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

Optimal. Leaf size=116 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{7/2}}+\frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}+\frac {2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac {2 x^{5/2}}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2159, 2162} \[ \frac {2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac {2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{7/2}}+\frac {2 x^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/ArcTanh[Tanh[a + b*x]],x]

[Out]

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

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

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Mathematica [A]  time = 0.13, size = 108, normalized size = 0.93 \[ \frac {2 \left (-35 b^{3/2} x^{3/2} \tanh ^{-1}(\tanh (a+b x))+15 \sqrt {b} \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-15 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )+23 b^{5/2} x^{5/2}\right )}{15 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(23*b^(5/2)*x^(5/2) - 35*b^(3/2)*x^(3/2)*ArcTanh[Tanh[a + b*x]] + 15*Sqrt[b]*Sqrt[x]*ArcTanh[Tanh[a + b*x]]
^2 - 15*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(5/2
)))/(15*b^(7/2))

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fricas [A]  time = 0.48, size = 132, normalized size = 1.14 \[ \left [\frac {15 \, a^{2} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, a^{2} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 59, normalized size = 0.51 \[ -\frac {2 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, b^{4} x^{\frac {5}{2}} - 5 \, a b^{3} x^{\frac {3}{2}} + 15 \, a^{2} b^{2} \sqrt {x}\right )}}{15 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.27, size = 330, normalized size = 2.84 \[ \frac {2 x^{\frac {5}{2}}}{5 b}-\frac {2 x^{\frac {3}{2}} a}{3 b^{2}}-\frac {2 x^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3 b^{2}}+\frac {2 a^{2} \sqrt {x}}{b^{3}}+\frac {4 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{3}}+\frac {2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{b^{3}}-\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{3}}{b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {6 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {6 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^(5/2)/b-2/3/b^2*x^(3/2)*a-2/3/b^2*x^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)+2/b^3*a^2*x^(1/2)+4/b^3*a*(arctan
h(tanh(b*x+a))-b*x-a)*x^(1/2)+2/b^3*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)-2/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^3-6/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*(arctanh(tanh(b*x+a))-b*x-a)-6/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)^2-2
/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))*(arctanh(tanh
(b*x+a))-b*x-a)^3

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maxima [A]  time = 0.42, size = 54, normalized size = 0.47 \[ -\frac {2 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, b^{2} x^{\frac {5}{2}} - 5 \, a b x^{\frac {3}{2}} + 15 \, a^{2} \sqrt {x}\right )}}{15 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*b^2*x^(5/2) - 5*a*b*x^(3/2) + 15*a^2*sqrt(x))/b^3

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mupad [B]  time = 1.39, size = 415, normalized size = 3.58 \[ \frac {2\,x^{5/2}}{5\,b}+\frac {x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{3\,b^2}+\frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,b^3}+\frac {\sqrt {2}\,\ln \left (\frac {16\,b^{15/2}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}+2\,\sqrt {2}\,b\,x\right )}{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^{5/2}}{8\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/atanh(tanh(a + b*x)),x)

[Out]

(2*x^(5/2))/(5*b) + (x^(3/2)*(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))/(3*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)^2)/(2*b^3) + (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)*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*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^(5/2))/(8*b^(7/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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