Integrand size = 10, antiderivative size = 80 \[ \int \sinh ^{\frac {5}{2}}(a+b x) \, dx=\frac {6 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{5 b \sqrt {i \sinh (a+b x)}}+\frac {2 \cosh (a+b x) \sinh ^{\frac {3}{2}}(a+b x)}{5 b} \]
2/5*cosh(b*x+a)*sinh(b*x+a)^(3/2)/b-6/5*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*x)^2 )^(1/2)/sin(1/2*I*a+1/4*Pi+1/2*I*b*x)*EllipticE(cos(1/2*I*a+1/4*Pi+1/2*I*b *x),2^(1/2))*sinh(b*x+a)^(1/2)/b/(I*sinh(b*x+a))^(1/2)
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \sinh ^{\frac {5}{2}}(a+b x) \, dx=\frac {-6 E\left (\left .\frac {1}{4} (-2 i a+\pi -2 i b x)\right |2\right ) \sqrt {i \sinh (a+b x)}+\sinh (a+b x) \sinh (2 (a+b x))}{5 b \sqrt {\sinh (a+b x)}} \]
(-6*EllipticE[((-2*I)*a + Pi - (2*I)*b*x)/4, 2]*Sqrt[I*Sinh[a + b*x]] + Si nh[a + b*x]*Sinh[2*(a + b*x)])/(5*b*Sqrt[Sinh[a + b*x]])
Time = 0.31 (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, 3119}
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.
\(\displaystyle \int \sinh ^{\frac {5}{2}}(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-i \sin (i a+i b x))^{5/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {2 \sinh ^{\frac {3}{2}}(a+b x) \cosh (a+b x)}{5 b}-\frac {3}{5} \int \sqrt {\sinh (a+b x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sinh ^{\frac {3}{2}}(a+b x) \cosh (a+b x)}{5 b}-\frac {3}{5} \int \sqrt {-i \sin (i a+i b x)}dx\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 \sinh ^{\frac {3}{2}}(a+b x) \cosh (a+b x)}{5 b}-\frac {3 \sqrt {\sinh (a+b x)} \int \sqrt {i \sinh (a+b x)}dx}{5 \sqrt {i \sinh (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sinh ^{\frac {3}{2}}(a+b x) \cosh (a+b x)}{5 b}-\frac {3 \sqrt {\sinh (a+b x)} \int \sqrt {\sin (i a+i b x)}dx}{5 \sqrt {i \sinh (a+b x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 \sinh ^{\frac {3}{2}}(a+b x) \cosh (a+b x)}{5 b}+\frac {6 i \sqrt {\sinh (a+b x)} E\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{5 b \sqrt {i \sinh (a+b x)}}\) |
(((6*I)/5)*EllipticE[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[Sinh[a + b*x]])/(b*Sq rt[I*Sinh[a + b*x]]) + (2*Cosh[a + b*x]*Sinh[a + b*x]^(3/2))/(5*b)
3.1.8.3.1 Defintions of rubi rules used
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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 = 0.75 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.05
method | result | size |
default | \(\frac {-\frac {6 \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {3 \sqrt {1-i \sinh \left (b x +a \right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (b x +a \right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (b x +a \right )}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {2 \cosh \left (b x +a \right )^{4}}{5}-\frac {2 \cosh \left (b x +a \right )^{2}}{5}}{\cosh \left (b x +a \right ) \sqrt {\sinh \left (b x +a \right )}\, b}\) | \(164\) |
(-6/5*(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)*EllipticE((1-I*sinh(b*x+a))^(1/2),1/2*2^(1/2))+3/5*(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))+2/5*cosh(b*x+a)^4-2/5*cosh(b*x+a)^2) /cosh(b*x+a)/sinh(b*x+a)^(1/2)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.52 \[ \int \sinh ^{\frac {5}{2}}(a+b x) \, dx=\frac {12 \, {\left (\sqrt {2} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt {2} \sinh \left (b x + a\right )^{2}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )\right ) + {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{2} + 12 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} + 6 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 1\right )} \sqrt {\sinh \left (b x + a\right )}}{10 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \]
1/10*(12*(sqrt(2)*cosh(b*x + a)^2 + 2*sqrt(2)*cosh(b*x + a)*sinh(b*x + a) + sqrt(2)*sinh(b*x + a)^2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(b*x + a) + sinh(b*x + a))) + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh (b*x + a)^3 + sinh(b*x + a)^4 + 6*(cosh(b*x + a)^2 + 2)*sinh(b*x + a)^2 + 12*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + 6*cosh(b*x + a))*sinh(b*x + a) - 1)*sqrt(sinh(b*x + a)))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)
\[ \int \sinh ^{\frac {5}{2}}(a+b x) \, dx=\int \sinh ^{\frac {5}{2}}{\left (a + b x \right )}\, dx \]
\[ \int \sinh ^{\frac {5}{2}}(a+b x) \, dx=\int { \sinh \left (b x + a\right )^{\frac {5}{2}} \,d x } \]
\[ \int \sinh ^{\frac {5}{2}}(a+b x) \, dx=\int { \sinh \left (b x + a\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \sinh ^{\frac {5}{2}}(a+b x) \, dx=\int {\mathrm {sinh}\left (a+b\,x\right )}^{5/2} \,d x \]