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

Optimal. Leaf size=177 \[ \frac{2 i \sqrt{\frac{a+b \cosh (x)}{a+b}} \text{EllipticF}\left (\frac{i x}{2},\frac{2 b}{a+b}\right )}{3 \left (a^2-b^2\right ) \sqrt{a+b \cosh (x)}}-\frac{8 a b \sinh (x)}{3 \left (a^2-b^2\right )^2 \sqrt{a+b \cosh (x)}}-\frac{2 b \sinh (x)}{3 \left (a^2-b^2\right ) (a+b \cosh (x))^{3/2}}-\frac{8 i a \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cosh (x)}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.206961, antiderivative size = 177, 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} \[ -\frac{8 a b \sinh (x)}{3 \left (a^2-b^2\right )^2 \sqrt{a+b \cosh (x)}}-\frac{2 b \sinh (x)}{3 \left (a^2-b^2\right ) (a+b \cosh (x))^{3/2}}+\frac{2 i \sqrt{\frac{a+b \cosh (x)}{a+b}} F\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{3 \left (a^2-b^2\right ) \sqrt{a+b \cosh (x)}}-\frac{8 i a \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cosh (x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[x])^(-5/2),x]

[Out]

(((-8*I)/3)*a*Sqrt[a + b*Cosh[x]]*EllipticE[(I/2)*x, (2*b)/(a + b)])/((a^2 - b^2)^2*Sqrt[(a + b*Cosh[x])/(a +
b)]) + (((2*I)/3)*Sqrt[(a + b*Cosh[x])/(a + b)]*EllipticF[(I/2)*x, (2*b)/(a + b)])/((a^2 - b^2)*Sqrt[a + b*Cos
h[x]]) - (2*b*Sinh[x])/(3*(a^2 - b^2)*(a + b*Cosh[x])^(3/2)) - (8*a*b*Sinh[x])/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos
h[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 \cosh (x))^{5/2}} \, dx &=-\frac{2 b \sinh (x)}{3 \left (a^2-b^2\right ) (a+b \cosh (x))^{3/2}}-\frac{2 \int \frac{-\frac{3 a}{2}+\frac{1}{2} b \cosh (x)}{(a+b \cosh (x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{2 b \sinh (x)}{3 \left (a^2-b^2\right ) (a+b \cosh (x))^{3/2}}-\frac{8 a b \sinh (x)}{3 \left (a^2-b^2\right )^2 \sqrt{a+b \cosh (x)}}+\frac{4 \int \frac{\frac{1}{4} \left (3 a^2+b^2\right )+a b \cosh (x)}{\sqrt{a+b \cosh (x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac{2 b \sinh (x)}{3 \left (a^2-b^2\right ) (a+b \cosh (x))^{3/2}}-\frac{8 a b \sinh (x)}{3 \left (a^2-b^2\right )^2 \sqrt{a+b \cosh (x)}}+\frac{(4 a) \int \sqrt{a+b \cosh (x)} \, dx}{3 \left (a^2-b^2\right )^2}-\frac{\int \frac{1}{\sqrt{a+b \cosh (x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{2 b \sinh (x)}{3 \left (a^2-b^2\right ) (a+b \cosh (x))^{3/2}}-\frac{8 a b \sinh (x)}{3 \left (a^2-b^2\right )^2 \sqrt{a+b \cosh (x)}}+\frac{\left (4 a \sqrt{a+b \cosh (x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cosh (x)}{a+b}} \, dx}{3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cosh (x)}{a+b}}}-\frac{\sqrt{\frac{a+b \cosh (x)}{a+b}} \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cosh (x)}{a+b}}} \, dx}{3 \left (a^2-b^2\right ) \sqrt{a+b \cosh (x)}}\\ &=-\frac{8 i a \sqrt{a+b \cosh (x)} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cosh (x)}{a+b}}}+\frac{2 i \sqrt{\frac{a+b \cosh (x)}{a+b}} F\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{3 \left (a^2-b^2\right ) \sqrt{a+b \cosh (x)}}-\frac{2 b \sinh (x)}{3 \left (a^2-b^2\right ) (a+b \cosh (x))^{3/2}}-\frac{8 a b \sinh (x)}{3 \left (a^2-b^2\right )^2 \sqrt{a+b \cosh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.526546, size = 135, normalized size = 0.76 \[ \frac{2 i (a-b) (a+b)^2 \left (\frac{a+b \cosh (x)}{a+b}\right )^{3/2} \text{EllipticF}\left (\frac{i x}{2},\frac{2 b}{a+b}\right )+2 b \sinh (x) \left (-5 a^2-4 a b \cosh (x)+b^2\right )-8 i a (a+b)^2 \left (\frac{a+b \cosh (x)}{a+b}\right )^{3/2} E\left (\frac{i x}{2}|\frac{2 b}{a+b}\right )}{3 (a-b)^2 (a+b)^2 (a+b \cosh (x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[x])^(-5/2),x]

[Out]

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

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Maple [B]  time = 0.147, size = 459, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*cosh(1/2*x)^2*b+a-b)*sinh(1/2*x)^2)^(1/2)*(-1/3/b/(a-b)/(a+b)*cosh(1/2*x)*(2*b*sinh(1/2*x)^4+(a+b)*sinh(1/
2*x)^2)^(1/2)/(cosh(1/2*x)^2+1/2*(a-b)/b)^2-16/3*b*sinh(1/2*x)^2/(a-b)^2/(a+b)^2*cosh(1/2*x)*a/((2*cosh(1/2*x)
^2*b+a-b)*sinh(1/2*x)^2)^(1/2)+2*(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)/(-2*b/(a-b))^(1/2)*((2*cosh(1/2*x)^2*b+
a-b)/(a-b))^(1/2)*(-sinh(1/2*x)^2)^(1/2)/(2*b*sinh(1/2*x)^4+(a+b)*sinh(1/2*x)^2)^(1/2)*EllipticF(cosh(1/2*x)*(
-2*b/(a-b))^(1/2),1/2*((-2*a+2*b)/b)^(1/2))-32/3*a*b/(a+b)^2/(a-b)^2*(-a+b)/(-2*b/(a-b))^(1/2)*((2*cosh(1/2*x)
^2*b+a-b)/(a-b))^(1/2)*(-sinh(1/2*x)^2)^(1/2)/(2*b*sinh(1/2*x)^4+(a+b)*sinh(1/2*x)^2)^(1/2)/(2*a-2*b)*(Ellipti
cF(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*((-2*a+2*b)/b)^(1/2))-EllipticE(cosh(1/2*x)*(-2*b/(a-b))^(1/2),1/2*((-2*
a+2*b)/b)^(1/2))))/sinh(1/2*x)/(2*sinh(1/2*x)^2*b+a+b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cosh \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*cosh(x))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cosh(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 \cosh \left (x\right ) + a}}{b^{3} \cosh \left (x\right )^{3} + 3 \, a b^{2} \cosh \left (x\right )^{2} + 3 \, a^{2} b \cosh \left (x\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*cosh(x) + a)/(b^3*cosh(x)^3 + 3*a*b^2*cosh(x)^2 + 3*a^2*b*cosh(x) + a^3), 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(1/(a+b*cosh(x))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cosh \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*cosh(x))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cosh(x) + a)^(-5/2), x)