3.137 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=113 \[ \frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 x}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2168, 2161} \[ -\frac {5 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{8 x}+\frac {5 b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac {5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^4,x]

[Out]

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

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Mathematica [A]  time = 0.08, size = 107, normalized size = 0.95 \[ \frac {1}{24} \left (-\frac {15 b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac {15 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x}-\frac {8 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^3}-\frac {10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^4,x]

[Out]

((-15*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/x - (10*b*ArcTanh[Tanh[a + b*x]]^(3/2))/x^2 - (8*ArcTanh[Tanh[a + b*x]
]^(5/2))/x^3 - (15*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/Sqrt[-(b*x
) + ArcTanh[Tanh[a + b*x]]])/24

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fricas [A]  time = 0.66, size = 146, normalized size = 1.29 \[ \left [\frac {15 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a x^{3}}, \frac {15 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,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*(33*a*b^2*x^2 + 26*a^2*b*x + 8*a^3)
*sqrt(b*x + a))/(a*x^3), 1/24*(15*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (33*a*b^2*x^2 + 26*a^2*b
*x + 8*a^3)*sqrt(b*x + a))/(a*x^3)]

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giac [A]  time = 0.19, size = 88, normalized size = 0.78 \[ \frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {2} {\left (33 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 15 \, \sqrt {b x + a} a^{2} b^{4}\right )}}{b^{3} x^{3}}\right )}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="giac")

[Out]

1/48*sqrt(2)*(15*sqrt(2)*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - sqrt(2)*(33*(b*x + a)^(5/2)*b^4 - 40*(b
*x + a)^(3/2)*a*b^4 + 15*sqrt(b*x + a)*a^2*b^4)/(b^3*x^3))/b

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maple [A]  time = 0.15, size = 144, normalized size = 1.27 \[ 2 b^{3} \left (\frac {-\frac {11 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16}+\left (\frac {5 \arctanh \left (\tanh \left (b x +a \right )\right )}{6}-\frac {5 b x}{6}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}+\left (-\frac {5 a^{2}}{16}-\frac {5 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{8}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{16}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{3} x^{3}}-\frac {5 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^4,x)

[Out]

2*b^3*((-11/16*arctanh(tanh(b*x+a))^(5/2)+(5/6*arctanh(tanh(b*x+a))-5/6*b*x)*arctanh(tanh(b*x+a))^(3/2)+(-5/16
*a^2-5/8*a*(arctanh(tanh(b*x+a))-b*x-a)-5/16*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^3/x
^3-5/16/(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 \[ \int \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^4, x)

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mupad [B]  time = 5.68, size = 669, normalized size = 5.92 \[ \frac {13\,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 )}^2}{12\,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 {\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^3\,\left (3\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+6\,b\,x\right )}-\frac {11\,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}}}{8\,x}+\frac {\sqrt {2}\,b^3\,\ln \left (\frac {\sqrt {\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}\,\left (\sqrt {2}\,b\,x-\sqrt {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 )+\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}+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}\,2{}\mathrm {i}\right )\,64{}\mathrm {i}}{x}\right )\,5{}\mathrm {i}}{16\,\sqrt {\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}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^4,x)

[Out]

(2^(1/2)*b^3*log(((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)*((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)*64i)/x)*5i)/(16*(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)) - ((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^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)) - (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))/(8*x) + (13*b*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(ex
p(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)/(12*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 \[ \int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2)/x**4,x)

[Out]

Integral(atanh(tanh(a + b*x))**(5/2)/x**4, x)

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