3.581 \(\int \frac{(d+e x)^3 (f+g x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=183 \[ \frac{(d+e x) (d g+e f) \left (32 d^2 g^2-11 d e f g+2 e^2 f^2\right )}{15 d^3 e^4 \sqrt{d^2-e^2 x^2}}+\frac{(d+e x)^2 (2 e f-13 d g) (d g+e f)^2}{15 d^2 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{(d+e x)^3 (d g+e f)^3}{5 d e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{g^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

((e*f + d*g)^3*(d + e*x)^3)/(5*d*e^4*(d^2 - e^2*x^2)^(5/2)) + ((2*e*f - 13*d*g)*(e*f + d*g)^2*(d + e*x)^2)/(15
*d^2*e^4*(d^2 - e^2*x^2)^(3/2)) + ((e*f + d*g)*(2*e^2*f^2 - 11*d*e*f*g + 32*d^2*g^2)*(d + e*x))/(15*d^3*e^4*Sq
rt[d^2 - e^2*x^2]) - (g^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^4

________________________________________________________________________________________

Rubi [A]  time = 0.40349, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1635, 778, 217, 203} \[ \frac{(d+e x) (d g+e f) \left (32 d^2 g^2-11 d e f g+2 e^2 f^2\right )}{15 d^3 e^4 \sqrt{d^2-e^2 x^2}}+\frac{(d+e x)^2 (2 e f-13 d g) (d g+e f)^2}{15 d^2 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{(d+e x)^3 (d g+e f)^3}{5 d e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{g^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(f + g*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((e*f + d*g)^3*(d + e*x)^3)/(5*d*e^4*(d^2 - e^2*x^2)^(5/2)) + ((2*e*f - 13*d*g)*(e*f + d*g)^2*(d + e*x)^2)/(15
*d^2*e^4*(d^2 - e^2*x^2)^(3/2)) + ((e*f + d*g)*(2*e^2*f^2 - 11*d*e*f*g + 32*d^2*g^2)*(d + e*x))/(15*d^3*e^4*Sq
rt[d^2 - e^2*x^2]) - (g^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^4

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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])

Rubi steps

\begin{align*} \int \frac{(d+e x)^3 (f+g x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{(e f+d g)^3 (d+e x)^3}{5 d e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{(d+e x)^2 \left (-\frac{2 e^3 f^3-9 d e^2 f^2 g-9 d^2 e f g^2-3 d^3 g^3}{e^3}+\frac{5 d g^2 (3 e f+d g) x}{e^2}+\frac{5 d g^3 x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac{(e f+d g)^3 (d+e x)^3}{5 d e^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{(2 e f-13 d g) (e f+d g)^2 (d+e x)^2}{15 d^2 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{(d+e x) \left (\frac{2 e^3 f^3-9 d e^2 f^2 g+21 d^2 e f g^2+17 d^3 g^3}{e^3}+\frac{15 d^2 g^3 x}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac{(e f+d g)^3 (d+e x)^3}{5 d e^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{(2 e f-13 d g) (e f+d g)^2 (d+e x)^2}{15 d^2 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{(e f+d g) \left (2 e^2 f^2-11 d e f g+32 d^2 g^2\right ) (d+e x)}{15 d^3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{g^3 \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{e^3}\\ &=\frac{(e f+d g)^3 (d+e x)^3}{5 d e^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{(2 e f-13 d g) (e f+d g)^2 (d+e x)^2}{15 d^2 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{(e f+d g) \left (2 e^2 f^2-11 d e f g+32 d^2 g^2\right ) (d+e x)}{15 d^3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{g^3 \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{e^3}\\ &=\frac{(e f+d g)^3 (d+e x)^3}{5 d e^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{(2 e f-13 d g) (e f+d g)^2 (d+e x)^2}{15 d^2 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{(e f+d g) \left (2 e^2 f^2-11 d e f g+32 d^2 g^2\right ) (d+e x)}{15 d^3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{g^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4}\\ \end{align*}

Mathematica [A]  time = 0.809311, size = 182, normalized size = 0.99 \[ \frac{(d+e x) \left (\sqrt{1-\frac{e^2 x^2}{d^2}} (d g+e f) \left (d^2 e^2 \left (7 f^2+33 f g x+32 g^2 x^2\right )-d^3 e g (16 f+51 g x)+22 d^4 g^2-d e^3 f x (6 f+11 g x)+2 e^4 f^2 x^2\right )-15 d^2 g^3 (d-e x)^3 \sin ^{-1}\left (\frac{e x}{d}\right )\right )}{15 d^3 e^4 (d-e x)^2 \sqrt{d^2-e^2 x^2} \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(f + g*x)^3)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*((e*f + d*g)*Sqrt[1 - (e^2*x^2)/d^2]*(22*d^4*g^2 + 2*e^4*f^2*x^2 - d*e^3*f*x*(6*f + 11*g*x) - d^3*e
*g*(16*f + 51*g*x) + d^2*e^2*(7*f^2 + 33*f*g*x + 32*g^2*x^2)) - 15*d^2*g^3*(d - e*x)^3*ArcSin[(e*x)/d]))/(15*d
^3*e^4*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

________________________________________________________________________________________

Maple [B]  time = 0.093, size = 713, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(g*x+f)^3/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/15/d*x/(-e^2*x^2+d^2)^(3/2)*f^3+2/15/d^3*x/(-e^2*x^2+d^2)^(1/2)*f^3+1/3*x^2*e/(-e^2*x^2+d^2)^(5/2)*f^3+1/5*e
*g^3*x^5/(-e^2*x^2+d^2)^(5/2)-1/3/e*g^3*x^3/(-e^2*x^2+d^2)^(3/2)+8/5/e^3*g^3*x/(-e^2*x^2+d^2)^(1/2)-1/e^3*g^3/
(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+3*x^4/(-e^2*x^2+d^2)^(5/2)*d*g^3+22/15*d^5/e^4/(-e^2*x^
2+d^2)^(5/2)*g^3+7/15*d^2/e/(-e^2*x^2+d^2)^(5/2)*f^3+4/5*x/(-e^2*x^2+d^2)^(5/2)*d*f^3+3*x^4*e/(-e^2*x^2+d^2)^(
5/2)*f*g^2-11/3*d^3/e^2*x^2/(-e^2*x^2+d^2)^(5/2)*g^3+2/5*d^4/e^3/(-e^2*x^2+d^2)^(5/2)*f*g^2+3/2*x^3/e/(-e^2*x^
2+d^2)^(5/2)*d^2*g^3+9/2*x^3/(-e^2*x^2+d^2)^(5/2)*d*f*g^2+3/2*x^3*e/(-e^2*x^2+d^2)^(5/2)*f^2*g-9/10*d^4/e^3*x/
(-e^2*x^2+d^2)^(5/2)*g^3+3/10/e^3*x/(-e^2*x^2+d^2)^(3/2)*d^2*g^3-3/10/e*x/(-e^2*x^2+d^2)^(3/2)*f^2*g+3*x^2/(-e
^2*x^2+d^2)^(5/2)*d*f^2*g-3/5*d^3/e^2/(-e^2*x^2+d^2)^(5/2)*f^2*g-21/10*d^3/e^2*x/(-e^2*x^2+d^2)^(5/2)*f*g^2+9/
10*d^2/e*x/(-e^2*x^2+d^2)^(5/2)*f^2*g+7/10/e^2*x/(-e^2*x^2+d^2)^(3/2)*d*f*g^2+7/5/d/e^2*x/(-e^2*x^2+d^2)^(1/2)
*f*g^2-3/5/d^2/e*x/(-e^2*x^2+d^2)^(1/2)*f^2*g-d^2/e*x^2/(-e^2*x^2+d^2)^(5/2)*f*g^2

________________________________________________________________________________________

Maxima [B]  time = 1.57726, size = 1220, normalized size = 6.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e^3*g^3*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x
^2 + d^2)^(5/2)*e^6)) - 1/3*e*g^3*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4))
+ 1/5*d*f^3*x/(-e^2*x^2 + d^2)^(5/2) + 3/5*d^2*f^3/((-e^2*x^2 + d^2)^(5/2)*e) + 3/5*d^3*f^2*g/((-e^2*x^2 + d^2
)^(5/2)*e^2) + 4/15*f^3*x/((-e^2*x^2 + d^2)^(3/2)*d) + 4/15*d^2*g^3*x/((-e^2*x^2 + d^2)^(3/2)*e^3) + 8/15*f^3*
x/(sqrt(-e^2*x^2 + d^2)*d^3) - 7/15*g^3*x/(sqrt(-e^2*x^2 + d^2)*e^3) + 3*(e^3*f*g^2 + d*e^2*g^3)*x^4/((-e^2*x^
2 + d^2)^(5/2)*e^2) - g^3*arcsin(e^2*x/sqrt(d^2*e^2))/(sqrt(e^2)*e^3) + 3/2*(e^3*f^2*g + 3*d*e^2*f*g^2 + d^2*e
*g^3)*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 4*(e^3*f*g^2 + d*e^2*g^3)*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 1/3*
(e^3*f^3 + 9*d*e^2*f^2*g + 9*d^2*e*f*g^2 + d^3*g^3)*x^2/((-e^2*x^2 + d^2)^(5/2)*e^2) - 9/10*(e^3*f^2*g + 3*d*e
^2*f*g^2 + d^2*e*g^3)*d^2*x/((-e^2*x^2 + d^2)^(5/2)*e^4) + 3/5*(d*e^2*f^3 + 3*d^2*e*f^2*g + d^3*f*g^2)*x/((-e^
2*x^2 + d^2)^(5/2)*e^2) + 8/5*(e^3*f*g^2 + d*e^2*g^3)*d^4/((-e^2*x^2 + d^2)^(5/2)*e^6) - 2/15*(e^3*f^3 + 9*d*e
^2*f^2*g + 9*d^2*e*f*g^2 + d^3*g^3)*d^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 3/10*(e^3*f^2*g + 3*d*e^2*f*g^2 + d^2*e
*g^3)*x/((-e^2*x^2 + d^2)^(3/2)*e^4) - 1/5*(d*e^2*f^3 + 3*d^2*e*f^2*g + d^3*f*g^2)*x/((-e^2*x^2 + d^2)^(3/2)*d
^2*e^2) + 3/5*(e^3*f^2*g + 3*d*e^2*f*g^2 + d^2*e*g^3)*x/(sqrt(-e^2*x^2 + d^2)*d^2*e^4) - 2/5*(d*e^2*f^3 + 3*d^
2*e*f^2*g + d^3*f*g^2)*x/(sqrt(-e^2*x^2 + d^2)*d^4*e^2)

________________________________________________________________________________________

Fricas [B]  time = 2.21653, size = 917, normalized size = 5.01 \begin{align*} -\frac{7 \, d^{3} e^{3} f^{3} - 9 \, d^{4} e^{2} f^{2} g + 6 \, d^{5} e f g^{2} + 22 \, d^{6} g^{3} -{\left (7 \, e^{6} f^{3} - 9 \, d e^{5} f^{2} g + 6 \, d^{2} e^{4} f g^{2} + 22 \, d^{3} e^{3} g^{3}\right )} x^{3} + 3 \,{\left (7 \, d e^{5} f^{3} - 9 \, d^{2} e^{4} f^{2} g + 6 \, d^{3} e^{3} f g^{2} + 22 \, d^{4} e^{2} g^{3}\right )} x^{2} - 3 \,{\left (7 \, d^{2} e^{4} f^{3} - 9 \, d^{3} e^{3} f^{2} g + 6 \, d^{4} e^{2} f g^{2} + 22 \, d^{5} e g^{3}\right )} x - 30 \,{\left (d^{3} e^{3} g^{3} x^{3} - 3 \, d^{4} e^{2} g^{3} x^{2} + 3 \, d^{5} e g^{3} x - d^{6} g^{3}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (7 \, d^{2} e^{3} f^{3} - 9 \, d^{3} e^{2} f^{2} g + 6 \, d^{4} e f g^{2} + 22 \, d^{5} g^{3} +{\left (2 \, e^{5} f^{3} - 9 \, d e^{4} f^{2} g + 21 \, d^{2} e^{3} f g^{2} + 32 \, d^{3} e^{2} g^{3}\right )} x^{2} - 3 \,{\left (2 \, d e^{4} f^{3} - 9 \, d^{2} e^{3} f^{2} g + 6 \, d^{3} e^{2} f g^{2} + 17 \, d^{4} e g^{3}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{3} e^{7} x^{3} - 3 \, d^{4} e^{6} x^{2} + 3 \, d^{5} e^{5} x - d^{6} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(7*d^3*e^3*f^3 - 9*d^4*e^2*f^2*g + 6*d^5*e*f*g^2 + 22*d^6*g^3 - (7*e^6*f^3 - 9*d*e^5*f^2*g + 6*d^2*e^4*f
*g^2 + 22*d^3*e^3*g^3)*x^3 + 3*(7*d*e^5*f^3 - 9*d^2*e^4*f^2*g + 6*d^3*e^3*f*g^2 + 22*d^4*e^2*g^3)*x^2 - 3*(7*d
^2*e^4*f^3 - 9*d^3*e^3*f^2*g + 6*d^4*e^2*f*g^2 + 22*d^5*e*g^3)*x - 30*(d^3*e^3*g^3*x^3 - 3*d^4*e^2*g^3*x^2 + 3
*d^5*e*g^3*x - d^6*g^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (7*d^2*e^3*f^3 - 9*d^3*e^2*f^2*g + 6*d^4*e
*f*g^2 + 22*d^5*g^3 + (2*e^5*f^3 - 9*d*e^4*f^2*g + 21*d^2*e^3*f*g^2 + 32*d^3*e^2*g^3)*x^2 - 3*(2*d*e^4*f^3 - 9
*d^2*e^3*f^2*g + 6*d^3*e^2*f*g^2 + 17*d^4*e*g^3)*x)*sqrt(-e^2*x^2 + d^2))/(d^3*e^7*x^3 - 3*d^4*e^6*x^2 + 3*d^5
*e^5*x - d^6*e^4)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{3} \left (f + g x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(g*x+f)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**3*(f + g*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.20298, size = 417, normalized size = 2.28 \begin{align*} -g^{3} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-4\right )} \mathrm{sgn}\left (d\right ) - \frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left ({\left (x{\left (\frac{{\left (32 \, d^{4} g^{3} e^{8} + 21 \, d^{3} f g^{2} e^{9} - 9 \, d^{2} f^{2} g e^{10} + 2 \, d f^{3} e^{11}\right )} x e^{\left (-7\right )}}{d^{4}} + \frac{45 \,{\left (d^{5} g^{3} e^{7} + d^{4} f g^{2} e^{8}\right )} e^{\left (-7\right )}}{d^{4}}\right )} - \frac{5 \,{\left (7 \, d^{6} g^{3} e^{6} - 3 \, d^{5} f g^{2} e^{7} - 9 \, d^{4} f^{2} g e^{8} + d^{3} f^{3} e^{9}\right )} e^{\left (-7\right )}}{d^{4}}\right )} x - \frac{5 \,{\left (11 \, d^{7} g^{3} e^{5} + 3 \, d^{6} f g^{2} e^{6} - 9 \, d^{5} f^{2} g e^{7} - d^{4} f^{3} e^{8}\right )} e^{\left (-7\right )}}{d^{4}}\right )} x + \frac{15 \,{\left (d^{8} g^{3} e^{4} + d^{5} f^{3} e^{7}\right )} e^{\left (-7\right )}}{d^{4}}\right )} x + \frac{{\left (22 \, d^{9} g^{3} e^{3} + 6 \, d^{8} f g^{2} e^{4} - 9 \, d^{7} f^{2} g e^{5} + 7 \, d^{6} f^{3} e^{6}\right )} e^{\left (-7\right )}}{d^{4}}\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-g^3*arcsin(x*e/d)*e^(-4)*sgn(d) - 1/15*sqrt(-x^2*e^2 + d^2)*((((x*((32*d^4*g^3*e^8 + 21*d^3*f*g^2*e^9 - 9*d^2
*f^2*g*e^10 + 2*d*f^3*e^11)*x*e^(-7)/d^4 + 45*(d^5*g^3*e^7 + d^4*f*g^2*e^8)*e^(-7)/d^4) - 5*(7*d^6*g^3*e^6 - 3
*d^5*f*g^2*e^7 - 9*d^4*f^2*g*e^8 + d^3*f^3*e^9)*e^(-7)/d^4)*x - 5*(11*d^7*g^3*e^5 + 3*d^6*f*g^2*e^6 - 9*d^5*f^
2*g*e^7 - d^4*f^3*e^8)*e^(-7)/d^4)*x + 15*(d^8*g^3*e^4 + d^5*f^3*e^7)*e^(-7)/d^4)*x + (22*d^9*g^3*e^3 + 6*d^8*
f*g^2*e^4 - 9*d^7*f^2*g*e^5 + 7*d^6*f^3*e^6)*e^(-7)/d^4)/(x^2*e^2 - d^2)^3