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

Optimal. Leaf size=108 \[ -\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{b^{7/2}}+\frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {5 x^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2159, 2162} \[ \frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}{b^{7/2}}-\frac {x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {5 x^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x^(3/2))/(3*b^2) + (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]]))/b^3 - (5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x -
ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2))/b^(7/2) - x^(5/2)/(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^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {5 \int \frac {x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b}\\ &=\frac {5 x^{3/2}}{3 b^2}-\frac {x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b^2}\\ &=\frac {5 x^{3/2}}{3 b^2}+\frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b^3}\\ &=\frac {5 x^{3/2}}{3 b^2}+\frac {5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {5 \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^{7/2}}-\frac {x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 119, normalized size = 1.10 \[ \frac {5 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{7/2}}-\frac {\sqrt {x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^3 \tanh ^{-1}(\tanh (a+b x))}-\frac {4 \sqrt {x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^3}+\frac {2 x^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 161, normalized size = 1.49 \[ \left [\frac {15 \, {\left (a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {15 \, {\left (a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b*x + a^2)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(2*b^2*x^2 - 10*a*b*x
- 15*a^2)*sqrt(x))/(b^4*x + a*b^3), 1/3*(15*(a*b*x + a^2)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (2*b^2*x^2
- 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x + a*b^3)]

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giac [A]  time = 0.19, size = 65, normalized size = 0.60 \[ \frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {a^{2} \sqrt {x}}{{\left (b x + a\right )} b^{3}} + \frac {2 \, {\left (b^{4} x^{\frac {3}{2}} - 6 \, a b^{3} \sqrt {x}\right )}}{3 \, b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - a^2*sqrt(x)/((b*x + a)*b^3) + 2/3*(b^4*x^(3/2) - 6*a*b^3*s
qrt(x))/b^6

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maple [B]  time = 0.27, size = 294, normalized size = 2.72 \[ \frac {2 x^{\frac {3}{2}}}{3 b^{2}}-\frac {4 a \sqrt {x}}{b^{3}}-\frac {4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{3}}-\frac {\sqrt {x}\, a^{2}}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \sqrt {x}\, a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {\sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{2}}{b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {10 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^(3/2)/b^2-4/b^3*a*x^(1/2)-4/b^3*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)-1/b^3*x^(1/2)/arctanh(tanh(b*x+a))*
a^2-2/b^3*x^(1/2)/arctanh(tanh(b*x+a))*a*(arctanh(tanh(b*x+a))-b*x-a)-1/b^3*x^(1/2)/arctanh(tanh(b*x+a))*(arct
anh(tanh(b*x+a))-b*x-a)^2+5/b^3/((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+10/b^3/((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*(arctanh(tanh(b*x+a))-b*x-a)+5/b^3/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(ta
nh(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 = 64, normalized size = 0.59 \[ \frac {2 \, b^{2} x^{\frac {5}{2}} - 10 \, a b x^{\frac {3}{2}} - 15 \, a^{2} \sqrt {x}}{3 \, {\left (b^{4} x + a b^{3}\right )}} + \frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*b^2*x^(5/2) - 10*a*b*x^(3/2) - 15*a^2*sqrt(x))/(b^4*x + a*b^3) + 5*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqr
t(a*b)*b^3)

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mupad [B]  time = 1.53, size = 463, normalized size = 4.29 \[ \frac {2\,x^{3/2}}{3\,b^2}+\frac {2\,\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^3}+\frac {5\,\sqrt {2}\,\ln \left (\frac {16\,b^{15/2}\,\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}}{8\,b^{7/2}}-\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 )}^2}{2\,b^3\,\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^(5/2)/atanh(tanh(a + b*x))^2,x)

[Out]

(2*x^(3/2))/(3*b^2) + (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^3 + (5*2^(1/2)*log((16*b^(15/2)*(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)*e
xp(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)) - lo
g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(3/2))/(8*b^(7/2)) - (x^(1/2)*(log(2/(exp(2*a)*e
xp(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)/(2*b^3*(log((2*exp(2*a)*e
xp(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 \[ \int \frac {x^{\frac {5}{2}}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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