3.170 \(\int \frac{a+b \tanh ^{-1}(\frac{c}{x^2})}{x^6} \, dx\)

Optimal. Leaf size=65 \[ -\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5}+\frac{b \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{5 c^{5/2}}+\frac{b \tanh ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{5 c^{5/2}}-\frac{2 b}{15 c x^3} \]

[Out]

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

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Rubi [A]  time = 0.0375865, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6097, 263, 325, 212, 206, 203} \[ -\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5}+\frac{b \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{5 c^{5/2}}+\frac{b \tanh ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{5 c^{5/2}}-\frac{2 b}{15 c x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x^2])/x^6,x]

[Out]

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

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{x^6} \, dx &=-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5}-\frac{1}{5} (2 b c) \int \frac{1}{\left (1-\frac{c^2}{x^4}\right ) x^8} \, dx\\ &=-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5}-\frac{1}{5} (2 b c) \int \frac{1}{x^4 \left (-c^2+x^4\right )} \, dx\\ &=-\frac{2 b}{15 c x^3}-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5}-\frac{(2 b) \int \frac{1}{-c^2+x^4} \, dx}{5 c}\\ &=-\frac{2 b}{15 c x^3}-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5}+\frac{b \int \frac{1}{c-x^2} \, dx}{5 c^2}+\frac{b \int \frac{1}{c+x^2} \, dx}{5 c^2}\\ &=-\frac{2 b}{15 c x^3}+\frac{b \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{5 c^{5/2}}-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5}+\frac{b \tanh ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{5 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.023624, size = 90, normalized size = 1.38 \[ -\frac{a}{5 x^5}-\frac{b \log \left (\sqrt{c}-x\right )}{10 c^{5/2}}+\frac{b \log \left (\sqrt{c}+x\right )}{10 c^{5/2}}+\frac{b \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{5 c^{5/2}}-\frac{2 b}{15 c x^3}-\frac{b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c/x^2])/x^6,x]

[Out]

-a/(5*x^5) - (2*b)/(15*c*x^3) + (b*ArcTan[x/Sqrt[c]])/(5*c^(5/2)) - (b*ArcTanh[c/x^2])/(5*x^5) - (b*Log[Sqrt[c
] - x])/(10*c^(5/2)) + (b*Log[Sqrt[c] + x])/(10*c^(5/2))

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Maple [A]  time = 0.011, size = 55, normalized size = 0.9 \begin{align*} -{\frac{a}{5\,{x}^{5}}}-{\frac{b}{5\,{x}^{5}}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) }-{\frac{2\,b}{15\,c{x}^{3}}}+{\frac{b}{5}\arctan \left ({x{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{5}{2}}}}+{\frac{b}{5}{\it Artanh} \left ({\frac{1}{x}\sqrt{c}} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c/x^2))/x^6,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7849, size = 468, normalized size = 7.2 \begin{align*} \left [\frac{6 \, b \sqrt{c} x^{5} \arctan \left (\frac{x}{\sqrt{c}}\right ) + 3 \, b \sqrt{c} x^{5} \log \left (\frac{x^{2} + 2 \, \sqrt{c} x + c}{x^{2} - c}\right ) - 4 \, b c^{2} x^{2} - 3 \, b c^{3} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) - 6 \, a c^{3}}{30 \, c^{3} x^{5}}, -\frac{6 \, b \sqrt{-c} x^{5} \arctan \left (\frac{\sqrt{-c} x}{c}\right ) + 3 \, b \sqrt{-c} x^{5} \log \left (\frac{x^{2} - 2 \, \sqrt{-c} x - c}{x^{2} + c}\right ) + 4 \, b c^{2} x^{2} + 3 \, b c^{3} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + 6 \, a c^{3}}{30 \, c^{3} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))/x^6,x, algorithm="fricas")

[Out]

[1/30*(6*b*sqrt(c)*x^5*arctan(x/sqrt(c)) + 3*b*sqrt(c)*x^5*log((x^2 + 2*sqrt(c)*x + c)/(x^2 - c)) - 4*b*c^2*x^
2 - 3*b*c^3*log((x^2 + c)/(x^2 - c)) - 6*a*c^3)/(c^3*x^5), -1/30*(6*b*sqrt(-c)*x^5*arctan(sqrt(-c)*x/c) + 3*b*
sqrt(-c)*x^5*log((x^2 - 2*sqrt(-c)*x - c)/(x^2 + c)) + 4*b*c^2*x^2 + 3*b*c^3*log((x^2 + c)/(x^2 - c)) + 6*a*c^
3)/(c^3*x^5)]

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Sympy [A]  time = 52.4155, size = 668, normalized size = 10.28 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c/x**2))/x**6,x)

[Out]

Piecewise((-a/(5*x**5), Eq(c, 0)), (-(a - oo*b)/(5*x**5), Eq(c, -x**2)), (-(a + oo*b)/(5*x**5), Eq(c, x**2)),
(6*a*c**71/(-30*c**71*x**5 + 30*c**69*x**9) - 6*a*c**69*x**4/(-30*c**71*x**5 + 30*c**69*x**9) + 6*b*c**(137/2)
*x**5*log(-sqrt(c) + x)/(-30*c**71*x**5 + 30*c**69*x**9) - 3*b*c**(137/2)*x**5*log(-I*sqrt(c) + x)/(-30*c**71*
x**5 + 30*c**69*x**9) + 3*I*b*c**(137/2)*x**5*log(-I*sqrt(c) + x)/(-30*c**71*x**5 + 30*c**69*x**9) - 3*b*c**(1
37/2)*x**5*log(I*sqrt(c) + x)/(-30*c**71*x**5 + 30*c**69*x**9) - 3*I*b*c**(137/2)*x**5*log(I*sqrt(c) + x)/(-30
*c**71*x**5 + 30*c**69*x**9) + 6*b*c**(137/2)*x**5*atanh(c/x**2)/(-30*c**71*x**5 + 30*c**69*x**9) - 6*b*c**(13
3/2)*x**9*log(-sqrt(c) + x)/(-30*c**71*x**5 + 30*c**69*x**9) + 3*b*c**(133/2)*x**9*log(-I*sqrt(c) + x)/(-30*c*
*71*x**5 + 30*c**69*x**9) - 3*I*b*c**(133/2)*x**9*log(-I*sqrt(c) + x)/(-30*c**71*x**5 + 30*c**69*x**9) + 3*b*c
**(133/2)*x**9*log(I*sqrt(c) + x)/(-30*c**71*x**5 + 30*c**69*x**9) + 3*I*b*c**(133/2)*x**9*log(I*sqrt(c) + x)/
(-30*c**71*x**5 + 30*c**69*x**9) - 6*b*c**(133/2)*x**9*atanh(c/x**2)/(-30*c**71*x**5 + 30*c**69*x**9) + 6*b*c*
*71*atanh(c/x**2)/(-30*c**71*x**5 + 30*c**69*x**9) + 4*b*c**70*x**2/(-30*c**71*x**5 + 30*c**69*x**9) - 6*b*c**
69*x**4*atanh(c/x**2)/(-30*c**71*x**5 + 30*c**69*x**9) - 4*b*c**68*x**6/(-30*c**71*x**5 + 30*c**69*x**9), True
))

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Giac [A]  time = 1.31428, size = 100, normalized size = 1.54 \begin{align*} -\frac{1}{5} \, b{\left (\frac{\arctan \left (\frac{x}{\sqrt{-c}}\right )}{\sqrt{-c} c^{2}} - \frac{\arctan \left (\frac{x}{\sqrt{c}}\right )}{c^{\frac{5}{2}}}\right )} - \frac{b \log \left (\frac{x^{2} + c}{x^{2} - c}\right )}{10 \, x^{5}} - \frac{2 \, b x^{2} + 3 \, a c}{15 \, c x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))/x^6,x, algorithm="giac")

[Out]

-1/5*b*(arctan(x/sqrt(-c))/(sqrt(-c)*c^2) - arctan(x/sqrt(c))/c^(5/2)) - 1/10*b*log((x^2 + c)/(x^2 - c))/x^5 -
1/15*(2*b*x^2 + 3*a*c)/(c*x^5)