3.210 \(\int \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=125 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {1}{4 b^2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{4 b^2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 125, 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, 2163, 2162} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {1}{4 b^2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{4 b^2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]/(4*b^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2))
- 1/(4*b^2*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])) - Sqrt[x]/(2*b*ArcTanh[Tanh[a + b*x]]^2) - 1/(4*b^2*Sqrt[x
]*ArcTanh[Tanh[a + b*x]])

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 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

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

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Mathematica [A]  time = 0.15, size = 107, normalized size = 0.86 \[ \frac {1}{4} \left (\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}+\frac {\sqrt {x}}{b \tanh ^{-1}(\tanh (a+b x))^2-b^2 x \tanh ^{-1}(\tanh (a+b x))}-\frac {2 \sqrt {x}}{b \tanh ^{-1}(\tanh (a+b x))^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

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

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fricas [A]  time = 0.52, size = 186, normalized size = 1.49 \[ \left [-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}, -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) - {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 52, normalized size = 0.42 \[ \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} + \frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.27, size = 98, normalized size = 0.78 \[ \frac {\frac {2 x^{\frac {3}{2}}}{8 \arctanh \left (\tanh \left (b x +a \right )\right )-8 b x}-\frac {\sqrt {x}}{4 b}}{\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/arctanh(tanh(b*x+a))^3,x)

[Out]

2*(1/8/(arctanh(tanh(b*x+a))-b*x)*x^(3/2)-1/8*x^(1/2)/b)/arctanh(tanh(b*x+a))^2+1/4/(arctanh(tanh(b*x+a))-b*x)
/b/((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))

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maxima [A]  time = 0.43, size = 64, normalized size = 0.51 \[ \frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.84, size = 580, normalized size = 4.64 \[ \frac {\sqrt {2}\,\ln \left (-\frac {4\,\sqrt {b}\,\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}\,\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 (b^3\,{\left (2\,a-\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 )+2\,b\,x\right )}^2+4\,a^2\,b^3-4\,a\,b^3\,\left (2\,a-\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 )+2\,b\,x\right )\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 )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}\right )}{4\,b^{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 )}^{3/2}}-\frac {2\,\sqrt {x}}{b\,{\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}-\frac {\sqrt {x}}{b\,\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 )\,\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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/atanh(tanh(a + b*x))^3,x)

[Out]

(2^(1/2)*log(-(4*b^(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^(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)*(b^3*(2*a - 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*b*x)^2 + 4*a^2*b^3 - 4*a*b^3*(2*a - 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*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)))))/(4*b^(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)^(3/2)) - (2*x^(1/2))/(b*(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) - x^(1/2)/(b*(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*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/atanh(tanh(b*x+a))**3,x)

[Out]

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

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