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

Optimal. Leaf size=83 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}{b^{5/2}}-\frac {x^{3/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \sqrt {x}}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2159, 2162} \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}{b^{5/2}}-\frac {x^{3/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \sqrt {x}}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

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))^2} \, dx &=-\frac {x^{3/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \int \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b}\\ &=\frac {3 \sqrt {x}}{b^2}-\frac {x^{3/2}}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b^2}\\ &=\frac {3 \sqrt {x}}{b^2}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}{b^{5/2}}-\frac {x^{3/2}}{b \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 81, normalized size = 0.98 \[ -\frac {3 \sqrt {\tanh ^{-1}(\tanh (a+b x))-b x} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{5/2}}-\frac {x^{3/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \sqrt {x}}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 134, normalized size = 1.61 \[ \left [\frac {3 \, {\left (b x + a\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (2 \, b x + 3 \, a\right )} \sqrt {x}}{2 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (b x + a\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (2 \, b x + 3 \, a\right )} \sqrt {x}}{b^{3} x + a b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(b*x + a)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(2*b*x + 3*a)*sqrt(x))/(b^3
*x + a*b^2), -(3*(b*x + a)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) - (2*b*x + 3*a)*sqrt(x))/(b^3*x + a*b^2)]

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giac [A]  time = 0.52, size = 46, normalized size = 0.55 \[ -\frac {3 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {a \sqrt {x}}{{\left (b x + a\right )} b^{2}} + \frac {2 \, \sqrt {x}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.26, size = 160, normalized size = 1.93 \[ \frac {2 \sqrt {x}}{b^{2}}+\frac {\sqrt {x}\, a}{b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {3 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a}{b^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {3 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 50, normalized size = 0.60 \[ \frac {2 \, b x^{\frac {3}{2}} + 3 \, a \sqrt {x}}{b^{3} x + a b^{2}} - \frac {3 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.77, size = 403, normalized size = 4.86 \[ \frac {2\,\sqrt {x}}{b^2}-\frac {\sqrt {x}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b^2\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}+\frac {3\,\sqrt {2}\,\ln \left (\frac {4\,b^{11/2}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}+2\,\sqrt {2}\,b\,x\right )}{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}\right )\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}{4\,b^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/atanh(tanh(a + b*x))^2,x)

[Out]

(2*x^(1/2))/b^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)))) + (3*2^(1/2)*log((4*b^(11/2)*(2^(1/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(e
xp(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))/(exp(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)*ex
p(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))*(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)))*(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))/(4*b^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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