3.486 \(\int \frac {\tanh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=156 \[ \frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

-(1/(1+sinh(f*x+e)^2))^(1/2)*(1+sinh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/
2))*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/(a-b)/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)+(1/(1+sinh(f*x+e
)^2))^(1/2)*(1+sinh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/2))*sech(f*x+e)*(
a+b*sinh(f*x+e)^2)^(1/2)/(a-b)/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3196, 471, 422, 418, 492, 411} \[ \frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 422

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

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 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

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]

Rubi steps

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

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Mathematica [C]  time = 0.41, size = 109, normalized size = 0.70 \[ \frac {\sqrt {2} \tanh (e+f x) (-2 a-b \cosh (2 (e+f x))+b)-2 i a \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )}{2 f (a-b) \sqrt {2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]^2/Sqrt[a + b*Sinh[e + f*x]^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]*(-2*a + b - b*Cosh[2*(
e + f*x)])*Tanh[e + f*x])/(2*(a - b)*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\tanh \left (f x + e\right )^{2}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.66, size = 374, normalized size = 2.40 \[ -\frac {2 \, {\left (\frac {\arctan \left (-\frac {\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b} + \sqrt {b}}{2 \, \sqrt {a - b}}\right ) e^{e}}{\sqrt {a - b}} - \frac {\arctan \left (-\frac {\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}}{\sqrt {-b}}\right ) e^{e}}{\sqrt {-b}} - \frac {2 \, {\left ({\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} e^{e} - \sqrt {b} e^{e}\right )}}{{\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} + 2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} \sqrt {b} + 4 \, a - 3 \, b}\right )}}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(arctan(-1/2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e)
+ b) + sqrt(b))/sqrt(a - b))*e^e/sqrt(a - b) - arctan(-(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a
*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))/sqrt(-b))*e^e/sqrt(-b) - 2*((sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(
4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*e^e - sqrt(b)*e^e)/((sqrt(b)*e^(2*f*x + 2*e) -
sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))^2 + 2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt
(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*sqrt(b) + 4*a - 3*b))/f^2

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maple [A]  time = 0.30, size = 239, normalized size = 1.53 \[ \frac {-\sqrt {-\frac {b}{a}}\, b \left (\sinh ^{3}\left (f x +e \right )\right )+a \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-b \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )+b \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\sqrt {-\frac {b}{a}}\, a \sinh \left (f x +e \right )}{\left (a -b \right ) \sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-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))+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))/(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \left (f x + e\right )^{2}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tanh}\left (e+f\,x\right )}^2}{\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(f*x+e)**2/(a+b*sinh(f*x+e)**2)**(1/2),x)

[Out]

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

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