3.2.70 \(\int \frac {a+b \tanh ^{-1}(\frac {c}{x^2})}{x^6} \, dx\) [170]

Optimal. Leaf size=65 \[ -\frac {2 b}{15 c x^3}+\frac {b \text {ArcTan}\left (\frac {x}{\sqrt {c}}\right )}{5 c^{5/2}}-\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{5 x^5}+\frac {b \tanh ^{-1}\left (\frac {x}{\sqrt {c}}\right )}{5 c^{5/2}} \]

[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 = 0.03, 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 = {6037, 269, 331, 218, 212, 209} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{5 x^5}+\frac {b \text {ArcTan}\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} \end {gather*}

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 269

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 331

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 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && 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 \begin {gather*} -\frac {a}{5 x^5}-\frac {2 b}{15 c x^3}+\frac {b \text {ArcTan}\left (\frac {x}{\sqrt {c}}\right )}{5 c^{5/2}}-\frac {b \tanh ^{-1}\left (\frac {c}{x^2}\right )}{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}} \end {gather*}

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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Maple [A]
time = 0.13, size = 57, normalized size = 0.88

method	result	size
derivativedivides	\(-\frac {a}{5 x^{5}}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{5 x^{5}}-\frac {2 b}{15 c \,x^{3}}-\frac {b \arctan \left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}+\frac {b \arctanh \left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}\)	\(57\)
default	\(-\frac {a}{5 x^{5}}-\frac {b \arctanh \left (\frac {c}{x^{2}}\right )}{5 x^{5}}-\frac {2 b}{15 c \,x^{3}}-\frac {b \arctan \left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}+\frac {b \arctanh \left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}\)	\(57\)
risch	\(-\frac {b \ln \left (x^{2}+c \right )}{10 x^{5}}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}}{20 x^{5}}+\frac {i b \pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}{20 x^{5}}+\frac {i b \pi \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{3}}{20 x^{5}}+\frac {i b \pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}}{20 x^{5}}-\frac {i b \pi \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{2}}{10 x^{5}}-\frac {i b \pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (-x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}{20 x^{5}}+\frac {i b \pi \mathrm {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )^{3}}{20 x^{5}}+\frac {i b \pi }{10 x^{5}}-\frac {i b \pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}}{20 x^{5}}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (x^{2}+c \right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )^{2}}{20 x^{5}}-\frac {a}{5 x^{5}}+\frac {b \ln \left (-x^{2}+c \right )}{10 x^{5}}+\frac {b \arctanh \left (\frac {x}{\sqrt {c}}\right )}{5 c^{\frac {5}{2}}}-\frac {2 b}{15 c \,x^{3}}+\frac {b \arctan \left (\frac {x}{\sqrt {c}}\right )}{5 c^{\frac {5}{2}}}\)	\(347\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c/x^2))/x^6,x,method=_RETURNVERBOSE)

[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(1/x*c^(1/2))+1/5*b/c^(5/2)*arctanh(1/x*c
^(1/2))

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Maxima [A]
time = 0.47, size = 65, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

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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Fricas [A] Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (49) = 98\).
time = 0.38, size = 196, normalized size = 3.02 \begin {gather*} \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 ] \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 994 vs. \(2 (61) = 122\).
time = 7.13, size = 994, normalized size = 15.29 \begin {gather*} \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 a c^{13} \sqrt {- c}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {6 a c^{11} x^{4} \sqrt {- c}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {6 b c^{\frac {21}{2}} x^{5} \sqrt {- c} \log {\left (- \sqrt {c} + x \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {3 b c^{\frac {21}{2}} x^{5} \sqrt {- c} \log {\left (x - \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {3 b c^{\frac {21}{2}} x^{5} \sqrt {- c} \log {\left (x + \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {6 b c^{\frac {21}{2}} x^{5} \sqrt {- c} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {6 b c^{\frac {17}{2}} x^{9} \sqrt {- c} \log {\left (- \sqrt {c} + x \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {3 b c^{\frac {17}{2}} x^{9} \sqrt {- c} \log {\left (x - \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {3 b c^{\frac {17}{2}} x^{9} \sqrt {- c} \log {\left (x + \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {6 b c^{\frac {17}{2}} x^{9} \sqrt {- c} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {6 b c^{13} \sqrt {- c} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {4 b c^{12} x^{2} \sqrt {- c}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {3 b c^{11} x^{5} \log {\left (x - \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {3 b c^{11} x^{5} \log {\left (x + \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {6 b c^{11} x^{4} \sqrt {- c} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {4 b c^{10} x^{6} \sqrt {- c}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} + \frac {3 b c^{9} x^{9} \log {\left (x - \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} - \frac {3 b c^{9} x^{9} \log {\left (x + \sqrt {- c} \right )}}{- 30 c^{13} x^{5} \sqrt {- c} + 30 c^{11} x^{9} \sqrt {- c}} & \text {otherwise} \end {cases} \end {gather*}

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*a*c**13*sqrt(-c)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*a*c**11*x**4*sqrt(-c)/(-30*c**13*x*
*5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) + 6*b*c**(21/2)*x**5*sqrt(-c)*log(-sqrt(c) + x)/(-30*c**13*x**5*sqrt(-c)
+ 30*c**11*x**9*sqrt(-c)) - 3*b*c**(21/2)*x**5*sqrt(-c)*log(x - sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11
*x**9*sqrt(-c)) - 3*b*c**(21/2)*x**5*sqrt(-c)*log(x + sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(
-c)) + 6*b*c**(21/2)*x**5*sqrt(-c)*atanh(c/x**2)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*b*c**(
17/2)*x**9*sqrt(-c)*log(-sqrt(c) + x)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) + 3*b*c**(17/2)*x**9*
sqrt(-c)*log(x - sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) + 3*b*c**(17/2)*x**9*sqrt(-c)*lo
g(x + sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*b*c**(17/2)*x**9*sqrt(-c)*atanh(c/x**2)
/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) + 6*b*c**13*sqrt(-c)*atanh(c/x**2)/(-30*c**13*x**5*sqrt(-c
) + 30*c**11*x**9*sqrt(-c)) + 4*b*c**12*x**2*sqrt(-c)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 3*b
*c**11*x**5*log(x - sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) + 3*b*c**11*x**5*log(x + sqrt
(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*b*c**11*x**4*sqrt(-c)*atanh(c/x**2)/(-30*c**13*x*
*5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 4*b*c**10*x**6*sqrt(-c)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(
-c)) + 3*b*c**9*x**9*log(x - sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 3*b*c**9*x**9*log(
x + sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)), True))

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Giac [A]
time = 0.49, size = 74, normalized size = 1.14 \begin {gather*} -\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}} \end {gather*}

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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Mupad [B]
time = 1.00, size = 69, normalized size = 1.06 \begin {gather*} \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}} \end {gather*}

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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