3.4.58 \(\int \frac {(f+g x)^2}{(d+e x)^4 (d^2-e^2 x^2)} \, dx\)

Optimal. Leaf size=139 \[ \frac {(d g+e f)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^5 e^3}-\frac {(d g+e f)^2}{16 d^4 e^3 (d+e x)}-\frac {(d g+e f)^2}{16 d^3 e^3 (d+e x)^2}-\frac {(3 d g+e f) (e f-d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4} \]

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Rubi [A]  time = 0.13, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {848, 88, 208} \begin {gather*} -\frac {(3 d g+e f) (e f-d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(d g+e f)^2}{16 d^4 e^3 (d+e x)}-\frac {(d g+e f)^2}{16 d^3 e^3 (d+e x)^2}+\frac {(d g+e f)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^5 e^3}-\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^2/((d + e*x)^4*(d^2 - e^2*x^2)),x]

[Out]

-(e*f - d*g)^2/(8*d*e^3*(d + e*x)^4) - ((e*f - d*g)*(e*f + 3*d*g))/(12*d^2*e^3*(d + e*x)^3) - (e*f + d*g)^2/(1
6*d^3*e^3*(d + e*x)^2) - (e*f + d*g)^2/(16*d^4*e^3*(d + e*x)) + ((e*f + d*g)^2*ArcTanh[(e*x)/d])/(16*d^5*e^3)

Rule 88

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 208

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

Rule 848

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*x)/e)^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 {(f+g x)^2}{(d+e x)^4 \left (d^2-e^2 x^2\right )} \, dx &=\int \frac {(f+g x)^2}{(d-e x) (d+e x)^5} \, dx\\ &=\int \left (\frac {(-e f+d g)^2}{2 d e^2 (d+e x)^5}+\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^2 (d+e x)^4}+\frac {(e f+d g)^2}{8 d^3 e^2 (d+e x)^3}+\frac {(e f+d g)^2}{16 d^4 e^2 (d+e x)^2}+\frac {(e f+d g)^2}{16 d^4 e^2 \left (d^2-e^2 x^2\right )}\right ) \, dx\\ &=-\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4}-\frac {(e f-d g) (e f+3 d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(e f+d g)^2}{16 d^3 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{16 d^4 e^3 (d+e x)}+\frac {(e f+d g)^2 \int \frac {1}{d^2-e^2 x^2} \, dx}{16 d^4 e^2}\\ &=-\frac {(e f-d g)^2}{8 d e^3 (d+e x)^4}-\frac {(e f-d g) (e f+3 d g)}{12 d^2 e^3 (d+e x)^3}-\frac {(e f+d g)^2}{16 d^3 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{16 d^4 e^3 (d+e x)}+\frac {(e f+d g)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^5 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 142, normalized size = 1.02 \begin {gather*} -\frac {\frac {12 d^4 (e f-d g)^2}{(d+e x)^4}+\frac {6 d^2 (d g+e f)^2}{(d+e x)^2}+\frac {8 d^3 \left (-3 d^2 g^2+2 d e f g+e^2 f^2\right )}{(d+e x)^3}+\frac {6 d (d g+e f)^2}{d+e x}+3 (d g+e f)^2 \log (d-e x)-3 (d g+e f)^2 \log (d+e x)}{96 d^5 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^2/((d + e*x)^4*(d^2 - e^2*x^2)),x]

[Out]

-1/96*((12*d^4*(e*f - d*g)^2)/(d + e*x)^4 + (8*d^3*(e^2*f^2 + 2*d*e*f*g - 3*d^2*g^2))/(d + e*x)^3 + (6*d^2*(e*
f + d*g)^2)/(d + e*x)^2 + (6*d*(e*f + d*g)^2)/(d + e*x) + 3*(e*f + d*g)^2*Log[d - e*x] - 3*(e*f + d*g)^2*Log[d
+ e*x])/(d^5*e^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(f+g x)^2}{(d+e x)^4 \left (d^2-e^2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(f + g*x)^2/((d + e*x)^4*(d^2 - e^2*x^2)),x]

[Out]

IntegrateAlgebraic[(f + g*x)^2/((d + e*x)^4*(d^2 - e^2*x^2)), x]

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fricas [B]  time = 0.42, size = 511, normalized size = 3.68 \begin {gather*} -\frac {32 \, d^{4} e^{2} f^{2} + 16 \, d^{5} e f g + 6 \, {\left (d e^{5} f^{2} + 2 \, d^{2} e^{4} f g + d^{3} e^{3} g^{2}\right )} x^{3} + 24 \, {\left (d^{2} e^{4} f^{2} + 2 \, d^{3} e^{3} f g + d^{4} e^{2} g^{2}\right )} x^{2} + 2 \, {\left (19 \, d^{3} e^{3} f^{2} + 38 \, d^{4} e^{2} f g + 3 \, d^{5} e g^{2}\right )} x - 3 \, {\left (d^{4} e^{2} f^{2} + 2 \, d^{5} e f g + d^{6} g^{2} + {\left (e^{6} f^{2} + 2 \, d e^{5} f g + d^{2} e^{4} g^{2}\right )} x^{4} + 4 \, {\left (d e^{5} f^{2} + 2 \, d^{2} e^{4} f g + d^{3} e^{3} g^{2}\right )} x^{3} + 6 \, {\left (d^{2} e^{4} f^{2} + 2 \, d^{3} e^{3} f g + d^{4} e^{2} g^{2}\right )} x^{2} + 4 \, {\left (d^{3} e^{3} f^{2} + 2 \, d^{4} e^{2} f g + d^{5} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + 3 \, {\left (d^{4} e^{2} f^{2} + 2 \, d^{5} e f g + d^{6} g^{2} + {\left (e^{6} f^{2} + 2 \, d e^{5} f g + d^{2} e^{4} g^{2}\right )} x^{4} + 4 \, {\left (d e^{5} f^{2} + 2 \, d^{2} e^{4} f g + d^{3} e^{3} g^{2}\right )} x^{3} + 6 \, {\left (d^{2} e^{4} f^{2} + 2 \, d^{3} e^{3} f g + d^{4} e^{2} g^{2}\right )} x^{2} + 4 \, {\left (d^{3} e^{3} f^{2} + 2 \, d^{4} e^{2} f g + d^{5} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{96 \, {\left (d^{5} e^{7} x^{4} + 4 \, d^{6} e^{6} x^{3} + 6 \, d^{7} e^{5} x^{2} + 4 \, d^{8} e^{4} x + d^{9} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(e*x+d)^4/(-e^2*x^2+d^2),x, algorithm="fricas")

[Out]

-1/96*(32*d^4*e^2*f^2 + 16*d^5*e*f*g + 6*(d*e^5*f^2 + 2*d^2*e^4*f*g + d^3*e^3*g^2)*x^3 + 24*(d^2*e^4*f^2 + 2*d
^3*e^3*f*g + d^4*e^2*g^2)*x^2 + 2*(19*d^3*e^3*f^2 + 38*d^4*e^2*f*g + 3*d^5*e*g^2)*x - 3*(d^4*e^2*f^2 + 2*d^5*e
*f*g + d^6*g^2 + (e^6*f^2 + 2*d*e^5*f*g + d^2*e^4*g^2)*x^4 + 4*(d*e^5*f^2 + 2*d^2*e^4*f*g + d^3*e^3*g^2)*x^3 +
6*(d^2*e^4*f^2 + 2*d^3*e^3*f*g + d^4*e^2*g^2)*x^2 + 4*(d^3*e^3*f^2 + 2*d^4*e^2*f*g + d^5*e*g^2)*x)*log(e*x +
d) + 3*(d^4*e^2*f^2 + 2*d^5*e*f*g + d^6*g^2 + (e^6*f^2 + 2*d*e^5*f*g + d^2*e^4*g^2)*x^4 + 4*(d*e^5*f^2 + 2*d^2
*e^4*f*g + d^3*e^3*g^2)*x^3 + 6*(d^2*e^4*f^2 + 2*d^3*e^3*f*g + d^4*e^2*g^2)*x^2 + 4*(d^3*e^3*f^2 + 2*d^4*e^2*f
*g + d^5*e*g^2)*x)*log(e*x - d))/(d^5*e^7*x^4 + 4*d^6*e^6*x^3 + 6*d^7*e^5*x^2 + 4*d^8*e^4*x + d^9*e^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(e*x+d)^4/(-e^2*x^2+d^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -(exp(2)^3*d*g*f-2*exp(2)^3*exp(1)*f^2-2
*exp(2)^2*d^2*exp(1)*g^2+6*exp(2)^2*d*exp(1)^2*g*f-2*exp(2)^2*exp(1)^3*f^2-2*exp(2)*d^2*exp(1)^3*g^2+exp(2)*d*
exp(1)^4*g*f)/(exp(2)^4*d^5-4*exp(2)^3*d^5*exp(1)^2+6*exp(2)^2*d^5*exp(1)^4-4*exp(2)*d^5*exp(1)^6+d^5*exp(1)^8
)*ln(abs(-x^2*exp(2)+d^2))-(-exp(2)^4*f^2-exp(2)^3*d^2*g^2+8*exp(2)^3*d*exp(1)*g*f-6*exp(2)^3*exp(1)^2*f^2-6*e
xp(2)^2*d^2*exp(1)^2*g^2+8*exp(2)^2*d*exp(1)^3*g*f-exp(2)^2*exp(1)^4*f^2-exp(2)*d^2*exp(1)^4*g^2)*1/2/(exp(2)^
4*d^4-4*exp(2)^3*d^4*exp(1)^2+6*exp(2)^2*d^4*exp(1)^4-4*exp(2)*d^4*exp(1)^6+d^4*exp(1)^8)/exp(1)/abs(d)*ln(abs
(-2*x*exp(2)-2*exp(1)*abs(d))/abs(-2*x*exp(2)+2*exp(1)*abs(d)))-(-2*exp(2)^3*d*exp(1)*g*f+4*exp(2)^3*exp(1)^2*
f^2+4*exp(2)^2*d^2*exp(1)^2*g^2-12*exp(2)^2*d*exp(1)^3*g*f+4*exp(2)^2*exp(1)^4*f^2+4*exp(2)*d^2*exp(1)^4*g^2-2
*exp(2)*d*exp(1)^5*g*f)/(exp(2)^4*d^5*exp(1)-4*exp(2)^3*d^5*exp(1)^3+6*exp(2)^2*d^5*exp(1)^5-4*exp(2)*d^5*exp(
1)^7+d^5*exp(1)^9)*ln(abs(x*exp(1)+d))-((6*exp(2)^3*d^2*exp(1)^3*g*f-9*exp(2)^3*d*exp(1)^4*f^2-9*exp(2)^2*d^3*
exp(1)^4*g^2+12*exp(2)^2*d^2*exp(1)^5*g*f+6*exp(2)^2*d*exp(1)^6*f^2+6*exp(2)*d^3*exp(1)^6*g^2-18*exp(2)*d^2*ex
p(1)^7*g*f+3*exp(2)*d*exp(1)^8*f^2+3*d^3*exp(1)^8*g^2)*x^2+(15*exp(2)^3*d^3*exp(1)^2*g*f-21*exp(2)^3*d^2*exp(1
)^3*f^2-21*exp(2)^2*d^4*exp(1)^3*g^2+21*exp(2)^2*d^3*exp(1)^4*g*f+18*exp(2)^2*d^2*exp(1)^5*f^2+18*exp(2)*d^4*e
xp(1)^5*g^2-39*exp(2)*d^3*exp(1)^6*g*f+3*exp(2)*d^2*exp(1)^7*f^2+3*d^4*exp(1)^7*g^2+3*d^3*exp(1)^8*g*f)*x-exp(
2)^3*d^5*g^2+11*exp(2)^3*d^4*exp(1)*g*f-13*exp(2)^3*d^3*exp(1)^2*f^2-9*exp(2)^2*d^5*exp(1)^2*g^2+3*exp(2)^2*d^
4*exp(1)^3*g*f+15*exp(2)^2*d^3*exp(1)^4*f^2+9*exp(2)*d^5*exp(1)^4*g^2-15*exp(2)*d^4*exp(1)^5*g*f-3*exp(2)*d^3*
exp(1)^6*f^2+d^5*exp(1)^6*g^2+d^4*exp(1)^7*g*f+d^3*exp(1)^8*f^2)/3/d^5/exp(1)/(exp(2)-exp(1)^2)^4/(x*exp(1)+d)
^3

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maple [B]  time = 0.01, size = 312, normalized size = 2.24 \begin {gather*} -\frac {d \,g^{2}}{8 \left (e x +d \right )^{4} e^{3}}-\frac {f^{2}}{8 \left (e x +d \right )^{4} d e}+\frac {f g}{4 \left (e x +d \right )^{4} e^{2}}-\frac {f g}{6 \left (e x +d \right )^{3} d \,e^{2}}-\frac {f^{2}}{12 \left (e x +d \right )^{3} d^{2} e}+\frac {g^{2}}{4 \left (e x +d \right )^{3} e^{3}}-\frac {g^{2}}{16 \left (e x +d \right )^{2} d \,e^{3}}-\frac {f g}{8 \left (e x +d \right )^{2} d^{2} e^{2}}-\frac {f^{2}}{16 \left (e x +d \right )^{2} d^{3} e}-\frac {g^{2}}{16 \left (e x +d \right ) d^{2} e^{3}}-\frac {f g}{8 \left (e x +d \right ) d^{3} e^{2}}-\frac {g^{2} \ln \left (e x -d \right )}{32 d^{3} e^{3}}+\frac {g^{2} \ln \left (e x +d \right )}{32 d^{3} e^{3}}-\frac {f^{2}}{16 \left (e x +d \right ) d^{4} e}-\frac {f g \ln \left (e x -d \right )}{16 d^{4} e^{2}}+\frac {f g \ln \left (e x +d \right )}{16 d^{4} e^{2}}-\frac {f^{2} \ln \left (e x -d \right )}{32 d^{5} e}+\frac {f^{2} \ln \left (e x +d \right )}{32 d^{5} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^2/(e*x+d)^4/(-e^2*x^2+d^2),x)

[Out]

-1/32/e^3/d^3*ln(e*x-d)*g^2-1/16/e^2/d^4*ln(e*x-d)*f*g-1/32/e/d^5*ln(e*x-d)*f^2+1/4/e^3/(e*x+d)^3*g^2-1/6/d/e^
2/(e*x+d)^3*f*g-1/12/d^2/e/(e*x+d)^3*f^2-1/8/e^3*d/(e*x+d)^4*g^2+1/4/e^2/(e*x+d)^4*f*g-1/8/e/d/(e*x+d)^4*f^2+1
/32/e^3/d^3*ln(e*x+d)*g^2+1/16/e^2/d^4*ln(e*x+d)*f*g+1/32/e/d^5*ln(e*x+d)*f^2-1/16/d^2/e^3/(e*x+d)*g^2-1/8/d^3
/e^2/(e*x+d)*f*g-1/16/d^4/e/(e*x+d)*f^2-1/16/d/e^3/(e*x+d)^2*g^2-1/8/d^2/e^2/(e*x+d)^2*f*g-1/16/d^3/e/(e*x+d)^
2*f^2

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maxima [A]  time = 0.49, size = 236, normalized size = 1.70 \begin {gather*} -\frac {16 \, d^{3} e f^{2} + 8 \, d^{4} f g + 3 \, {\left (e^{4} f^{2} + 2 \, d e^{3} f g + d^{2} e^{2} g^{2}\right )} x^{3} + 12 \, {\left (d e^{3} f^{2} + 2 \, d^{2} e^{2} f g + d^{3} e g^{2}\right )} x^{2} + {\left (19 \, d^{2} e^{2} f^{2} + 38 \, d^{3} e f g + 3 \, d^{4} g^{2}\right )} x}{48 \, {\left (d^{4} e^{6} x^{4} + 4 \, d^{5} e^{5} x^{3} + 6 \, d^{6} e^{4} x^{2} + 4 \, d^{7} e^{3} x + d^{8} e^{2}\right )}} + \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{32 \, d^{5} e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{32 \, d^{5} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(e*x+d)^4/(-e^2*x^2+d^2),x, algorithm="maxima")

[Out]

-1/48*(16*d^3*e*f^2 + 8*d^4*f*g + 3*(e^4*f^2 + 2*d*e^3*f*g + d^2*e^2*g^2)*x^3 + 12*(d*e^3*f^2 + 2*d^2*e^2*f*g
+ d^3*e*g^2)*x^2 + (19*d^2*e^2*f^2 + 38*d^3*e*f*g + 3*d^4*g^2)*x)/(d^4*e^6*x^4 + 4*d^5*e^5*x^3 + 6*d^6*e^4*x^2
+ 4*d^7*e^3*x + d^8*e^2) + 1/32*(e^2*f^2 + 2*d*e*f*g + d^2*g^2)*log(e*x + d)/(d^5*e^3) - 1/32*(e^2*f^2 + 2*d*
e*f*g + d^2*g^2)*log(e*x - d)/(d^5*e^3)

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mupad [B]  time = 0.15, size = 180, normalized size = 1.29 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,{\left (d\,g+e\,f\right )}^2}{16\,d^5\,e^3}-\frac {\frac {x^3\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{16\,d^4}+\frac {2\,e\,f^2+d\,g\,f}{6\,d\,e^2}+\frac {x\,\left (3\,d^2\,g^2+38\,d\,e\,f\,g+19\,e^2\,f^2\right )}{48\,d^2\,e^2}+\frac {x^2\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^3\,e}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)^2/((d^2 - e^2*x^2)*(d + e*x)^4),x)

[Out]

(atanh((e*x)/d)*(d*g + e*f)^2)/(16*d^5*e^3) - ((x^3*(d^2*g^2 + e^2*f^2 + 2*d*e*f*g))/(16*d^4) + (2*e*f^2 + d*f
*g)/(6*d*e^2) + (x*(3*d^2*g^2 + 19*e^2*f^2 + 38*d*e*f*g))/(48*d^2*e^2) + (x^2*(d^2*g^2 + e^2*f^2 + 2*d*e*f*g))
/(4*d^3*e))/(d^4 + e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x)

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sympy [B]  time = 1.93, size = 282, normalized size = 2.03 \begin {gather*} - \frac {8 d^{4} f g + 16 d^{3} e f^{2} + x^{3} \left (3 d^{2} e^{2} g^{2} + 6 d e^{3} f g + 3 e^{4} f^{2}\right ) + x^{2} \left (12 d^{3} e g^{2} + 24 d^{2} e^{2} f g + 12 d e^{3} f^{2}\right ) + x \left (3 d^{4} g^{2} + 38 d^{3} e f g + 19 d^{2} e^{2} f^{2}\right )}{48 d^{8} e^{2} + 192 d^{7} e^{3} x + 288 d^{6} e^{4} x^{2} + 192 d^{5} e^{5} x^{3} + 48 d^{4} e^{6} x^{4}} - \frac {\left (d g + e f\right )^{2} \log {\left (- \frac {d \left (d g + e f\right )^{2}}{e \left (d^{2} g^{2} + 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{32 d^{5} e^{3}} + \frac {\left (d g + e f\right )^{2} \log {\left (\frac {d \left (d g + e f\right )^{2}}{e \left (d^{2} g^{2} + 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{32 d^{5} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2/(e*x+d)**4/(-e**2*x**2+d**2),x)

[Out]

-(8*d**4*f*g + 16*d**3*e*f**2 + x**3*(3*d**2*e**2*g**2 + 6*d*e**3*f*g + 3*e**4*f**2) + x**2*(12*d**3*e*g**2 +
24*d**2*e**2*f*g + 12*d*e**3*f**2) + x*(3*d**4*g**2 + 38*d**3*e*f*g + 19*d**2*e**2*f**2))/(48*d**8*e**2 + 192*
d**7*e**3*x + 288*d**6*e**4*x**2 + 192*d**5*e**5*x**3 + 48*d**4*e**6*x**4) - (d*g + e*f)**2*log(-d*(d*g + e*f)
**2/(e*(d**2*g**2 + 2*d*e*f*g + e**2*f**2)) + x)/(32*d**5*e**3) + (d*g + e*f)**2*log(d*(d*g + e*f)**2/(e*(d**2
*g**2 + 2*d*e*f*g + e**2*f**2)) + x)/(32*d**5*e**3)

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