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]

-5/2*arctan(tanh(x)/(-1-tanh(x)^2)^(1/2))+2*arctan(2^(1/2)*tanh(x)/(-1-tanh(x)^2)^(1/2))*2^(1/2)+1/2*(-1-tanh(
x)^2)^(1/2)*tanh(x)

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Rubi [A]  time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.07, 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]

-1/2*((-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))/Cosh[2*x]^(3/2)

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fricas [C]  time = 0.41, size = 361, normalized size = 5.39 \[ \frac {2 \, {\left (\sqrt {-2} e^{\left (4 \, x\right )} + 2 \, \sqrt {-2} e^{\left (2 \, x\right )} + \sqrt {-2}\right )} \log \left (2 \, {\left (\sqrt {-2} \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} + 2 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )}\right ) - 2 \, {\left (\sqrt {-2} e^{\left (4 \, x\right )} + 2 \, \sqrt {-2} e^{\left (2 \, x\right )} + \sqrt {-2}\right )} \log \left (-2 \, {\left (\sqrt {-2} \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} - 2 \, e^{\left (2 \, x\right )} - 2\right )} e^{\left (-2 \, x\right )}\right ) + {\left (5 i \, e^{\left (4 \, x\right )} + 10 i \, e^{\left (2 \, x\right )} + 5 i\right )} \log \left ({\left (4 i \, \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} - 4 \, e^{\left (2 \, x\right )} + 4\right )} e^{\left (-2 \, x\right )}\right ) + {\left (-5 i \, e^{\left (4 \, x\right )} - 10 i \, e^{\left (2 \, x\right )} - 5 i\right )} \log \left ({\left (-4 i \, \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} - 4 \, e^{\left (2 \, x\right )} + 4\right )} e^{\left (-2 \, x\right )}\right ) - 2 \, {\left (\sqrt {-2} e^{\left (4 \, x\right )} + 2 \, \sqrt {-2} e^{\left (2 \, x\right )} + \sqrt {-2}\right )} \log \left (4 \, {\left (\sqrt {-2 \, e^{\left (4 \, x\right )} - 2} {\left (e^{\left (2 \, x\right )} - 2\right )} + \sqrt {-2} e^{\left (4 \, x\right )} - \sqrt {-2} e^{\left (2 \, x\right )} + 2 \, \sqrt {-2}\right )} e^{\left (-4 \, x\right )}\right ) + 2 \, {\left (\sqrt {-2} e^{\left (4 \, x\right )} + 2 \, \sqrt {-2} e^{\left (2 \, x\right )} + \sqrt {-2}\right )} \log \left (4 \, {\left (\sqrt {-2 \, e^{\left (4 \, x\right )} - 2} {\left (e^{\left (2 \, x\right )} - 2\right )} - \sqrt {-2} e^{\left (4 \, x\right )} + \sqrt {-2} e^{\left (2 \, x\right )} - 2 \, \sqrt {-2}\right )} e^{\left (-4 \, x\right )}\right ) + 2 \, \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} {\left (e^{\left (2 \, x\right )} - 1\right )}}{4 \, {\left (e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1\right )}} \]

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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giac [C]  time = 0.17, size = 204, normalized size = 3.04 \[ -\frac {1}{4} \, \sqrt {2} {\left (-5 i \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1}{\sqrt {2} + \sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} - 1}\right ) - \frac {4 \, {\left (-3 i \, {\left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{3} + i \, {\left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} + i \, \sqrt {e^{\left (4 \, x\right )} + 1} - i \, e^{\left (2 \, x\right )} + i\right )}}{{\left ({\left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} - 2 \, \sqrt {e^{\left (4 \, x\right )} + 1} + 2 \, e^{\left (2 \, x\right )} - 1\right )}^{2}} - 4 i \, \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) - 4 i \, \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) + 4 i \, \log \left (-\sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right )\right )} \]

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

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maple [B]  time = 0.11, size = 211, normalized size = 3.15 \[ -\frac {\left (-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )\right )^{\frac {3}{2}}}{6}+\frac {\tanh \relax (x ) \sqrt {-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )}}{4}-\frac {5 \arctan \left (\frac {\tanh \relax (x )}{\sqrt {-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )}}\right )}{4}+\sqrt {-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )}-\sqrt {2}\, \arctan \left (\frac {\left (-2-2 \tanh \relax (x )\right ) \sqrt {2}}{4 \sqrt {-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )}}\right )+\frac {\left (-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )\right )^{\frac {3}{2}}}{6}+\frac {\tanh \relax (x ) \sqrt {-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )}}{4}-\frac {5 \arctan \left (\frac {\tanh \relax (x )}{\sqrt {-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )}}\right )}{4}-\sqrt {-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )}+\sqrt {2}\, \arctan \left (\frac {\left (-2+2 \tanh \relax (x )\right ) \sqrt {2}}{4 \sqrt {-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*arctan(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))+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+2*ta
nh(x))*2^(1/2)/(-(1+tanh(x))^2+2*tanh(x))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-\tanh \relax (x)^{2} - 1\right )}^{\frac {3}{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (-{\mathrm {tanh}\relax (x)}^2-1\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((- tanh(x)^2 - 1)^(3/2),x)

[Out]

int((- tanh(x)^2 - 1)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- \tanh ^{2}{\relax (x )} - 1\right )^{\frac {3}{2}}\, dx \]

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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