3.138 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^5} \, dx\)

Optimal. Leaf size=167 \[ \frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^4 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3} \]

[Out]

5/64*b^4*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)-
5/24*b*arctanh(tanh(b*x+a))^(3/2)/x^3-1/4*arctanh(tanh(b*x+a))^(5/2)/x^4-5/64*b^3/x/arctanh(tanh(b*x+a))^(1/2)
+5/64*b^4/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(1/2)-5/32*b^2*arctanh(tanh(b*x+a))^(1/2)/x^2

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Rubi [A]  time = 0.12, antiderivative size = 167, 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} \[ -\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}+\frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^4 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^5,x]

[Out]

(5*b^4*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(64*(b*x - ArcTanh[Tanh[a + b*
x]])^(3/2)) - (5*b^3)/(64*x*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*b^4)/(64*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[Ar
cTanh[Tanh[a + b*x]]]) - (5*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(32*x^2) - (5*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(2
4*x^3) - ArcTanh[Tanh[a + b*x]]^(5/2)/(4*x^4)

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))^{5/2}}{x^5} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac {1}{8} (5 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^4} \, dx\\ &=-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac {1}{16} \left (5 b^2\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx\\ &=-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac {1}{64} \left (5 b^3\right ) \int \frac {1}{x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac {1}{128} \left (5 b^4\right ) \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac {\left (5 b^4\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{128 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {5 b^4 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {5 b^3}{64 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 134, normalized size = 0.80 \[ \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{64 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (10 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+8 b x \tanh ^{-1}(\tanh (a+b x))^2-48 \tanh ^{-1}(\tanh (a+b x))^3+15 b^3 x^3\right )}{192 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^5,x]

[Out]

(5*b^4*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(64*(-(b*x) + ArcTanh[Tanh
[a + b*x]])^(3/2)) - (Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^3*x^3 + 10*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 8*b*x*Arc
Tanh[Tanh[a + b*x]]^2 - 48*ArcTanh[Tanh[a + b*x]]^3))/(192*x^4*(b*x - ArcTanh[Tanh[a + b*x]]))

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fricas [A]  time = 0.62, size = 167, normalized size = 1.00 \[ \left [\frac {15 \, \sqrt {a} b^{4} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, -\frac {15 \, \sqrt {-a} b^{4} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^5,x, algorithm="fricas")

[Out]

[1/384*(15*sqrt(a)*b^4*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^3*x^3 + 118*a^2*b^2*x^2 +
136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4), -1/192*(15*sqrt(-a)*b^4*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) +
(15*a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4)]

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giac [A]  time = 0.39, size = 108, normalized size = 0.65 \[ -\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {2} {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x + a} a^{3} b^{5}\right )}}{a b^{4} x^{4}}\right )}}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^5,x, algorithm="giac")

[Out]

-1/384*sqrt(2)*(15*sqrt(2)*b^5*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + sqrt(2)*(15*(b*x + a)^(7/2)*b^5 +
73*(b*x + a)^(5/2)*a*b^5 - 55*(b*x + a)^(3/2)*a^2*b^5 + 15*sqrt(b*x + a)*a^3*b^5)/(a*b^4*x^4))/b

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maple [A]  time = 0.15, size = 169, normalized size = 1.01 \[ 2 b^{4} \left (\frac {-\frac {5 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{128 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {73 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{384}+\left (\frac {55 \arctanh \left (\tanh \left (b x +a \right )\right )}{384}-\frac {55 b x}{384}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {5 a^{2}}{128}-\frac {5 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{64}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{128}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{4} x^{4}}+\frac {5 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{128 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {3}{2}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^5,x)

[Out]

2*b^4*((-5/128/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(7/2)-73/384*arctanh(tanh(b*x+a))^(5/2)+(55/384
*arctanh(tanh(b*x+a))-55/384*b*x)*arctanh(tanh(b*x+a))^(3/2)+(-5/128*a^2-5/64*a*(arctanh(tanh(b*x+a))-b*x-a)-5
/128*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^4/x^4+5/128/(arctanh(tanh(b*x+a))-b*x)^(3/2
)*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 \[ \int \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^5,x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^5, x)

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mupad [B]  time = 5.85, size = 1069, normalized size = 6.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^5,x)

[Out]

(5*b^3*(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))/
(32*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)^3)/(4*x^4*(4*log(2
/(exp(2*a)*exp(2*b*x) + 1)) - 4*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 8*b*x)) + (2^(1/2)*b^
4*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))*1024i)/(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)/(64*(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)) - (59*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)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*e
xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(48*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)) + (17*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)*(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)/(16*x^3*(3*log(2/(exp(2*a)*exp(2*b*x) + 1)) - 3*log((2*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2)/x**5,x)

[Out]

Integral(atanh(tanh(a + b*x))**(5/2)/x**5, x)

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