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]

2/5*B*cosh(x)*(a+I*a*sinh(x))^(3/2)+8/15*a^2*(5*I*A+3*B)*cosh(x)/(a+I*a*sinh(x))^(1/2)+2/15*a*(5*I*A+3*B)*cosh
(x)*(a+I*a*sinh(x))^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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]

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 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)]

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*}

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Mathematica [A]  time = 0.22, 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]

-1/15*(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]))/(Cosh[x/2] + I*Sinh[x/2])

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fricas [A]  time = 0.67, size = 80, normalized size = 0.99 \[ \frac {1}{30} \, {\left (3 i \, B a e^{\left (5 \, x\right )} - 5 \, {\left (-2 i \, A - 3 \, B\right )} a e^{\left (4 \, x\right )} + {\left (90 \, A - 60 i \, B\right )} a e^{\left (3 \, x\right )} - 30 \, {\left (-3 i \, A - 2 \, B\right )} a e^{\left (2 \, x\right )} + {\left (10 \, A - 15 i \, B\right )} a e^{x} - 3 \, B a\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} e^{\left (-2 \, x\right )} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sinh \relax (x) + A\right )} {\left (i \, a \sinh \relax (x) + a\right )}^{\frac {3}{2}}\,{d x} \]

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)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (a +i a \sinh \relax (x )\right )^{\frac {3}{2}} \left (A +B \sinh \relax (x )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sinh \relax (x) + A\right )} {\left (i \, a \sinh \relax (x) + a\right )}^{\frac {3}{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\mathrm {sinh}\relax (x)\right )\,{\left (a+a\,\mathrm {sinh}\relax (x)\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sinh(x))*(a + a*sinh(x)*1i)^(3/2),x)

[Out]

int((A + B*sinh(x))*(a + a*sinh(x)*1i)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\sinh {\relax (x )} - i\right )\right )^{\frac {3}{2}} \left (A + B \sinh {\relax (x )}\right )\, dx \]

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]

Integral((I*a*(sinh(x) - I))**(3/2)*(A + B*sinh(x)), x)

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