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

Optimal. Leaf size=212 \[ \frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

5/8*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)))^(7/2)-1
/8*b^2/x/arctanh(tanh(b*x+a))^(5/2)+1/8*b^3/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(5/2)+1/12*b/x^2/a
rctanh(tanh(b*x+a))^(3/2)-5/24*b^3/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(3/2)-1/3/x^3/arctanh(tan
h(b*x+a))^(1/2)+5/8*b^3/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ \frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(5*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
]])^(7/2)) - b^2/(8*x*ArcTanh[Tanh[a + b*x]]^(5/2)) + b^3/(8*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b
*x]]^(5/2)) + b/(12*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)) - (5*b^3)/(24*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[T
anh[a + b*x]]^(3/2)) - 1/(3*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^3*Sq
rt[ArcTanh[Tanh[a + b*x]]])

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 {1}{x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {1}{6} b \int \frac {1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {1}{8} b^2 \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {1}{16} \left (5 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {\left (5 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (5 b^3\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (5 b^3\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {5 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 )^{7/2}}-\frac {b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac {b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {1}{3 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 117, normalized size = 0.55 \[ \frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}}+\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-26 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+33 b^2 x^2\right )}{24 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(5*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]])^(7/2)) + (Sqrt[ArcTanh[Tanh[a + b*x]]]*(33*b^2*x^2 - 26*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh
[a + b*x]]^2))/(24*x^3*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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fricas [A]  time = 0.56, size = 145, normalized size = 0.68 \[ \left [\frac {15 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{4} x^{3}}, -\frac {15 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{4} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*sqrt(a)*b^3*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*x^2 - 10*a^2*b*x + 8*a^3)
*sqrt(b*x + a))/(a^4*x^3), -1/24*(15*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 - 10*a^
2*b*x + 8*a^3)*sqrt(b*x + a))/(a^4*x^3)]

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giac [A]  time = 0.21, size = 84, normalized size = 0.40 \[ -\frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 33 \, \sqrt {b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(15*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*x + a)^(5/2)*b^4 - 40*(b*x + a)^(3/2)*a*b
^4 + 33*sqrt(b*x + a)*a^2*b^4)/(a^3*b^3*x^3))/b

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maple [A]  time = 0.18, size = 200, normalized size = 0.94 \[ 2 b^{3} \left (\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) b^{3} x^{3}}+\frac {\frac {10 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) b^{2} x^{2}}+\frac {10 \left (\frac {6 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) b x}-\frac {6 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{3 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right )}}{-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*(2/3*arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/b^3/x^3+10/3/(-4*arctanh(tanh(b*x+a))+4*
b*x)*(arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/b^2/x^2+3/(-4*arctanh(tanh(b*x+a))+4*b*x)*(2*
arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/b/x-2/(-4*arctanh(tanh(b*x+a))+4*b*x)/(arctanh(tanh
(b*x+a))-b*x)^(1/2)*arctanh(arctanh(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 \[ \int \frac {1}{x^{4} \sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^4*sqrt(arctanh(tanh(b*x + a)))), x)

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mupad [B]  time = 5.75, size = 1086, normalized size = 5.12 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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))/(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)) + (5
*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))/(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) + (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)^7 + 84*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)*e
xp(2*b*x) + 1)) + 2*b*x)^5 - 280*a^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(ex
p(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4 + 560*a^4*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + l
og(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3 - 672*a^5*(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 - 128*a^7 - 14*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)^6 + 448*a^6*(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))*1i)/(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)))*5i)/(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)^(7/2)) + (10*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*(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*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 \[ \int \frac {1}{x^{4} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**4*sqrt(atanh(tanh(a + b*x)))), x)

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