3.136 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=110 \[ \frac {15}{4} b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {15}{4} b^2 \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))} \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2159, 2161} \[ \frac {15}{4} b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}-\frac {15}{4} b^2 \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))} \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^3,x]

[Out]

(-15*b^2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Tanh[a + b
*x]]])/4 + (15*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/4 - (5*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(4*x) - ArcTanh[Tanh[a
+ b*x]]^(5/2)/(2*x^2)

Rule 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 108, normalized size = 0.98 \[ -\frac {-15 b^2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}+15 b^2 x^2 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}+2 \tanh ^{-1}(\tanh (a+b x))^{5/2}+5 b x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^3,x]

[Out]

-1/4*(-15*b^2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]] + 5*b*x*ArcTanh[Tanh[a + b*x]]^(3/2) + 2*ArcTanh[Tanh[a + b*x]]
^(5/2) + 15*b^2*x^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*Sqrt[-(b*x) +
ArcTanh[Tanh[a + b*x]]])/x^2

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 133, normalized size = 1.21 \[ \left [\frac {15 \, \sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, x^{2}}, \frac {15 \, \sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^3,x, algorithm="fricas")

[Out]

[1/8*(15*sqrt(a)*b^2*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b
*x + a))/x^2, 1/4*(15*sqrt(-a)*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (8*b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b
*x + a))/x^2]

________________________________________________________________________________________

giac [A]  time = 0.19, size = 92, normalized size = 0.84 \[ \frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} a b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 8 \, \sqrt {2} \sqrt {b x + a} b^{3} - \frac {\sqrt {2} {\left (9 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} - 7 \, \sqrt {b x + a} a^{2} b^{3}\right )}}{b^{2} x^{2}}\right )}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^3,x, algorithm="giac")

[Out]

1/8*sqrt(2)*(15*sqrt(2)*a*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 8*sqrt(2)*sqrt(b*x + a)*b^3 - sqrt(2)*
(9*(b*x + a)^(3/2)*a*b^3 - 7*sqrt(b*x + a)*a^2*b^3)/(b^2*x^2))/b

________________________________________________________________________________________

maple [A]  time = 0.16, size = 142, normalized size = 1.29 \[ 2 b^{2} \left (\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\left (-\frac {9 \arctanh \left (\tanh \left (b x +a \right )\right )}{8}+\frac {9 b x}{8}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (\frac {7 a^{2}}{8}+\frac {7 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{4}+\frac {7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{8}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{2} x^{2}}-\frac {15 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\, \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^3,x)

[Out]

2*b^2*(arctanh(tanh(b*x+a))^(1/2)+((-9/8*arctanh(tanh(b*x+a))+9/8*b*x)*arctanh(tanh(b*x+a))^(3/2)+(7/8*a^2+7/4
*a*(arctanh(tanh(b*x+a))-b*x-a)+7/8*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2-15/8*(
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)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^3,x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^3, x)

________________________________________________________________________________________

mupad [B]  time = 2.07, size = 614, normalized size = 5.58 \[ 2\,b^2\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}+b^2\,\ln \left (\frac {64\,\left (2\,\sqrt {2}\,a-2\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\sqrt {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,b\,x}-\sqrt {2}\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )+\sqrt {2}\,b\,x\right )}{x\,\sqrt {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,b\,x}}\right )\,\sqrt {\frac {225\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{128}-\frac {225\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{128}-\frac {225\,b\,x}{64}}-\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,x^2\,\left (2\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+4\,b\,x\right )}+\frac {9\,b\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{8\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^3,x)

[Out]

2*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) +
b^2*log((64*(2*2^(1/2)*a - 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))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)
) - 2*b*x)^(1/2) - 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))/(x*(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)^(1/2)))*((225*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/128 - (2
25*log(2/(exp(2*a)*exp(2*b*x) + 1)))/128 - (225*b*x)/64)^(1/2) - ((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)^3)/(4*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)) + (9*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))/(8*x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2)/x**3,x)

[Out]

Integral(atanh(tanh(a + b*x))**(5/2)/x**3, x)

________________________________________________________________________________________