Integrand size = 12, antiderivative size = 88 \[ \int (b \sinh (c+d x))^{3/2} \, dx=\frac {2 i b^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{3 d \sqrt {b \sinh (c+d x)}}+\frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d} \] Output:
2/3*I*b^2*InverseJacobiAM(1/2*I*c-1/4*Pi+1/2*I*d*x,2^(1/2))*(I*sinh(d*x+c) )^(1/2)/d/(b*sinh(d*x+c))^(1/2)+2/3*b*cosh(d*x+c)*(b*sinh(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int (b \sinh (c+d x))^{3/2} \, dx=\frac {b^2 \left (\sinh (2 (c+d x))-2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh (2 (c+d x))+\sinh (2 (c+d x))\right ) \sqrt {1-\cosh (2 c+2 d x)-\sinh (2 c+2 d x)}\right )}{3 d \sqrt {b \sinh (c+d x)}} \] Input:
Integrate[(b*Sinh[c + d*x])^(3/2),x]
Output:
(b^2*(Sinh[2*(c + d*x)] - 2*Hypergeometric2F1[1/4, 1/2, 5/4, Cosh[2*(c + d *x)] + Sinh[2*(c + d*x)]]*Sqrt[1 - Cosh[2*c + 2*d*x] - Sinh[2*c + 2*d*x]]) )/(3*d*Sqrt[b*Sinh[c + d*x]])
Time = 0.34 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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.
\(\displaystyle \int (b \sinh (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-i b \sin (i c+i d x))^{3/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}-\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sinh (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}-\frac {1}{3} b^2 \int \frac {1}{\sqrt {-i b \sin (i c+i d x)}}dx\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}-\frac {b^2 \sqrt {i \sinh (c+d x)} \int \frac {1}{\sqrt {i \sinh (c+d x)}}dx}{3 \sqrt {b \sinh (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}-\frac {b^2 \sqrt {i \sinh (c+d x)} \int \frac {1}{\sqrt {\sin (i c+i d x)}}dx}{3 \sqrt {b \sinh (c+d x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 b \cosh (c+d x) \sqrt {b \sinh (c+d x)}}{3 d}+\frac {2 i b^2 \sqrt {i \sinh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{3 d \sqrt {b \sinh (c+d x)}}\) |
Input:
Int[(b*Sinh[c + d*x])^(3/2),x]
Output:
(((2*I)/3)*b^2*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2]*Sqrt[I*Sinh[c + d*x]]) /(d*Sqrt[b*Sinh[c + d*x]]) + (2*b*Cosh[c + d*x]*Sqrt[b*Sinh[c + d*x]])/(3* d)
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 = 0.22 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.20
method | result | size |
default | \(-\frac {b^{2} \left (i \sqrt {-i \sinh \left (d x +c \right )+1}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-i \sinh \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{2}\right )}{3 \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\) | \(106\) |
Input:
int((b*sinh(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/3*b^2*(I*(-I*sinh(d*x+c)+1)^(1/2)*2^(1/2)*(1+I*sinh(d*x+c))^(1/2)*(I*si nh(d*x+c))^(1/2)*EllipticF((-I*sinh(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*sinh(d* x+c)*cosh(d*x+c)^2)/cosh(d*x+c)/(b*sinh(d*x+c))^(1/2)/d
Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.27 \[ \int (b \sinh (c+d x))^{3/2} \, dx=-\frac {4 \, \sqrt {\frac {1}{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b\right )} \sqrt {b \sinh \left (d x + c\right )}}{3 \, {\left (d \cosh \left (d x + c\right ) + d \sinh \left (d x + c\right )\right )}} \] Input:
integrate((b*sinh(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-1/3*(4*sqrt(1/2)*(b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt(b)*weierstrassP Inverse(4, 0, cosh(d*x + c) + sinh(d*x + c)) - (b*cosh(d*x + c)^2 + 2*b*co sh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)*sqrt(b*sinh(d*x + c)))/ (d*cosh(d*x + c) + d*sinh(d*x + c))
\[ \int (b \sinh (c+d x))^{3/2} \, dx=\int \left (b \sinh {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((b*sinh(d*x+c))**(3/2),x)
Output:
Integral((b*sinh(c + d*x))**(3/2), x)
\[ \int (b \sinh (c+d x))^{3/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((b*sinh(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((b*sinh(d*x + c))^(3/2), x)
\[ \int (b \sinh (c+d x))^{3/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((b*sinh(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((b*sinh(d*x + c))^(3/2), x)
Timed out. \[ \int (b \sinh (c+d x))^{3/2} \, dx=\int {\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:
int((b*sinh(c + d*x))^(3/2),x)
Output:
int((b*sinh(c + d*x))^(3/2), x)
\[ \int (b \sinh (c+d x))^{3/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sinh \left (d x +c \right )}\, \sinh \left (d x +c \right )d x \right ) b \] Input:
int((b*sinh(d*x+c))^(3/2),x)
Output:
sqrt(b)*int(sqrt(sinh(c + d*x))*sinh(c + d*x),x)*b