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

Optimal. Leaf size=104 \[ \frac{64 i a^3 \cosh (c+d x)}{15 d \sqrt{a+i a \sinh (c+d x)}}+\frac{16 i a^2 \cosh (c+d x) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d} \]

[Out]

(((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

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Rubi [A]  time = 0.0531309, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2647, 2646} \[ \frac{64 i a^3 \cosh (c+d x)}{15 d \sqrt{a+i a \sinh (c+d x)}}+\frac{16 i a^2 \cosh (c+d x) \sqrt{a+i a \sinh (c+d x)}}{15 d}+\frac{2 i a \cosh (c+d x) (a+i a \sinh (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[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 2647

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] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

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]

Rubi steps

\begin{align*} \int (a+i a \sinh (c+d x))^{5/2} \, dx &=\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]  time = 0.452801, size = 145, normalized size = 1.39 \[ \frac{a^2 (\sinh (c+d x)-i)^2 \sqrt{a+i a \sinh (c+d x)} \left (-150 \sinh \left (\frac{1}{2} (c+d x)\right )+25 \sinh \left (\frac{3}{2} (c+d x)\right )+3 \sinh \left (\frac{5}{2} (c+d x)\right )-150 i \cosh \left (\frac{1}{2} (c+d x)\right )-25 i \cosh \left (\frac{3}{2} (c+d x)\right )+3 i \cosh \left (\frac{5}{2} (c+d x)\right )\right )}{30 d \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.14, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\sinh \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.98379, size = 360, normalized size = 3.46 \begin{align*} -\frac{\sqrt{\frac{1}{2}}{\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{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )}}{30 \, d e^{\left (3 \, d x + 3 \, c\right )} - 30 i \, d e^{\left (2 \, d x + 2 \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(1/2)*(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(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - I*a)*e^(-1/2*d*x - 1/2*c)/(3
0*d*e^(3*d*x + 3*c) - 30*I*d*e^(2*d*x + 2*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)