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

Optimal. Leaf size=217 \[ \frac{(a+b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{a f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\tanh (e+f x)}{f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}-\frac{2 \sqrt{a} \sqrt{b} \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{f (a-b)^2 \sqrt{a+b \sinh ^2(e+f x)} \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}}} \]

[Out]

(-2*Sqrt[a]*Sqrt[b]*Cosh[e + f*x]*EllipticE[ArcTan[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a]], 1 - a/b])/((a - b)^2*f*Sq
rt[(a*Cosh[e + f*x]^2)/(a + b*Sinh[e + f*x]^2)]*Sqrt[a + b*Sinh[e + f*x]^2]) + ((a + b)*EllipticF[ArcTan[Sinh[
e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(a*(a - b)^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sin
h[e + f*x]^2))/a]) - Tanh[e + f*x]/((a - b)*f*Sqrt[a + b*Sinh[e + f*x]^2])

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Rubi [A]  time = 0.212858, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3196, 471, 525, 418, 411} \[ -\frac{\tanh (e+f x)}{f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}-\frac{2 \sqrt{a} \sqrt{b} \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{f (a-b)^2 \sqrt{a+b \sinh ^2(e+f x)} \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}}}+\frac{(a+b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{a f (a-b)^2 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a]*Sqrt[b]*Cosh[e + f*x]*EllipticE[ArcTan[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a]], 1 - a/b])/((a - b)^2*f*Sq
rt[(a*Cosh[e + f*x]^2)/(a + b*Sinh[e + f*x]^2)]*Sqrt[a + b*Sinh[e + f*x]^2]) + ((a + b)*EllipticF[ArcTan[Sinh[
e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(a*(a - b)^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sin
h[e + f*x]^2))/a]) - Tanh[e + f*x]/((a - b)*f*Sqrt[a + b*Sinh[e + f*x]^2])

Rule 3196

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/(1 - ff^2*x^2)^((m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\tanh ^2(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{\tanh (e+f x)}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{a-b x^2}{\sqrt{1+x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{(-a+b) f}\\ &=-\frac{\tanh (e+f x)}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\left (2 a b \sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x^2}}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{(a-b) (-a+b) f}-\frac{\left ((a+b) \sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{(a-b) (-a+b) f}\\ &=-\frac{2 \sqrt{a} \sqrt{b} \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{(a-b)^2 f \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}} \sqrt{a+b \sinh ^2(e+f x)}}+\frac{(a+b) F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a (a-b)^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\tanh (e+f x)}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.28214, size = 158, normalized size = 0.73 \[ \frac{i \sqrt{2} (a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-2 \tanh (e+f x) (a+b \cosh (2 (e+f x)))-2 i \sqrt{2} a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{f (a-b)^2 \sqrt{4 a+2 b \cosh (2 (e+f x))-2 b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[2]*a*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] + I*Sqrt[2]*(a - b)*Sqrt
[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] - 2*(a + b*Cosh[2*(e + f*x)])*Tanh[e + f*x])/(
(a - b)^2*f*Sqrt[4*a - 2*b + 2*b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.185, size = 256, normalized size = 1.2 \begin{align*} -{\frac{1}{ \left ( a-b \right ) ^{2}\cosh \left ( fx+e \right ) f} \left ( 2\,\sqrt{-{\frac{b}{a}}}b \left ( \sinh \left ( fx+e \right ) \right ) ^{3}-a\sqrt{{\frac{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{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{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{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 ) -2\,b\sqrt{{\frac{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{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 ) +\sqrt{-{\frac{b}{a}}}a\sinh \left ( fx+e \right ) +b\sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}} \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(tanh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

-(2*(-1/a*b)^(1/2)*b*sinh(f*x+e)^3-a*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)
*(-1/a*b)^(1/2),(a/b)^(1/2))+b*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a
*b)^(1/2),(a/b)^(1/2))-2*b*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^
(1/2),(a/b)^(1/2))+(-1/a*b)^(1/2)*a*sinh(f*x+e)+b*sinh(f*x+e)*(-1/a*b)^(1/2))/(a-b)^2/(-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{\tanh \left (f x + e\right )^{2}}{{\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(tanh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(tanh(f*x + e)^2/(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} \tanh \left (f x + e\right )^{2}}{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(tanh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(e + f*x)**2/(a + b*sinh(e + f*x)**2)**(3/2), x)

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

[Out]

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