Integrand size = 10, antiderivative size = 150 \[ \int \sinh ^3\left (a+b x^n\right ) \, dx=-\frac {3^{-1/n} e^{3 a} x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},b x^n\right )}{8 n}+\frac {3^{-1/n} e^{-3 a} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},3 b x^n\right )}{8 n} \] Output:
-1/8*exp(3*a)*x*GAMMA(1/n,-3*b*x^n)/(3^(1/n))/n/((-b*x^n)^(1/n))+3/8*exp(a )*x*GAMMA(1/n,-b*x^n)/n/((-b*x^n)^(1/n))-3/8*x*GAMMA(1/n,b*x^n)/exp(a)/n/( (b*x^n)^(1/n))+1/8*x*GAMMA(1/n,3*b*x^n)/(3^(1/n))/exp(3*a)/n/((b*x^n)^(1/n ))
Time = 0.07 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93 \[ \int \sinh ^3\left (a+b x^n\right ) \, dx=\frac {3^{-1/n} e^{-3 a} x \left (-b^2 x^{2 n}\right )^{-1/n} \left (-e^{6 a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},-3 b x^n\right )+3^{1+\frac {1}{n}} e^{4 a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (\frac {1}{n},-b x^n\right )+\left (-b x^n\right )^{\frac {1}{n}} \left (-3^{1+\frac {1}{n}} e^{2 a} \Gamma \left (\frac {1}{n},b x^n\right )+\Gamma \left (\frac {1}{n},3 b x^n\right )\right )\right )}{8 n} \] Input:
Integrate[Sinh[a + b*x^n]^3,x]
Output:
(x*(-(E^(6*a)*(b*x^n)^n^(-1)*Gamma[n^(-1), -3*b*x^n]) + 3^(1 + n^(-1))*E^( 4*a)*(b*x^n)^n^(-1)*Gamma[n^(-1), -(b*x^n)] + (-(b*x^n))^n^(-1)*(-(3^(1 + n^(-1))*E^(2*a)*Gamma[n^(-1), b*x^n]) + Gamma[n^(-1), 3*b*x^n])))/(8*3^n^( -1)*E^(3*a)*n*(-(b^2*x^(2*n)))^n^(-1))
Time = 0.35 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5831, 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.
\(\displaystyle \int \sinh ^3\left (a+b x^n\right ) \, dx\) |
\(\Big \downarrow \) 5831 |
\(\displaystyle \int \left (\frac {1}{4} \sinh \left (3 a+3 b x^n\right )-\frac {3}{4} \sinh \left (a+b x^n\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{3 a} 3^{-1/n} x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-3 b x^n\right )}{8 n}+\frac {3 e^a x \left (-b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},-b x^n\right )}{8 n}-\frac {3 e^{-a} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},b x^n\right )}{8 n}+\frac {e^{-3 a} 3^{-1/n} x \left (b x^n\right )^{-1/n} \Gamma \left (\frac {1}{n},3 b x^n\right )}{8 n}\) |
Input:
Int[Sinh[a + b*x^n]^3,x]
Output:
-1/8*(E^(3*a)*x*Gamma[n^(-1), -3*b*x^n])/(3^n^(-1)*n*(-(b*x^n))^n^(-1)) + (3*E^a*x*Gamma[n^(-1), -(b*x^n)])/(8*n*(-(b*x^n))^n^(-1)) - (3*x*Gamma[n^( -1), b*x^n])/(8*E^a*n*(b*x^n)^n^(-1)) + (x*Gamma[n^(-1), 3*b*x^n])/(8*3^n^ (-1)*E^(3*a)*n*(b*x^n)^n^(-1))
Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[Ex pandTrigReduce[(a + b*Sinh[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
\[\int \sinh \left (a +b \,x^{n}\right )^{3}d x\]
Input:
int(sinh(a+b*x^n)^3,x)
Output:
int(sinh(a+b*x^n)^3,x)
\[ \int \sinh ^3\left (a+b x^n\right ) \, dx=\int { \sinh \left (b x^{n} + a\right )^{3} \,d x } \] Input:
integrate(sinh(a+b*x^n)^3,x, algorithm="fricas")
Output:
integral(sinh(b*x^n + a)^3, x)
\[ \int \sinh ^3\left (a+b x^n\right ) \, dx=\int \sinh ^{3}{\left (a + b x^{n} \right )}\, dx \] Input:
integrate(sinh(a+b*x**n)**3,x)
Output:
Integral(sinh(a + b*x**n)**3, x)
Time = 0.17 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \sinh ^3\left (a+b x^n\right ) \, dx=\frac {x e^{\left (-3 \, a\right )} \Gamma \left (\frac {1}{n}, 3 \, b x^{n}\right )}{8 \, \left (3 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} n} - \frac {3 \, x e^{\left (-a\right )} \Gamma \left (\frac {1}{n}, b x^{n}\right )}{8 \, \left (b x^{n}\right )^{\left (\frac {1}{n}\right )} n} + \frac {3 \, x e^{a} \Gamma \left (\frac {1}{n}, -b x^{n}\right )}{8 \, \left (-b x^{n}\right )^{\left (\frac {1}{n}\right )} n} - \frac {x e^{\left (3 \, a\right )} \Gamma \left (\frac {1}{n}, -3 \, b x^{n}\right )}{8 \, \left (-3 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} n} \] Input:
integrate(sinh(a+b*x^n)^3,x, algorithm="maxima")
Output:
1/8*x*e^(-3*a)*gamma(1/n, 3*b*x^n)/((3*b*x^n)^(1/n)*n) - 3/8*x*e^(-a)*gamm a(1/n, b*x^n)/((b*x^n)^(1/n)*n) + 3/8*x*e^a*gamma(1/n, -b*x^n)/((-b*x^n)^( 1/n)*n) - 1/8*x*e^(3*a)*gamma(1/n, -3*b*x^n)/((-3*b*x^n)^(1/n)*n)
\[ \int \sinh ^3\left (a+b x^n\right ) \, dx=\int { \sinh \left (b x^{n} + a\right )^{3} \,d x } \] Input:
integrate(sinh(a+b*x^n)^3,x, algorithm="giac")
Output:
integrate(sinh(b*x^n + a)^3, x)
Timed out. \[ \int \sinh ^3\left (a+b x^n\right ) \, dx=\int {\mathrm {sinh}\left (a+b\,x^n\right )}^3 \,d x \] Input:
int(sinh(a + b*x^n)^3,x)
Output:
int(sinh(a + b*x^n)^3, x)
\[ \int \sinh ^3\left (a+b x^n\right ) \, dx=\int \sinh \left (x^{n} b +a \right )^{3}d x \] Input:
int(sinh(a+b*x^n)^3,x)
Output:
int(sinh(x**n*b + a)**3,x)