = 0. \[ \int x^3 \sin \left (a+b (c+d x)^n\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{x}^{3}\sin \left ( a+b \left ( dx+c \right ) ^{n} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*sin((d*x + c)^n*b + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*sin(a + b*(c + d*x)**n), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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