3.241 \(\int \frac {x^{5/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\)

Optimal. Leaf size=145 \[ \frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{7/2}}+\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{8 b^3}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{12 b^2}+\frac {x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b} \]

[Out]

5/8*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^3/b^(7/2)+1/3*x^(5/2)*arcta
nh(tanh(b*x+a))^(1/2)/b+5/12*x^(3/2)*(b*x-arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(1/2)/b^2+5/8*(b*x-arctan
h(tanh(b*x+a)))^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)/b^3

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Rubi [A]  time = 0.08, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ \frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{12 b^2}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{7/2}}+\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{8 b^3}+\frac {x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/(8*b^(7/2)) + (x^
(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]])/(3*b) + (5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*
x]]])/(12*b^2) + (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(8*b^3)

Rule 2165

Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTanh[(
Rt[a*b, 2]*Sqrt[u])/(a*Sqrt[v])])/Rt[a*b, 2], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[u, v,
x]

Rule 2169

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 + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

Rubi steps

\begin {align*} \int \frac {x^{5/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b}-\frac {\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {x^{3/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{6 b}\\ &=\frac {x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b}+\frac {5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^2}+\frac {\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 b^2}\\ &=\frac {x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b}+\frac {5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^2}+\frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 b^3}-\frac {\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 b^3}\\ &=\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 b^{7/2}}+\frac {x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b}+\frac {5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{12 b^2}+\frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 105, normalized size = 0.72 \[ \frac {5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{8 b^{7/2}}+\frac {\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-40 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2+33 b^2 x^2\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(33*b^2*x^2 - 40*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh[a + b*x]]^
2))/(24*b^3) + (5*(b*x - ArcTanh[Tanh[a + b*x]])^3*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*b
^(7/2))

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fricas [A]  time = 0.64, size = 140, normalized size = 0.97 \[ \left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{4}}, \frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*b^3*x^2 - 10*a*b^2*x + 15*a^2*b)
*sqrt(b*x + a)*sqrt(x))/b^4, 1/24*(15*a^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*b^3*x^2 - 1
0*a*b^2*x + 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/b^4]

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giac [A]  time = 0.14, size = 64, normalized size = 0.44 \[ \frac {1}{24} \, \sqrt {b x + a} {\left (2 \, x {\left (\frac {4 \, x}{b} - \frac {5 \, a}{b^{2}}\right )} + \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x} + \frac {5 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(b*x + a)*(2*x*(4*x/b - 5*a/b^2) + 15*a^2/b^3)*sqrt(x) + 5/8*a^3*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x
+ a)))/b^(7/2)

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maple [B]  time = 0.30, size = 304, normalized size = 2.10 \[ \frac {x^{\frac {5}{2}} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 b}-\frac {5 a \,x^{\frac {3}{2}} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{12 b^{2}}+\frac {5 a^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{8 b^{3}}-\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) a^{3}}{8 b^{\frac {7}{2}}}-\frac {15 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8 b^{\frac {7}{2}}}+\frac {5 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{4 b^{3}}-\frac {15 a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{8 b^{\frac {7}{2}}}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x^{\frac {3}{2}} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{12 b^{2}}+\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{8 b^{3}}-\frac {5 \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{8 b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x)

[Out]

1/3*x^(5/2)*arctanh(tanh(b*x+a))^(1/2)/b-5/12/b^2*a*x^(3/2)*arctanh(tanh(b*x+a))^(1/2)+5/8/b^3*a^2*x^(1/2)*arc
tanh(tanh(b*x+a))^(1/2)-5/8/b^(7/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a^3-15/8/b^(7/2)*a^2*ln(b^(
1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+5/4/b^3*a*(arctanh(tanh(b*x+a))-b*x-a)*x
^(1/2)*arctanh(tanh(b*x+a))^(1/2)-15/8/b^(7/2)*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(
b*x+a))-b*x-a)^2-5/12/b^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(3/2)*arctanh(tanh(b*x+a))^(1/2)+5/8/b^3*(arctanh(tan
h(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)-5/8/b^(7/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2
))*(arctanh(tanh(b*x+a))-b*x-a)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {5}{2}}}{\sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/sqrt(arctanh(tanh(b*x + a))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{5/2}}{\sqrt {\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/atanh(tanh(a + b*x))^(1/2),x)

[Out]

int(x^(5/2)/atanh(tanh(a + b*x))^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/atanh(tanh(b*x+a))**(1/2),x)

[Out]

Timed out

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