\(\int x^2 (a+b \sin (c+d (f+g x)^n))^2 \, dx\) [272]

Optimal result

Integrand size = 22, antiderivative size = 856 \[ \int x^2 \left (a+b \sin \left (c+d (f+g x)^n\right )\right )^2 \, dx=\frac {\left (2 a^2+b^2\right ) f^2 x}{2 g^2}-\frac {\left (2 a^2+b^2\right ) f (f+g x)^2}{2 g^3}+\frac {\left (2 a^2+b^2\right ) (f+g x)^3}{6 g^3}+\frac {i a b e^{i c} f^2 (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g^3 n}-\frac {i a b e^{-i c} f^2 (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g^3 n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} f^2 (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g^3 n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} f^2 (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g^3 n}-\frac {2 i a b e^{i c} f (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )}{g^3 n}+\frac {2 i a b e^{-i c} f (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )}{g^3 n}-\frac {2^{-1-\frac {2}{n}} b^2 e^{2 i c} f (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )}{g^3 n}-\frac {2^{-1-\frac {2}{n}} b^2 e^{-2 i c} f (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )}{g^3 n}+\frac {i a b e^{i c} (f+g x)^3 \left (-i d (f+g x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-i d (f+g x)^n\right )}{g^3 n}-\frac {i a b e^{-i c} (f+g x)^3 \left (i d (f+g x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},i d (f+g x)^n\right )}{g^3 n}+\frac {2^{-2-\frac {3}{n}} b^2 e^{2 i c} (f+g x)^3 \left (-i d (f+g x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-2 i d (f+g x)^n\right )}{g^3 n}+\frac {2^{-2-\frac {3}{n}} b^2 e^{-2 i c} (f+g x)^3 \left (i d (f+g x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},2 i d (f+g x)^n\right )}{g^3 n} \] Output:

1/2*(2*a^2+b^2)*f^2*x/g^2-1/2*(2*a^2+b^2)*f*(g*x+f)^2/g^3+1/6*(2*a^2+b^2)* 
(g*x+f)^3/g^3+I*a*b*exp(I*c)*f^2*(g*x+f)*GAMMA(1/n,-I*d*(g*x+f)^n)/g^3/n/( 
(-I*d*(g*x+f)^n)^(1/n))-I*a*b*f^2*(g*x+f)*GAMMA(1/n,I*d*(g*x+f)^n)/exp(I*c 
)/g^3/n/((I*d*(g*x+f)^n)^(1/n))+2^(-2-1/n)*b^2*exp(2*I*c)*f^2*(g*x+f)*GAMM 
A(1/n,-2*I*d*(g*x+f)^n)/g^3/n/((-I*d*(g*x+f)^n)^(1/n))+2^(-2-1/n)*b^2*f^2* 
(g*x+f)*GAMMA(1/n,2*I*d*(g*x+f)^n)/exp(2*I*c)/g^3/n/((I*d*(g*x+f)^n)^(1/n) 
)-2*I*a*b*exp(I*c)*f*(g*x+f)^2*GAMMA(2/n,-I*d*(g*x+f)^n)/g^3/n/((-I*d*(g*x 
+f)^n)^(2/n))+2*I*a*b*f*(g*x+f)^2*GAMMA(2/n,I*d*(g*x+f)^n)/exp(I*c)/g^3/n/ 
((I*d*(g*x+f)^n)^(2/n))-2^(-1-2/n)*b^2*exp(2*I*c)*f*(g*x+f)^2*GAMMA(2/n,-2 
*I*d*(g*x+f)^n)/g^3/n/((-I*d*(g*x+f)^n)^(2/n))-2^(-1-2/n)*b^2*f*(g*x+f)^2* 
GAMMA(2/n,2*I*d*(g*x+f)^n)/exp(2*I*c)/g^3/n/((I*d*(g*x+f)^n)^(2/n))+I*a*b* 
exp(I*c)*(g*x+f)^3*GAMMA(3/n,-I*d*(g*x+f)^n)/g^3/n/((-I*d*(g*x+f)^n)^(3/n) 
)-I*a*b*(g*x+f)^3*GAMMA(3/n,I*d*(g*x+f)^n)/exp(I*c)/g^3/n/((I*d*(g*x+f)^n) 
^(3/n))+2^(-2-3/n)*b^2*exp(2*I*c)*(g*x+f)^3*GAMMA(3/n,-2*I*d*(g*x+f)^n)/g^ 
3/n/((-I*d*(g*x+f)^n)^(3/n))+2^(-2-3/n)*b^2*(g*x+f)^3*GAMMA(3/n,2*I*d*(g*x 
+f)^n)/exp(2*I*c)/g^3/n/((I*d*(g*x+f)^n)^(3/n))
 

Mathematica [A] (verified)

Time = 1.91 (sec) , antiderivative size = 648, normalized size of antiderivative = 0.76 \[ \int x^2 \left (a+b \sin \left (c+d (f+g x)^n\right )\right )^2 \, dx=\frac {\left (2 a^2+b^2\right ) g^3 n x^3+6 i a b e^{i c} (f+g x) \left (-i d (f+g x)^n\right )^{-3/n} \left (f^2 \left (-i d (f+g x)^n\right )^{2/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )-(f+g x) \left (2 f \left (-i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {3}{n},-i d (f+g x)^n\right )\right )\right )-6 i a b e^{-i c} (f+g x) \left (i d (f+g x)^n\right )^{-3/n} \left (f^2 \left (i d (f+g x)^n\right )^{2/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )-(f+g x) \left (2 f \left (i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {3}{n},i d (f+g x)^n\right )\right )\right )+3\ 2^{-\frac {3+n}{n}} b^2 e^{2 i c} (f+g x) \left (-i d (f+g x)^n\right )^{-3/n} \left (4^{\frac {1}{n}} f^2 \left (-i d (f+g x)^n\right )^{2/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )-(f+g x) \left (2^{1+\frac {1}{n}} f \left (-i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {3}{n},-2 i d (f+g x)^n\right )\right )\right )+3\ 2^{-\frac {3+n}{n}} b^2 e^{-2 i c} (f+g x) \left (i d (f+g x)^n\right )^{-3/n} \left (4^{\frac {1}{n}} f^2 \left (i d (f+g x)^n\right )^{2/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )-(f+g x) \left (2^{1+\frac {1}{n}} f \left (i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {3}{n},2 i d (f+g x)^n\right )\right )\right )}{6 g^3 n} \] Input:

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

Output:

((2*a^2 + b^2)*g^3*n*x^3 + ((6*I)*a*b*E^(I*c)*(f + g*x)*(f^2*((-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)^n^(-1)*Gamma[2/n, (-I)*d*(f + g*x)^n] - (f + g*x)*Gamma[3/n, ( 
-I)*d*(f + g*x)^n])))/((-I)*d*(f + g*x)^n)^(3/n) - ((6*I)*a*b*(f + g*x)*(f 
^2*(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)^n^(-1)*Gamma[2/n, I*d*(f + g*x)^n] - (f + g*x)*Gamma[3/ 
n, I*d*(f + g*x)^n])))/(E^(I*c)*(I*d*(f + g*x)^n)^(3/n)) + (3*b^2*E^((2*I) 
*c)*(f + g*x)*(4^n^(-1)*f^2*((-I)*d*(f + g*x)^n)^(2/n)*Gamma[n^(-1), (-2*I 
)*d*(f + g*x)^n] - (f + g*x)*(2^(1 + n^(-1))*f*((-I)*d*(f + g*x)^n)^n^(-1) 
*Gamma[2/n, (-2*I)*d*(f + g*x)^n] - (f + g*x)*Gamma[3/n, (-2*I)*d*(f + g*x 
)^n])))/(2^((3 + n)/n)*((-I)*d*(f + g*x)^n)^(3/n)) + (3*b^2*(f + g*x)*(4^n 
^(-1)*f^2*(I*d*(f + g*x)^n)^(2/n)*Gamma[n^(-1), (2*I)*d*(f + g*x)^n] - (f 
+ g*x)*(2^(1 + n^(-1))*f*(I*d*(f + g*x)^n)^n^(-1)*Gamma[2/n, (2*I)*d*(f + 
g*x)^n] - (f + g*x)*Gamma[3/n, (2*I)*d*(f + g*x)^n])))/(2^((3 + n)/n)*E^(( 
2*I)*c)*(I*d*(f + g*x)^n)^(3/n)))/(6*g^3*n)
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 819, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 3914

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {i a b e^{i c} (f+g x)^3 \Gamma \left (\frac {3}{n},-i d (f+g x)^n\right ) \left (-i d (f+g x)^n\right )^{-3/n}}{n}+\frac {2^{-2-\frac {3}{n}} b^2 e^{2 i c} (f+g x)^3 \Gamma \left (\frac {3}{n},-2 i d (f+g x)^n\right ) \left (-i d (f+g x)^n\right )^{-3/n}}{n}-\frac {2 i a b e^{i c} f (f+g x)^2 \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right ) \left (-i d (f+g x)^n\right )^{-2/n}}{n}-\frac {2^{-1-\frac {2}{n}} b^2 e^{2 i c} f (f+g x)^2 \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right ) \left (-i d (f+g x)^n\right )^{-2/n}}{n}+\frac {i a b e^{i c} f^2 (f+g x) \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right ) \left (-i d (f+g x)^n\right )^{-1/n}}{n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} f^2 (f+g x) \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right ) \left (-i d (f+g x)^n\right )^{-1/n}}{n}+\frac {1}{6} \left (2 a^2+b^2\right ) (f+g x)^3-\frac {1}{2} \left (2 a^2+b^2\right ) f (f+g x)^2+\frac {1}{2} \left (2 a^2+b^2\right ) f^2 (f+g x)-\frac {i a b e^{-i c} f^2 (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{n}+\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} f^2 (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{n}+\frac {2 i a b e^{-i c} f (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )}{n}-\frac {2^{-1-\frac {2}{n}} b^2 e^{-2 i c} f (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )}{n}-\frac {i a b e^{-i c} (f+g x)^3 \left (i d (f+g x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},i d (f+g x)^n\right )}{n}+\frac {2^{-2-\frac {3}{n}} b^2 e^{-2 i c} (f+g x)^3 \left (i d (f+g x)^n\right )^{-3/n} \Gamma \left (\frac {3}{n},2 i d (f+g x)^n\right )}{n}}{g^3}\)

Input:

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

Output:

(((2*a^2 + b^2)*f^2*(f + g*x))/2 - ((2*a^2 + b^2)*f*(f + g*x)^2)/2 + ((2*a 
^2 + b^2)*(f + g*x)^3)/6 + (I*a*b*E^(I*c)*f^2*(f + g*x)*Gamma[n^(-1), (-I) 
*d*(f + g*x)^n])/(n*((-I)*d*(f + g*x)^n)^n^(-1)) - (I*a*b*f^2*(f + g*x)*Ga 
mma[n^(-1), I*d*(f + g*x)^n])/(E^(I*c)*n*(I*d*(f + g*x)^n)^n^(-1)) + (2^(- 
2 - n^(-1))*b^2*E^((2*I)*c)*f^2*(f + g*x)*Gamma[n^(-1), (-2*I)*d*(f + g*x) 
^n])/(n*((-I)*d*(f + g*x)^n)^n^(-1)) + (2^(-2 - n^(-1))*b^2*f^2*(f + g*x)* 
Gamma[n^(-1), (2*I)*d*(f + g*x)^n])/(E^((2*I)*c)*n*(I*d*(f + g*x)^n)^n^(-1 
)) - ((2*I)*a*b*E^(I*c)*f*(f + g*x)^2*Gamma[2/n, (-I)*d*(f + g*x)^n])/(n*( 
(-I)*d*(f + g*x)^n)^(2/n)) + ((2*I)*a*b*f*(f + g*x)^2*Gamma[2/n, I*d*(f + 
g*x)^n])/(E^(I*c)*n*(I*d*(f + g*x)^n)^(2/n)) - (2^(-1 - 2/n)*b^2*E^((2*I)* 
c)*f*(f + g*x)^2*Gamma[2/n, (-2*I)*d*(f + g*x)^n])/(n*((-I)*d*(f + g*x)^n) 
^(2/n)) - (2^(-1 - 2/n)*b^2*f*(f + g*x)^2*Gamma[2/n, (2*I)*d*(f + g*x)^n]) 
/(E^((2*I)*c)*n*(I*d*(f + g*x)^n)^(2/n)) + (I*a*b*E^(I*c)*(f + g*x)^3*Gamm 
a[3/n, (-I)*d*(f + g*x)^n])/(n*((-I)*d*(f + g*x)^n)^(3/n)) - (I*a*b*(f + g 
*x)^3*Gamma[3/n, I*d*(f + g*x)^n])/(E^(I*c)*n*(I*d*(f + g*x)^n)^(3/n)) + ( 
2^(-2 - 3/n)*b^2*E^((2*I)*c)*(f + g*x)^3*Gamma[3/n, (-2*I)*d*(f + g*x)^n]) 
/(n*((-I)*d*(f + g*x)^n)^(3/n)) + (2^(-2 - 3/n)*b^2*(f + g*x)^3*Gamma[3/n, 
 (2*I)*d*(f + g*x)^n])/(E^((2*I)*c)*n*(I*d*(f + g*x)^n)^(3/n)))/g^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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/3*a^2*x^3 + 1/6*b^2*x^3 - 1/2*b^2*integrate(x^2*cos(2*(g*x + f)^n*d + 2* 
c), x) + 2*a*b*integrate(x^2*sin((g*x + f)^n*d + c), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^2 \left (a+b \sin \left (c+d (f+g x)^n\right )\right )^2 \, dx=\left (\int \sin \left (\left (g x +f \right )^{n} d +c \right )^{2} x^{2}d x \right ) b^{2}+2 \left (\int \sin \left (\left (g x +f \right )^{n} d +c \right ) x^{2}d x \right ) a b +\frac {a^{2} x^{3}}{3} \] Input:

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

Output:

(3*int(sin((f + g*x)**n*d + c)**2*x**2,x)*b**2 + 6*int(sin((f + g*x)**n*d 
+ c)*x**2,x)*a*b + a**2*x**3)/3