3.434 \(\int \coth ^2(e+f x) \sqrt{a+a \sinh ^2(e+f x)} \, dx\)

Optimal. Leaf size=56 \[ \frac{\tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{f}-\frac{\text{csch}(e+f x) \text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]  time = 0.113688, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3176, 3207, 2590, 14} \[ \frac{\tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{f}-\frac{\text{csch}(e+f x) \text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0754151, size = 35, normalized size = 0.62 \[ -\frac{\tanh (e+f x) \left (\text{csch}^2(e+f x)-1\right ) \sqrt{a \cosh ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a*Cosh[e + f*x]^2]*(-1 + Csch[e + f*x]^2)*Tanh[e + f*x])/f)

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Maple [A]  time = 0.085, size = 42, normalized size = 0.8 \begin{align*}{\frac{\cosh \left ( fx+e \right ) a \left ( -1+ \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \right ) }{\sinh \left ( fx+e \right ) f}{\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(coth(f*x+e)^2*(a+a*sinh(f*x+e)^2)^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.73373, size = 169, normalized size = 3.02 \begin{align*} \frac{\sqrt{a} e^{\left (-f x - e\right )}}{f{\left (e^{\left (-2 \, f x - 2 \, e\right )} - 1\right )}} - \frac{2 \, \sqrt{a} e^{\left (-2 \, f x - 2 \, e\right )} - \sqrt{a}}{2 \, f{\left (e^{\left (-f x - e\right )} - e^{\left (-3 \, f x - 3 \, e\right )}\right )}} + \frac{2 \, \sqrt{a} e^{\left (-f x - e\right )} - \sqrt{a} e^{\left (-3 \, f x - 3 \, e\right )}}{2 \, f{\left (e^{\left (-2 \, f x - 2 \, e\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.78849, size = 829, normalized size = 14.8 \begin{align*} \frac{{\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} + 6 \,{\left (\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} - 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} - 6 \, \cosh \left (f x + e\right )^{2} + 1\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 )}}{2 \,{\left (f \cosh \left (f x + e\right )^{3} +{\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )} \sinh \left (f x + e\right )^{3} + 3 \,{\left (f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{2} - f \cosh \left (f x + e\right ) +{\left (f \cosh \left (f x + e\right )^{3} - f \cosh \left (f x + e\right )\right )} e^{\left (2 \, f x + 2 \, e\right )} +{\left (3 \, f \cosh \left (f x + e\right )^{2} +{\left (3 \, f \cosh \left (f x + e\right )^{2} - f\right )} e^{\left (2 \, f x + 2 \, e\right )} - f\right )} \sinh \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^3 + e^(f*x + e)*sinh(f*x + e)^4 + 6*(cosh(f*x + e)^2 - 1)*e^(f*
x + e)*sinh(f*x + e)^2 + 4*(cosh(f*x + e)^3 - 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) + (cosh(f*x + e)^4 -
6*cosh(f*x + e)^2 + 1)*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)/(f*cosh(f*x
+ e)^3 + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e)^3 + 3*(f*cosh(f*x + e)*e^(2*f*x + 2*e) + f*cosh(f*x + e))*sinh
(f*x + e)^2 - f*cosh(f*x + e) + (f*cosh(f*x + e)^3 - f*cosh(f*x + e))*e^(2*f*x + 2*e) + (3*f*cosh(f*x + e)^2 +
(3*f*cosh(f*x + e)^2 - f)*e^(2*f*x + 2*e) - f)*sinh(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29183, size = 77, normalized size = 1.38 \begin{align*} -\frac{\sqrt{a}{\left (\frac{{\left (5 \, e^{\left (2 \, f x + 2 \, e\right )} - 1\right )} e^{\left (-e\right )}}{e^{\left (3 \, f x + 2 \, e\right )} - e^{\left (f x\right )}} - e^{\left (f x + e\right )}\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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