3.250 \(\int \frac {x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=128 \[ \frac {15 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{7/2}}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

15/4*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))^2/b^(7/2)-2*x^(5/2)/b/arct
anh(tanh(b*x+a))^(1/2)+5/2*x^(3/2)*arctanh(tanh(b*x+a))^(1/2)/b^2+15/4*(b*x-arctanh(tanh(b*x+a)))*x^(1/2)*arct
anh(tanh(b*x+a))^(1/2)/b^3

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Rubi [A]  time = 0.08, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ \frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3}+\frac {15 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{7/2}}-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^2)/(4*b^(7/2)) - (2
*x^(5/2))/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*x^(3/2)*Sqrt[ArcTanh[Tanh[a + b*x]]])/(2*b^2) + (15*Sqrt[x]*(b
*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])/(4*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 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]))

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}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 \int \frac {x^{3/2}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}-\frac {\left (15 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{4 b^2}\\ &=-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3}+\frac {\left (15 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 b^3}\\ &=\frac {15 \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 )^2}{4 b^{7/2}}-\frac {2 x^{5/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 b^2}+\frac {15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 104, normalized size = 0.81 \[ \frac {15 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{4 b^{7/2}}-\frac {\sqrt {x} \left (-25 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right )}{4 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[x]*(8*b^2*x^2 - 25*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh[a + b*x]]^2))/(b^3*Sqrt[ArcTanh[Tan
h[a + b*x]]]) + (15*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])
/(4*b^(7/2))

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fricas [A]  time = 0.57, size = 175, normalized size = 1.37 \[ \left [\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*(a^2*b*x + a^3)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^3*x^2 - 5*a*b^2*x -
15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^4), -1/4*(15*(a^2*b*x + a^3)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt
(-b)/(b*sqrt(x))) - (2*b^3*x^2 - 5*a*b^2*x - 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^4)]

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giac [A]  time = 0.16, size = 63, normalized size = 0.49 \[ \frac {{\left (x {\left (\frac {2 \, x}{b} - \frac {5 \, a}{b^{2}}\right )} - \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x}}{4 \, \sqrt {b x + a}} - \frac {15 \, a^{2} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{4 \, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.25, size = 261, normalized size = 2.04 \[ \frac {x^{\frac {5}{2}}}{2 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {5 a \,x^{\frac {3}{2}}}{4 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {15 a^{2} \sqrt {x}}{4 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 a^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{4 b^{\frac {7}{2}}}-\frac {15 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{2 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{2 b^{\frac {7}{2}}}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x^{\frac {3}{2}}}{4 b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{4 b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{4 b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

1/2*x^(5/2)/b/arctanh(tanh(b*x+a))^(1/2)-5/4/b^2*a*x^(3/2)/arctanh(tanh(b*x+a))^(1/2)-15/4/b^3*a^2*x^(1/2)/arc
tanh(tanh(b*x+a))^(1/2)+15/4/b^(7/2)*a^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))-15/2/b^3*a*(arctanh(ta
nh(b*x+a))-b*x-a)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)+15/2/b^(7/2)*a*(arctanh(tanh(b*x+a))-b*x-a)*ln(b^(1/2)*x^
(1/2)+arctanh(tanh(b*x+a))^(1/2))-5/4/b^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(3/2)/arctanh(tanh(b*x+a))^(1/2)-15/4
/b^3*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)+15/4/b^(7/2)*(arctanh(tanh(b*x+a))-b*x-
a)^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {5}{2}}}{\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/arctanh(tanh(b*x + a))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{5/2}}{{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/atanh(tanh(a + b*x))^(3/2),x)

[Out]

int(x^(5/2)/atanh(tanh(a + b*x))^(3/2), 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))**(3/2),x)

[Out]

Timed out

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