3.151 \(\int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=55 \[ -\frac {16 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^3}+\frac {8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {2 x^2}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

-16/3*arctanh(tanh(b*x+a))^(3/2)/b^3-2*x^2/b/arctanh(tanh(b*x+a))^(1/2)+8*x*arctanh(tanh(b*x+a))^(1/2)/b^2

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2157, 30} \[ \frac {8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {16 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^3}-\frac {2 x^2}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2)/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (8*x*Sqrt[ArcTanh[Tanh[a + b*x]]])/b^2 - (16*ArcTanh[Tanh[a + b*x]
]^(3/2))/(3*b^3)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {2 x^2}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {4 \int \frac {x}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac {2 x^2}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {8 \int \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=-\frac {2 x^2}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {8 \operatorname {Subst}\left (\int \sqrt {x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}\\ &=-\frac {2 x^2}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {16 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 49, normalized size = 0.89 \[ -\frac {2 \left (-12 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+3 b^2 x^2\right )}{3 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*b^2*x^2 - 12*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/(3*b^3*Sqrt[ArcTanh[Tanh[a + b*x
]]])

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fricas [A]  time = 0.74, size = 40, normalized size = 0.73 \[ \frac {2 \, {\left (b^{2} x^{2} - 4 \, a b x - 8 \, a^{2}\right )} \sqrt {b x + a}}{3 \, {\left (b^{4} x + a b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b^2*x^2 - 4*a*b*x - 8*a^2)*sqrt(b*x + a)/(b^4*x + a*b^3)

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giac [A]  time = 0.15, size = 46, normalized size = 0.84 \[ -\frac {2 \, a^{2}}{\sqrt {b x + a} b^{3}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} b^{6} - 6 \, \sqrt {b x + a} a b^{6}\right )}}{3 \, b^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^2/(sqrt(b*x + a)*b^3) + 2/3*((b*x + a)^(3/2)*b^6 - 6*sqrt(b*x + a)*a*b^6)/b^9

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maple [B]  time = 0.18, size = 106, normalized size = 1.93 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}-4 a \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^3*(1/3*arctanh(tanh(b*x+a))^(3/2)-2*a*arctanh(tanh(b*x+a))^(1/2)-2*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(ta
nh(b*x+a))^(1/2)-(a^2+2*a*(arctanh(tanh(b*x+a))-b*x-a)+(arctanh(tanh(b*x+a))-b*x-a)^2)/arctanh(tanh(b*x+a))^(1
/2))

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maxima [A]  time = 0.53, size = 41, normalized size = 0.75 \[ \frac {2 \, {\left (b^{3} x^{3} - 3 \, a b^{2} x^{2} - 12 \, a^{2} b x - 8 \, a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b^3*x^3 - 3*a*b^2*x^2 - 12*a^2*b*x - 8*a^3)/((b*x + a)^(3/2)*b^3)

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mupad [B]  time = 1.29, size = 259, normalized size = 4.71 \[ -\frac {4\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (3\,b^2\,x^2-6\,b\,x\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+6\,b\,x\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-4\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\right )}{3\,b^3\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/atanh(tanh(a + b*x))^(3/2),x)

[Out]

-(4*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*(2*l
og(2/(exp(2*a)*exp(2*b*x) + 1))^2 - 4*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log(2/(exp(2*a)*e
xp(2*b*x) + 1)) + 2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 + 3*b^2*x^2 - 6*b*x*log((2*exp(2*
a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x*log(2/(exp(2*a)*exp(2*b*x) + 1))))/(3*b^3*(log((2*exp(2*a)*e
xp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(x**2/atanh(tanh(a + b*x))**(3/2), x)

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