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

Optimal. Leaf size=62 \[ \frac{\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{2 f \sqrt{a \cosh ^2(e+f x)}}-\frac{\tanh (e+f x)}{2 f \sqrt{a \cosh ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.120652, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3176, 3207, 2611, 3770} \[ \frac{\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))}{2 f \sqrt{a \cosh ^2(e+f x)}}-\frac{\tanh (e+f x)}{2 f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\tanh ^2(e+f x)}{\sqrt{a+a \sinh ^2(e+f x)}} \, dx &=\int \frac{\tanh ^2(e+f x)}{\sqrt{a \cosh ^2(e+f x)}} \, dx\\ &=\frac{\cosh (e+f x) \int \text{sech}(e+f x) \tanh ^2(e+f x) \, dx}{\sqrt{a \cosh ^2(e+f x)}}\\ &=-\frac{\tanh (e+f x)}{2 f \sqrt{a \cosh ^2(e+f x)}}+\frac{\cosh (e+f x) \int \text{sech}(e+f x) \, dx}{2 \sqrt{a \cosh ^2(e+f x)}}\\ &=\frac{\tan ^{-1}(\sinh (e+f x)) \cosh (e+f x)}{2 f \sqrt{a \cosh ^2(e+f x)}}-\frac{\tanh (e+f x)}{2 f \sqrt{a \cosh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0574041, size = 44, normalized size = 0.71 \[ \frac{\cosh (e+f x) \tan ^{-1}(\sinh (e+f x))-\tanh (e+f x)}{2 f \sqrt{a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.224, size = 51, normalized size = 0.8 \begin{align*}{\frac{1}{f\cosh \left ( fx+e \right ) } \left ({\frac{\arctan \left ( \sinh \left ( fx+e \right ) \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{2}}-{\frac{\sinh \left ( fx+e \right ) }{2}} \right ){\frac{1}{\sqrt{a \left ( \cosh \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+a*sinh(f*x+e)^2)^(1/2),x)

[Out]

(1/2*arctan(sinh(f*x+e))*cosh(f*x+e)^2-1/2*sinh(f*x+e))/cosh(f*x+e)/(a*cosh(f*x+e)^2)^(1/2)/f

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Maxima [B]  time = 1.80806, size = 293, normalized size = 4.73 \begin{align*} \frac{\frac{\arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt{a}} - \frac{e^{\left (-f x - e\right )} - e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} + \sqrt{a}}}{2 \, f} - \frac{3 \, \arctan \left (e^{\left (-f x - e\right )}\right )}{2 \, \sqrt{a} f} - \frac{5 \, e^{\left (-f x - e\right )} + 3 \, e^{\left (-3 \, f x - 3 \, e\right )}}{4 \,{\left (2 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} + \sqrt{a}\right )} f} + \frac{3 \, e^{\left (-f x - e\right )} + 5 \, e^{\left (-3 \, f x - 3 \, e\right )}}{4 \,{\left (2 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} + \sqrt{a} e^{\left (-4 \, f x - 4 \, e\right )} + \sqrt{a}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(arctan(e^(-f*x - e))/sqrt(a) - (e^(-f*x - e) - e^(-3*f*x - 3*e))/(2*sqrt(a)*e^(-2*f*x - 2*e) + sqrt(a)*e^
(-4*f*x - 4*e) + sqrt(a)))/f - 3/2*arctan(e^(-f*x - e))/(sqrt(a)*f) - 1/4*(5*e^(-f*x - e) + 3*e^(-3*f*x - 3*e)
)/((2*sqrt(a)*e^(-2*f*x - 2*e) + sqrt(a)*e^(-4*f*x - 4*e) + sqrt(a))*f) + 1/4*(3*e^(-f*x - e) + 5*e^(-3*f*x -
3*e))/((2*sqrt(a)*e^(-2*f*x - 2*e) + sqrt(a)*e^(-4*f*x - 4*e) + sqrt(a))*f)

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Fricas [B]  time = 1.84082, size = 1343, normalized size = 21.66 \begin{align*} -\frac{{\left (3 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{3} +{\left (3 \, \cosh \left (f x + e\right )^{2} - 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right ) -{\left (4 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{3} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{4} + 2 \,{\left (3 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} + 4 \,{\left (\cosh \left (f x + e\right )^{3} + \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )} \sinh \left (f x + e\right ) +{\left (\cosh \left (f x + e\right )^{4} + 2 \, \cosh \left (f x + e\right )^{2} + 1\right )} e^{\left (f x + e\right )}\right )} \arctan \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) +{\left (\cosh \left (f x + e\right )^{3} - \cosh \left (f x + e\right )\right )} e^{\left (f x + e\right )}\right )} \sqrt{a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} e^{\left (-f x - e\right )}}{a f \cosh \left (f x + e\right )^{4} +{\left (a f e^{\left (2 \, f x + 2 \, e\right )} + a f\right )} \sinh \left (f x + e\right )^{4} + 2 \, a f \cosh \left (f x + e\right )^{2} + 4 \,{\left (a f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{3} + 2 \,{\left (3 \, a f \cosh \left (f x + e\right )^{2} + a f +{\left (3 \, a f \cosh \left (f x + e\right )^{2} + a f\right )} e^{\left (2 \, f x + 2 \, e\right )}\right )} \sinh \left (f x + e\right )^{2} + a f +{\left (a f \cosh \left (f x + e\right )^{4} + 2 \, a f \cosh \left (f x + e\right )^{2} + a f\right )} e^{\left (2 \, f x + 2 \, e\right )} + 4 \,{\left (a f \cosh \left (f x + e\right )^{3} + a f \cosh \left (f x + e\right ) +{\left (a f \cosh \left (f x + e\right )^{3} + a f \cosh \left (f x + e\right )\right )} e^{\left (2 \, f x + 2 \, e\right )}\right )} \sinh \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^2 + e^(f*x + e)*sinh(f*x + e)^3 + (3*cosh(f*x + e)^2 - 1)*e^(f*x +
e)*sinh(f*x + e) - (4*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^3 + e^(f*x + e)*sinh(f*x + e)^4 + 2*(3*cosh(f*x
+ e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^2 + 4*(cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) + (co
sh(f*x + e)^4 + 2*cosh(f*x + e)^2 + 1)*e^(f*x + e))*arctan(cosh(f*x + e) + sinh(f*x + e)) + (cosh(f*x + e)^3 -
cosh(f*x + e))*e^(f*x + e))*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*e^(-f*x - e)/(a*f*cosh(f*x + e)
^4 + (a*f*e^(2*f*x + 2*e) + a*f)*sinh(f*x + e)^4 + 2*a*f*cosh(f*x + e)^2 + 4*(a*f*cosh(f*x + e)*e^(2*f*x + 2*e
) + a*f*cosh(f*x + e))*sinh(f*x + e)^3 + 2*(3*a*f*cosh(f*x + e)^2 + a*f + (3*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f
*x + 2*e))*sinh(f*x + e)^2 + a*f + (a*f*cosh(f*x + e)^4 + 2*a*f*cosh(f*x + e)^2 + a*f)*e^(2*f*x + 2*e) + 4*(a*
f*cosh(f*x + e)^3 + a*f*cosh(f*x + e) + (a*f*cosh(f*x + e)^3 + a*f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f*x +
e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{2}{\left (e + f x \right )}}{\sqrt{a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(e + f*x)**2/sqrt(a*(sinh(e + f*x)**2 + 1)), x)

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Giac [A]  time = 1.31437, size = 85, normalized size = 1.37 \begin{align*} \frac{\frac{\arctan \left (e^{\left (f x + e\right )}\right )}{\sqrt{a}} - \frac{\sqrt{a} e^{\left (3 \, f x + 3 \, e\right )} - \sqrt{a} e^{\left (f x + e\right )}}{a{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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