\(\int x^2 \sin (a+b (c+d x)^n) \, dx\) [261]

Optimal result

Integrand size = 16, antiderivative size = 369 \[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\frac {i c^2 e^{i a} (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i b (c+d x)^n\right )}{2 d^3 n}-\frac {i c^2 e^{-i a} (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i b (c+d x)^n\right )}{2 d^3 n}-\frac {i c e^{i a} (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i b (c+d x)^n\right )}{d^3 n}+\frac {i c e^{-i a} (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i b (c+d x)^n\right )}{d^3 n}+\frac {i e^{i a} (c+d x)^3 \left (-i b (c+d x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-i b (c+d x)^n\right )}{2 d^3 n}-\frac {i e^{-i a} (c+d x)^3 \left (i b (c+d x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},i b (c+d x)^n\right )}{2 d^3 n} \] Output:

1/2*I*c^2*exp(I*a)*(d*x+c)*GAMMA(1/n,-I*b*(d*x+c)^n)/d^3/n/((-I*b*(d*x+c)^ 
n)^(1/n))-1/2*I*c^2*(d*x+c)*GAMMA(1/n,I*b*(d*x+c)^n)/d^3/exp(I*a)/n/((I*b* 
(d*x+c)^n)^(1/n))-I*c*exp(I*a)*(d*x+c)^2*GAMMA(2/n,-I*b*(d*x+c)^n)/d^3/n/( 
(-I*b*(d*x+c)^n)^(2/n))+I*c*(d*x+c)^2*GAMMA(2/n,I*b*(d*x+c)^n)/d^3/exp(I*a 
)/n/((I*b*(d*x+c)^n)^(2/n))+1/2*I*exp(I*a)*(d*x+c)^3*GAMMA(3/n,-I*b*(d*x+c 
)^n)/d^3/n/((-I*b*(d*x+c)^n)^(3/n))-1/2*I*(d*x+c)^3*GAMMA(3/n,I*b*(d*x+c)^ 
n)/d^3/exp(I*a)/n/((I*b*(d*x+c)^n)^(3/n))
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.78 \[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\frac {i e^{-i a} (c+d x) \left (e^{2 i a} \left (-i b (c+d x)^n\right )^{-3/n} \left (c^2 \left (-i b (c+d x)^n\right )^{2/n} \Gamma \left (\frac {1}{n},-i b (c+d x)^n\right )-(c+d x) \left (2 c \left (-i b (c+d x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {2}{n},-i b (c+d x)^n\right )-(c+d x) \Gamma \left (\frac {3}{n},-i b (c+d x)^n\right )\right )\right )+\left (i b (c+d x)^n\right )^{-3/n} \left (-c^2 \left (i b (c+d x)^n\right )^{2/n} \Gamma \left (\frac {1}{n},i b (c+d x)^n\right )+(c+d x) \left (2 c \left (i b (c+d x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {2}{n},i b (c+d x)^n\right )-(c+d x) \Gamma \left (\frac {3}{n},i b (c+d x)^n\right )\right )\right )\right )}{2 d^3 n} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 355, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3914, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx\)

\(\Big \downarrow \) 3914

\(\displaystyle \frac {\int \left (\sin \left (b (c+d x)^n+a\right ) c^2-2 (c+d x) \sin \left (b (c+d x)^n+a\right ) c+(c+d x)^2 \sin \left (b (c+d x)^n+a\right )\right )d(c+d x)}{d^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {i e^{i a} c^2 (c+d x) \left (-i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i b (c+d x)^n\right )}{2 n}-\frac {i e^{-i a} c^2 (c+d x) \left (i b (c+d x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i b (c+d x)^n\right )}{2 n}+\frac {i e^{i a} (c+d x)^3 \left (-i b (c+d x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-i b (c+d x)^n\right )}{2 n}-\frac {i e^{i a} c (c+d x)^2 \left (-i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i b (c+d x)^n\right )}{n}+\frac {i e^{-i a} c (c+d x)^2 \left (i b (c+d x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i b (c+d x)^n\right )}{n}-\frac {i e^{-i a} (c+d x)^3 \left (i b (c+d x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},i b (c+d x)^n\right )}{2 n}}{d^3}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3914

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat 
or[n], 1]}, Simp[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]
 

Maple [F]

\[\int x^{2} \sin \left (a +b \left (d x +c \right )^{n}\right )d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\int { x^{2} \sin \left ({\left (d x + c\right )}^{n} b + a\right ) \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\int x^{2} \sin {\left (a + b \left (c + d x\right )^{n} \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\int { x^{2} \sin \left ({\left (d x + c\right )}^{n} b + a\right ) \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\int { x^{2} \sin \left ({\left (d x + c\right )}^{n} b + a\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\int x^2\,\sin \left (a+b\,{\left (c+d\,x\right )}^n\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^2 \sin \left (a+b (c+d x)^n\right ) \, dx=\int \sin \left (\left (d x +c \right )^{n} b +a \right ) x^{2}d x \] Input:

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

Output:

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