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

Optimal. Leaf size=89 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{b^{5/2}}+\frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac {2 x^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 89, 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 = {2159, 2162} \[ \frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{b^{5/2}}+\frac {2 x^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/ArcTanh[Tanh[a + b*x]],x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 86, normalized size = 0.97 \[ \frac {2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{5/2}}-\frac {2 \sqrt {x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^2}+\frac {2 x^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 103, normalized size = 1.16 \[ \left [\frac {3 \, a \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (b x - 3 \, a\right )} \sqrt {x}}{3 \, b^{2}}, \frac {2 \, {\left (3 \, a \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (b x - 3 \, a\right )} \sqrt {x}\right )}}{3 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 45, normalized size = 0.51 \[ \frac {2 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (b^{2} x^{\frac {3}{2}} - 3 \, a b \sqrt {x}\right )}}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.27, size = 207, normalized size = 2.33 \[ \frac {2 x^{\frac {3}{2}}}{3 b}-\frac {2 a \sqrt {x}}{b^{2}}-\frac {2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{2}}+\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{2}}{b^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {4 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 42, normalized size = 0.47 \[ \frac {2 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (b x^{\frac {3}{2}} - 3 \, a \sqrt {x}\right )}}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.85, size = 354, normalized size = 3.98 \[ \frac {2\,x^{3/2}}{3\,b}+\frac {\sqrt {x}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b^2}+\frac {\sqrt {2}\,\ln \left (\frac {4\,b^{11/2}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}+2\,\sqrt {2}\,b\,x\right )}{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^{3/2}}{4\,b^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/atanh(tanh(a + b*x)),x)

[Out]

(2*x^(3/2))/(3*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))/b^2 + (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))/(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))/(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)*exp(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*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(3/2))/(4*b^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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