3.267 \(\int x^2 (a+b \sin (c+d (f+g x)^n)) \, dx\)

Optimal. Leaf size=383 \[ \frac{i b e^{i c} f^2 (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-i d (f+g x)^n\right )}{2 g^3 n}-\frac{i b e^{-i c} f^2 (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},i d (f+g x)^n\right )}{2 g^3 n}+\frac{i b e^{i c} (f+g x)^3 \left (-i d (f+g x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-i d (f+g x)^n\right )}{2 g^3 n}-\frac{i b e^{i c} f (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-i d (f+g x)^n\right )}{g^3 n}+\frac{i b e^{-i c} f (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},i d (f+g x)^n\right )}{g^3 n}-\frac{i b e^{-i c} (f+g x)^3 \left (i d (f+g x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},i d (f+g x)^n\right )}{2 g^3 n}+\frac{a x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.344696, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {14, 3433, 3365, 2208, 3423, 2218} \[ \frac{i b e^{i c} f^2 (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},-i d (f+g x)^n\right )}{2 g^3 n}-\frac{i b e^{-i c} f^2 (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \text{Gamma}\left (\frac{1}{n},i d (f+g x)^n\right )}{2 g^3 n}+\frac{i b e^{i c} (f+g x)^3 \left (-i d (f+g x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-i d (f+g x)^n\right )}{2 g^3 n}-\frac{i b e^{i c} f (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-i d (f+g x)^n\right )}{g^3 n}+\frac{i b e^{-i c} f (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},i d (f+g x)^n\right )}{g^3 n}-\frac{i b e^{-i c} (f+g x)^3 \left (i d (f+g x)^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},i d (f+g x)^n\right )}{2 g^3 n}+\frac{a x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

Mathematica [A]  time = 12.5848, size = 403, normalized size = 1.05 \[ \frac{a x^3}{3}+\frac{i b (\cos (c)-i \sin (c)) (f+g x) \left (d^2 (f+g x)^{2 n}\right )^{-3/n} \left (f^2 (\cos (c)+i \sin (c))^2 \left (i d (f+g x)^n\right )^{3/n} \left (-i d (f+g x)^n\right )^{2/n} \text{Gamma}\left (\frac{1}{n},-i d (f+g x)^n\right )-(f+g x) \left (2 f (\cos (c)+i \sin (c))^2 \left (i d (f+g x)^n\right )^{3/n} \left (-i d (f+g x)^n\right )^{\frac{1}{n}} \text{Gamma}\left (\frac{2}{n},-i d (f+g x)^n\right )-(f+g x) \left ((\cos (c)+i \sin (c))^2 \left (i d (f+g x)^n\right )^{3/n} \text{Gamma}\left (\frac{3}{n},-i d (f+g x)^n\right )-\left (-i d (f+g x)^n\right )^{3/n} \text{Gamma}\left (\frac{3}{n},i d (f+g x)^n\right )\right )-2 f \left (i d (f+g x)^n\right )^{\frac{1}{n}} \left (-i d (f+g x)^n\right )^{3/n} \text{Gamma}\left (\frac{2}{n},i d (f+g x)^n\right )\right )-f^2 \left (i d (f+g x)^n\right )^{2/n} \left (-i d (f+g x)^n\right )^{3/n} \text{Gamma}\left (\frac{1}{n},i d (f+g x)^n\right )\right )}{2 g^3 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.118, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b\sin \left ( c+d \left ( gx+f \right ) ^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*sin(c+d*(g*x+f)^n)),x)

[Out]

int(x^2*(a+b*sin(c+d*(g*x+f)^n)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*sin(c+d*(g*x+f)^n)),x, algorithm="maxima")

[Out]

1/3*a*x^3 + b*integrate(x^2*sin((g*x + f)^n*d + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b x^{2} \sin \left ({\left (g x + f\right )}^{n} d + c\right ) + a x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*sin(c+d*(g*x+f)^n)),x, algorithm="fricas")

[Out]

integral(b*x^2*sin((g*x + f)^n*d + c) + a*x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*sin(c+d*(g*x+f)**n)),x)

[Out]

Integral(x**2*(a + b*sin(c + d*(f + g*x)**n)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*sin(c+d*(g*x+f)^n)),x, algorithm="giac")

[Out]

integrate((b*sin((g*x + f)^n*d + c) + a)*x^2, x)