3.110 \(\int \frac{1}{(a+b \sinh (x))^{5/2}} \, dx\)

Optimal. Leaf size=197 \[ -\frac{2 i \sqrt{\frac{a+b \sinh (x)}{a-i b}} \text{EllipticF}\left (\frac{\pi }{4}-\frac{i x}{2},\frac{2 b}{b+i a}\right )}{3 \left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}-\frac{8 a b \cosh (x)}{3 \left (a^2+b^2\right )^2 \sqrt{a+b \sinh (x)}}-\frac{2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac{8 i a \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 \left (a^2+b^2\right )^2 \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

[Out]

(-2*b*Cosh[x])/(3*(a^2 + b^2)*(a + b*Sinh[x])^(3/2)) - (8*a*b*Cosh[x])/(3*(a^2 + b^2)^2*Sqrt[a + b*Sinh[x]]) +
(((8*I)/3)*a*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/((a^2 + b^2)^2*Sqrt[(a + b*Sinh[
x])/(a - I*b)]) - (((2*I)/3)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[(a + b*Sinh[x])/(a - I*b)])/((a^2
+ b^2)*Sqrt[a + b*Sinh[x]])

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Rubi [A]  time = 0.211873, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {2664, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{8 a b \cosh (x)}{3 \left (a^2+b^2\right )^2 \sqrt{a+b \sinh (x)}}-\frac{2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}-\frac{2 i \sqrt{\frac{a+b \sinh (x)}{a-i b}} F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 \left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}+\frac{8 i a \sqrt{a+b \sinh (x)} E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right )}{3 \left (a^2+b^2\right )^2 \sqrt{\frac{a+b \sinh (x)}{a-i b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[x])^(-5/2),x]

[Out]

(-2*b*Cosh[x])/(3*(a^2 + b^2)*(a + b*Sinh[x])^(3/2)) - (8*a*b*Cosh[x])/(3*(a^2 + b^2)^2*Sqrt[a + b*Sinh[x]]) +
(((8*I)/3)*a*EllipticE[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/((a^2 + b^2)^2*Sqrt[(a + b*Sinh[
x])/(a - I*b)]) - (((2*I)/3)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[(a + b*Sinh[x])/(a - I*b)])/((a^2
+ b^2)*Sqrt[a + b*Sinh[x]])

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 2754

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

Rule 2752

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/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 2661

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

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))^{5/2}} \, dx &=-\frac{2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}-\frac{2 \int \frac{-\frac{3 a}{2}+\frac{1}{2} b \sinh (x)}{(a+b \sinh (x))^{3/2}} \, dx}{3 \left (a^2+b^2\right )}\\ &=-\frac{2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}-\frac{8 a b \cosh (x)}{3 \left (a^2+b^2\right )^2 \sqrt{a+b \sinh (x)}}+\frac{4 \int \frac{\frac{1}{4} \left (3 a^2-b^2\right )+a b \sinh (x)}{\sqrt{a+b \sinh (x)}} \, dx}{3 \left (a^2+b^2\right )^2}\\ &=-\frac{2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}-\frac{8 a b \cosh (x)}{3 \left (a^2+b^2\right )^2 \sqrt{a+b \sinh (x)}}+\frac{(4 a) \int \sqrt{a+b \sinh (x)} \, dx}{3 \left (a^2+b^2\right )^2}-\frac{\int \frac{1}{\sqrt{a+b \sinh (x)}} \, dx}{3 \left (a^2+b^2\right )}\\ &=-\frac{2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}-\frac{8 a b \cosh (x)}{3 \left (a^2+b^2\right )^2 \sqrt{a+b \sinh (x)}}+\frac{\left (4 a \sqrt{a+b \sinh (x)}\right ) \int \sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}} \, dx}{3 \left (a^2+b^2\right )^2 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{\sqrt{\frac{a+b \sinh (x)}{a-i b}} \int \frac{1}{\sqrt{\frac{a}{a-i b}+\frac{b \sinh (x)}{a-i b}}} \, dx}{3 \left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}\\ &=-\frac{2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}-\frac{8 a b \cosh (x)}{3 \left (a^2+b^2\right )^2 \sqrt{a+b \sinh (x)}}+\frac{8 i a E\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{a+b \sinh (x)}}{3 \left (a^2+b^2\right )^2 \sqrt{\frac{a+b \sinh (x)}{a-i b}}}-\frac{2 i F\left (\frac{\pi }{4}-\frac{i x}{2}|\frac{2 b}{i a+b}\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}}}{3 \left (a^2+b^2\right ) \sqrt{a+b \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.61847, size = 166, normalized size = 0.84 \[ \frac{-2 i \left (a^2+b^2\right ) \sqrt{\frac{a+b \sinh (x)}{a-i b}} (a+b \sinh (x)) \text{EllipticF}\left (\frac{1}{4} (\pi -2 i x),-\frac{2 i b}{a-i b}\right )-2 b \cosh (x) \left (5 a^2+4 a b \sinh (x)+b^2\right )+\frac{8 i a (a+b \sinh (x))^2 E\left (\frac{1}{4} (\pi -2 i x)|-\frac{2 i b}{a-i b}\right )}{\sqrt{\frac{a+b \sinh (x)}{a-i b}}}}{3 \left (a^2+b^2\right )^2 (a+b \sinh (x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[x])^(-5/2),x]

[Out]

(((8*I)*a*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*(a + b*Sinh[x])^2)/Sqrt[(a + b*Sinh[x])/(a - I*b)]
- (2*I)*(a^2 + b^2)*EllipticF[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*(a + b*Sinh[x])*Sqrt[(a + b*Sinh[x])/(a
- I*b)] - 2*b*Cosh[x]*(5*a^2 + b^2 + 4*a*b*Sinh[x]))/(3*(a^2 + b^2)^2*(a + b*Sinh[x])^(3/2))

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Maple [A]  time = 0.147, size = 438, normalized size = 2.2 \begin{align*}{\frac{1}{\cosh \left ( x \right ) }\sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}} \left ( -{\frac{2}{3\,b \left ({a}^{2}+{b}^{2} \right ) }\sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}} \left ( \sinh \left ( x \right ) +{\frac{a}{b}} \right ) ^{-2}}-{\frac{8\,b \left ( \cosh \left ( x \right ) \right ) ^{2}a}{3\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}}}}}+2\,{\frac{3\,{a}^{2}-{b}^{2}}{ \left ( 3\,{a}^{4}+6\,{a}^{2}{b}^{2}+3\,{b}^{4} \right ) \sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}}} \left ({\frac{a}{b}}-i \right ) \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{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{-b\sinh \left ( x \right ) -a}{ib-a}}},\sqrt{{\frac{a-ib}{ib+a}}} \right ) }+{\frac{8\,ab}{3\, \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ({\frac{a}{b}}-i \right ) \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{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}}} \left ( \left ( -{\frac{a}{b}}-i \right ){\it EllipticE} \left ( \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}},\sqrt{{\frac{a-ib}{ib+a}}} \right ) +i{\it EllipticF} \left ( \sqrt{{\frac{-b\sinh \left ( x \right ) -a}{ib-a}}},\sqrt{{\frac{a-ib}{ib+a}}} \right ) \right ){\frac{1}{\sqrt{ \left ( a+b\sinh \left ( x \right ) \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}}}}} \right ){\frac{1}{\sqrt{a+b\sinh \left ( x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sinh(x))^(5/2),x)

[Out]

((a+b*sinh(x))*cosh(x)^2)^(1/2)*(-2/3/b/(a^2+b^2)*((a+b*sinh(x))*cosh(x)^2)^(1/2)/(sinh(x)+a/b)^2-8/3*b*cosh(x
)^2/(a^2+b^2)^2*a/((a+b*sinh(x))*cosh(x)^2)^(1/2)+2*(3*a^2-b^2)/(3*a^4+6*a^2*b^2+3*b^4)*(a/b-I)*((-b*sinh(x)-a
)/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)*E
llipticF(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+8/3*a*b/(a^2+b^2)^2*(a/b-I)*((-b*sinh(x)-a)/(
I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*((I+sinh(x))*b/(I*b-a))^(1/2)/((a+b*sinh(x))*cosh(x)^2)^(1/2)*((-a
/b-I)*EllipticE(((-b*sinh(x)-a)/(I*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+I*EllipticF(((-b*sinh(x)-a)/(I*b-a))^(
1/2),((a-I*b)/(I*b+a))^(1/2))))/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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sinh(x) + a)^(-5/2), 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^{3} \sinh \left (x\right )^{3} + 3 \, a b^{2} \sinh \left (x\right )^{2} + 3 \, a^{2} b \sinh \left (x\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sinh(x) + a)/(b^3*sinh(x)^3 + 3*a*b^2*sinh(x)^2 + 3*a^2*b*sinh(x) + a^3), 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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x))**(5/2),x)

[Out]

Integral((a + b*sinh(x))**(-5/2), 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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sinh(x) + a)^(-5/2), x)