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

Optimal. Leaf size=63 \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{5 x^5}+\frac{1}{5} b c^{5/2} \tan ^{-1}\left (\sqrt{c} x\right )+\frac{1}{5} b c^{5/2} \tanh ^{-1}\left (\sqrt{c} x\right )-\frac{2 b c}{15 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0319471, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6097, 325, 212, 206, 203} \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{5 x^5}+\frac{1}{5} b c^{5/2} \tan ^{-1}\left (\sqrt{c} x\right )+\frac{1}{5} b c^{5/2} \tanh ^{-1}\left (\sqrt{c} x\right )-\frac{2 b c}{15 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])/x^6,x]

[Out]

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

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

Mathematica [A]  time = 0.0272086, size = 91, normalized size = 1.44 \[ -\frac{a}{5 x^5}-\frac{1}{10} b c^{5/2} \log \left (1-\sqrt{c} x\right )+\frac{1}{10} b c^{5/2} \log \left (\sqrt{c} x+1\right )+\frac{1}{5} b c^{5/2} \tan ^{-1}\left (\sqrt{c} x\right )-\frac{2 b c}{15 x^3}-\frac{b \tanh ^{-1}\left (c x^2\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(5*x^5) - (2*b*c)/(15*x^3) + (b*c^(5/2)*ArcTan[Sqrt[c]*x])/5 - (b*ArcTanh[c*x^2])/(5*x^5) - (b*c^(5/2)*Log[
1 - Sqrt[c]*x])/10 + (b*c^(5/2)*Log[1 + Sqrt[c]*x])/10

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Maple [A]  time = 0.013, size = 51, normalized size = 0.8 \begin{align*} -{\frac{a}{5\,{x}^{5}}}-{\frac{b{\it Artanh} \left ( c{x}^{2} \right ) }{5\,{x}^{5}}}+{\frac{b}{5}{c}^{{\frac{5}{2}}}\arctan \left ( x\sqrt{c} \right ) }+{\frac{b}{5}{c}^{{\frac{5}{2}}}{\it Artanh} \left ( x\sqrt{c} \right ) }-{\frac{2\,bc}{15\,{x}^{3}}} \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*b/x^5*arctanh(c*x^2)+1/5*b*c^(5/2)*arctan(x*c^(1/2))+1/5*b*c^(5/2)*arctanh(x*c^(1/2))-2/15*b*c/
x^3

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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 = 2.14109, size = 458, normalized size = 7.27 \begin{align*} \left [\frac{6 \, b c^{\frac{5}{2}} x^{5} \arctan \left (\sqrt{c} x\right ) + 3 \, b c^{\frac{5}{2}} x^{5} \log \left (\frac{c x^{2} + 2 \, \sqrt{c} x + 1}{c x^{2} - 1}\right ) - 4 \, b c x^{2} - 3 \, b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) - 6 \, a}{30 \, x^{5}}, -\frac{6 \, b \sqrt{-c} c^{2} x^{5} \arctan \left (\sqrt{-c} x\right ) - 3 \, b \sqrt{-c} c^{2} x^{5} \log \left (\frac{c x^{2} + 2 \, \sqrt{-c} x - 1}{c x^{2} + 1}\right ) + 4 \, b c x^{2} + 3 \, b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a}{30 \, 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*c^(5/2)*x^5*arctan(sqrt(c)*x) + 3*b*c^(5/2)*x^5*log((c*x^2 + 2*sqrt(c)*x + 1)/(c*x^2 - 1)) - 4*b*c*
x^2 - 3*b*log(-(c*x^2 + 1)/(c*x^2 - 1)) - 6*a)/x^5, -1/30*(6*b*sqrt(-c)*c^2*x^5*arctan(sqrt(-c)*x) - 3*b*sqrt(
-c)*c^2*x^5*log((c*x^2 + 2*sqrt(-c)*x - 1)/(c*x^2 + 1)) + 4*b*c*x^2 + 3*b*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 6*a)
/x^5]

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Sympy [A]  time = 30.8841, size = 833, normalized size = 13.22 \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, -1/x**2)), (-(a + oo*b)/(5*x**5), Eq(c, x**(-2
))), (-12*a*x**4/(60*x**9 - 60*x**5/c**2) + 12*a/(60*c**2*x**9 - 60*x**5) - 3*b*c**4*x**9*(1/c)**(3/2)*log(x +
I*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) - 3*I*b*c**4*x**9*(1/c)**(3/2)*log(x + I*sqrt(1/c))/(60*x**9 - 60*x**5/
c**2) + 6*b*c**3*x**9*sqrt(1/c)*log(x - I*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) - 6*I*b*c**3*x**9*sqrt(1/c)*log(
x - I*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) + 9*b*c**3*x**9*sqrt(1/c)*log(x + I*sqrt(1/c))/(60*x**9 - 60*x**5/c*
*2) + 9*I*b*c**3*x**9*sqrt(1/c)*log(x + I*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) - 12*b*c**3*x**9*sqrt(1/c)*log(x
- sqrt(1/c))/(60*x**9 - 60*x**5/c**2) - 12*b*c**3*x**9*sqrt(1/c)*atanh(c*x**2)/(60*x**9 - 60*x**5/c**2) + 3*b
*c**2*x**5*(1/c)**(3/2)*log(x + I*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) + 3*I*b*c**2*x**5*(1/c)**(3/2)*log(x + I
*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) - 8*b*c*x**6/(60*x**9 - 60*x**5/c**2) - 6*b*c*x**5*sqrt(1/c)*log(x - I*sq
rt(1/c))/(60*x**9 - 60*x**5/c**2) + 6*I*b*c*x**5*sqrt(1/c)*log(x - I*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) - 9*b
*c*x**5*sqrt(1/c)*log(x + I*sqrt(1/c))/(60*x**9 - 60*x**5/c**2) - 9*I*b*c*x**5*sqrt(1/c)*log(x + I*sqrt(1/c))/
(60*x**9 - 60*x**5/c**2) + 12*b*c*x**5*sqrt(1/c)*log(x - sqrt(1/c))/(60*x**9 - 60*x**5/c**2) + 12*b*c*x**5*sqr
t(1/c)*atanh(c*x**2)/(60*x**9 - 60*x**5/c**2) - 12*b*x**4*atanh(c*x**2)/(60*x**9 - 60*x**5/c**2) + 8*b*x**2/(6
0*c*x**9 - 60*x**5/c) + 12*b*atanh(c*x**2)/(60*c**2*x**9 - 60*x**5), True))

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Giac [A]  time = 1.53314, size = 123, normalized size = 1.95 \begin{align*} \frac{1}{10} \, b c^{3}{\left (\frac{2 \, \arctan \left (x \sqrt{{\left | c \right |}}\right )}{\sqrt{{\left | c \right |}}} + \frac{\log \left ({\left | x + \frac{1}{\sqrt{{\left | c \right |}}} \right |}\right )}{\sqrt{{\left | c \right |}}} - \frac{\log \left ({\left | x - \frac{1}{\sqrt{{\left | c \right |}}} \right |}\right )}{\sqrt{{\left | c \right |}}}\right )} - \frac{b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{10 \, x^{5}} - \frac{2 \, b c x^{2} + 3 \, a}{15 \, 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]

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