Integrand size = 17, antiderivative size = 104 \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\frac {64 i a^3 \cosh (c+d x)}{15 d \sqrt {a+i a \sinh (c+d x)}}+\frac {16 i a^2 \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d} \] Output:
64/15*I*a^3*cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))^(1/2)+16/15*I*a^2*cosh(d*x+c )*(a+I*a*sinh(d*x+c))^(1/2)/d+2/5*I*a*cosh(d*x+c)*(a+I*a*sinh(d*x+c))^(3/2 )/d
Time = 5.96 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.39 \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\frac {a^2 (-i+\sinh (c+d x))^2 \sqrt {a+i a \sinh (c+d x)} \left (-150 i \cosh \left (\frac {1}{2} (c+d x)\right )-25 i \cosh \left (\frac {3}{2} (c+d x)\right )+3 i \cosh \left (\frac {5}{2} (c+d x)\right )-150 \sinh \left (\frac {1}{2} (c+d x)\right )+25 \sinh \left (\frac {3}{2} (c+d x)\right )+3 \sinh \left (\frac {5}{2} (c+d x)\right )\right )}{30 d \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \] Input:
Integrate[(a + I*a*Sinh[c + d*x])^(5/2),x]
Output:
(a^2*(-I + Sinh[c + d*x])^2*Sqrt[a + I*a*Sinh[c + d*x]]*((-150*I)*Cosh[(c + d*x)/2] - (25*I)*Cosh[(3*(c + d*x))/2] + (3*I)*Cosh[(5*(c + d*x))/2] - 1 50*Sinh[(c + d*x)/2] + 25*Sinh[(3*(c + d*x))/2] + 3*Sinh[(5*(c + d*x))/2]) )/(30*d*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^5)
Time = 0.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 3126, 3042, 3126, 3042, 3125}
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.
\(\displaystyle \int (a+i a \sinh (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+a \sin (i c+i d x))^{5/2}dx\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {8}{5} a \int (i \sinh (c+d x) a+a)^{3/2}dx+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} a \int (\sin (i c+i d x) a+a)^{3/2}dx+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {i \sinh (c+d x) a+a}dx+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}\right )+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (i c+i d x) a+a}dx+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}\right )+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {8}{5} a \left (\frac {8 i a^2 \cosh (c+d x)}{3 d \sqrt {a+i a \sinh (c+d x)}}+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}\right )+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d}\) |
Input:
Int[(a + I*a*Sinh[c + d*x])^(5/2),x]
Output:
(((2*I)/5)*a*Cosh[c + d*x]*(a + I*a*Sinh[c + d*x])^(3/2))/d + (8*a*((((8*I )/3)*a^2*Cosh[c + d*x])/(d*Sqrt[a + I*a*Sinh[c + d*x]]) + (((2*I)/3)*a*Cos h[c + d*x]*Sqrt[a + I*a*Sinh[c + d*x]])/d))/5
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {8 a^{3} \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \left (3 \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{4}+4 \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2}+8\right ) \sqrt {2}}{15 \sqrt {a \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2}}\, d}\) | \(93\) |
Input:
int((a+I*a*sinh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
8/15*a^3*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*sinh(1/2*c+1/4*I*Pi+1/2*d*x)*(3*cosh (1/2*c+1/4*I*Pi+1/2*d*x)^4+4*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^2+8)*2^(1/2)/(a* cosh(1/2*c+1/4*I*Pi+1/2*d*x)^2)^(1/2)/d
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97 \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=-\frac {{\left (3 \, a^{2} e^{\left (5 \, d x + 5 \, c\right )} - 25 i \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 150 \, a^{2} e^{\left (3 \, d x + 3 \, c\right )} - 150 i \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 25 \, a^{2} e^{\left (d x + c\right )} + 3 i \, a^{2}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} e^{\left (-2 \, d x - 2 \, c\right )}}{30 \, d} \] Input:
integrate((a+I*a*sinh(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-1/30*(3*a^2*e^(5*d*x + 5*c) - 25*I*a^2*e^(4*d*x + 4*c) - 150*a^2*e^(3*d*x + 3*c) - 150*I*a^2*e^(2*d*x + 2*c) - 25*a^2*e^(d*x + c) + 3*I*a^2)*sqrt(1 /2*I*a*e^(-d*x - c))*e^(-2*d*x - 2*c)/d
Timed out. \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*sinh(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\int { {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+I*a*sinh(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((I*a*sinh(d*x + c) + a)^(5/2), x)
\[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\int { {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+I*a*sinh(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate((I*a*sinh(d*x + c) + a)^(5/2), x)
Timed out. \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\int {\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \] Input:
int((a + a*sinh(c + d*x)*1i)^(5/2),x)
Output:
int((a + a*sinh(c + d*x)*1i)^(5/2), x)
\[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sinh \left (d x +c \right ) i +1}d x -\left (\int \sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )^{2}d x \right )+2 \left (\int \sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )d x \right ) i \right ) \] Input:
int((a+I*a*sinh(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sinh(c + d*x)*i + 1),x) - int(sqrt(sinh(c + d*x)*i + 1)*sinh(c + d*x)**2,x) + 2*int(sqrt(sinh(c + d*x)*i + 1)*sinh(c + d*x),x )*i)