3.5.86 \(\int \frac {\tanh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\) [486]

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

[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 = 0.13, 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 = {3275, 482, 433, 429, 506, 422} \begin {gather*} \frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} F\left (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\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 (\text {ArcTan}(\sinh (e+f x))\left |1-\frac {b}{a}\right .\right )}{f (a-b) \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \end {gather*}

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 422

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

Rule 429

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

Rule 433

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 482

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 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[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 3275

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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] Result contains complex when optimal does not.
time = 0.32, size = 109, normalized size = 0.70 \begin {gather*} \frac {-2 i a \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )+\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))}} \end {gather*}

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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Maple [A]
time = 1.51, size = 239, normalized size = 1.53

method	result	size
default	\(\frac {-\sqrt {-\frac {b}{a}}\, b \left (\sinh ^{3}\left (f x +e \right )\right )+a \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-b \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )+b \sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\sqrt {-\frac {b}{a}}\, a \sinh \left (f x +e \right )}{\left (a -b \right ) \sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f}\)	\(239\)

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,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (176) = 352\).
time = 0.12, size = 703, normalized size = 4.51 \begin {gather*} \frac {{\left ({\left (2 \, a b - b^{2}\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (2 \, a b - b^{2}\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (2 \, a b - b^{2}\right )} \sinh \left (f x + e\right )^{2} + 2 \, a b - b^{2} - 2 \, {\left (b^{2} \cosh \left (f x + e\right )^{2} + 2 \, b^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + b^{2} \sinh \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {\frac {a^{2} - a b}{b^{2}}}\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} - a b}{b^{2}}} - 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} - a b}{b^{2}}} - 2 \, a + b}{b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} - 8 \, a b + b^{2} + 4 \, {\left (2 \, a b - b^{2}\right )} \sqrt {\frac {a^{2} - a b}{b^{2}}}}{b^{2}}) - 2 \, {\left ({\left (2 \, a^{2} - a b\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (2 \, a^{2} - a b\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (2 \, a^{2} - a b\right )} \sinh \left (f x + e\right )^{2} + 2 \, a^{2} - a b + 2 \, {\left ({\left (a b - b^{2}\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (a b - b^{2}\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (a b - b^{2}\right )} \sinh \left (f x + e\right )^{2} + a b - b^{2}\right )} \sqrt {\frac {a^{2} - a b}{b^{2}}}\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} - a b}{b^{2}}} - 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} - a b}{b^{2}}} - 2 \, a + b}{b}} {\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} - 8 \, a b + b^{2} + 4 \, {\left (2 \, a b - b^{2}\right )} \sqrt {\frac {a^{2} - a b}{b^{2}}}}{b^{2}}) - \sqrt {2} {\left (b^{2} \cosh \left (f x + e\right ) + b^{2} \sinh \left (f x + e\right )\right )} \sqrt {\frac {b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{{\left (a b^{2} - b^{3}\right )} f \cosh \left (f x + e\right )^{2} + 2 \, {\left (a b^{2} - b^{3}\right )} f \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (a b^{2} - b^{3}\right )} f \sinh \left (f x + e\right )^{2} + {\left (a b^{2} - b^{3}\right )} f} \end {gather*}

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]

(((2*a*b - b^2)*cosh(f*x + e)^2 + 2*(2*a*b - b^2)*cosh(f*x + e)*sinh(f*x + e) + (2*a*b - b^2)*sinh(f*x + e)^2
+ 2*a*b - b^2 - 2*(b^2*cosh(f*x + e)^2 + 2*b^2*cosh(f*x + e)*sinh(f*x + e) + b^2*sinh(f*x + e)^2 + b^2)*sqrt((
a^2 - a*b)/b^2))*sqrt(b)*sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 -
a*b)/b^2) - 2*a + b)/b)*(cosh(f*x + e) + sinh(f*x + e))), (8*a^2 - 8*a*b + b^2 + 4*(2*a*b - b^2)*sqrt((a^2 -
a*b)/b^2))/b^2) - 2*((2*a^2 - a*b)*cosh(f*x + e)^2 + 2*(2*a^2 - a*b)*cosh(f*x + e)*sinh(f*x + e) + (2*a^2 - a*
b)*sinh(f*x + e)^2 + 2*a^2 - a*b + 2*((a*b - b^2)*cosh(f*x + e)^2 + 2*(a*b - b^2)*cosh(f*x + e)*sinh(f*x + e)
+ (a*b - b^2)*sinh(f*x + e)^2 + a*b - b^2)*sqrt((a^2 - a*b)/b^2))*sqrt(b)*sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*
a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 - a*b)/b^2) - 2*a + b)/b)*(cosh(f*x + e) + sinh(f*x + e))), (8
*a^2 - 8*a*b + b^2 + 4*(2*a*b - b^2)*sqrt((a^2 - a*b)/b^2))/b^2) - sqrt(2)*(b^2*cosh(f*x + e) + b^2*sinh(f*x +
e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) +
sinh(f*x + e)^2)))/((a*b^2 - b^3)*f*cosh(f*x + e)^2 + 2*(a*b^2 - b^3)*f*cosh(f*x + e)*sinh(f*x + e) + (a*b^2
- b^3)*f*sinh(f*x + e)^2 + (a*b^2 - b^3)*f)

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

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage3:=type(sage2):;OUTPUT:Error
: Bad Argument Type

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

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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