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

Optimal. Leaf size=98 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{5/2} \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac {3 \sqrt {x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

-1/2*x^(3/2)/b/arctanh(tanh(b*x+a))^2-3/4*x^(1/2)/b^2/arctanh(tanh(b*x+a))-3/4*arctanh(b^(1/2)*x^(1/2)/(b*x-ar
ctanh(tanh(b*x+a)))^(1/2))/b^(5/2)/(b*x-arctanh(tanh(b*x+a)))^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2168, 2162} \[ -\frac {3 \sqrt {x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{5/2} \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac {x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*b^(5/2)*Sqrt[b*x - ArcTanh[Tanh[a + b*x]
]]) - x^(3/2)/(2*b*ArcTanh[Tanh[a + b*x]]^2) - (3*Sqrt[x])/(4*b^2*ArcTanh[Tanh[a + b*x]])

Rule 2162

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-2*ArcTanh[Sqrt
[v]/Rt[-((b*u - a*v)/a), 2]])/(a*Rt[-((b*u - a*v)/a), 2]), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /;
PiecewiseLinearQ[u, v, 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^{3/2}}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 \int \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac {x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {3 \sqrt {x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{5/2} \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac {x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {3 \sqrt {x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 96, normalized size = 0.98 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 b^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac {3 \sqrt {x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x^(3/2)/(b*ArcTanh[Tanh[a + b*x]]^2) - (3*Sqrt[x])/(4*b^2*ArcTanh[Tanh[a + b*x]]) + (3*ArcTan[(Sqrt[b]*Sq
rt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(4*b^(5/2)*Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]])

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fricas [A]  time = 0.50, size = 185, normalized size = 1.89 \[ \left [-\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (5 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {x}}{8 \, {\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}, -\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (5 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {x}}{4 \, {\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(5*a*b^2*x +
3*a^2*b)*sqrt(x))/(a*b^5*x^2 + 2*a^2*b^4*x + a^3*b^3), -1/4*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a*b)*arctan(sqr
t(a*b)/(b*sqrt(x))) + (5*a*b^2*x + 3*a^2*b)*sqrt(x))/(a*b^5*x^2 + 2*a^2*b^4*x + a^3*b^3)]

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giac [A]  time = 0.17, size = 47, normalized size = 0.48 \[ \frac {3 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{2}} - \frac {5 \, b x^{\frac {3}{2}} + 3 \, a \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2) - 1/4*(5*b*x^(3/2) + 3*a*sqrt(x))/((b*x + a)^2*b^2)

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maple [A]  time = 0.27, size = 85, normalized size = 0.87 \[ \frac {-\frac {5 x^{\frac {3}{2}}}{4 b}-\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}{4 b^{2}}}{\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 b^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-5/8*x^(3/2)/b-3/8*(arctanh(tanh(b*x+a))-b*x)/b^2*x^(1/2))/arctanh(tanh(b*x+a))^2+3/4/b^2/((arctanh(tanh(b*
x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))

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maxima [A]  time = 0.43, size = 61, normalized size = 0.62 \[ -\frac {5 \, b x^{\frac {3}{2}} + 3 \, a \sqrt {x}}{4 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} + \frac {3 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(5*b*x^(3/2) + 3*a*sqrt(x))/(b^4*x^2 + 2*a*b^3*x + a^2*b^2) + 3/4*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*
b^2)

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mupad [B]  time = 1.87, size = 667, normalized size = 6.81 \[ \frac {3\,\sqrt {2}\,\ln \left (\frac {16\,b^{11/2}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x\right )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x}+2\,\sqrt {2}\,b\,x\right )}{\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 )}{8\,b^{5/2}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x}}-\frac {\sqrt {x}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x\right )}{b^2\,{\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 )}^2}-\frac {\sqrt {x}\,\left (\frac {1}{b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x\right )}+\frac {8\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-8\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+16\,b\,x}{2\,b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x\right )}^2}\right )\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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\,b\,x\right )}{2\,b\,\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^(3/2)/atanh(tanh(a + b*x))^3,x)

[Out]

(3*2^(1/2)*log((16*b^(11/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*
x) + 1)) + 2*b*x)^(1/2)*(2^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp
(2*b*x) + 1)) + 2*b*x) - 4*b^(1/2)*x^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2) + 2*2^(1/2)*b*x))/(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1
)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))))/(8*b^(5/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*
b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(1/2)) - (x^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a
)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(b^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)
) - log(2/(exp(2*a)*exp(2*b*x) + 1)))^2) - (x^(1/2)*(1/(b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*
exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)) + (8*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 8*log((2*exp(2*a)*exp
(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 16*b*x)/(2*b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b
*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2))*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(
exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(2*b*(log((2*exp(2*a)*exp(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 \[ \int \frac {x^{\frac {3}{2}}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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