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]

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

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 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 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 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^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*}

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Mathematica [A]  time = 0.02, 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]

-1/5*a/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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fricas [A]  time = 0.53, size = 196, normalized size = 3.02 \[ \left [\frac {6 \, b \sqrt {c} x^{5} \arctan \left (\frac {x}{\sqrt {c}}\right ) + 3 \, b \sqrt {c} x^{5} \log \left (\frac {x^{2} + 2 \, \sqrt {c} x + c}{x^{2} - c}\right ) - 4 \, b c^{2} x^{2} - 3 \, b c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) - 6 \, a c^{3}}{30 \, c^{3} x^{5}}, -\frac {6 \, b \sqrt {-c} x^{5} \arctan \left (\frac {\sqrt {-c} x}{c}\right ) + 3 \, b \sqrt {-c} x^{5} \log \left (\frac {x^{2} - 2 \, \sqrt {-c} x - c}{x^{2} + c}\right ) + 4 \, b c^{2} x^{2} + 3 \, b c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + 6 \, a c^{3}}{30 \, c^{3} x^{5}}\right ] \]

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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giac [A]  time = 0.18, size = 74, normalized size = 1.14 \[ -\frac {1}{5} \, b {\left (\frac {\arctan \left (\frac {x}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {\arctan \left (\frac {x}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )} - \frac {b \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{10 \, x^{5}} - \frac {2 \, b x^{2} + 3 \, a c}{15 \, c x^{5}} \]

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)

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maple [A]  time = 0.04, size = 55, normalized size = 0.85 \[ -\frac {a}{5 x^{5}}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{5 x^{5}}-\frac {2 b}{15 c \,x^{3}}+\frac {b \arctan \left (\frac {x}{\sqrt {c}}\right )}{5 c^{\frac {5}{2}}}+\frac {b \arctanh \left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}} \]

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)-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 [A]  time = 0.42, size = 65, normalized size = 1.00 \[ \frac {1}{30} \, {\left (c {\left (\frac {6 \, \arctan \left (\frac {x}{\sqrt {c}}\right )}{c^{\frac {7}{2}}} - \frac {3 \, \log \left (\frac {x - \sqrt {c}}{x + \sqrt {c}}\right )}{c^{\frac {7}{2}}} - \frac {4}{c^{2} x^{3}}\right )} - \frac {6 \, \operatorname {artanh}\left (\frac {c}{x^{2}}\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(c*(6*arctan(x/sqrt(c))/c^(7/2) - 3*log((x - sqrt(c))/(x + sqrt(c)))/c^(7/2) - 4/(c^2*x^3)) - 6*arctanh(c
/x^2)/x^5)*b - 1/5*a/x^5

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mupad [B]  time = 1.00, size = 69, normalized size = 1.06 \[ \frac {b\,\mathrm {atan}\left (\frac {x}{\sqrt {c}}\right )}{5\,c^{5/2}}-\frac {2\,b}{15\,c\,x^3}-\frac {a}{5\,x^5}-\frac {b\,\ln \left (x^2+c\right )}{10\,x^5}+\frac {b\,\ln \left (x^2-c\right )}{10\,x^5}-\frac {b\,\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{5\,c^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c/x^2))/x^6,x)

[Out]

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

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sympy [A]  time = 20.04, size = 797, normalized size = 12.26 \[ \begin {cases} - \frac {a}{5 x^{5}} & \text {for}\: c = 0 \\- \frac {a - \infty b}{5 x^{5}} & \text {for}\: c = - x^{2} \\- \frac {a + \infty b}{5 x^{5}} & \text {for}\: c = x^{2} \\\frac {6 i a c^{\frac {27}{2}}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {6 i a c^{\frac {23}{2}} x^{4}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {6 i b c^{\frac {27}{2}} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {4 i b c^{\frac {25}{2}} x^{2}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {6 i b c^{\frac {23}{2}} x^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {4 i b c^{\frac {21}{2}} x^{6}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {6 i b c^{11} x^{5} \log {\left (- \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {3 b c^{11} x^{5} \log {\left (- i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {3 i b c^{11} x^{5} \log {\left (- i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {3 b c^{11} x^{5} \log {\left (i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {3 i b c^{11} x^{5} \log {\left (i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {6 i b c^{11} x^{5} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {6 i b c^{9} x^{9} \log {\left (- \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {3 b c^{9} x^{9} \log {\left (- i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {3 i b c^{9} x^{9} \log {\left (- i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {3 b c^{9} x^{9} \log {\left (i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} + \frac {3 i b c^{9} x^{9} \log {\left (i \sqrt {c} + x \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} - \frac {6 i b c^{9} x^{9} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 i c^{\frac {27}{2}} x^{5} + 30 i c^{\frac {23}{2}} x^{9}} & \text {otherwise} \end {cases} \]

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*I*a*c**(27/2)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) - 6*I*a*c**(23/2)*x**4/(-30*I*c**(27/2)*x**5 + 3
0*I*c**(23/2)*x**9) + 6*I*b*c**(27/2)*atanh(c/x**2)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) + 4*I*b*c**(2
5/2)*x**2/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) - 6*I*b*c**(23/2)*x**4*atanh(c/x**2)/(-30*I*c**(27/2)*x
**5 + 30*I*c**(23/2)*x**9) - 4*I*b*c**(21/2)*x**6/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) + 6*I*b*c**11*x
**5*log(-sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) - 3*b*c**11*x**5*log(-I*sqrt(c) + x)/(-30*I
*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) - 3*I*b*c**11*x**5*log(-I*sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**
(23/2)*x**9) + 3*b*c**11*x**5*log(I*sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) - 3*I*b*c**11*x*
*5*log(I*sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) + 6*I*b*c**11*x**5*atanh(c/x**2)/(-30*I*c**
(27/2)*x**5 + 30*I*c**(23/2)*x**9) - 6*I*b*c**9*x**9*log(-sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*
x**9) + 3*b*c**9*x**9*log(-I*sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) + 3*I*b*c**9*x**9*log(-
I*sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9) - 3*b*c**9*x**9*log(I*sqrt(c) + x)/(-30*I*c**(27/2
)*x**5 + 30*I*c**(23/2)*x**9) + 3*I*b*c**9*x**9*log(I*sqrt(c) + x)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9
) - 6*I*b*c**9*x**9*atanh(c/x**2)/(-30*I*c**(27/2)*x**5 + 30*I*c**(23/2)*x**9), True))

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