3.6.58 \(\int \frac {(d+e x)^6 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\) [558]

Optimal. Leaf size=177 \[ \frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right ) x}{e^2}+\frac {d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x^2}{e}+\frac {1}{3} \left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^3+\frac {1}{2} e g (e f+3 d g) x^4+\frac {1}{5} e^2 g^2 x^5+\frac {16 d^4 (e f+d g)^2}{e^3 (d-e x)}+\frac {32 d^3 (e f+d g) (e f+2 d g) \log (d-e x)}{e^3} \]

[Out]

d^2*(48*d^2*g^2+64*d*e*f*g+17*e^2*f^2)*x/e^2+d*(16*d^2*g^2+17*d*e*f*g+3*e^2*f^2)*x^2/e+1/3*(17*d^2*g^2+12*d*e*
f*g+e^2*f^2)*x^3+1/2*e*g*(3*d*g+e*f)*x^4+1/5*e^2*g^2*x^5+16*d^4*(d*g+e*f)^2/e^3/(-e*x+d)+32*d^3*(d*g+e*f)*(2*d
*g+e*f)*ln(-e*x+d)/e^3

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Rubi [A]
time = 0.15, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {862, 90} \begin {gather*} \frac {16 d^4 (d g+e f)^2}{e^3 (d-e x)}+\frac {32 d^3 (d g+e f) (2 d g+e f) \log (d-e x)}{e^3}+\frac {1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac {d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac {d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac {1}{2} e g x^4 (3 d g+e f)+\frac {1}{5} e^2 g^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^6*(f + g*x)^2)/(d^2 - e^2*x^2)^2,x]

[Out]

(d^2*(17*e^2*f^2 + 64*d*e*f*g + 48*d^2*g^2)*x)/e^2 + (d*(3*e^2*f^2 + 17*d*e*f*g + 16*d^2*g^2)*x^2)/e + ((e^2*f
^2 + 12*d*e*f*g + 17*d^2*g^2)*x^3)/3 + (e*g*(e*f + 3*d*g)*x^4)/2 + (e^2*g^2*x^5)/5 + (16*d^4*(e*f + d*g)^2)/(e
^3*(d - e*x)) + (32*d^3*(e*f + d*g)*(e*f + 2*d*g)*Log[d - e*x])/e^3

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 862

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^4 (f+g x)^2}{(d-e x)^2} \, dx\\ &=\int \left (\frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right )}{e^2}+\frac {2 d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x}{e}+\left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^2+2 e g (e f+3 d g) x^3+e^2 g^2 x^4+\frac {32 d^3 (-e f-2 d g) (e f+d g)}{e^2 (d-e x)}+\frac {16 d^4 (e f+d g)^2}{e^2 (-d+e x)^2}\right ) \, dx\\ &=\frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right ) x}{e^2}+\frac {d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x^2}{e}+\frac {1}{3} \left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^3+\frac {1}{2} e g (e f+3 d g) x^4+\frac {1}{5} e^2 g^2 x^5+\frac {16 d^4 (e f+d g)^2}{e^3 (d-e x)}+\frac {32 d^3 (e f+d g) (e f+2 d g) \log (d-e x)}{e^3}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 185, normalized size = 1.05 \begin {gather*} \frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right ) x}{e^2}+\frac {d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x^2}{e}+\frac {1}{3} \left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^3+\frac {1}{2} e g (e f+3 d g) x^4+\frac {1}{5} e^2 g^2 x^5-\frac {16 d^4 (e f+d g)^2}{e^3 (-d+e x)}+\frac {32 d^3 \left (e^2 f^2+3 d e f g+2 d^2 g^2\right ) \log (d-e x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^6*(f + g*x)^2)/(d^2 - e^2*x^2)^2,x]

[Out]

(d^2*(17*e^2*f^2 + 64*d*e*f*g + 48*d^2*g^2)*x)/e^2 + (d*(3*e^2*f^2 + 17*d*e*f*g + 16*d^2*g^2)*x^2)/e + ((e^2*f
^2 + 12*d*e*f*g + 17*d^2*g^2)*x^3)/3 + (e*g*(e*f + 3*d*g)*x^4)/2 + (e^2*g^2*x^5)/5 - (16*d^4*(e*f + d*g)^2)/(e
^3*(-d + e*x)) + (32*d^3*(e^2*f^2 + 3*d*e*f*g + 2*d^2*g^2)*Log[d - e*x])/e^3

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Maple [A]
time = 0.07, size = 217, normalized size = 1.23

method	result	size
default	\(\frac {\frac {1}{5} g^{2} e^{4} x^{5}+\frac {3}{2} d \,e^{3} g^{2} x^{4}+\frac {1}{2} e^{4} f g \,x^{4}+\frac {17}{3} d^{2} e^{2} g^{2} x^{3}+4 d \,e^{3} f g \,x^{3}+\frac {1}{3} e^{4} f^{2} x^{3}+16 d^{3} e \,g^{2} x^{2}+17 d^{2} e^{2} f g \,x^{2}+3 d \,e^{3} f^{2} x^{2}+48 d^{4} g^{2} x +64 d^{3} e f g x +17 d^{2} e^{2} f^{2} x}{e^{2}}+\frac {16 d^{4} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}+\frac {32 d^{3} \left (2 d^{2} g^{2}+3 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(217\)
risch	\(\frac {e^{2} g^{2} x^{5}}{5}+\frac {3 e d \,g^{2} x^{4}}{2}+\frac {e^{2} f g \,x^{4}}{2}+\frac {17 d^{2} g^{2} x^{3}}{3}+4 e d f g \,x^{3}+\frac {e^{2} f^{2} x^{3}}{3}+\frac {16 d^{3} g^{2} x^{2}}{e}+17 d^{2} f g \,x^{2}+3 e d \,f^{2} x^{2}+\frac {48 d^{4} g^{2} x}{e^{2}}+\frac {64 d^{3} f g x}{e}+17 d^{2} f^{2} x +\frac {16 d^{6} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {32 d^{5} f g}{e^{2} \left (-e x +d \right )}+\frac {16 d^{4} f^{2}}{e \left (-e x +d \right )}+\frac {64 d^{5} \ln \left (-e x +d \right ) g^{2}}{e^{3}}+\frac {96 d^{4} \ln \left (-e x +d \right ) f g}{e^{2}}+\frac {32 d^{3} \ln \left (-e x +d \right ) f^{2}}{e}\)	\(239\)
norman	\(\frac {\left (-\frac {127}{3} d^{4} g^{2}-60 f g \,d^{3} e -\frac {50}{3} d^{2} e^{2} f^{2}\right ) x^{3}+\left (-\frac {82}{15} g^{2} d^{2} e^{2}-4 f g d \,e^{3}-\frac {1}{3} f^{2} e^{4}\right ) x^{5}+\left (-\frac {29}{2} g^{2} d^{3} e -\frac {33}{2} f g \,d^{2} e^{2}-3 f^{2} d \,e^{3}\right ) x^{4}+\frac {d^{2} \left (32 d^{5} g^{2}+49 d^{4} e f g +19 d^{3} e^{2} f^{2}\right )}{e^{3}}+\frac {d^{4} \left (64 d^{2} g^{2}+96 d e f g +33 e^{2} f^{2}\right ) x}{e^{2}}-\frac {g^{2} e^{4} x^{7}}{5}-\frac {e^{3} g \left (3 d g +e f \right ) x^{6}}{2}}{-e^{2} x^{2}+d^{2}}+\frac {32 d^{3} \left (2 d^{2} g^{2}+3 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(246\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/e^2*(1/5*g^2*e^4*x^5+3/2*d*e^3*g^2*x^4+1/2*e^4*f*g*x^4+17/3*d^2*e^2*g^2*x^3+4*d*e^3*f*g*x^3+1/3*e^4*f^2*x^3+
16*d^3*e*g^2*x^2+17*d^2*e^2*f*g*x^2+3*d*e^3*f^2*x^2+48*d^4*g^2*x+64*d^3*e*f*g*x+17*d^2*e^2*f^2*x)+16*d^4*(d^2*
g^2+2*d*e*f*g+e^2*f^2)/e^3/(-e*x+d)+32*d^3/e^3*(2*d^2*g^2+3*d*e*f*g+e^2*f^2)*ln(-e*x+d)

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Maxima [A]
time = 0.29, size = 208, normalized size = 1.18 \begin {gather*} 32 \, {\left (2 \, d^{5} g^{2} + 3 \, d^{4} f g e + d^{3} f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right ) + \frac {1}{30} \, {\left (6 \, g^{2} x^{5} e^{4} + 15 \, {\left (3 \, d g^{2} e^{3} + f g e^{4}\right )} x^{4} + 10 \, {\left (17 \, d^{2} g^{2} e^{2} + 12 \, d f g e^{3} + f^{2} e^{4}\right )} x^{3} + 30 \, {\left (16 \, d^{3} g^{2} e + 17 \, d^{2} f g e^{2} + 3 \, d f^{2} e^{3}\right )} x^{2} + 30 \, {\left (48 \, d^{4} g^{2} + 64 \, d^{3} f g e + 17 \, d^{2} f^{2} e^{2}\right )} x\right )} e^{\left (-2\right )} - \frac {16 \, {\left (d^{6} g^{2} + 2 \, d^{5} f g e + d^{4} f^{2} e^{2}\right )}}{x e^{4} - d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="maxima")

[Out]

32*(2*d^5*g^2 + 3*d^4*f*g*e + d^3*f^2*e^2)*e^(-3)*log(x*e - d) + 1/30*(6*g^2*x^5*e^4 + 15*(3*d*g^2*e^3 + f*g*e
^4)*x^4 + 10*(17*d^2*g^2*e^2 + 12*d*f*g*e^3 + f^2*e^4)*x^3 + 30*(16*d^3*g^2*e + 17*d^2*f*g*e^2 + 3*d*f^2*e^3)*
x^2 + 30*(48*d^4*g^2 + 64*d^3*f*g*e + 17*d^2*f^2*e^2)*x)*e^(-2) - 16*(d^6*g^2 + 2*d^5*f*g*e + d^4*f^2*e^2)/(x*
e^4 - d*e^3)

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Fricas [A]
time = 2.21, size = 280, normalized size = 1.58 \begin {gather*} -\frac {480 \, d^{6} g^{2} - {\left (6 \, g^{2} x^{6} + 15 \, f g x^{5} + 10 \, f^{2} x^{4}\right )} e^{6} - {\left (39 \, d g^{2} x^{5} + 105 \, d f g x^{4} + 80 \, d f^{2} x^{3}\right )} e^{5} - 5 \, {\left (25 \, d^{2} g^{2} x^{4} + 78 \, d^{2} f g x^{3} + 84 \, d^{2} f^{2} x^{2}\right )} e^{4} - 10 \, {\left (31 \, d^{3} g^{2} x^{3} + 141 \, d^{3} f g x^{2} - 51 \, d^{3} f^{2} x\right )} e^{3} - 480 \, {\left (2 \, d^{4} g^{2} x^{2} - 4 \, d^{4} f g x - d^{4} f^{2}\right )} e^{2} + 480 \, {\left (3 \, d^{5} g^{2} x + 2 \, d^{5} f g\right )} e + 960 \, {\left (2 \, d^{6} g^{2} - d^{3} f^{2} x e^{3} - {\left (3 \, d^{4} f g x - d^{4} f^{2}\right )} e^{2} - {\left (2 \, d^{5} g^{2} x - 3 \, d^{5} f g\right )} e\right )} \log \left (x e - d\right )}{30 \, {\left (x e^{4} - d e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="fricas")

[Out]

-1/30*(480*d^6*g^2 - (6*g^2*x^6 + 15*f*g*x^5 + 10*f^2*x^4)*e^6 - (39*d*g^2*x^5 + 105*d*f*g*x^4 + 80*d*f^2*x^3)
*e^5 - 5*(25*d^2*g^2*x^4 + 78*d^2*f*g*x^3 + 84*d^2*f^2*x^2)*e^4 - 10*(31*d^3*g^2*x^3 + 141*d^3*f*g*x^2 - 51*d^
3*f^2*x)*e^3 - 480*(2*d^4*g^2*x^2 - 4*d^4*f*g*x - d^4*f^2)*e^2 + 480*(3*d^5*g^2*x + 2*d^5*f*g)*e + 960*(2*d^6*
g^2 - d^3*f^2*x*e^3 - (3*d^4*f*g*x - d^4*f^2)*e^2 - (2*d^5*g^2*x - 3*d^5*f*g)*e)*log(x*e - d))/(x*e^4 - d*e^3)

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Sympy [A]
time = 0.47, size = 199, normalized size = 1.12 \begin {gather*} \frac {32 d^{3} \left (d g + e f\right ) \left (2 d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} + \frac {e^{2} g^{2} x^{5}}{5} + x^{4} \cdot \left (\frac {3 d e g^{2}}{2} + \frac {e^{2} f g}{2}\right ) + x^{3} \cdot \left (\frac {17 d^{2} g^{2}}{3} + 4 d e f g + \frac {e^{2} f^{2}}{3}\right ) + x^{2} \cdot \left (\frac {16 d^{3} g^{2}}{e} + 17 d^{2} f g + 3 d e f^{2}\right ) + x \left (\frac {48 d^{4} g^{2}}{e^{2}} + \frac {64 d^{3} f g}{e} + 17 d^{2} f^{2}\right ) + \frac {- 16 d^{6} g^{2} - 32 d^{5} e f g - 16 d^{4} e^{2} f^{2}}{- d e^{3} + e^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**6*(g*x+f)**2/(-e**2*x**2+d**2)**2,x)

[Out]

32*d**3*(d*g + e*f)*(2*d*g + e*f)*log(-d + e*x)/e**3 + e**2*g**2*x**5/5 + x**4*(3*d*e*g**2/2 + e**2*f*g/2) + x
**3*(17*d**2*g**2/3 + 4*d*e*f*g + e**2*f**2/3) + x**2*(16*d**3*g**2/e + 17*d**2*f*g + 3*d*e*f**2) + x*(48*d**4
*g**2/e**2 + 64*d**3*f*g/e + 17*d**2*f**2) + (-16*d**6*g**2 - 32*d**5*e*f*g - 16*d**4*e**2*f**2)/(-d*e**3 + e*
*4*x)

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Giac [A]
time = 1.22, size = 218, normalized size = 1.23 \begin {gather*} 32 \, {\left (2 \, d^{5} g^{2} + 3 \, d^{4} f g e + d^{3} f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) + \frac {1}{30} \, {\left (6 \, g^{2} x^{5} e^{12} + 45 \, d g^{2} x^{4} e^{11} + 170 \, d^{2} g^{2} x^{3} e^{10} + 480 \, d^{3} g^{2} x^{2} e^{9} + 1440 \, d^{4} g^{2} x e^{8} + 15 \, f g x^{4} e^{12} + 120 \, d f g x^{3} e^{11} + 510 \, d^{2} f g x^{2} e^{10} + 1920 \, d^{3} f g x e^{9} + 10 \, f^{2} x^{3} e^{12} + 90 \, d f^{2} x^{2} e^{11} + 510 \, d^{2} f^{2} x e^{10}\right )} e^{\left (-10\right )} - \frac {16 \, {\left (d^{6} g^{2} + 2 \, d^{5} f g e + d^{4} f^{2} e^{2}\right )} e^{\left (-3\right )}}{x e - d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="giac")

[Out]

32*(2*d^5*g^2 + 3*d^4*f*g*e + d^3*f^2*e^2)*e^(-3)*log(abs(x*e - d)) + 1/30*(6*g^2*x^5*e^12 + 45*d*g^2*x^4*e^11
+ 170*d^2*g^2*x^3*e^10 + 480*d^3*g^2*x^2*e^9 + 1440*d^4*g^2*x*e^8 + 15*f*g*x^4*e^12 + 120*d*f*g*x^3*e^11 + 51
0*d^2*f*g*x^2*e^10 + 1920*d^3*f*g*x*e^9 + 10*f^2*x^3*e^12 + 90*d*f^2*x^2*e^11 + 510*d^2*f^2*x*e^10)*e^(-10) -
16*(d^6*g^2 + 2*d^5*f*g*e + d^4*f^2*e^2)*e^(-3)/(x*e - d)

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Mupad [B]
time = 2.61, size = 565, normalized size = 3.19 \begin {gather*} x^2\,\left (\frac {2\,d\,\left (d^2\,g^2+3\,d\,e\,f\,g+e^2\,f^2\right )}{e}-\frac {d^2\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{2\,e^2}+\frac {d\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e}\right )+x^4\,\left (\frac {e\,g\,\left (2\,d\,g+e\,f\right )}{2}+\frac {d\,e\,g^2}{2}\right )+x\,\left (\frac {d^4\,g^2+8\,d^3\,e\,f\,g+6\,d^2\,e^2\,f^2}{e^2}-\frac {d^2\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e^2}+\frac {2\,d\,\left (\frac {4\,d\,\left (d^2\,g^2+3\,d\,e\,f\,g+e^2\,f^2\right )}{e}-\frac {d^2\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e^2}+\frac {2\,d\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e}\right )}{e}\right )+x^3\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{3\,e^2}-\frac {d^2\,g^2}{3}+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{3\,e}\right )+\frac {\ln \left (e\,x-d\right )\,\left (64\,d^5\,g^2+96\,d^4\,e\,f\,g+32\,d^3\,e^2\,f^2\right )}{e^3}+\frac {16\,\left (d^6\,g^2+2\,d^5\,e\,f\,g+d^4\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {e^2\,g^2\,x^5}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^2*(d + e*x)^6)/(d^2 - e^2*x^2)^2,x)

[Out]

x^2*((2*d*(d^2*g^2 + e^2*f^2 + 3*d*e*f*g))/e - (d^2*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/(2*e^2) + (d*((e^4*f^2
+ 6*d^2*e^2*g^2 + 8*d*e^3*f*g)/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e) + x^4*((e*g*(2*d
*g + e*f))/2 + (d*e*g^2)/2) + x*((d^4*g^2 + 6*d^2*e^2*f^2 + 8*d^3*e*f*g)/e^2 - (d^2*((e^4*f^2 + 6*d^2*e^2*g^2
+ 8*d*e^3*f*g)/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e^2 + (2*d*((4*d*(d^2*g^2 + e^2*f^2
+ 3*d*e*f*g))/e - (d^2*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e^2 + (2*d*((e^4*f^2 + 6*d^2*e^2*g^2 + 8*d*e^3*f*g)
/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e))/e) + x^3*((e^4*f^2 + 6*d^2*e^2*g^2 + 8*d*e^3*
f*g)/(3*e^2) - (d^2*g^2)/3 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/(3*e)) + (log(e*x - d)*(64*d^5*g^2 + 32*d
^3*e^2*f^2 + 96*d^4*e*f*g))/e^3 + (16*(d^6*g^2 + d^4*e^2*f^2 + 2*d^5*e*f*g))/(e*(d*e^2 - e^3*x)) + (e^2*g^2*x^
5)/5

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