\(\int x^m \sin ^3(a+b x^n) \, dx\) [144]

Optimal result

Integrand size = 14, antiderivative size = 237 \[ \int x^m \sin ^3\left (a+b x^n\right ) \, dx=\frac {3 i e^{i a} x^{1+m} \left (-i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-i b x^n\right )}{8 n}-\frac {3 i e^{-i a} x^{1+m} \left (i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},i b x^n\right )}{8 n}-\frac {i 3^{-\frac {1+m}{n}} e^{3 i a} x^{1+m} \left (-i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-3 i b x^n\right )}{8 n}+\frac {i 3^{-\frac {1+m}{n}} e^{-3 i a} x^{1+m} \left (i b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},3 i b x^n\right )}{8 n} \] Output:

3/8*I*exp(I*a)*x^(1+m)*GAMMA((1+m)/n,-I*b*x^n)/n/((-I*b*x^n)^((1+m)/n))-3/ 
8*I*x^(1+m)*GAMMA((1+m)/n,I*b*x^n)/exp(I*a)/n/((I*b*x^n)^((1+m)/n))-1/8*I* 
exp(3*I*a)*x^(1+m)*GAMMA((1+m)/n,-3*I*b*x^n)/(3^((1+m)/n))/n/((-I*b*x^n)^( 
(1+m)/n))+1/8*I*x^(1+m)*GAMMA((1+m)/n,3*I*b*x^n)/(3^((1+m)/n))/exp(3*I*a)/ 
n/((I*b*x^n)^((1+m)/n))
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.95 \[ \int x^m \sin ^3\left (a+b x^n\right ) \, dx=\frac {i 3^{-\frac {1+m}{n}} e^{-3 i a} x^{1+m} \left (b^2 x^{2 n}\right )^{-\frac {1+m}{n}} \left (3^{\frac {1+m+n}{n}} e^{4 i a} \left (i b x^n\right )^{\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-i b x^n\right )-3^{\frac {1+m+n}{n}} e^{2 i a} \left (-i b x^n\right )^{\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},i b x^n\right )-e^{6 i a} \left (i b x^n\right )^{\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-3 i b x^n\right )+\left (-i b x^n\right )^{\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},3 i b x^n\right )\right )}{8 n} \] Input:

Integrate[x^m*Sin[a + b*x^n]^3,x]
 

Output:

((I/8)*x^(1 + m)*(3^((1 + m + n)/n)*E^((4*I)*a)*(I*b*x^n)^((1 + m)/n)*Gamm 
a[(1 + m)/n, (-I)*b*x^n] - 3^((1 + m + n)/n)*E^((2*I)*a)*((-I)*b*x^n)^((1 
+ m)/n)*Gamma[(1 + m)/n, I*b*x^n] - E^((6*I)*a)*(I*b*x^n)^((1 + m)/n)*Gamm 
a[(1 + m)/n, (-3*I)*b*x^n] + ((-I)*b*x^n)^((1 + m)/n)*Gamma[(1 + m)/n, (3* 
I)*b*x^n]))/(3^((1 + m)/n)*E^((3*I)*a)*n*(b^2*x^(2*n))^((1 + m)/n))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3906, 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.

\(\displaystyle \int x^m \sin ^3\left (a+b x^n\right ) \, dx\)

\(\Big \downarrow \) 3906

\(\displaystyle \int \left (\frac {3}{4} x^m \sin \left (a+b x^n\right )-\frac {1}{4} x^m \sin \left (3 a+3 b x^n\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 i e^{i a} x^{m+1} \left (-i b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-i b x^n\right )}{8 n}-\frac {3 i e^{-i a} x^{m+1} \left (i b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},i b x^n\right )}{8 n}-\frac {i e^{3 i a} 3^{-\frac {m+1}{n}} x^{m+1} \left (-i b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-3 i b x^n\right )}{8 n}+\frac {i e^{-3 i a} 3^{-\frac {m+1}{n}} x^{m+1} \left (i b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},3 i b x^n\right )}{8 n}\)

Input:

Int[x^m*Sin[a + b*x^n]^3,x]
 

Output:

(((3*I)/8)*E^(I*a)*x^(1 + m)*Gamma[(1 + m)/n, (-I)*b*x^n])/(n*((-I)*b*x^n) 
^((1 + m)/n)) - (((3*I)/8)*x^(1 + m)*Gamma[(1 + m)/n, I*b*x^n])/(E^(I*a)*n 
*(I*b*x^n)^((1 + m)/n)) - ((I/8)*E^((3*I)*a)*x^(1 + m)*Gamma[(1 + m)/n, (- 
3*I)*b*x^n])/(3^((1 + m)/n)*n*((-I)*b*x^n)^((1 + m)/n)) + ((I/8)*x^(1 + m) 
*Gamma[(1 + m)/n, (3*I)*b*x^n])/(3^((1 + m)/n)*E^((3*I)*a)*n*(I*b*x^n)^((1 
 + m)/n))
 

Defintions of rubi rules used

rule 2009

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

rule 3906

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x 
_Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

Maple [F]

\[\int x^{m} \sin \left (a +b \,x^{n}\right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(x^m*cos(b*x^n + a)^2 - x^m)*sin(b*x^n + a), x)
 

Sympy [F]

\[ \int x^m \sin ^3\left (a+b x^n\right ) \, dx=\int x^{m} \sin ^{3}{\left (a + b x^{n} \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^m \sin ^3\left (a+b x^n\right ) \, dx=\int x^m\,{\sin \left (a+b\,x^n\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int(x**m*sin(x**n*b + a)**3,x)