3.227 \(\int (1+\tanh ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=50 \[ 2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{\tanh ^2(x)+1}}\right )-\frac{1}{2} \tanh (x) \sqrt{\tanh ^2(x)+1}-\frac{5}{2} \sinh ^{-1}(\tanh (x)) \]

[Out]

(-5*ArcSinh[Tanh[x]])/2 + 2*Sqrt[2]*ArcTanh[(Sqrt[2]*Tanh[x])/Sqrt[1 + Tanh[x]^2]] - (Tanh[x]*Sqrt[1 + Tanh[x]
^2])/2

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Rubi [A]  time = 0.0388703, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3661, 416, 523, 215, 377, 206} \[ 2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{\tanh ^2(x)+1}}\right )-\frac{1}{2} \tanh (x) \sqrt{\tanh ^2(x)+1}-\frac{5}{2} \sinh ^{-1}(\tanh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tanh[x]^2)^(3/2),x]

[Out]

(-5*ArcSinh[Tanh[x]])/2 + 2*Sqrt[2]*ArcTanh[(Sqrt[2]*Tanh[x])/Sqrt[1 + Tanh[x]^2]] - (Tanh[x]*Sqrt[1 + Tanh[x]
^2])/2

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 215

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

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

Rubi steps

\begin{align*} \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^{3/2}}{1-x^2} \, dx,x,\tanh (x)\right )\\ &=-\frac{1}{2} \tanh (x) \sqrt{1+\tanh ^2(x)}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{-3-5 x^2}{\left (1-x^2\right ) \sqrt{1+x^2}} \, dx,x,\tanh (x)\right )\\ &=-\frac{1}{2} \tanh (x) \sqrt{1+\tanh ^2(x)}-\frac{5}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,\tanh (x)\right )+4 \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{1+x^2}} \, dx,x,\tanh (x)\right )\\ &=-\frac{5}{2} \sinh ^{-1}(\tanh (x))-\frac{1}{2} \tanh (x) \sqrt{1+\tanh ^2(x)}+4 \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\frac{\tanh (x)}{\sqrt{1+\tanh ^2(x)}}\right )\\ &=-\frac{5}{2} \sinh ^{-1}(\tanh (x))+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{1+\tanh ^2(x)}}\right )-\frac{1}{2} \tanh (x) \sqrt{1+\tanh ^2(x)}\\ \end{align*}

Mathematica [A]  time = 0.133756, size = 74, normalized size = 1.48 \[ -\frac{\left (\tanh ^2(x)+1\right )^{3/2} \left (-4 \sqrt{2} \sinh ^{-1}\left (\sqrt{2} \sinh (x)\right ) \cosh ^3(x)+\sinh (x) \sqrt{\cosh (2 x)} \cosh (x)+5 \cosh ^3(x) \tanh ^{-1}\left (\frac{\sinh (x)}{\sqrt{\cosh (2 x)}}\right )\right )}{2 \cosh ^{\frac{3}{2}}(2 x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tanh[x]^2)^(3/2),x]

[Out]

-((-4*Sqrt[2]*ArcSinh[Sqrt[2]*Sinh[x]]*Cosh[x]^3 + 5*ArcTanh[Sinh[x]/Sqrt[Cosh[2*x]]]*Cosh[x]^3 + Cosh[x]*Sqrt
[Cosh[2*x]]*Sinh[x])*(1 + Tanh[x]^2)^(3/2))/(2*Cosh[2*x]^(3/2))

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Maple [B]  time = 0.025, size = 158, normalized size = 3.2 \begin{align*}{\frac{1}{6} \left ( \left ( 1+\tanh \left ( x \right ) \right ) ^{2}-2\,\tanh \left ( x \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{\tanh \left ( x \right ) }{4}\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}-2\,\tanh \left ( x \right ) }}-{\frac{5\,{\it Arcsinh} \left ( \tanh \left ( x \right ) \right ) }{2}}+\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}-2\,\tanh \left ( x \right ) }-\sqrt{2}{\it Artanh} \left ({\frac{ \left ( 2-2\,\tanh \left ( x \right ) \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ( 1+\tanh \left ( x \right ) \right ) ^{2}-2\,\tanh \left ( x \right ) }}}} \right ) -{\frac{1}{6} \left ( \left ( \tanh \left ( x \right ) -1 \right ) ^{2}+2\,\tanh \left ( x \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{\tanh \left ( x \right ) }{4}\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}+2\,\tanh \left ( x \right ) }}-\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}+2\,\tanh \left ( x \right ) }+\sqrt{2}{\it Artanh} \left ({\frac{ \left ( 2\,\tanh \left ( x \right ) +2 \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{ \left ( \tanh \left ( x \right ) -1 \right ) ^{2}+2\,\tanh \left ( x \right ) }}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+tanh(x)^2)^(3/2),x)

[Out]

1/6*((1+tanh(x))^2-2*tanh(x))^(3/2)-1/4*tanh(x)*((1+tanh(x))^2-2*tanh(x))^(1/2)-5/2*arcsinh(tanh(x))+((1+tanh(
x))^2-2*tanh(x))^(1/2)-2^(1/2)*arctanh(1/4*(2-2*tanh(x))*2^(1/2)/((1+tanh(x))^2-2*tanh(x))^(1/2))-1/6*((tanh(x
)-1)^2+2*tanh(x))^(3/2)-1/4*tanh(x)*((tanh(x)-1)^2+2*tanh(x))^(1/2)-((tanh(x)-1)^2+2*tanh(x))^(1/2)+2^(1/2)*ar
ctanh(1/4*(2*tanh(x)+2)*2^(1/2)/((tanh(x)-1)^2+2*tanh(x))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\tanh \left (x\right )^{2} + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tanh(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((tanh(x)^2 + 1)^(3/2), x)

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Fricas [B]  time = 2.0444, size = 3526, normalized size = 70.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tanh(x)^2)^(3/2),x, algorithm="fricas")

[Out]

1/4*(2*(sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + 2*(3*sqrt(2)*cosh(x)^2 + sqrt(2)
)*sinh(x)^2 + 2*sqrt(2)*cosh(x)^2 + 4*(sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(-2*(cosh(x)
^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2 - 3)*sinh(x)^6 - 3*cosh(x)^6 + 2*(28*cosh(x)^3 - 9*cosh(x
))*sinh(x)^5 + 5*(14*cosh(x)^4 - 9*cosh(x)^2 + 1)*sinh(x)^4 + 5*cosh(x)^4 + 4*(14*cosh(x)^5 - 15*cosh(x)^3 + 5
*cosh(x))*sinh(x)^3 + (28*cosh(x)^6 - 45*cosh(x)^4 + 30*cosh(x)^2 - 4)*sinh(x)^2 - 4*cosh(x)^2 + 2*(4*cosh(x)^
7 - 9*cosh(x)^5 + 10*cosh(x)^3 - 4*cosh(x))*sinh(x) + (sqrt(2)*cosh(x)^6 + 6*sqrt(2)*cosh(x)*sinh(x)^5 + sqrt(
2)*sinh(x)^6 + 3*(5*sqrt(2)*cosh(x)^2 - sqrt(2))*sinh(x)^4 - 3*sqrt(2)*cosh(x)^4 + 4*(5*sqrt(2)*cosh(x)^3 - 3*
sqrt(2)*cosh(x))*sinh(x)^3 + (15*sqrt(2)*cosh(x)^4 - 18*sqrt(2)*cosh(x)^2 + 4*sqrt(2))*sinh(x)^2 + 4*sqrt(2)*c
osh(x)^2 + 2*(3*sqrt(2)*cosh(x)^5 - 6*sqrt(2)*cosh(x)^3 + 4*sqrt(2)*cosh(x))*sinh(x) - 4*sqrt(2))*sqrt((cosh(x
)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4)/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(
x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6)) + 2*(sqrt
(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + 2*(3*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2
+ 2*sqrt(2)*cosh(x)^2 + 4*(sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(2*(cosh(x)^4 + 4*cosh(x
)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(2*cosh(x)^3 + cosh(x))*sinh(x) + (sqrt(
2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 + sqrt(2))*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^
2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) - 5*(cosh(x)^4 + 4*cosh(
x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)*
log((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 2*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x)
+ sinh(x)^2)) - 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) + 5*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x
)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)*log((cosh(x)^2 + 2*co
sh(x)*sinh(x) + sinh(x)^2 - 2*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) - 1)/(
cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) - 4*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt((cosh(x)^
2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*
(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\tanh ^{2}{\left (x \right )} + 1\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tanh(x)**2)**(3/2),x)

[Out]

Integral((tanh(x)**2 + 1)**(3/2), x)

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Giac [B]  time = 1.25641, size = 273, normalized size = 5.46 \begin{align*} -\frac{1}{4} \, \sqrt{2}{\left (5 \, \sqrt{2} \log \left (\frac{\sqrt{2} - \sqrt{e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1}{\sqrt{2} + \sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} - 1}\right ) - \frac{4 \,{\left (3 \,{\left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{3} -{\left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} - \sqrt{e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} - 1\right )}}{{\left ({\left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} - 2 \, \sqrt{e^{\left (4 \, x\right )} + 1} + 2 \, e^{\left (2 \, x\right )} - 1\right )}^{2}} + 4 \, \log \left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) + 4 \, \log \left (\sqrt{e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) - 4 \, \log \left (-\sqrt{e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tanh(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/4*sqrt(2)*(5*sqrt(2)*log((sqrt(2) - sqrt(e^(4*x) + 1) + e^(2*x) + 1)/(sqrt(2) + sqrt(e^(4*x) + 1) - e^(2*x)
- 1)) - 4*(3*(sqrt(e^(4*x) + 1) - e^(2*x))^3 - (sqrt(e^(4*x) + 1) - e^(2*x))^2 - sqrt(e^(4*x) + 1) + e^(2*x)
- 1)/((sqrt(e^(4*x) + 1) - e^(2*x))^2 - 2*sqrt(e^(4*x) + 1) + 2*e^(2*x) - 1)^2 + 4*log(sqrt(e^(4*x) + 1) - e^(
2*x) + 1) + 4*log(sqrt(e^(4*x) + 1) - e^(2*x)) - 4*log(-sqrt(e^(4*x) + 1) + e^(2*x) + 1))