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]
-2/15*b*c/x^3+1/5*b*c^(5/2)*arctan(x*c^(1/2))+1/5*(-a-b*arctanh(c*x^2))/x^5+1/5*b*c^(5/2)*arctanh(x*c^(1/2))
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Rubi [A] time = 0.03, antiderivative size = 63, 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, 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 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 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 (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*}
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Mathematica [A] time = 0.03, 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]
-1/5*a/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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fricas [A] time = 1.02, size = 187, normalized size = 2.97 \[ \left [\frac {6 \, b c^{\frac {5}{2}} x^{5} \arctan \left (\sqrt {c} x\right ) + 3 \, b c^{\frac {5}{2}} x^{5} \log \left (\frac {c x^{2} + 2 \, \sqrt {c} x + 1}{c x^{2} - 1}\right ) - 4 \, b c x^{2} - 3 \, b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) - 6 \, a}{30 \, x^{5}}, -\frac {6 \, b \sqrt {-c} c^{2} x^{5} \arctan \left (\sqrt {-c} x\right ) - 3 \, b \sqrt {-c} c^{2} x^{5} \log \left (\frac {c x^{2} + 2 \, \sqrt {-c} x - 1}{c x^{2} + 1}\right ) + 4 \, b c x^{2} + 3 \, b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a}{30 \, x^{5}}\right ] \]
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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giac [A] time = 0.22, size = 91, normalized size = 1.44 \[ \frac {1}{10} \, b c^{3} {\left (\frac {2 \, \arctan \left (x \sqrt {{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} + \frac {\log \left ({\left | x + \frac {1}{\sqrt {{\left | c \right |}}} \right |}\right )}{\sqrt {{\left | c \right |}}} - \frac {\log \left ({\left | x - \frac {1}{\sqrt {{\left | c \right |}}} \right |}\right )}{\sqrt {{\left | c \right |}}}\right )} - \frac {b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{10 \, x^{5}} - \frac {2 \, b c x^{2} + 3 \, a}{15 \, x^{5}} \]
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
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maple [A] time = 0.03, size = 51, normalized size = 0.81 \[ -\frac {a}{5 x^{5}}-\frac {b \arctanh \left (c \,x^{2}\right )}{5 x^{5}}-\frac {2 b c}{15 x^{3}}+\frac {b \,c^{\frac {5}{2}} \arctan \left (x \sqrt {c}\right )}{5}+\frac {b \,c^{\frac {5}{2}} \arctanh \left (x \sqrt {c}\right )}{5} \]
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*c^(5/2)*arctan(x*c^(1/2))+1/5*b*c^(5/2)*arctanh(x*c^(1/
2))
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maxima [A] time = 0.40, size = 66, normalized size = 1.05 \[ \frac {1}{30} \, {\left ({\left (6 \, c^{\frac {3}{2}} \arctan \left (\sqrt {c} x\right ) - 3 \, c^{\frac {3}{2}} \log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right ) - \frac {4}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \]
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*((6*c^(3/2)*arctan(sqrt(c)*x) - 3*c^(3/2)*log((c*x - sqrt(c))/(c*x + sqrt(c))) - 4/x^3)*c - 6*arctanh(c*x
^2)/x^5)*b - 1/5*a/x^5
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mupad [B] time = 1.03, size = 71, normalized size = 1.13 \[ \frac {b\,c^{5/2}\,\mathrm {atan}\left (\sqrt {c}\,x\right )}{5}-\frac {\frac {2\,b\,c\,x^2}{3}+a}{5\,x^5}-\frac {b\,\ln \left (c\,x^2+1\right )}{10\,x^5}+\frac {b\,\ln \left (1-c\,x^2\right )}{10\,x^5}-\frac {b\,c^{5/2}\,\mathrm {atan}\left (\sqrt {c}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{5} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*atanh(c*x^2))/x^6,x)
[Out]
(b*c^(5/2)*atan(c^(1/2)*x))/5 - (a + (2*b*c*x^2)/3)/(5*x^5) - (b*c^(5/2)*atan(c^(1/2)*x*1i)*1i)/5 - (b*log(c*x
^2 + 1))/(10*x^5) + (b*log(1 - c*x^2))/(10*x^5)
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sympy [A] time = 20.47, size = 1867, normalized size = 29.63 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atanh(c*x**2))/x**6,x)
[Out]
Piecewise((-(a - oo*b)/(5*x**5), Eq(c, -1/x**2)), (-(a + oo*b)/(5*x**5), Eq(c, x**(-2))), (-a/(5*x**5), Eq(c,
0)), (-3*a*c*x**4*sqrt(1/c)/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**5*sqr
t(1/c)/c) - 3*I*a*c*x**4*sqrt(1/c)/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x
**5*sqrt(1/c)/c) + 3*a*sqrt(1/c)/(15*c**2*x**9*sqrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I
*x**5*sqrt(1/c)) + 3*I*a*sqrt(1/c)/(15*c**2*x**9*sqrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15
*I*x**5*sqrt(1/c)) + 3*I*b*c**4*x**9*log(x + I*sqrt(1/c))/(15*c**2*x**9*sqrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) -
15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)) - 3*I*b*c**4*x**9*log(x - sqrt(1/c))/(15*c**2*x**9*sqrt(1/c) + 15*I*
c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)) - 3*I*b*c**4*x**9*atanh(c*x**2)/(15*c**2*x**9*s
qrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)) + 3*b*c**3*x**9*log(x - I*sqrt(
1/c))/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**5*sqrt(1/c)/c) - 3*b*c**3*x
**9*log(x - sqrt(1/c))/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**5*sqrt(1/c
)/c) - 3*b*c**3*x**9*atanh(c*x**2)/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x
**5*sqrt(1/c)/c) - 2*b*c**2*x**6*sqrt(1/c)/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c
- 15*I*x**5*sqrt(1/c)/c) - 2*I*b*c**2*x**6*sqrt(1/c)/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sq
rt(1/c)/c - 15*I*x**5*sqrt(1/c)/c) - 3*I*b*c**2*x**5*log(x + I*sqrt(1/c))/(15*c**2*x**9*sqrt(1/c) + 15*I*c**2*
x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)) + 3*I*b*c**2*x**5*log(x - sqrt(1/c))/(15*c**2*x**9*s
qrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)) + 3*I*b*c**2*x**5*atanh(c*x**2)
/(15*c**2*x**9*sqrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)) - 3*b*c*x**5*lo
g(x - I*sqrt(1/c))/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**5*sqrt(1/c)/c)
+ 3*b*c*x**5*log(x - sqrt(1/c))/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**
5*sqrt(1/c)/c) + 3*b*c*x**5*atanh(c*x**2)/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c -
15*I*x**5*sqrt(1/c)/c) - 3*b*c*x**4*sqrt(1/c)*atanh(c*x**2)/(15*c*x**9*sqrt(1/c) + 15*I*c*x**9*sqrt(1/c) - 15
*x**5*sqrt(1/c)/c - 15*I*x**5*sqrt(1/c)/c) - 3*I*b*c*x**4*sqrt(1/c)*atanh(c*x**2)/(15*c*x**9*sqrt(1/c) + 15*I*
c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**5*sqrt(1/c)/c) + 2*b*x**2*sqrt(1/c)/(15*c*x**9*sqrt(1/c) + 15
*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**5*sqrt(1/c)/c) + 2*I*b*x**2*sqrt(1/c)/(15*c*x**9*sqrt(1/c)
+ 15*I*c*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c)/c - 15*I*x**5*sqrt(1/c)/c) + 3*b*sqrt(1/c)*atanh(c*x**2)/(15*c**2
*x**9*sqrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)) + 3*I*b*sqrt(1/c)*atanh(
c*x**2)/(15*c**2*x**9*sqrt(1/c) + 15*I*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(1/c) - 15*I*x**5*sqrt(1/c)), True))
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