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

Optimal. Leaf size=35 \[ \frac {1}{2} \tanh ^{-1}\left (\frac {\tanh (x)}{\sqrt {-\text {sech}^2(x)}}\right )-\frac {1}{2} \tanh (x) \sqrt {-\text {sech}^2(x)} \]

[Out]

1/2*arctanh(tanh(x)/(-sech(x)^2)^(1/2))-1/2*(-sech(x)^2)^(1/2)*tanh(x)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3657, 4122, 195, 217, 206} \[ \frac {1}{2} \tanh ^{-1}\left (\frac {\tanh (x)}{\sqrt {-\text {sech}^2(x)}}\right )-\frac {1}{2} \tanh (x) \sqrt {-\text {sech}^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + Tanh[x]^2)^(3/2),x]

[Out]

ArcTanh[Tanh[x]/Sqrt[-Sech[x]^2]]/2 - (Sqrt[-Sech[x]^2]*Tanh[x])/2

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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])

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 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 28, normalized size = 0.80 \[ -\frac {1}{2} \sqrt {-\text {sech}^2(x)} \left (\tanh (x)+2 \cosh (x) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Tanh[x]^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 1, normalized size = 0.03 \[ 0 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

________________________________________________________________________________________

giac [A]  time = 0.12, size = 41, normalized size = 1.17 \[ \frac {\sqrt {-e^{\left (2 \, x\right )}} + \frac {1}{\sqrt {-e^{\left (2 \, x\right )}}}}{{\left (\sqrt {-e^{\left (2 \, x\right )}} + \frac {1}{\sqrt {-e^{\left (2 \, x\right )}}}\right )}^{2} - 4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.12, size = 28, normalized size = 0.80 \[ -\frac {\tanh \relax (x ) \sqrt {-1+\tanh ^{2}\relax (x )}}{2}+\frac {\ln \left (\tanh \relax (x )+\sqrt {-1+\tanh ^{2}\relax (x )}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*tanh(x)*(-1+tanh(x)^2)^(1/2)+1/2*ln(tanh(x)+(-1+tanh(x)^2)^(1/2))

________________________________________________________________________________________

maxima [C]  time = 0.46, size = 32, normalized size = 0.91 \[ \frac {-i \, e^{\left (3 \, x\right )} + i \, e^{x}}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1} - i \, \arctan \left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.17, size = 27, normalized size = 0.77 \[ \frac {\ln \left (\mathrm {tanh}\relax (x)+\sqrt {{\mathrm {tanh}\relax (x)}^2-1}\right )}{2}-\frac {\mathrm {tanh}\relax (x)\,\sqrt {{\mathrm {tanh}\relax (x)}^2-1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x)^2 - 1)^(3/2),x)

[Out]

log(tanh(x) + (tanh(x)^2 - 1)^(1/2))/2 - (tanh(x)*(tanh(x)^2 - 1)^(1/2))/2

________________________________________________________________________________________

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)

________________________________________________________________________________________