### 3.228 $$\int (-1-\tanh ^2(x))^{3/2} \, dx$$

Optimal. Leaf size=67 $\frac{1}{2} \tanh (x) \sqrt{-\tanh ^2(x)-1}-\frac{5}{2} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{-\tanh ^2(x)-1}}\right )+2 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{-\tanh ^2(x)-1}}\right )$

[Out]

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

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Rubi [A]  time = 0.0473687, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {3661, 416, 523, 217, 203, 377} $\frac{1}{2} \tanh (x) \sqrt{-\tanh ^2(x)-1}-\frac{5}{2} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{-\tanh ^2(x)-1}}\right )+2 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{-\tanh ^2(x)-1}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[(-1 - Tanh[x]^2)^(3/2),x]

[Out]

(-5*ArcTan[Tanh[x]/Sqrt[-1 - Tanh[x]^2]])/2 + 2*Sqrt[2]*ArcTan[(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 217

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

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])

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]

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}{\sqrt{-1-x^2} \left (1-x^2\right )} \, 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}{\sqrt{-1-x^2} \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{2} \tanh (x) \sqrt{-1-\tanh ^2(x)}-\frac{5}{2} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\tanh (x)}{\sqrt{-1-\tanh ^2(x)}}\right )+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} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{-1-\tanh ^2(x)}}\right )+2 \sqrt{2} \tan ^{-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 = 0.0671395, size = 76, normalized size = 1.13 $-\frac{\left (-\tanh ^2(x)-1\right )^{3/2} \left (-4 \sqrt{2} \sinh ^{-1}\left (\sqrt{2} \sinh (x)\right ) \cosh ^3(x)+\sinh (x) \sqrt{\cosh (2 x)} \cosh (x)+5 \cosh ^3(x) \tanh ^{-1}\left (\frac{\sinh (x)}{\sqrt{\cosh (2 x)}}\right )\right )}{2 \cosh ^{\frac{3}{2}}(2 x)}$

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 = 0.027, size = 211, 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}{4}\arctan \left ({\tanh \left ( x \right ){\frac{1}{\sqrt{- \left ( 1+\tanh \left ( x \right ) \right ) ^{2}+2\,\tanh \left ( x \right ) }}}} \right ) }-\sqrt{- \left ( 1+\tanh \left ( x \right ) \right ) ^{2}+2\,\tanh \left ( x \right ) }+\sqrt{2}\arctan \left ({\frac{ \left ( 2\,\tanh \left ( x \right ) -2 \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 ) }}-{\frac{5}{4}\arctan \left ({\tanh \left ( x \right ){\frac{1}{\sqrt{- \left ( \tanh \left ( x \right ) -1 \right ) ^{2}-2\,\tanh \left ( x \right ) }}}} \right ) }+\sqrt{- \left ( \tanh \left ( x \right ) -1 \right ) ^{2}-2\,\tanh \left ( x \right ) }-\sqrt{2}\arctan \left ({\frac{ \left ( -2-2\,\tanh \left ( x \right ) \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)-5/4*arctan(tanh(x)/(-(1+tanh
(x))^2+2*tanh(x))^(1/2))-(-(1+tanh(x))^2+2*tanh(x))^(1/2)+2^(1/2)*arctan(1/4*(2*tanh(x)-2)*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)-5/4*a
rctan(tanh(x)/(-(tanh(x)-1)^2-2*tanh(x))^(1/2))+(-(tanh(x)-1)^2-2*tanh(x))^(1/2)-2^(1/2)*arctan(1/4*(-2-2*tanh
(x))*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 [C]  time = 1.84119, size = 1103, normalized size = 16.46 \begin{align*} \frac{2 \,{\left (\sqrt{-2} e^{\left (4 \, x\right )} + 2 \, \sqrt{-2} e^{\left (2 \, x\right )} + \sqrt{-2}\right )} \log \left (2 \,{\left (\sqrt{-2} \sqrt{-2 \, e^{\left (4 \, x\right )} - 2} + 2 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )}\right ) - 2 \,{\left (\sqrt{-2} e^{\left (4 \, x\right )} + 2 \, \sqrt{-2} e^{\left (2 \, x\right )} + \sqrt{-2}\right )} \log \left (-2 \,{\left (\sqrt{-2} \sqrt{-2 \, e^{\left (4 \, x\right )} - 2} - 2 \, e^{\left (2 \, x\right )} - 2\right )} e^{\left (-2 \, x\right )}\right ) +{\left (5 i \, e^{\left (4 \, x\right )} + 10 i \, e^{\left (2 \, x\right )} + 5 i\right )} \log \left ({\left (4 i \, \sqrt{-2 \, e^{\left (4 \, x\right )} - 2} - 4 \, e^{\left (2 \, x\right )} + 4\right )} e^{\left (-2 \, x\right )}\right ) +{\left (-5 i \, e^{\left (4 \, x\right )} - 10 i \, e^{\left (2 \, x\right )} - 5 i\right )} \log \left ({\left (-4 i \, \sqrt{-2 \, e^{\left (4 \, x\right )} - 2} - 4 \, e^{\left (2 \, x\right )} + 4\right )} e^{\left (-2 \, x\right )}\right ) - 2 \,{\left (\sqrt{-2} e^{\left (4 \, x\right )} + 2 \, \sqrt{-2} e^{\left (2 \, x\right )} + \sqrt{-2}\right )} \log \left (4 \,{\left (\sqrt{-2 \, e^{\left (4 \, x\right )} - 2}{\left (e^{\left (2 \, x\right )} - 2\right )} + \sqrt{-2} e^{\left (4 \, x\right )} - \sqrt{-2} e^{\left (2 \, x\right )} + 2 \, \sqrt{-2}\right )} e^{\left (-4 \, x\right )}\right ) + 2 \,{\left (\sqrt{-2} e^{\left (4 \, x\right )} + 2 \, \sqrt{-2} e^{\left (2 \, x\right )} + \sqrt{-2}\right )} \log \left (4 \,{\left (\sqrt{-2 \, e^{\left (4 \, x\right )} - 2}{\left (e^{\left (2 \, x\right )} - 2\right )} - \sqrt{-2} e^{\left (4 \, x\right )} + \sqrt{-2} e^{\left (2 \, x\right )} - 2 \, \sqrt{-2}\right )} e^{\left (-4 \, x\right )}\right ) + 2 \, \sqrt{-2 \, e^{\left (4 \, x\right )} - 2}{\left (e^{\left (2 \, x\right )} - 1\right )}}{4 \,{\left (e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1\right )}} \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)*e^(4*x) + 2*sqrt(-2)*e^(2*x) + sqrt(-2))*log(2*(sqrt(-2)*sqrt(-2*e^(4*x) - 2) + 2*e^(2*x) + 2
)*e^(-2*x)) - 2*(sqrt(-2)*e^(4*x) + 2*sqrt(-2)*e^(2*x) + sqrt(-2))*log(-2*(sqrt(-2)*sqrt(-2*e^(4*x) - 2) - 2*e
^(2*x) - 2)*e^(-2*x)) + (5*I*e^(4*x) + 10*I*e^(2*x) + 5*I)*log((4*I*sqrt(-2*e^(4*x) - 2) - 4*e^(2*x) + 4)*e^(-
2*x)) + (-5*I*e^(4*x) - 10*I*e^(2*x) - 5*I)*log((-4*I*sqrt(-2*e^(4*x) - 2) - 4*e^(2*x) + 4)*e^(-2*x)) - 2*(sqr
t(-2)*e^(4*x) + 2*sqrt(-2)*e^(2*x) + sqrt(-2))*log(4*(sqrt(-2*e^(4*x) - 2)*(e^(2*x) - 2) + sqrt(-2)*e^(4*x) -
sqrt(-2)*e^(2*x) + 2*sqrt(-2))*e^(-4*x)) + 2*(sqrt(-2)*e^(4*x) + 2*sqrt(-2)*e^(2*x) + sqrt(-2))*log(4*(sqrt(-2
*e^(4*x) - 2)*(e^(2*x) - 2) - sqrt(-2)*e^(4*x) + sqrt(-2)*e^(2*x) - 2*sqrt(-2))*e^(-4*x)) + 2*sqrt(-2*e^(4*x)
- 2)*(e^(2*x) - 1))/(e^(4*x) + 2*e^(2*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 [C]  time = 1.21645, size = 273, normalized size = 4.07 \begin{align*} \frac{1}{2} \, \sqrt{2}{\left (5 \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left ({\left (-i \, \sqrt{e^{\left (4 \, x\right )} + 1} - i\right )} e^{\left (-2 \, x\right )} - i\right )}\right ) + \frac{2 \,{\left (3 \,{\left (i \, \sqrt{e^{\left (4 \, x\right )} + 1} + i\right )}^{3} e^{\left (-6 \, x\right )} + i \,{\left (i \, \sqrt{e^{\left (4 \, x\right )} + 1} + i\right )}^{2} e^{\left (-4 \, x\right )} + i \,{\left (\sqrt{e^{\left (4 \, x\right )} + 1} + 1\right )} e^{\left (-2 \, x\right )} - i\right )}}{{\left (i \,{\left (i \, \sqrt{e^{\left (4 \, x\right )} + 1} + i\right )}^{2} e^{\left (-4 \, x\right )} +{\left (-2 i \, \sqrt{e^{\left (4 \, x\right )} + 1} - 2 i\right )} e^{\left (-2 \, x\right )} + i\right )}^{2}} + 2 i \, \log \left (-{\left (\sqrt{e^{\left (4 \, x\right )} + 1} + 1\right )} e^{\left (-2 \, x\right )}\right ) - 2 i \, \log \left (-{\left (-i \, \sqrt{e^{\left (4 \, x\right )} + 1} - i\right )} e^{\left (-2 \, x\right )} + i\right ) + 2 i \, \log \left (-{\left (-i \, \sqrt{e^{\left (4 \, x\right )} + 1} - i\right )} e^{\left (-2 \, x\right )} - i\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]

1/2*sqrt(2)*(5*sqrt(2)*arctan(-1/2*sqrt(2)*((-I*sqrt(e^(4*x) + 1) - I)*e^(-2*x) - I)) + 2*(3*(I*sqrt(e^(4*x) +
1) + I)^3*e^(-6*x) + I*(I*sqrt(e^(4*x) + 1) + I)^2*e^(-4*x) + I*(sqrt(e^(4*x) + 1) + 1)*e^(-2*x) - I)/(I*(I*s
qrt(e^(4*x) + 1) + I)^2*e^(-4*x) + (-2*I*sqrt(e^(4*x) + 1) - 2*I)*e^(-2*x) + I)^2 + 2*I*log(-(sqrt(e^(4*x) + 1
) + 1)*e^(-2*x)) - 2*I*log(-(-I*sqrt(e^(4*x) + 1) - I)*e^(-2*x) + I) + 2*I*log(-(-I*sqrt(e^(4*x) + 1) - I)*e^(
-2*x) - I))