3.24 \(\int (a \tanh ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=35 \[ a \coth (x) \sqrt {a \tanh ^2(x)} \log (\cosh (x))-\frac {1}{2} a \tanh (x) \sqrt {a \tanh ^2(x)} \]

[Out]

a*coth(x)*ln(cosh(x))*(a*tanh(x)^2)^(1/2)-1/2*a*(a*tanh(x)^2)^(1/2)*tanh(x)

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3658, 3473, 3475} \[ a \coth (x) \sqrt {a \tanh ^2(x)} \log (\cosh (x))-\frac {1}{2} a \tanh (x) \sqrt {a \tanh ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Tanh[x]^2)^(3/2),x]

[Out]

a*Coth[x]*Log[Cosh[x]]*Sqrt[a*Tanh[x]^2] - (a*Tanh[x]*Sqrt[a*Tanh[x]^2])/2

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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
]])

Rubi steps

\begin {align*} \int \left (a \tanh ^2(x)\right )^{3/2} \, dx &=\left (a \coth (x) \sqrt {a \tanh ^2(x)}\right ) \int \tanh ^3(x) \, dx\\ &=-\frac {1}{2} a \tanh (x) \sqrt {a \tanh ^2(x)}+\left (a \coth (x) \sqrt {a \tanh ^2(x)}\right ) \int \tanh (x) \, dx\\ &=a \coth (x) \log (\cosh (x)) \sqrt {a \tanh ^2(x)}-\frac {1}{2} a \tanh (x) \sqrt {a \tanh ^2(x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.80 \[ \frac {1}{2} a \sqrt {a \tanh ^2(x)} (\text {csch}(x) \text {sech}(x)+2 \coth (x) \log (\cosh (x))) \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Tanh[x]^2)^(3/2),x]

[Out]

(a*(2*Coth[x]*Log[Cosh[x]] + Csch[x]*Sech[x])*Sqrt[a*Tanh[x]^2])/2

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fricas [B]  time = 0.51, size = 467, normalized size = 13.34 \[ -\frac {{\left (a x \cosh \relax (x)^{4} + {\left (a x e^{\left (2 \, x\right )} + a x\right )} \sinh \relax (x)^{4} + 4 \, {\left (a x \cosh \relax (x) e^{\left (2 \, x\right )} + a x \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 2 \, {\left (a x - a\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, a x \cosh \relax (x)^{2} + a x + {\left (3 \, a x \cosh \relax (x)^{2} + a x - a\right )} e^{\left (2 \, x\right )} - a\right )} \sinh \relax (x)^{2} + a x + {\left (a x \cosh \relax (x)^{4} + 2 \, {\left (a x - a\right )} \cosh \relax (x)^{2} + a x\right )} e^{\left (2 \, x\right )} - {\left (a \cosh \relax (x)^{4} + {\left (a e^{\left (2 \, x\right )} + a\right )} \sinh \relax (x)^{4} + 4 \, {\left (a \cosh \relax (x) e^{\left (2 \, x\right )} + a \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 2 \, a \cosh \relax (x)^{2} + 2 \, {\left (3 \, a \cosh \relax (x)^{2} + {\left (3 \, a \cosh \relax (x)^{2} + a\right )} e^{\left (2 \, x\right )} + a\right )} \sinh \relax (x)^{2} + {\left (a \cosh \relax (x)^{4} + 2 \, a \cosh \relax (x)^{2} + a\right )} e^{\left (2 \, x\right )} + 4 \, {\left (a \cosh \relax (x)^{3} + a \cosh \relax (x) + {\left (a \cosh \relax (x)^{3} + a \cosh \relax (x)\right )} e^{\left (2 \, x\right )}\right )} \sinh \relax (x) + a\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 4 \, {\left (a x \cosh \relax (x)^{3} + {\left (a x - a\right )} \cosh \relax (x) + {\left (a x \cosh \relax (x)^{3} + {\left (a x - a\right )} \cosh \relax (x)\right )} e^{\left (2 \, x\right )}\right )} \sinh \relax (x)\right )} \sqrt {\frac {a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1}}}{{\left (e^{\left (2 \, x\right )} - 1\right )} \sinh \relax (x)^{4} - \cosh \relax (x)^{4} + 4 \, {\left (\cosh \relax (x) e^{\left (2 \, x\right )} - \cosh \relax (x)\right )} \sinh \relax (x)^{3} - 2 \, {\left (3 \, \cosh \relax (x)^{2} - {\left (3 \, \cosh \relax (x)^{2} + 1\right )} e^{\left (2 \, x\right )} + 1\right )} \sinh \relax (x)^{2} - 2 \, \cosh \relax (x)^{2} + {\left (\cosh \relax (x)^{4} + 2 \, \cosh \relax (x)^{2} + 1\right )} e^{\left (2 \, x\right )} - 4 \, {\left (\cosh \relax (x)^{3} - {\left (\cosh \relax (x)^{3} + \cosh \relax (x)\right )} e^{\left (2 \, x\right )} + \cosh \relax (x)\right )} \sinh \relax (x) - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tanh(x)^2)^(3/2),x, algorithm="fricas")

[Out]

-(a*x*cosh(x)^4 + (a*x*e^(2*x) + a*x)*sinh(x)^4 + 4*(a*x*cosh(x)*e^(2*x) + a*x*cosh(x))*sinh(x)^3 + 2*(a*x - a
)*cosh(x)^2 + 2*(3*a*x*cosh(x)^2 + a*x + (3*a*x*cosh(x)^2 + a*x - a)*e^(2*x) - a)*sinh(x)^2 + a*x + (a*x*cosh(
x)^4 + 2*(a*x - a)*cosh(x)^2 + a*x)*e^(2*x) - (a*cosh(x)^4 + (a*e^(2*x) + a)*sinh(x)^4 + 4*(a*cosh(x)*e^(2*x)
+ a*cosh(x))*sinh(x)^3 + 2*a*cosh(x)^2 + 2*(3*a*cosh(x)^2 + (3*a*cosh(x)^2 + a)*e^(2*x) + a)*sinh(x)^2 + (a*co
sh(x)^4 + 2*a*cosh(x)^2 + a)*e^(2*x) + 4*(a*cosh(x)^3 + a*cosh(x) + (a*cosh(x)^3 + a*cosh(x))*e^(2*x))*sinh(x)
+ a)*log(2*cosh(x)/(cosh(x) - sinh(x))) + 4*(a*x*cosh(x)^3 + (a*x - a)*cosh(x) + (a*x*cosh(x)^3 + (a*x - a)*c
osh(x))*e^(2*x))*sinh(x))*sqrt((a*e^(4*x) - 2*a*e^(2*x) + a)/(e^(4*x) + 2*e^(2*x) + 1))/((e^(2*x) - 1)*sinh(x)
^4 - cosh(x)^4 + 4*(cosh(x)*e^(2*x) - cosh(x))*sinh(x)^3 - 2*(3*cosh(x)^2 - (3*cosh(x)^2 + 1)*e^(2*x) + 1)*sin
h(x)^2 - 2*cosh(x)^2 + (cosh(x)^4 + 2*cosh(x)^2 + 1)*e^(2*x) - 4*(cosh(x)^3 - (cosh(x)^3 + cosh(x))*e^(2*x) +
cosh(x))*sinh(x) - 1)

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giac [A]  time = 5.01, size = 52, normalized size = 1.49 \[ -{\left (x \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \log \left (e^{\left (2 \, x\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) - \frac {2 \, e^{\left (2 \, x\right )} \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right )}{{\left (e^{\left (2 \, x\right )} + 1\right )}^{2}}\right )} a^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tanh(x)^2)^(3/2),x, algorithm="giac")

[Out]

-(x*sgn(e^(4*x) - 1) - log(e^(2*x) + 1)*sgn(e^(4*x) - 1) - 2*e^(2*x)*sgn(e^(4*x) - 1)/(e^(2*x) + 1)^2)*a^(3/2)

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maple [A]  time = 0.08, size = 30, normalized size = 0.86 \[ -\frac {\left (a \left (\tanh ^{2}\relax (x )\right )\right )^{\frac {3}{2}} \left (\tanh ^{2}\relax (x )+\ln \left (\tanh \relax (x )-1\right )+\ln \left (1+\tanh \relax (x )\right )\right )}{2 \tanh \relax (x )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*tanh(x)^2)^(3/2),x)

[Out]

-1/2*(a*tanh(x)^2)^(3/2)*(tanh(x)^2+ln(tanh(x)-1)+ln(1+tanh(x)))/tanh(x)^3

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maxima [A]  time = 0.50, size = 42, normalized size = 1.20 \[ -a^{\frac {3}{2}} x - a^{\frac {3}{2}} \log \left (e^{\left (-2 \, x\right )} + 1\right ) - \frac {2 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tanh(x)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (a\,{\mathrm {tanh}\relax (x)}^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*tanh(x)^2)^(3/2),x)

[Out]

int((a*tanh(x)^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*tanh(x)**2)**(3/2),x)

[Out]

Integral((a*tanh(x)**2)**(3/2), x)

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