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

Optimal. Leaf size=57 \[ \frac{\tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{f}-\frac{\text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \tan ^{-1}(\sinh (e+f x))}{f} \]

[Out]

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

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Rubi [A]  time = 0.108551, antiderivative size = 57, 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 = {3176, 3207, 2592, 321, 203} \[ \frac{\tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{f}-\frac{\text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \tan ^{-1}(\sinh (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((ArcTan[Sinh[e + f*x]]*Sqrt[a*Cosh[e + f*x]^2]*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 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \sqrt{a+a \sinh ^2(e+f x)} \tanh ^2(e+f x) \, dx &=\int \sqrt{a \cosh ^2(e+f x)} \tanh ^2(e+f x) \, dx\\ &=\left (\sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \int \sinh (e+f x) \tanh (e+f x) \, dx\\ &=\frac{\left (\sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\sqrt{a \cosh ^2(e+f x)} \tanh (e+f x)}{f}-\frac{\left (\sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{\tan ^{-1}(\sinh (e+f x)) \sqrt{a \cosh ^2(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 = 0.0526466, size = 40, normalized size = 0.7 \[ \frac{\text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \left (\sinh (e+f x)-\tan ^{-1}(\sinh (e+f x))\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.088, size = 41, normalized size = 0.7 \begin{align*} -{\frac{a\cosh \left ( fx+e \right ) \left ( -\sinh \left ( fx+e \right ) +\arctan \left ( \sinh \left ( fx+e \right ) \right ) \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((a+a*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)^2,x)

[Out]

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

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Maxima [A]  time = 1.76125, size = 68, normalized size = 1.19 \begin{align*} \frac{2 \, \sqrt{a} \arctan \left (e^{\left (-f x - e\right )}\right )}{f} + \frac{\sqrt{a} e^{\left (f x + e\right )}}{2 \, f} - \frac{\sqrt{a} e^{\left (-f x - e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.97095, size = 497, normalized size = 8.72 \begin{align*} \frac{{\left (2 \, \cosh \left (f x + e\right ) e^{\left (f x + e\right )} \sinh \left (f x + e\right ) + e^{\left (f x + e\right )} \sinh \left (f x + e\right )^{2} - 4 \,{\left (\cosh \left (f x + e\right ) e^{\left (f x + e\right )} + e^{\left (f x + e\right )} \sinh \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 )^{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 ) e^{\left (2 \, f x + 2 \, e\right )} + f \cosh \left (f x + e\right ) +{\left (f 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((a+a*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)^2,x, algorithm="fricas")

[Out]

1/2*(2*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e) + e^(f*x + e)*sinh(f*x + e)^2 - 4*(cosh(f*x + e)*e^(f*x + e) +
e^(f*x + e)*sinh(f*x + e))*arctan(cosh(f*x + e) + sinh(f*x + e)) + (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)*e^(2*f*x + 2*e) + f*cosh(f*x + e) + (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 )} \tanh ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.26655, size = 51, normalized size = 0.89 \begin{align*} -\frac{\sqrt{a}{\left (4 \, \arctan \left (e^{\left (f x + e\right )}\right ) - e^{\left (f x + e\right )} + e^{\left (-f x - e\right )}\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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