3.113 \(\int (a+i a \sinh (x))^{3/2} (A+B \sinh (x)) \, dx\)

Optimal. Leaf size=81 \[ \frac{8 a^2 (3 B+5 i A) \cosh (x)}{15 \sqrt{a+i a \sinh (x)}}+\frac{2}{15} a (3 B+5 i A) \cosh (x) \sqrt{a+i a \sinh (x)}+\frac{2}{5} B \cosh (x) (a+i a \sinh (x))^{3/2} \]

[Out]

(8*a^2*((5*I)*A + 3*B)*Cosh[x])/(15*Sqrt[a + I*a*Sinh[x]]) + (2*a*((5*I)*A + 3*B)*Cosh[x]*Sqrt[a + I*a*Sinh[x]
])/15 + (2*B*Cosh[x]*(a + I*a*Sinh[x])^(3/2))/5

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Rubi [A]  time = 0.0808424, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2751, 2647, 2646} \[ \frac{8 a^2 (3 B+5 i A) \cosh (x)}{15 \sqrt{a+i a \sinh (x)}}+\frac{2}{15} a (3 B+5 i A) \cosh (x) \sqrt{a+i a \sinh (x)}+\frac{2}{5} B \cosh (x) (a+i a \sinh (x))^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Sinh[x])^(3/2)*(A + B*Sinh[x]),x]

[Out]

(8*a^2*((5*I)*A + 3*B)*Cosh[x])/(15*Sqrt[a + I*a*Sinh[x]]) + (2*a*((5*I)*A + 3*B)*Cosh[x]*Sqrt[a + I*a*Sinh[x]
])/15 + (2*B*Cosh[x]*(a + I*a*Sinh[x])^(3/2))/5

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

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 (x))^{3/2} (A+B \sinh (x)) \, dx &=\frac{2}{5} B \cosh (x) (a+i a \sinh (x))^{3/2}+\frac{1}{5} (5 A-3 i B) \int (a+i a \sinh (x))^{3/2} \, dx\\ &=\frac{2}{15} a (5 i A+3 B) \cosh (x) \sqrt{a+i a \sinh (x)}+\frac{2}{5} B \cosh (x) (a+i a \sinh (x))^{3/2}+\frac{1}{15} (4 a (5 A-3 i B)) \int \sqrt{a+i a \sinh (x)} \, dx\\ &=\frac{8 a^2 (5 i A+3 B) \cosh (x)}{15 \sqrt{a+i a \sinh (x)}}+\frac{2}{15} a (5 i A+3 B) \cosh (x) \sqrt{a+i a \sinh (x)}+\frac{2}{5} B \cosh (x) (a+i a \sinh (x))^{3/2}\\ \end{align*}

Mathematica [A]  time = 0.214721, size = 83, normalized size = 1.02 \[ -\frac{a \sqrt{a+i a \sinh (x)} \left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right ) (2 (5 A-9 i B) \sinh (x)-50 i A+3 B \cosh (2 x)-39 B)}{15 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Sinh[x])^(3/2)*(A + B*Sinh[x]),x]

[Out]

-(a*(Cosh[x/2] - I*Sinh[x/2])*Sqrt[a + I*a*Sinh[x]]*((-50*I)*A - 39*B + 3*B*Cosh[2*x] + 2*(5*A - (9*I)*B)*Sinh
[x]))/(15*(Cosh[x/2] + I*Sinh[x/2]))

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Maple [F]  time = 0.108, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\sinh \left ( x \right ) \right ) ^{{\frac{3}{2}}} \left ( A+B\sinh \left ( x \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*sinh(x))^(3/2)*(A+B*sinh(x)),x)

[Out]

int((a+I*a*sinh(x))^(3/2)*(A+B*sinh(x)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sinh \left (x\right ) + A\right )}{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="maxima")

[Out]

integrate((B*sinh(x) + A)*(I*a*sinh(x) + a)^(3/2), x)

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Fricas [A]  time = 1.76496, size = 294, normalized size = 3.63 \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left (3 i \, B a e^{\left (5 \, x\right )} +{\left (10 i \, A + 15 \, B\right )} a e^{\left (4 \, x\right )} + 30 \,{\left (3 \, A - 2 i \, B\right )} a e^{\left (3 \, x\right )} +{\left (90 i \, A + 60 \, B\right )} a e^{\left (2 \, x\right )} + 5 \,{\left (2 \, A - 3 i \, B\right )} a e^{x} - 3 \, B a\right )} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a} e^{\left (-\frac{1}{2} \, x\right )}}{30 \,{\left (e^{\left (3 \, x\right )} - i \, e^{\left (2 \, x\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="fricas")

[Out]

1/30*sqrt(1/2)*(3*I*B*a*e^(5*x) + (10*I*A + 15*B)*a*e^(4*x) + 30*(3*A - 2*I*B)*a*e^(3*x) + (90*I*A + 60*B)*a*e
^(2*x) + 5*(2*A - 3*I*B)*a*e^x - 3*B*a)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*e^(-1/2*x)/(e^(3*x) - I*e^(2*x))

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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(x))**(3/2)*(A+B*sinh(x)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sinh \left (x\right ) + A\right )}{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(x))^(3/2)*(A+B*sinh(x)),x, algorithm="giac")

[Out]

integrate((B*sinh(x) + A)*(I*a*sinh(x) + a)^(3/2), x)