3.128 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=146 \[ \frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {b^2}{8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}-\frac {b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 x^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 146, 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, 2161} \[ \frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^2}{8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^4,x]

[Out]

(b^3*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(8*(b*x - ArcTanh[Tanh[a + b*x]]
)^(3/2)) - b^2/(8*x*Sqrt[ArcTanh[Tanh[a + b*x]]]) + b^3/(8*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a
+ b*x]]]) - (b*Sqrt[ArcTanh[Tanh[a + b*x]]])/(4*x^2) - ArcTanh[Tanh[a + b*x]]^(3/2)/(3*x^3)

Rule 2161

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTan[Sqrt[v
]/Rt[(b*u - a*v)/a, 2]])/(a*Rt[(b*u - a*v)/a, 2]), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; Piecewis
eLinearQ[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 {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^4} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}+\frac {1}{2} b \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx\\ &=-\frac {b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}+\frac {1}{8} b^2 \int \frac {1}{x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {b^2}{8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}-\frac {1}{16} b^3 \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=-\frac {b^2}{8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}-\frac {b^3 \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {b^2}{8 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 117, normalized size = 0.80 \[ \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}+\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-\frac {b^2}{8 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}-\frac {\tanh ^{-1}(\tanh (a+b x))-b x}{3 x^3}-\frac {7 b}{12 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^4,x]

[Out]

(b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(8*(-(b*x) + ArcTanh[Tanh[a
+ b*x]])^(3/2)) + Sqrt[ArcTanh[Tanh[a + b*x]]]*((-7*b)/(12*x^2) - b^2/(8*x*(-(b*x) + ArcTanh[Tanh[a + b*x]]))
- (-(b*x) + ArcTanh[Tanh[a + b*x]])/(3*x^3))

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fricas [A]  time = 0.46, size = 145, normalized size = 0.99 \[ \left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{2} x^{3}}, -\frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{2} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/48*(3*sqrt(a)*b^3*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(3*a*b^2*x^2 + 14*a^2*b*x + 8*a^3)*s
qrt(b*x + a))/(a^2*x^3), -1/24*(3*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*a*b^2*x^2 + 14*a^2*b*
x + 8*a^3)*sqrt(b*x + a))/(a^2*x^3)]

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giac [A]  time = 0.21, size = 93, normalized size = 0.64 \[ -\frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x + a} a^{2} b^{4}\right )}}{a b^{3} x^{3}}\right )}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^4,x, algorithm="giac")

[Out]

-1/48*sqrt(2)*(3*sqrt(2)*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + sqrt(2)*(3*(b*x + a)^(5/2)*b^4 + 8*
(b*x + a)^(3/2)*a*b^4 - 3*sqrt(b*x + a)*a^2*b^4)/(a*b^3*x^3))/b

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maple [A]  time = 0.15, size = 116, normalized size = 0.79 \[ 2 b^{3} \left (\frac {-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{6}+\left (\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{16}-\frac {b x}{16}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{3} x^{3}}+\frac {\arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {3}{2}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(3/2)/x^4,x)

[Out]

2*b^3*((-1/16/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(5/2)-1/6*arctanh(tanh(b*x+a))^(3/2)+(1/16*arcta
nh(tanh(b*x+a))-1/16*b*x)*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3+1/16/(arctanh(tanh(b*x+a))-b*x)^(3/2)*arctanh(ar
ctanh(tanh(b*x+a))^(1/2)/(arctanh(tanh(b*x+a))-b*x)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(3/2)/x^4, x)

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mupad [B]  time = 5.42, size = 1019, normalized size = 6.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(3/2)/x^4,x)

[Out]

(11*b^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))
/(12*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)) +
((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/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)^2)/(2*x^3*(3*log(
2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x)) - (2*b^2*(lo
g((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(x*(3*log(
2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x)) + (2^(1/2)*b
^3*log(((2*2^(1/2)*a + (log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x)
+ 1))/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)^(1/2)*2i - 2^(1/2)*(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) + 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)^3 - 8*a^3 - 6*a*(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 + 12*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))*16i)/(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)^(1/2)))*1i)/(8*(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)^(3/2)) - (3*b*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex
p(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((7*log(2/(exp(2*a)*exp(2*b*x) + 1)))/18 - (7*log
((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/18 + (7*b*x)/9))/(x^2*(2*log(2/(exp(2*a)*exp(2*b*x) + 1))
- 2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 4*b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(3/2)/x**4,x)

[Out]

Integral(atanh(tanh(a + b*x))**(3/2)/x**4, x)

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