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

Optimal. Leaf size=176 \[ \frac {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {15}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2162} \[ \frac {3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac {15 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {15}{4 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

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

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Mathematica [A]  time = 0.18, size = 141, normalized size = 0.80 \[ -\frac {b \sqrt {x}}{2 \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac {7 b \sqrt {x}}{4 \tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac {2}{\sqrt {x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

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

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fricas [A]  time = 0.47, size = 214, normalized size = 1.22 \[ \left [\frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac {15 \, {\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.30, size = 59, normalized size = 0.34 \[ -\frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2}{a^{3} \sqrt {x}} - \frac {7 \, b^{2} x^{\frac {3}{2}} + 9 \, a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 181, normalized size = 1.03 \[ -\frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}-\frac {7 b^{2} x^{\frac {3}{2}}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {9 b a \sqrt {x}}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {9 b \sqrt {x}\, \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {15 b \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(arctanh(tanh(b*x+a))-b*x)^3/x^(1/2)-7/4/(arctanh(tanh(b*x+a))-b*x)^3*b^2/arctanh(tanh(b*x+a))^2*x^(3/2)-9/
4/(arctanh(tanh(b*x+a))-b*x)^3*b/arctanh(tanh(b*x+a))^2*a*x^(1/2)-9/4/(arctanh(tanh(b*x+a))-b*x)^3*b/arctanh(t
anh(b*x+a))^2*x^(1/2)*(arctanh(tanh(b*x+a))-b*x-a)-15/4/(arctanh(tanh(b*x+a))-b*x)^3*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 = 73, normalized size = 0.41 \[ -\frac {15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} + 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} - \frac {15 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.17, size = 1077, normalized size = 6.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*((12*b)/(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 + (3*b*(16*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 16*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 3
2*b*x))/(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)
- (16*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 16*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 32*b*x)/
(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)/(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)))) + (15*2^(1/2)*b
^(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)*e
xp(2*b*x) + 1)) + 2*b*x)^(1/2) + 2*2^(1/2)*b*x)*((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)^6 + 60*a^2*(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 - 160*a^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e
xp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3 + 240*a^4*(2*a - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 + 64*a^6 - 12*a*(2*a - 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)) + 2*b*x)^5 - 192*a^5*(2*a - log((2*ex
p(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))))))/(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)^(7/2) - (8*b*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)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {3}{2}} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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