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

Optimal. Leaf size=110 \[ -\frac {15 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}{4 b^{7/2}}-\frac {5 x^{3/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {15 \sqrt {x}}{4 b^3} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 110, 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} \[ -\frac {5 x^{3/2}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {15 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}{4 b^{7/2}}-\frac {x^{5/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {15 \sqrt {x}}{4 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, size = 104, normalized size = 0.95 \[ \frac {1}{4} \left (-\frac {15 \sqrt {\tanh ^{-1}(\tanh (a+b x))-b x} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{7/2}}-\frac {5 x^{3/2}}{b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {2 x^{5/2}}{b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {15 \sqrt {x}}{b^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 200, normalized size = 1.82 \[ \left [\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(8*b^2*x^2
+ 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a/b)*a
rctan(b*sqrt(x)*sqrt(a/b)/a) - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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giac [A]  time = 0.21, size = 59, normalized size = 0.54 \[ -\frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} + \frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3 + 1/4*(9*a*b*x^(3/2) + 7*a^2*sqrt(x))/((b*
x + a)^2*b^3)

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maple [B]  time = 0.26, size = 249, normalized size = 2.26 \[ \frac {2 \sqrt {x}}{b^{3}}+\frac {9 x^{\frac {3}{2}} a}{4 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {9 x^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {7 a^{2} \sqrt {x}}{4 b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {7 \sqrt {x}\, a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{2 b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {7 \sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{4 b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) a}{4 b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}-\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4 b^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^(1/2)/b^3+9/4/b^2/arctanh(tanh(b*x+a))^2*x^(3/2)*a+9/4/b^2/arctanh(tanh(b*x+a))^2*x^(3/2)*(arctanh(tanh(b*
x+a))-b*x-a)+7/4/b^3/arctanh(tanh(b*x+a))^2*a^2*x^(1/2)+7/2/b^3/arctanh(tanh(b*x+a))^2*x^(1/2)*a*(arctanh(tanh
(b*x+a))-b*x-a)+7/4/b^3/arctanh(tanh(b*x+a))^2*x^(1/2)*(arctanh(tanh(b*x+a))-b*x-a)^2-15/4/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-15/4/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)

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maxima [A]  time = 0.44, size = 73, normalized size = 0.66 \[ \frac {8 \, b^{2} x^{\frac {5}{2}} + 25 \, a b x^{\frac {3}{2}} + 15 \, a^{2} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*b^2*x^(5/2) + 25*a*b*x^(3/2) + 15*a^2*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 15/4*a*arctan(b*sqrt(x
)/sqrt(a*b))/(sqrt(a*b)*b^3)

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mupad [B]  time = 1.69, size = 511, normalized size = 4.65 \[ \frac {2\,\sqrt {x}}{b^3}-\frac {9\,\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 )}{4\,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 )}+\frac {15\,\sqrt {2}\,\ln \left (\frac {64\,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 )\,\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}}{16\,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 )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/atanh(tanh(a + b*x))^3,x)

[Out]

(2*x^(1/2))/b^3 - (9*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))/(4*b^3*(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)))) + (15*2^(1/2)*log((64*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)*e
xp(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)^(1/2))/(16*b^(7/2)) - (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*(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)))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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