Integrand size = 12, antiderivative size = 88 \[ \int (b \sinh (c+d x))^{5/2} \, dx=\frac {6 i b^2 E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{5 d \sqrt {i \sinh (c+d x)}}+\frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d} \] Output:
-6/5*I*b^2*EllipticE(cos(1/2*I*c+1/4*Pi+1/2*I*d*x),2^(1/2))*(b*sinh(d*x+c) )^(1/2)/d/(I*sinh(d*x+c))^(1/2)+2/5*b*cosh(d*x+c)*(b*sinh(d*x+c))^(3/2)/d
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int (b \sinh (c+d x))^{5/2} \, dx=\frac {b^2 \sqrt {b \sinh (c+d x)} \left (-\frac {6 i E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right )}{\sqrt {i \sinh (c+d x)}}+\sinh (2 (c+d x))\right )}{5 d} \] Input:
Integrate[(b*Sinh[c + d*x])^(5/2),x]
Output:
(b^2*Sqrt[b*Sinh[c + d*x]]*(((-6*I)*EllipticE[((-2*I)*c + Pi - (2*I)*d*x)/ 4, 2])/Sqrt[I*Sinh[c + d*x]] + Sinh[2*(c + d*x)]))/(5*d)
Time = 0.33 (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, 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.
\(\displaystyle \int (b \sinh (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (-i b \sin (i c+i d x))^{5/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d}-\frac {3}{5} b^2 \int \sqrt {b \sinh (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d}-\frac {3}{5} b^2 \int \sqrt {-i b \sin (i c+i d x)}dx\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d}-\frac {3 b^2 \sqrt {b \sinh (c+d x)} \int \sqrt {i \sinh (c+d x)}dx}{5 \sqrt {i \sinh (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d}-\frac {3 b^2 \sqrt {b \sinh (c+d x)} \int \sqrt {\sin (i c+i d x)}dx}{5 \sqrt {i \sinh (c+d x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 b \cosh (c+d x) (b \sinh (c+d x))^{3/2}}{5 d}+\frac {6 i b^2 E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{5 d \sqrt {i \sinh (c+d x)}}\) |
Input:
Int[(b*Sinh[c + d*x])^(5/2),x]
Output:
(((6*I)/5)*b^2*EllipticE[(I*c - Pi/2 + I*d*x)/2, 2]*Sqrt[b*Sinh[c + d*x]]) /(d*Sqrt[I*Sinh[c + d*x]]) + (2*b*Cosh[c + d*x]*(b*Sinh[c + d*x])^(3/2))/( 5*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[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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (71 ) = 142\).
Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.93
method | result | size |
default | \(-\frac {b^{3} \left (6 \sqrt {-i \sinh \left (d x +c \right )+1}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-i \sinh \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-i \sinh \left (d x +c \right )+1}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-i \sinh \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cosh \left (d x +c \right )^{4}+2 \cosh \left (d x +c \right )^{2}\right )}{5 \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\) | \(170\) |
Input:
int((b*sinh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/5*b^3*(6*(-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)*EllipticE((-I*sinh(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*(-I*sin h(d*x+c)+1)^(1/2)*2^(1/2)*(1+I*sinh(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)*El lipticF((-I*sinh(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*cosh(d*x+c)^4+2*cosh(d*x+c )^2)/cosh(d*x+c)/(b*sinh(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (67) = 134\).
Time = 0.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.76 \[ \int (b \sinh (c+d x))^{5/2} \, dx=\frac {24 \, \sqrt {\frac {1}{2}} {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2}\right )} \sqrt {b} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + {\left (b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} + 12 \, b^{2} \cosh \left (d x + c\right )^{2} + 6 \, {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} - b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 6 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b \sinh \left (d x + c\right )}}{10 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2}\right )}} \] Input:
integrate((b*sinh(d*x+c))^(5/2),x, algorithm="fricas")
Output:
1/10*(24*sqrt(1/2)*(b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c ) + b^2*sinh(d*x + c)^2)*sqrt(b)*weierstrassZeta(4, 0, weierstrassPInverse (4, 0, cosh(d*x + c) + sinh(d*x + c))) + (b^2*cosh(d*x + c)^4 + 4*b^2*cosh (d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 12*b^2*cosh(d*x + c)^2 + 6*(b^2*cosh(d*x + c)^2 + 2*b^2)*sinh(d*x + c)^2 - b^2 + 4*(b^2*cosh(d*x + c)^3 + 6*b^2*cosh(d*x + c))*sinh(d*x + c))*sqrt(b*sinh(d*x + c)))/(d*cosh (d*x + c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2)
\[ \int (b \sinh (c+d x))^{5/2} \, dx=\int \left (b \sinh {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((b*sinh(d*x+c))**(5/2),x)
Output:
Integral((b*sinh(c + d*x))**(5/2), x)
\[ \int (b \sinh (c+d x))^{5/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((b*sinh(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((b*sinh(d*x + c))^(5/2), x)
\[ \int (b \sinh (c+d x))^{5/2} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((b*sinh(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((b*sinh(d*x + c))^(5/2), x)
Timed out. \[ \int (b \sinh (c+d x))^{5/2} \, dx=\int {\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int((b*sinh(c + d*x))^(5/2),x)
Output:
int((b*sinh(c + d*x))^(5/2), x)
\[ \int (b \sinh (c+d x))^{5/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sinh \left (d x +c \right )}\, \sinh \left (d x +c \right )^{2}d x \right ) b^{2} \] Input:
int((b*sinh(d*x+c))^(5/2),x)
Output:
sqrt(b)*int(sqrt(sinh(c + d*x))*sinh(c + d*x)**2,x)*b**2