3.119 \(\int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^4} \, dx\)

Optimal. Leaf size=179 \[ \frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[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)))^(5/2)+1
/24*b^2/x/arctanh(tanh(b*x+a))^(3/2)-1/24*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(3/2)-1/12*b/x^2
/arctanh(tanh(b*x+a))^(1/2)+1/8*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(1/2)-1/3*arctanh(tanh(b
*x+a))^(1/2)/x^3

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 179, 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, 2161} \[ \frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/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]]
)^(5/2)) + b^2/(24*x*ArcTanh[Tanh[a + b*x]]^(3/2)) - b^3/(24*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b
*x]]^(3/2)) - b/(12*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + b^3/(8*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[T
anh[a + b*x]]]) - Sqrt[ArcTanh[Tanh[a + b*x]]]/(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 {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^4} \, dx &=-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}+\frac {1}{6} b \int \frac {1}{x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac {1}{24} b^2 \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}+\frac {1}{16} b^3 \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac {b^3 \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{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 ) \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 )^{5/2}}+\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 115, normalized size = 0.64 \[ \frac {1}{24} \left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (14 b x \tanh ^{-1}(\tanh (a+b x))-8 \tanh ^{-1}(\tanh (a+b x))^2-3 b^2 x^2\right )}{x^3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac {3 b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^4,x]

[Out]

((-3*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a
+ b*x]])^(5/2) + (Sqrt[ArcTanh[Tanh[a + b*x]]]*(-3*b^2*x^2 + 14*b*x*ArcTanh[Tanh[a + b*x]] - 8*ArcTanh[Tanh[a
+ b*x]]^2))/(x^3*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2))/24

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 145, normalized size = 0.81 \[ \left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{3} x^{3}}, \frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{3} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/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 - 2*a^2*b*x - 8*a^3)*sq
rt(b*x + a))/(a^3*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 - 2*a^2*b*x -
8*a^3)*sqrt(b*x + a))/(a^3*x^3)]

________________________________________________________________________________________

giac [A]  time = 0.92, size = 93, normalized size = 0.52 \[ \frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {\sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x + a} a^{2} b^{4}\right )}}{a^{2} b^{3} x^{3}}\right )}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maple [A]  time = 0.18, size = 185, normalized size = 1.03 \[ 2 b^{3} \left (\frac {\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16 a^{2}+32 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+16 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{16}}{b^{3} x^{3}}-\frac {\arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(1/2)/x^4,x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.51, size = 964, normalized size = 5.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(1/2)/x^4,x)

[Out]

(b*(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))/(3*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)) +
(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))/(2
*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)^2) + (2
^(1/2)*b^3*log((((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)*(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^(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)^5 + 40*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)^3 - 80*a^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(e
xp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 - 32*a^5 - 10*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)^4 + 80*a^4*(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))*4i)/(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)/(4*(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)) - ((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))/(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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(1/2)/x**4,x)

[Out]

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

________________________________________________________________________________________