3.109 \(\int \frac{1}{(a+b \sinh (x))^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ -\frac{2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{2 i \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{\left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

[Out]

(-2*b*Cosh[x])/((a^2 + b^2)*Sqrt[a + b*Sinh[x]]) + ((2*I)*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a +
b*Sinh[x]])/((a^2 + b^2)*Sqrt[(a + b*Sinh[x])/(a - I*b)])

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Rubi [A]  time = 0.0597764, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2664, 21, 2655, 2653} \[ -\frac{2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{2 i \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{\left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[x])^(-3/2),x]

[Out]

(-2*b*Cosh[x])/((a^2 + b^2)*Sqrt[a + b*Sinh[x]]) + ((2*I)*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a +
b*Sinh[x]])/((a^2 + b^2)*Sqrt[(a + b*Sinh[x])/(a - I*b)])

Rule 2664

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 + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+b \sinh (x))^{3/2}} \, dx &=-\frac{2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}-\frac{2 \int \frac{-\frac{a}{2}-\frac{1}{2} b \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx}{a^2+b^2}\\ &=-\frac{2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{\int \sqrt{a+b \sinh (x)} \, dx}{a^2+b^2}\\ &=-\frac{2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{\sqrt{a+b \sinh (x)} \int \sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}} \, dx}{\left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}}\\ &=-\frac{2 b \cosh (x)}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{2 i E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{a+b \sinh (x)}}{\left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}}\\ \end{align*}

Mathematica [A]  time = 0.15025, size = 81, normalized size = 0.86 \[ \frac{-2 b \cosh (x)+2 (b+i a) \sqrt{\frac{a+b \sinh (x)}{a-i b}} E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )}{\left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[x])^(-3/2),x]

[Out]

(-2*b*Cosh[x] + 2*(I*a + b)*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*Sqrt[(a + b*Sinh[x])/(a - I*b)])
/((a^2 + b^2)*Sqrt[a + b*Sinh[x]])

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Maple [B]  time = 0.095, size = 456, normalized size = 4.9 \begin{align*} 2\,{\frac{1}{ \left ({a}^{2}+{b}^{2} \right ) b\cosh \left ( x \right ) \sqrt{a+b\sinh \left ( x \right ) }} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}}{\it EllipticF} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ){a}^{2}+\sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}}{\it EllipticF} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ){b}^{2}-\sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}}{\it EllipticE} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ){a}^{2}-\sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}}\sqrt{{\frac{ \left ( i-\sinh \left ( x \right ) \right ) b}{ib+a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( x \right ) \right ) b}{ib-a}}}{\it EllipticE} \left ( \sqrt{-{\frac{a+b\sinh \left ( x \right ) }{ib-a}}},\sqrt{-{\frac{ib-a}{ib+a}}} \right ){b}^{2}-{b}^{2} \left ( \sinh \left ( x \right ) \right ) ^{2}-{b}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sinh(x))^(3/2),x)

[Out]

2*((-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticF((-(a+b
*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^2+(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^
(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticF((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*b^2-(-(
a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticE((-(a+b*sinh(
x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*a^2-(-(a+b*sinh(x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*
((I+sinh(x))*b/(I*b-a))^(1/2)*EllipticE((-(a+b*sinh(x))/(I*b-a))^(1/2),(-(I*b-a)/(I*b+a))^(1/2))*b^2-b^2*sinh(
x)^2-b^2)/(a^2+b^2)/b/cosh(x)/(a+b*sinh(x))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \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(1/(a+b*sinh(x))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sinh(x) + a)^(-3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sinh \left (x\right ) + a}}{b^{2} \sinh \left (x\right )^{2} + 2 \, a b \sinh \left (x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sinh(x) + a)/(b^2*sinh(x)^2 + 2*a*b*sinh(x) + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \sinh{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x))**(3/2),x)

[Out]

Integral((a + b*sinh(x))**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \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(1/(a+b*sinh(x))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sinh(x) + a)^(-3/2), x)