\(\int x (a+b \sin (c+d (f+g x)^n))^2 \, dx\) [273]

Optimal result

Integrand size = 20, antiderivative size = 556 \[ \int x \left (a+b \sin \left (c+d (f+g x)^n\right )\right )^2 \, dx=-\frac {\left (2 a^2+b^2\right ) f x}{2 g}+\frac {\left (2 a^2+b^2\right ) (f+g x)^2}{4 g^2}-\frac {i a b e^{i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{-i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{2 i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{g^2 n}-\frac {2^{-2-\frac {1}{n}} b^2 e^{-2 i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{g^2 n}+\frac {i a b e^{i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )}{g^2 n}-\frac {i a b e^{-i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{2 i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )}{g^2 n}+\frac {4^{-1-\frac {1}{n}} b^2 e^{-2 i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )}{g^2 n} \] Output:

-1/2*(2*a^2+b^2)*f*x/g+1/4*(2*a^2+b^2)*(g*x+f)^2/g^2-I*a*b*exp(I*c)*f*(g*x 
+f)*GAMMA(1/n,-I*d*(g*x+f)^n)/g^2/n/((-I*d*(g*x+f)^n)^(1/n))+I*a*b*f*(g*x+ 
f)*GAMMA(1/n,I*d*(g*x+f)^n)/exp(I*c)/g^2/n/((I*d*(g*x+f)^n)^(1/n))-2^(-2-1 
/n)*b^2*exp(2*I*c)*f*(g*x+f)*GAMMA(1/n,-2*I*d*(g*x+f)^n)/g^2/n/((-I*d*(g*x 
+f)^n)^(1/n))-2^(-2-1/n)*b^2*f*(g*x+f)*GAMMA(1/n,2*I*d*(g*x+f)^n)/exp(2*I* 
c)/g^2/n/((I*d*(g*x+f)^n)^(1/n))+I*a*b*exp(I*c)*(g*x+f)^2*GAMMA(2/n,-I*d*( 
g*x+f)^n)/g^2/n/((-I*d*(g*x+f)^n)^(2/n))-I*a*b*(g*x+f)^2*GAMMA(2/n,I*d*(g* 
x+f)^n)/exp(I*c)/g^2/n/((I*d*(g*x+f)^n)^(2/n))+4^(-1-1/n)*b^2*exp(2*I*c)*( 
g*x+f)^2*GAMMA(2/n,-2*I*d*(g*x+f)^n)/g^2/n/((-I*d*(g*x+f)^n)^(2/n))+4^(-1- 
1/n)*b^2*(g*x+f)^2*GAMMA(2/n,2*I*d*(g*x+f)^n)/exp(2*I*c)/g^2/n/((I*d*(g*x+ 
f)^n)^(2/n))
 

Mathematica [A] (verified)

Time = 1.75 (sec) , antiderivative size = 434, normalized size of antiderivative = 0.78 \[ \int x \left (a+b \sin \left (c+d (f+g x)^n\right )\right )^2 \, dx=\frac {\left (2 a^2+b^2\right ) g^2 n x^2-4^{-1/n} b^2 e^{2 i c} (f+g x) \left (-i d (f+g x)^n\right )^{-2/n} \left (2^{\frac {1}{n}} f \left (-i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )\right )-4^{-1/n} b^2 e^{-2 i c} (f+g x) \left (i d (f+g x)^n\right )^{-2/n} \left (2^{\frac {1}{n}} f \left (i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )\right )+4 a b (f+g x) \left (-i d (f+g x)^n\right )^{-2/n} \left (f \left (-i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )\right ) (-i \cos (c)+\sin (c))+4 a b (f+g x) \left (i d (f+g x)^n\right )^{-2/n} \left (f \left (i d (f+g x)^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )-(f+g x) \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )\right ) (i \cos (c)+\sin (c))}{4 g^2 n} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 534, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.

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

\(\Big \downarrow \) 3914

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{4} \left (2 a^2+b^2\right ) (f+g x)^2-\frac {1}{2} f \left (2 a^2+b^2\right ) (f+g x)+\frac {i a b e^{i c} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-i d (f+g x)^n\right )}{n}-\frac {i a b e^{i c} f (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-i d (f+g x)^n\right )}{n}+\frac {i a b e^{-i c} f (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},i d (f+g x)^n\right )}{n}-\frac {i a b e^{-i c} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},i d (f+g x)^n\right )}{n}+\frac {b^2 e^{2 i c} 4^{-\frac {1}{n}-1} (f+g x)^2 \left (-i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},-2 i d (f+g x)^n\right )}{n}-\frac {b^2 e^{2 i c} f 2^{-\frac {1}{n}-2} (f+g x) \left (-i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-2 i d (f+g x)^n\right )}{n}-\frac {b^2 e^{-2 i c} f 2^{-\frac {1}{n}-2} (f+g x) \left (i d (f+g x)^n\right )^{-1/n} \Gamma \left (\frac {1}{n},2 i d (f+g x)^n\right )}{n}+\frac {b^2 e^{-2 i c} 4^{-\frac {1}{n}-1} (f+g x)^2 \left (i d (f+g x)^n\right )^{-2/n} \Gamma \left (\frac {2}{n},2 i d (f+g x)^n\right )}{n}}{g^2}\)

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*a^2*x^2 + 1/4*b^2*x^2 - 1/2*b^2*integrate(x*cos(2*(g*x + f)^n*d + 2*c) 
, x) + 2*a*b*integrate(x*sin((g*x + f)^n*d + c), x)
 

Giac [F]

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

integrate(x*(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, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*int(sin((f + g*x)**n*d + c)**2*x,x)*b**2 + 4*int(sin((f + g*x)**n*d + c 
)*x,x)*a*b + a**2*x**2)/2