\(\int (a+i a \sinh (c+d x))^{5/2} \, dx\) [66]

Optimal result

Integrand size = 17, antiderivative size = 104 \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\frac {64 i a^3 \cosh (c+d x)}{15 d \sqrt {a+i a \sinh (c+d x)}}+\frac {16 i a^2 \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d} \]

[Out]

2/5*I*a*cosh(d*x+c)*(a+I*a*sinh(d*x+c))^(3/2)/d+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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2726, 2725} \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\frac {64 i a^3 \cosh (c+d x)}{15 d \sqrt {a+i a \sinh (c+d x)}}+\frac {16 i a^2 \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{15 d}+\frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d} \]

[In]

Int[(a + I*a*Sinh[c + d*x])^(5/2),x]

[Out]

(((64*I)/15)*a^3*Cosh[c + d*x])/(d*Sqrt[a + I*a*Sinh[c + d*x]]) + (((16*I)/15)*a^2*Cosh[c + d*x]*Sqrt[a + I*a*
Sinh[c + d*x]])/d + (((2*I)/5)*a*Cosh[c + d*x]*(a + I*a*Sinh[c + d*x])^(3/2))/d

Rule 2725

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] && EqQ[a^2 - b^2, 0]

Rule 2726

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d}+\frac {1}{5} (8 a) \int (a+i a \sinh (c+d x))^{3/2} \, dx \\ & = \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}+\frac {1}{15} \left (32 a^2\right ) \int \sqrt {a+i a \sinh (c+d x)} \, 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.04 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.39 \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\frac {a^2 (-i+\sinh (c+d x))^2 \sqrt {a+i a \sinh (c+d x)} \left (-150 i \cosh \left (\frac {1}{2} (c+d x)\right )-25 i \cosh \left (\frac {3}{2} (c+d x)\right )+3 i \cosh \left (\frac {5}{2} (c+d x)\right )-150 \sinh \left (\frac {1}{2} (c+d x)\right )+25 \sinh \left (\frac {3}{2} (c+d x)\right )+3 \sinh \left (\frac {5}{2} (c+d x)\right )\right )}{30 d \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

[In]

Integrate[(a + I*a*Sinh[c + d*x])^(5/2),x]

[Out]

(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] - 150*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)

Maple [F]

\[\int \left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}d x\]

[In]

int((a+I*a*sinh(d*x+c))^(5/2),x)

[Out]

int((a+I*a*sinh(d*x+c))^(5/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97 \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=-\frac {{\left (3 \, a^{2} e^{\left (5 \, d x + 5 \, c\right )} - 25 i \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 150 \, a^{2} e^{\left (3 \, d x + 3 \, c\right )} - 150 i \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 25 \, a^{2} e^{\left (d x + c\right )} + 3 i \, a^{2}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} e^{\left (-2 \, d x - 2 \, c\right )}}{30 \, d} \]

[In]

integrate((a+I*a*sinh(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-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

Sympy [F(-1)]

Timed out. \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*sinh(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\int { {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+I*a*sinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((I*a*sinh(d*x + c) + a)^(5/2), x)

Giac [F]

\[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\int { {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+I*a*sinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*sinh(d*x + c) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+i a \sinh (c+d x))^{5/2} \, dx=\int {\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

[In]

int((a + a*sinh(c + d*x)*1i)^(5/2),x)

[Out]

int((a + a*sinh(c + d*x)*1i)^(5/2), x)