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

Optimal. Leaf size=46 \[ -\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{x}+\frac {b \tan ^{-1}\left (\frac {x}{\sqrt {c}}\right )}{\sqrt {c}}+\frac {b \tanh ^{-1}\left (\frac {x}{\sqrt {c}}\right )}{\sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, 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, 263, 212, 206, 203} \[ -\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{x}+\frac {b \tan ^{-1}\left (\frac {x}{\sqrt {c}}\right )}{\sqrt {c}}+\frac {b \tanh ^{-1}\left (\frac {x}{\sqrt {c}}\right )}{\sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x^2])/x^2,x]

[Out]

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

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])

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 72, normalized size = 1.57 \[ -\frac {a}{x}-\frac {b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{x}-\frac {b \log \left (\sqrt {c}-x\right )}{2 \sqrt {c}}+\frac {b \log \left (\sqrt {c}+x\right )}{2 \sqrt {c}}+\frac {b \tan ^{-1}\left (\frac {x}{\sqrt {c}}\right )}{\sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 159, normalized size = 3.46 \[ \left [\frac {2 \, b \sqrt {c} x \arctan \left (\frac {x}{\sqrt {c}}\right ) + b \sqrt {c} x \log \left (\frac {x^{2} + 2 \, \sqrt {c} x + c}{x^{2} - c}\right ) - b c \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) - 2 \, a c}{2 \, c x}, -\frac {2 \, b \sqrt {-c} x \arctan \left (\frac {\sqrt {-c} x}{c}\right ) + b \sqrt {-c} x \log \left (\frac {x^{2} - 2 \, \sqrt {-c} x - c}{x^{2} + c}\right ) + b c \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + 2 \, a c}{2 \, c x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 62, normalized size = 1.35 \[ -b c {\left (\frac {\arctan \left (\frac {x}{\sqrt {-c}}\right )}{\sqrt {-c} c} - \frac {\arctan \left (\frac {x}{\sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} - \frac {b \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{2 \, x} - \frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 44, normalized size = 0.96 \[ -\frac {a}{x}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{x}+\frac {b \arctan \left (\frac {x}{\sqrt {c}}\right )}{\sqrt {c}}+\frac {b \arctanh \left (\frac {\sqrt {c}}{x}\right )}{\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 57, normalized size = 1.24 \[ \frac {1}{2} \, {\left (c {\left (\frac {2 \, \arctan \left (\frac {x}{\sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {\log \left (\frac {x - \sqrt {c}}{x + \sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} - \frac {2 \, \operatorname {artanh}\left (\frac {c}{x^{2}}\right )}{x}\right )} b - \frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.96, size = 59, normalized size = 1.28 \[ \frac {b\,\mathrm {atan}\left (\frac {x}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {a}{x}-\frac {b\,\ln \left (x^2+c\right )}{2\,x}+\frac {b\,\ln \left (x^2-c\right )}{2\,x}-\frac {b\,\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c/x^2))/x^2,x)

[Out]

(b*atan(x/c^(1/2)))/c^(1/2) - a/x - (b*atan((x*1i)/c^(1/2))*1i)/c^(1/2) - (b*log(c + x^2))/(2*x) + (b*log(x^2
- c))/(2*x)

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sympy [A]  time = 9.31, size = 620, normalized size = 13.48 \[ \begin {cases} - \frac {a}{x} & \text {for}\: c = 0 \\- \frac {a - \infty b}{x} & \text {for}\: c = - x^{2} \\- \frac {a + \infty b}{x} & \text {for}\: c = x^{2} \\\frac {2 i a c^{4}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {2 i a c^{2} x^{4}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} + \frac {2 i b c^{\frac {7}{2}} x \log {\left (- \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {b c^{\frac {7}{2}} x \log {\left (- i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {i b c^{\frac {7}{2}} x \log {\left (- i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} + \frac {b c^{\frac {7}{2}} x \log {\left (i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {i b c^{\frac {7}{2}} x \log {\left (i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} + \frac {2 i b c^{\frac {7}{2}} x \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {2 i b c^{\frac {3}{2}} x^{5} \log {\left (- \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} + \frac {b c^{\frac {3}{2}} x^{5} \log {\left (- i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} + \frac {i b c^{\frac {3}{2}} x^{5} \log {\left (- i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {b c^{\frac {3}{2}} x^{5} \log {\left (i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} + \frac {i b c^{\frac {3}{2}} x^{5} \log {\left (i \sqrt {c} + x \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {2 i b c^{\frac {3}{2}} x^{5} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} + \frac {2 i b c^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} - \frac {2 i b c^{2} x^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 2 i c^{4} x + 2 i c^{2} x^{5}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a/x, Eq(c, 0)), (-(a - oo*b)/x, Eq(c, -x**2)), (-(a + oo*b)/x, Eq(c, x**2)), (2*I*a*c**4/(-2*I*c**
4*x + 2*I*c**2*x**5) - 2*I*a*c**2*x**4/(-2*I*c**4*x + 2*I*c**2*x**5) + 2*I*b*c**(7/2)*x*log(-sqrt(c) + x)/(-2*
I*c**4*x + 2*I*c**2*x**5) - b*c**(7/2)*x*log(-I*sqrt(c) + x)/(-2*I*c**4*x + 2*I*c**2*x**5) - I*b*c**(7/2)*x*lo
g(-I*sqrt(c) + x)/(-2*I*c**4*x + 2*I*c**2*x**5) + b*c**(7/2)*x*log(I*sqrt(c) + x)/(-2*I*c**4*x + 2*I*c**2*x**5
) - I*b*c**(7/2)*x*log(I*sqrt(c) + x)/(-2*I*c**4*x + 2*I*c**2*x**5) + 2*I*b*c**(7/2)*x*atanh(c/x**2)/(-2*I*c**
4*x + 2*I*c**2*x**5) - 2*I*b*c**(3/2)*x**5*log(-sqrt(c) + x)/(-2*I*c**4*x + 2*I*c**2*x**5) + b*c**(3/2)*x**5*l
og(-I*sqrt(c) + x)/(-2*I*c**4*x + 2*I*c**2*x**5) + I*b*c**(3/2)*x**5*log(-I*sqrt(c) + x)/(-2*I*c**4*x + 2*I*c*
*2*x**5) - b*c**(3/2)*x**5*log(I*sqrt(c) + x)/(-2*I*c**4*x + 2*I*c**2*x**5) + I*b*c**(3/2)*x**5*log(I*sqrt(c)
+ x)/(-2*I*c**4*x + 2*I*c**2*x**5) - 2*I*b*c**(3/2)*x**5*atanh(c/x**2)/(-2*I*c**4*x + 2*I*c**2*x**5) + 2*I*b*c
**4*atanh(c/x**2)/(-2*I*c**4*x + 2*I*c**2*x**5) - 2*I*b*c**2*x**4*atanh(c/x**2)/(-2*I*c**4*x + 2*I*c**2*x**5),
True))

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