Integrand size = 10, antiderivative size = 80 \[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{3 b \sqrt {\sinh (a+b x)}}+\frac {2 \cosh (a+b x) \sqrt {\sinh (a+b x)}}{3 b} \] Output:
2/3*I*InverseJacobiAM(1/2*I*a-1/4*Pi+1/2*I*b*x,2^(1/2))*(I*sinh(b*x+a))^(1 /2)/b/sinh(b*x+a)^(1/2)+2/3*cosh(b*x+a)*sinh(b*x+a)^(1/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=\frac {\sinh (2 (a+b x))-2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 (a+b x))+\sinh (2 (a+b x))\right ) \sqrt {1-\cosh (2 a+2 b x)-\sinh (2 a+2 b x)}}{3 b \sqrt {\sinh (a+b x)}} \] Input:
Integrate[Sinh[a + b*x]^(3/2),x]
Output:
(Sinh[2*(a + b*x)] - 2*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(a + b*x)] + Sinh[2*(a + b*x)]]*Sqrt[1 - Cosh[2*a + 2*b*x] - Sinh[2*a + 2*b*x]])/(3*b *Sqrt[Sinh[a + b*x]])
Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3115, 3042, 3121, 3042, 3120}
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.
\(\displaystyle \int \sinh ^{\frac {3}{2}}(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-i \sin (i a+i b x))^{3/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}-\frac {1}{3} \int \frac {1}{\sqrt {\sinh (a+b x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}-\frac {1}{3} \int \frac {1}{\sqrt {-i \sin (i a+i b x)}}dx\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {i \sinh (a+b x)}}dx}{3 \sqrt {\sinh (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {\sin (i a+i b x)}}dx}{3 \sqrt {\sinh (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sqrt {\sinh (a+b x)} \cosh (a+b x)}{3 b}+\frac {2 i \sqrt {i \sinh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{3 b \sqrt {\sinh (a+b x)}}\) |
Input:
Int[Sinh[a + b*x]^(3/2),x]
Output:
(((2*I)/3)*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/(b* Sqrt[Sinh[a + b*x]]) + (2*Cosh[a + b*x]*Sqrt[Sinh[a + b*x]])/(3*b)
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25
method | result | size |
default | \(\frac {-\frac {i \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{3}}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\) | \(100\) |
Input:
int(sinh(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
(-1/3*I*(1-I*sinh(b*x+a))^(1/2)*2^(1/2)*(1+I*sinh(b*x+a))^(1/2)*(I*sinh(b* x+a))^(1/2)*EllipticF((1-I*sinh(b*x+a))^(1/2),1/2*2^(1/2))+2/3*cosh(b*x+a) ^2*sinh(b*x+a))/cosh(b*x+a)/sinh(b*x+a)^(1/2)/b
Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.29 \[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=-\frac {2 \, {\left (\sqrt {2} \cosh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \sqrt {\sinh \left (b x + a\right )}}{3 \, {\left (b \cosh \left (b x + a\right ) + b \sinh \left (b x + a\right )\right )}} \] Input:
integrate(sinh(b*x+a)^(3/2),x, algorithm="fricas")
Output:
-1/3*(2*(sqrt(2)*cosh(b*x + a) + sqrt(2)*sinh(b*x + a))*weierstrassPInvers e(4, 0, cosh(b*x + a) + sinh(b*x + a)) - (cosh(b*x + a)^2 + 2*cosh(b*x + a )*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*sqrt(sinh(b*x + a)))/(b*cosh(b*x + a) + b*sinh(b*x + a))
\[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int \sinh ^{\frac {3}{2}}{\left (a + b x \right )}\, dx \] Input:
integrate(sinh(b*x+a)**(3/2),x)
Output:
Integral(sinh(a + b*x)**(3/2), x)
\[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int { \sinh \left (b x + a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(sinh(b*x+a)^(3/2),x, algorithm="maxima")
Output:
integrate(sinh(b*x + a)^(3/2), x)
\[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int { \sinh \left (b x + a\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(sinh(b*x+a)^(3/2),x, algorithm="giac")
Output:
integrate(sinh(b*x + a)^(3/2), x)
Timed out. \[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int {\mathrm {sinh}\left (a+b\,x\right )}^{3/2} \,d x \] Input:
int(sinh(a + b*x)^(3/2),x)
Output:
int(sinh(a + b*x)^(3/2), x)
\[ \int \sinh ^{\frac {3}{2}}(a+b x) \, dx=\int \sqrt {\sinh \left (b x +a \right )}\, \sinh \left (b x +a \right )d x \] Input:
int(sinh(b*x+a)^(3/2),x)
Output:
int(sqrt(sinh(a + b*x))*sinh(a + b*x),x)