3.169 \(\int \frac{a+b \tanh ^{-1}(\frac{c}{x^2})}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0371315, 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, 298, 203, 206} \[ -\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{3 x^3}-\frac{b \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{3 c^{3/2}}+\frac{b \tanh ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{3 c^{3/2}}-\frac{2 b}{3 c x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x^2])/x^4,x]

[Out]

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

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 298

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0305724, size = 90, normalized size = 1.38 \[ -\frac{a}{3 x^3}-\frac{b \log \left (\sqrt{c}-x\right )}{6 c^{3/2}}+\frac{b \log \left (\sqrt{c}+x\right )}{6 c^{3/2}}-\frac{b \tan ^{-1}\left (\frac{x}{\sqrt{c}}\right )}{3 c^{3/2}}-\frac{b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{3 x^3}-\frac{2 b}{3 c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(3*x^3) - (2*b)/(3*c*x) - (b*ArcTan[x/Sqrt[c]])/(3*c^(3/2)) - (b*ArcTanh[c/x^2])/(3*x^3) - (b*Log[Sqrt[c] -
x])/(6*c^(3/2)) + (b*Log[Sqrt[c] + x])/(6*c^(3/2))

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Maple [A]  time = 0.012, size = 55, normalized size = 0.9 \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b}{3\,{x}^{3}}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) }-{\frac{2\,b}{3\,cx}}-{\frac{b}{3}\arctan \left ({x{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}}+{\frac{b}{3}{\it Artanh} \left ({\frac{1}{x}\sqrt{c}} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a/x^3-1/3*b/x^3*arctanh(c/x^2)-2/3*b/c/x-1/3*b*arctan(x/c^(1/2))/c^(3/2)+1/3*b/c^(3/2)*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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84796, size = 451, normalized size = 6.94 \begin{align*} \left [-\frac{2 \, b \sqrt{c} x^{3} \arctan \left (\frac{x}{\sqrt{c}}\right ) - b \sqrt{c} x^{3} \log \left (\frac{x^{2} + 2 \, \sqrt{c} x + c}{x^{2} - c}\right ) + 4 \, b c x^{2} + b c^{2} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + 2 \, a c^{2}}{6 \, c^{2} x^{3}}, -\frac{2 \, b \sqrt{-c} x^{3} \arctan \left (\frac{\sqrt{-c} x}{c}\right ) + b \sqrt{-c} x^{3} \log \left (\frac{x^{2} + 2 \, \sqrt{-c} x - c}{x^{2} + c}\right ) + 4 \, b c x^{2} + b c^{2} \log \left (\frac{x^{2} + c}{x^{2} - c}\right ) + 2 \, a c^{2}}{6 \, c^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 25.6988, size = 706, normalized size = 10.86 \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**4,x)

[Out]

Piecewise((-a/(3*x**3), Eq(c, 0)), (-(a - oo*b)/(3*x**3), Eq(c, -x**2)), (-(a + oo*b)/(3*x**3), Eq(c, x**2)),
(2*a*c**(65/2)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) - 2*a*c**(61/2)*x**4/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x*
*7) + 2*b*c**(65/2)*atanh(c/x**2)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) + 4*b*c**(63/2)*x**2/(-6*c**(65/2)*x*
*3 + 6*c**(61/2)*x**7) - 2*b*c**(61/2)*x**4*atanh(c/x**2)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) - 4*b*c**(59/
2)*x**6/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) + 2*b*c**31*x**3*log(-sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**(6
1/2)*x**7) - b*c**31*x**3*log(-I*sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) - I*b*c**31*x**3*log(-I*s
qrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) - b*c**31*x**3*log(I*sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c*
*(61/2)*x**7) + I*b*c**31*x**3*log(I*sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) + 2*b*c**31*x**3*atan
h(c/x**2)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) - 2*b*c**29*x**7*log(-sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**
(61/2)*x**7) + b*c**29*x**7*log(-I*sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) + I*b*c**29*x**7*log(-I
*sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) + b*c**29*x**7*log(I*sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*
c**(61/2)*x**7) - I*b*c**29*x**7*log(I*sqrt(c) + x)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7) - 2*b*c**29*x**7*at
anh(c/x**2)/(-6*c**(65/2)*x**3 + 6*c**(61/2)*x**7), True))

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Giac [A]  time = 1.30144, size = 97, normalized size = 1.49 \begin{align*} -\frac{1}{3} \, b{\left (\frac{\arctan \left (\frac{x}{\sqrt{-c}}\right )}{\sqrt{-c} c} + \frac{\arctan \left (\frac{x}{\sqrt{c}}\right )}{c^{\frac{3}{2}}}\right )} - \frac{b \log \left (\frac{x^{2} + c}{x^{2} - c}\right )}{6 \, x^{3}} - \frac{2 \, b x^{2} + a c}{3 \, c x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b*(arctan(x/sqrt(-c))/(sqrt(-c)*c) + arctan(x/sqrt(c))/c^(3/2)) - 1/6*b*log((x^2 + c)/(x^2 - c))/x^3 - 1/
3*(2*b*x^2 + a*c)/(c*x^3)