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

Optimal. Leaf size=152 \[ \frac {1}{4 b^2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {1}{4 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \sqrt {b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {3}{4 b \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2162} \[ \frac {1}{4 b^2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {1}{4 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \sqrt {b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {3}{4 b \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

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

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Mathematica [A]  time = 0.09, size = 118, normalized size = 0.78 \[ \frac {\sqrt {x}}{2 \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}+\frac {3 \sqrt {x}}{4 \tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 \sqrt {b} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

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

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fricas [A]  time = 0.51, size = 186, normalized size = 1.22 \[ \left [-\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (3 \, a b^{2} x + 5 \, a^{2} b\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}}, -\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) - {\left (3 \, a b^{2} x + 5 \, a^{2} b\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 47, normalized size = 0.31 \[ \frac {3 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{2}} + \frac {3 \, b x^{\frac {3}{2}} + 5 \, a \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.27, size = 112, normalized size = 0.74 \[ \frac {\sqrt {x}}{2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 \sqrt {x}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {3 \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^2+3/4/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)/arctanh
(tanh(b*x+a))+3/4/(arctanh(tanh(b*x+a))-b*x)^2/((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 = 60, normalized size = 0.39 \[ \frac {3 \, b x^{\frac {3}{2}} + 5 \, a \sqrt {x}}{4 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} + \frac {3 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.90, size = 741, normalized size = 4.88 \[ \frac {6\,\sqrt {x}}{\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 )}^2}-\frac {4\,\sqrt {x}}{{\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\,\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 )}+\frac {3\,\sqrt {2}\,\ln \left (\frac {\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 (16\,a^4\,b+b\,{\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 )}^4-8\,a\,b\,{\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 )}^3-32\,a^3\,b\,\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 )+24\,a^2\,b\,{\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\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 )}{2\,\sqrt {b}\,{\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}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*x^(1/2))/((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)^2) - (4*x^(1/2)
)/((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*(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^(1/2)*log((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)*(16*a^4*b + b*(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)^4 - 8*a*b*(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)^3 - 32*a^3*b*(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) + 24*a^2*b*(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))/(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^(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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(x)*atanh(tanh(a + b*x))**3), x)

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