Integrand size = 14, antiderivative size = 91 \[ \int (i \sinh (c+d x))^{7/2} \, dx=-\frac {10 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right )}{21 d}+\frac {10 i \cosh (c+d x) \sqrt {i \sinh (c+d x)}}{21 d}+\frac {2 i \cosh (c+d x) (i \sinh (c+d x))^{5/2}}{7 d} \] Output:
-10/21*I*InverseJacobiAM(1/2*I*c-1/4*Pi+1/2*I*d*x,2^(1/2))/d+10/21*I*cosh( d*x+c)*(I*sinh(d*x+c))^(1/2)/d+2/7*I*cosh(d*x+c)*(I*sinh(d*x+c))^(5/2)/d
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71 \[ \int (i \sinh (c+d x))^{7/2} \, dx=\frac {i \left (20 \operatorname {EllipticF}\left (\frac {1}{4} (-2 i c+\pi -2 i d x),2\right )+(23 \cosh (c+d x)-3 \cosh (3 (c+d x))) \sqrt {i \sinh (c+d x)}\right )}{42 d} \] Input:
Integrate[(I*Sinh[c + d*x])^(7/2),x]
Output:
((I/42)*(20*EllipticF[((-2*I)*c + Pi - (2*I)*d*x)/4, 2] + (23*Cosh[c + d*x ] - 3*Cosh[3*(c + d*x)])*Sqrt[I*Sinh[c + d*x]]))/d
Time = 0.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3115, 3042, 3115, 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.
\(\displaystyle \int (i \sinh (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (i c+i d x)^{7/2}dx\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{7} \int (i \sinh (c+d x))^{3/2}dx+\frac {2 i (i \sinh (c+d x))^{5/2} \cosh (c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} \int \sin (i c+i d x)^{3/2}dx+\frac {2 i (i \sinh (c+d x))^{5/2} \cosh (c+d x)}{7 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {i \sinh (c+d x)}}dx+\frac {2 i \sqrt {i \sinh (c+d x)} \cosh (c+d x)}{3 d}\right )+\frac {2 i (i \sinh (c+d x))^{5/2} \cosh (c+d x)}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (i c+i d x)}}dx+\frac {2 i \sqrt {i \sinh (c+d x)} \cosh (c+d x)}{3 d}\right )+\frac {2 i (i \sinh (c+d x))^{5/2} \cosh (c+d x)}{7 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 i (i \sinh (c+d x))^{5/2} \cosh (c+d x)}{7 d}+\frac {5}{7} \left (\frac {2 i \sqrt {i \sinh (c+d x)} \cosh (c+d x)}{3 d}-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right )}{3 d}\right )\) |
Input:
Int[(I*Sinh[c + d*x])^(7/2),x]
Output:
(5*((((-2*I)/3)*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2])/d + (((2*I)/3)*Cosh[ c + d*x]*Sqrt[I*Sinh[c + d*x]])/d))/7 + (((2*I)/7)*Cosh[c + d*x]*(I*Sinh[c + d*x])^(5/2))/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]
Time = 0.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.34
method | result | size |
default | \(-\frac {i \left (6 i \cosh \left (d x +c \right )^{4} \sinh \left (d x +c \right )-5 \sqrt {-i \sinh \left (d x +c \right )+1}\, \sqrt {2}\, \sqrt {1+i \sinh \left (d x +c \right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-i \sinh \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-16 i \cosh \left (d x +c \right )^{2} \sinh \left (d x +c \right )\right )}{21 \cosh \left (d x +c \right ) \sqrt {i \sinh \left (d x +c \right )}\, d}\) | \(122\) |
Input:
int((I*sinh(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/21*I*(6*I*cosh(d*x+c)^4*sinh(d*x+c)-5*(-I*sinh(d*x+c)+1)^(1/2)*2^(1/2)* (1+I*sinh(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)*EllipticF((-I*sinh(d*x+c)+1) ^(1/2),1/2*2^(1/2))-16*I*cosh(d*x+c)^2*sinh(d*x+c))/cosh(d*x+c)/(I*sinh(d* x+c))^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int (i \sinh (c+d x))^{7/2} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} {\left (-3 i \, e^{\left (6 \, d x + 6 \, c\right )} + 23 i \, e^{\left (4 \, d x + 4 \, c\right )} + 23 i \, e^{\left (2 \, d x + 2 \, c\right )} - 3 i\right )} \sqrt {i \, e^{\left (2 \, d x + 2 \, c\right )} - i} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )} - 80 i \, \sqrt {\frac {1}{2} i} e^{\left (3 \, d x + 3 \, c\right )} {\rm weierstrassPInverse}\left (4, 0, e^{\left (d x + c\right )}\right )\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{84 \, d} \] Input:
integrate((I*sinh(d*x+c))^(7/2),x, algorithm="fricas")
Output:
1/84*(sqrt(1/2)*(-3*I*e^(6*d*x + 6*c) + 23*I*e^(4*d*x + 4*c) + 23*I*e^(2*d *x + 2*c) - 3*I)*sqrt(I*e^(2*d*x + 2*c) - I)*e^(-1/2*d*x - 1/2*c) - 80*I*s qrt(1/2*I)*e^(3*d*x + 3*c)*weierstrassPInverse(4, 0, e^(d*x + c)))*e^(-3*d *x - 3*c)/d
Timed out. \[ \int (i \sinh (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate((I*sinh(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int (i \sinh (c+d x))^{7/2} \, dx=\int { \left (i \, \sinh \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((I*sinh(d*x+c))^(7/2),x, algorithm="maxima")
Output:
integrate((I*sinh(d*x + c))^(7/2), x)
\[ \int (i \sinh (c+d x))^{7/2} \, dx=\int { \left (i \, \sinh \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((I*sinh(d*x+c))^(7/2),x, algorithm="giac")
Output:
integrate((I*sinh(d*x + c))^(7/2), x)
Timed out. \[ \int (i \sinh (c+d x))^{7/2} \, dx=\int {\left (\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \] Input:
int((sinh(c + d*x)*1i)^(7/2),x)
Output:
int((sinh(c + d*x)*1i)^(7/2), x)
\[ \int (i \sinh (c+d x))^{7/2} \, dx=-\sqrt {i}\, \left (\int \sqrt {\sinh \left (d x +c \right )}\, \sinh \left (d x +c \right )^{3}d x \right ) i \] Input:
int((I*sinh(d*x+c))^(7/2),x)
Output:
- sqrt(i)*int(sqrt(sinh(c + d*x))*sinh(c + d*x)**3,x)*i