3.3.36 \(\int \frac {1}{x^5 (d+e x^2) (a+c x^4)} \, dx\) [236]

Optimal. Leaf size=156 \[ -\frac {1}{4 a d x^4}+\frac {e}{2 a d^2 x^2}+\frac {c^{3/2} e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )}-\frac {\left (c d^2-a e^2\right ) \log (x)}{a^2 d^3}-\frac {e^4 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )}+\frac {c^2 d \log \left (a+c x^4\right )}{4 a^2 \left (c d^2+a e^2\right )} \]

[Out]

-1/4/a/d/x^4+1/2*e/a/d^2/x^2+1/2*c^(3/2)*e*arctan(x^2*c^(1/2)/a^(1/2))/a^(3/2)/(a*e^2+c*d^2)-(-a*e^2+c*d^2)*ln
(x)/a^2/d^3-1/2*e^4*ln(e*x^2+d)/d^3/(a*e^2+c*d^2)+1/4*c^2*d*ln(c*x^4+a)/a^2/(a*e^2+c*d^2)

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Rubi [A]
time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1266, 908, 649, 211, 266} \begin {gather*} \frac {c^{3/2} e \text {ArcTan}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )}+\frac {c^2 d \log \left (a+c x^4\right )}{4 a^2 \left (a e^2+c d^2\right )}-\frac {\log (x) \left (c d^2-a e^2\right )}{a^2 d^3}-\frac {e^4 \log \left (d+e x^2\right )}{2 d^3 \left (a e^2+c d^2\right )}+\frac {e}{2 a d^2 x^2}-\frac {1}{4 a d x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(d + e*x^2)*(a + c*x^4)),x]

[Out]

-1/4*1/(a*d*x^4) + e/(2*a*d^2*x^2) + (c^(3/2)*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(3/2)*(c*d^2 + a*e^2)) - (
(c*d^2 - a*e^2)*Log[x])/(a^2*d^3) - (e^4*Log[d + e*x^2])/(2*d^3*(c*d^2 + a*e^2)) + (c^2*d*Log[a + c*x^4])/(4*a
^2*(c*d^2 + a*e^2))

Rule 211

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1266

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
 - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rubi steps

\begin {align*} \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^3 (d+e x) \left (a+c x^2\right )} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{a d x^3}-\frac {e}{a d^2 x^2}+\frac {-c d^2+a e^2}{a^2 d^3 x}-\frac {e^5}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {c^2 (a e+c d x)}{a^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{4 a d x^4}+\frac {e}{2 a d^2 x^2}-\frac {\left (c d^2-a e^2\right ) \log (x)}{a^2 d^3}-\frac {e^4 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )}+\frac {c^2 \text {Subst}\left (\int \frac {a e+c d x}{a+c x^2} \, dx,x,x^2\right )}{2 a^2 \left (c d^2+a e^2\right )}\\ &=-\frac {1}{4 a d x^4}+\frac {e}{2 a d^2 x^2}-\frac {\left (c d^2-a e^2\right ) \log (x)}{a^2 d^3}-\frac {e^4 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )}+\frac {\left (c^3 d\right ) \text {Subst}\left (\int \frac {x}{a+c x^2} \, dx,x,x^2\right )}{2 a^2 \left (c d^2+a e^2\right )}+\frac {\left (c^2 e\right ) \text {Subst}\left (\int \frac {1}{a+c x^2} \, dx,x,x^2\right )}{2 a \left (c d^2+a e^2\right )}\\ &=-\frac {1}{4 a d x^4}+\frac {e}{2 a d^2 x^2}+\frac {c^{3/2} e \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )}-\frac {\left (c d^2-a e^2\right ) \log (x)}{a^2 d^3}-\frac {e^4 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )}+\frac {c^2 d \log \left (a+c x^4\right )}{4 a^2 \left (c d^2+a e^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 209, normalized size = 1.34 \begin {gather*} -\frac {a c d^4+a^2 d^2 e^2-2 a c d^3 e x^2-2 a^2 d e^3 x^2+2 \sqrt {a} c^{3/2} d^3 e x^4 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 \sqrt {a} c^{3/2} d^3 e x^4 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+4 c^2 d^4 x^4 \log (x)-4 a^2 e^4 x^4 \log (x)+2 a^2 e^4 x^4 \log \left (d+e x^2\right )-c^2 d^4 x^4 \log \left (a+c x^4\right )}{4 a^2 d^3 \left (c d^2+a e^2\right ) x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(d + e*x^2)*(a + c*x^4)),x]

[Out]

-1/4*(a*c*d^4 + a^2*d^2*e^2 - 2*a*c*d^3*e*x^2 - 2*a^2*d*e^3*x^2 + 2*Sqrt[a]*c^(3/2)*d^3*e*x^4*ArcTan[1 - (Sqrt
[2]*c^(1/4)*x)/a^(1/4)] + 2*Sqrt[a]*c^(3/2)*d^3*e*x^4*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)] + 4*c^2*d^4*x^4*
Log[x] - 4*a^2*e^4*x^4*Log[x] + 2*a^2*e^4*x^4*Log[d + e*x^2] - c^2*d^4*x^4*Log[a + c*x^4])/(a^2*d^3*(c*d^2 + a
*e^2)*x^4)

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Maple [A]
time = 0.18, size = 127, normalized size = 0.81

method result size
default \(-\frac {e^{4} \ln \left (e \,x^{2}+d \right )}{2 d^{3} \left (a \,e^{2}+c \,d^{2}\right )}+\frac {c^{2} \left (\frac {d \ln \left (c \,x^{4}+a \right )}{2}+\frac {a e \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) a^{2}}-\frac {1}{4 a d \,x^{4}}+\frac {\left (a \,e^{2}-c \,d^{2}\right ) \ln \left (x \right )}{d^{3} a^{2}}+\frac {e}{2 a \,d^{2} x^{2}}\) \(127\)
risch \(\frac {\frac {e \,x^{2}}{2 d^{2} a}-\frac {1}{4 d a}}{x^{4}}+\frac {\ln \left (x \right ) e^{2}}{d^{3} a}-\frac {\ln \left (x \right ) c}{d \,a^{2}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\left (a^{5} e^{2}+d^{2} a^{4} c \right ) \textit {\_Z}^{2}-2 a^{2} c^{2} d \textit {\_Z} +c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-6 a^{6} d^{4} e^{4}-7 a^{5} c \,d^{6} e^{2}-5 a^{4} c^{2} d^{8}\right ) \textit {\_R}^{3}+\left (-18 a^{4} c \,d^{3} e^{4}+16 a^{3} c^{2} d^{5} e^{2}+5 a^{2} c^{3} d^{7}\right ) \textit {\_R}^{2}+\left (-8 a^{3} c \,e^{6}+24 a^{2} c^{2} d^{2} e^{4}-8 a \,c^{3} d^{4} e^{2}\right ) \textit {\_R} -8 c^{3} d \,e^{4}\right ) x^{2}+\left (-2 a^{6} d^{5} e^{3}+2 a^{5} c \,d^{7} e \right ) \textit {\_R}^{3}+\left (8 a^{5} d^{2} e^{5}-12 a^{4} c \,d^{4} e^{3}+7 a^{3} c^{2} d^{6} e \right ) \textit {\_R}^{2}+\left (-16 a^{3} c d \,e^{5}+18 a^{2} c^{2} d^{3} e^{3}-4 a \,c^{3} d^{5} e \right ) \textit {\_R} +8 a \,c^{2} e^{5}-8 c^{3} d^{2} e^{3}\right )\right )}{4}-\frac {e^{4} \ln \left (e \,x^{2}+d \right )}{2 d^{3} \left (a \,e^{2}+c \,d^{2}\right )}\) \(379\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(e*x^2+d)/(c*x^4+a),x,method=_RETURNVERBOSE)

[Out]

-1/2*e^4*ln(e*x^2+d)/d^3/(a*e^2+c*d^2)+1/2*c^2/(a*e^2+c*d^2)/a^2*(1/2*d*ln(c*x^4+a)+a*e/(a*c)^(1/2)*arctan(c*x
^2/(a*c)^(1/2)))-1/4/a/d/x^4+(a*e^2-c*d^2)/d^3/a^2*ln(x)+1/2*e/a/d^2/x^2

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Maxima [A]
time = 0.49, size = 143, normalized size = 0.92 \begin {gather*} \frac {c^{2} d \log \left (c x^{4} + a\right )}{4 \, {\left (a^{2} c d^{2} + a^{3} e^{2}\right )}} + \frac {c^{2} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right ) e}{2 \, {\left (a c d^{2} + a^{2} e^{2}\right )} \sqrt {a c}} - \frac {e^{4} \log \left (x^{2} e + d\right )}{2 \, {\left (c d^{5} + a d^{3} e^{2}\right )}} - \frac {{\left (c d^{2} - a e^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{2} d^{3}} + \frac {2 \, x^{2} e - d}{4 \, a d^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(e*x^2+d)/(c*x^4+a),x, algorithm="maxima")

[Out]

1/4*c^2*d*log(c*x^4 + a)/(a^2*c*d^2 + a^3*e^2) + 1/2*c^2*arctan(c*x^2/sqrt(a*c))*e/((a*c*d^2 + a^2*e^2)*sqrt(a
*c)) - 1/2*e^4*log(x^2*e + d)/(c*d^5 + a*d^3*e^2) - 1/2*(c*d^2 - a*e^2)*log(x^2)/(a^2*d^3) + 1/4*(2*x^2*e - d)
/(a*d^2*x^4)

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Fricas [A]
time = 54.53, size = 350, normalized size = 2.24 \begin {gather*} \left [\frac {a c d^{3} x^{4} \sqrt {-\frac {c}{a}} e \log \left (\frac {c x^{4} + 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) + c^{2} d^{4} x^{4} \log \left (c x^{4} + a\right ) + 2 \, a c d^{3} x^{2} e - 2 \, a^{2} x^{4} e^{4} \log \left (x^{2} e + d\right ) - a c d^{4} + 2 \, a^{2} d x^{2} e^{3} - a^{2} d^{2} e^{2} - 4 \, {\left (c^{2} d^{4} x^{4} - a^{2} x^{4} e^{4}\right )} \log \left (x\right )}{4 \, {\left (a^{2} c d^{5} x^{4} + a^{3} d^{3} x^{4} e^{2}\right )}}, -\frac {2 \, a c d^{3} x^{4} \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) e - c^{2} d^{4} x^{4} \log \left (c x^{4} + a\right ) - 2 \, a c d^{3} x^{2} e + 2 \, a^{2} x^{4} e^{4} \log \left (x^{2} e + d\right ) + a c d^{4} - 2 \, a^{2} d x^{2} e^{3} + a^{2} d^{2} e^{2} + 4 \, {\left (c^{2} d^{4} x^{4} - a^{2} x^{4} e^{4}\right )} \log \left (x\right )}{4 \, {\left (a^{2} c d^{5} x^{4} + a^{3} d^{3} x^{4} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(e*x^2+d)/(c*x^4+a),x, algorithm="fricas")

[Out]

[1/4*(a*c*d^3*x^4*sqrt(-c/a)*e*log((c*x^4 + 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) + c^2*d^4*x^4*log(c*x^4 + a)
+ 2*a*c*d^3*x^2*e - 2*a^2*x^4*e^4*log(x^2*e + d) - a*c*d^4 + 2*a^2*d*x^2*e^3 - a^2*d^2*e^2 - 4*(c^2*d^4*x^4 -
a^2*x^4*e^4)*log(x))/(a^2*c*d^5*x^4 + a^3*d^3*x^4*e^2), -1/4*(2*a*c*d^3*x^4*sqrt(c/a)*arctan(a*sqrt(c/a)/(c*x^
2))*e - c^2*d^4*x^4*log(c*x^4 + a) - 2*a*c*d^3*x^2*e + 2*a^2*x^4*e^4*log(x^2*e + d) + a*c*d^4 - 2*a^2*d*x^2*e^
3 + a^2*d^2*e^2 + 4*(c^2*d^4*x^4 - a^2*x^4*e^4)*log(x))/(a^2*c*d^5*x^4 + a^3*d^3*x^4*e^2)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(e*x**2+d)/(c*x**4+a),x)

[Out]

Timed out

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Giac [A]
time = 5.42, size = 168, normalized size = 1.08 \begin {gather*} \frac {c^{2} d \log \left (c x^{4} + a\right )}{4 \, {\left (a^{2} c d^{2} + a^{3} e^{2}\right )}} + \frac {c^{2} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right ) e}{2 \, {\left (a c d^{2} + a^{2} e^{2}\right )} \sqrt {a c}} - \frac {e^{5} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \, {\left (c d^{5} e + a d^{3} e^{3}\right )}} - \frac {{\left (c d^{2} - a e^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{2} d^{3}} + \frac {3 \, c d^{2} x^{4} - 3 \, a x^{4} e^{2} + 2 \, a d x^{2} e - a d^{2}}{4 \, a^{2} d^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(e*x^2+d)/(c*x^4+a),x, algorithm="giac")

[Out]

1/4*c^2*d*log(c*x^4 + a)/(a^2*c*d^2 + a^3*e^2) + 1/2*c^2*arctan(c*x^2/sqrt(a*c))*e/((a*c*d^2 + a^2*e^2)*sqrt(a
*c)) - 1/2*e^5*log(abs(x^2*e + d))/(c*d^5*e + a*d^3*e^3) - 1/2*(c*d^2 - a*e^2)*log(x^2)/(a^2*d^3) + 1/4*(3*c*d
^2*x^4 - 3*a*x^4*e^2 + 2*a*d*x^2*e - a*d^2)/(a^2*d^3*x^4)

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Mupad [B]
time = 1.87, size = 1017, normalized size = 6.52 \begin {gather*} \frac {\ln \left (25\,a^2\,c^9\,d^{20}\,{\left (-a^5\,c^3\right )}^{3/2}-64\,a^{19}\,c^4\,e^{20}\,x^2-25\,a^9\,c^{14}\,d^{20}\,x^2-64\,a^{17}\,c^2\,e^{20}\,\sqrt {-a^5\,c^3}+100\,a^3\,d^8\,e^{12}\,{\left (-a^5\,c^3\right )}^{5/2}+128\,a^{11}\,d^2\,e^{18}\,{\left (-a^5\,c^3\right )}^{3/2}-112\,c^3\,d^{14}\,e^6\,{\left (-a^5\,c^3\right )}^{5/2}-76\,a^{10}\,c^{13}\,d^{18}\,e^2\,x^2-138\,a^{11}\,c^{12}\,d^{16}\,e^4\,x^2-112\,a^{12}\,c^{11}\,d^{14}\,e^6\,x^2+55\,a^{13}\,c^{10}\,d^{12}\,e^8\,x^2+104\,a^{14}\,c^9\,d^{10}\,e^{10}\,x^2+100\,a^{15}\,c^8\,d^8\,e^{12}\,x^2+172\,a^{16}\,c^7\,d^6\,e^{14}\,x^2+32\,a^{17}\,c^6\,d^4\,e^{16}\,x^2-128\,a^{18}\,c^5\,d^2\,e^{18}\,x^2+55\,a\,c^2\,d^{12}\,e^8\,{\left (-a^5\,c^3\right )}^{5/2}+104\,a^2\,c\,d^{10}\,e^{10}\,{\left (-a^5\,c^3\right )}^{5/2}-32\,a^{10}\,c\,d^4\,e^{16}\,{\left (-a^5\,c^3\right )}^{3/2}+76\,a^3\,c^8\,d^{18}\,e^2\,{\left (-a^5\,c^3\right )}^{3/2}+138\,a^4\,c^7\,d^{16}\,e^4\,{\left (-a^5\,c^3\right )}^{3/2}-172\,a^9\,c^2\,d^6\,e^{14}\,{\left (-a^5\,c^3\right )}^{3/2}\right )\,\left (e\,\sqrt {-a^5\,c^3}+a^2\,c^2\,d\right )}{4\,a^5\,e^2+4\,c\,a^4\,d^2}-\frac {e^4\,\ln \left (e\,x^2+d\right )}{2\,\left (c\,d^5+a\,d^3\,e^2\right )}-\frac {\ln \left (25\,a^9\,c^{14}\,d^{20}\,x^2+64\,a^{19}\,c^4\,e^{20}\,x^2+25\,a^2\,c^9\,d^{20}\,{\left (-a^5\,c^3\right )}^{3/2}-64\,a^{17}\,c^2\,e^{20}\,\sqrt {-a^5\,c^3}+100\,a^3\,d^8\,e^{12}\,{\left (-a^5\,c^3\right )}^{5/2}+128\,a^{11}\,d^2\,e^{18}\,{\left (-a^5\,c^3\right )}^{3/2}-112\,c^3\,d^{14}\,e^6\,{\left (-a^5\,c^3\right )}^{5/2}+76\,a^{10}\,c^{13}\,d^{18}\,e^2\,x^2+138\,a^{11}\,c^{12}\,d^{16}\,e^4\,x^2+112\,a^{12}\,c^{11}\,d^{14}\,e^6\,x^2-55\,a^{13}\,c^{10}\,d^{12}\,e^8\,x^2-104\,a^{14}\,c^9\,d^{10}\,e^{10}\,x^2-100\,a^{15}\,c^8\,d^8\,e^{12}\,x^2-172\,a^{16}\,c^7\,d^6\,e^{14}\,x^2-32\,a^{17}\,c^6\,d^4\,e^{16}\,x^2+128\,a^{18}\,c^5\,d^2\,e^{18}\,x^2+55\,a\,c^2\,d^{12}\,e^8\,{\left (-a^5\,c^3\right )}^{5/2}+104\,a^2\,c\,d^{10}\,e^{10}\,{\left (-a^5\,c^3\right )}^{5/2}-32\,a^{10}\,c\,d^4\,e^{16}\,{\left (-a^5\,c^3\right )}^{3/2}+76\,a^3\,c^8\,d^{18}\,e^2\,{\left (-a^5\,c^3\right )}^{3/2}+138\,a^4\,c^7\,d^{16}\,e^4\,{\left (-a^5\,c^3\right )}^{3/2}-172\,a^9\,c^2\,d^6\,e^{14}\,{\left (-a^5\,c^3\right )}^{3/2}\right )\,\left (e\,\sqrt {-a^5\,c^3}-a^2\,c^2\,d\right )}{4\,\left (a^5\,e^2+c\,a^4\,d^2\right )}-\frac {\frac {1}{4\,a\,d}-\frac {e\,x^2}{2\,a\,d^2}}{x^4}+\frac {\ln \left (x\right )\,\left (a\,e^2-c\,d^2\right )}{a^2\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + c*x^4)*(d + e*x^2)),x)

[Out]

(log(25*a^2*c^9*d^20*(-a^5*c^3)^(3/2) - 64*a^19*c^4*e^20*x^2 - 25*a^9*c^14*d^20*x^2 - 64*a^17*c^2*e^20*(-a^5*c
^3)^(1/2) + 100*a^3*d^8*e^12*(-a^5*c^3)^(5/2) + 128*a^11*d^2*e^18*(-a^5*c^3)^(3/2) - 112*c^3*d^14*e^6*(-a^5*c^
3)^(5/2) - 76*a^10*c^13*d^18*e^2*x^2 - 138*a^11*c^12*d^16*e^4*x^2 - 112*a^12*c^11*d^14*e^6*x^2 + 55*a^13*c^10*
d^12*e^8*x^2 + 104*a^14*c^9*d^10*e^10*x^2 + 100*a^15*c^8*d^8*e^12*x^2 + 172*a^16*c^7*d^6*e^14*x^2 + 32*a^17*c^
6*d^4*e^16*x^2 - 128*a^18*c^5*d^2*e^18*x^2 + 55*a*c^2*d^12*e^8*(-a^5*c^3)^(5/2) + 104*a^2*c*d^10*e^10*(-a^5*c^
3)^(5/2) - 32*a^10*c*d^4*e^16*(-a^5*c^3)^(3/2) + 76*a^3*c^8*d^18*e^2*(-a^5*c^3)^(3/2) + 138*a^4*c^7*d^16*e^4*(
-a^5*c^3)^(3/2) - 172*a^9*c^2*d^6*e^14*(-a^5*c^3)^(3/2))*(e*(-a^5*c^3)^(1/2) + a^2*c^2*d))/(4*a^5*e^2 + 4*a^4*
c*d^2) - (e^4*log(d + e*x^2))/(2*(c*d^5 + a*d^3*e^2)) - (log(25*a^9*c^14*d^20*x^2 + 64*a^19*c^4*e^20*x^2 + 25*
a^2*c^9*d^20*(-a^5*c^3)^(3/2) - 64*a^17*c^2*e^20*(-a^5*c^3)^(1/2) + 100*a^3*d^8*e^12*(-a^5*c^3)^(5/2) + 128*a^
11*d^2*e^18*(-a^5*c^3)^(3/2) - 112*c^3*d^14*e^6*(-a^5*c^3)^(5/2) + 76*a^10*c^13*d^18*e^2*x^2 + 138*a^11*c^12*d
^16*e^4*x^2 + 112*a^12*c^11*d^14*e^6*x^2 - 55*a^13*c^10*d^12*e^8*x^2 - 104*a^14*c^9*d^10*e^10*x^2 - 100*a^15*c
^8*d^8*e^12*x^2 - 172*a^16*c^7*d^6*e^14*x^2 - 32*a^17*c^6*d^4*e^16*x^2 + 128*a^18*c^5*d^2*e^18*x^2 + 55*a*c^2*
d^12*e^8*(-a^5*c^3)^(5/2) + 104*a^2*c*d^10*e^10*(-a^5*c^3)^(5/2) - 32*a^10*c*d^4*e^16*(-a^5*c^3)^(3/2) + 76*a^
3*c^8*d^18*e^2*(-a^5*c^3)^(3/2) + 138*a^4*c^7*d^16*e^4*(-a^5*c^3)^(3/2) - 172*a^9*c^2*d^6*e^14*(-a^5*c^3)^(3/2
))*(e*(-a^5*c^3)^(1/2) - a^2*c^2*d))/(4*(a^5*e^2 + a^4*c*d^2)) - (1/(4*a*d) - (e*x^2)/(2*a*d^2))/x^4 + (log(x)
*(a*e^2 - c*d^2))/(a^2*d^3)

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