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

Optimal. Leaf size=115 \[ -\frac{b \sinh (e+f x) \cosh (e+f x)}{a f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}-\frac{i \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{a f (a-b) \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

[Out]

-((b*Cosh[e + f*x]*Sinh[e + f*x])/(a*(a - b)*f*Sqrt[a + b*Sinh[e + f*x]^2])) - (I*EllipticE[I*e + I*f*x, b/a]*
Sqrt[a + b*Sinh[e + f*x]^2])/(a*(a - b)*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a])

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Rubi [A]  time = 0.067648, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3184, 21, 3178, 3177} \[ -\frac{b \sinh (e+f x) \cosh (e+f x)}{a f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}-\frac{i \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{a f (a-b) \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[e + f*x]^2)^(-3/2),x]

[Out]

-((b*Cosh[e + f*x]*Sinh[e + f*x])/(a*(a - b)*f*Sqrt[a + b*Sinh[e + f*x]^2])) - (I*EllipticE[I*e + I*f*x, b/a]*
Sqrt[a + b*Sinh[e + f*x]^2])/(a*(a - b)*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a])

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
 + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

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 3178

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

Rule 3177

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[e + f*x, -(b/a)])/f, x]
 /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rubi steps

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

Mathematica [A]  time = 0.149781, size = 100, normalized size = 0.87 \[ \frac{-\sqrt{2} b \sinh (2 (e+f x))-2 i a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{2 a f (a-b) \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[e + f*x]^2)^(-3/2),x]

[Out]

((-2*I)*a*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] - Sqrt[2]*b*Sinh[2*(e + f*x)])/(
2*a*(a - b)*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0., size = 252, normalized size = 2.2 \begin{align*}{\frac{1}{a \left ( a-b \right ) \cosh \left ( fx+e \right ) f} \left ( -\sqrt{-{\frac{b}{a}}}b\sinh \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}+a\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) b+\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) b \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-1/a*b)^(1/2)*b*sinh(f*x+e)*cosh(f*x+e)^2+a*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*Ellipti
cF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(s
inh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b+(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sin
h(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b)/a/(a-b)/(-1/a*b)^(1/2)/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/f

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

[Out]

integrate((b*sinh(f*x + e)^2 + 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 (f x + e\right )^{2} + a}}{b^{2} \sinh \left (f x + e\right )^{4} + 2 \, a b \sinh \left (f x + e\right )^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)/(b^2*sinh(f*x + e)^4 + 2*a*b*sinh(f*x + e)^2 + a^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(1/(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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

[Out]

integrate((b*sinh(f*x + e)^2 + a)^(-3/2), x)