3.1788 \(\int \frac{(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx\)

Optimal. Leaf size=227 \[ -\frac{2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )\right )}{f^3 \sqrt{e+f x} (d e-c f)^3}+\frac{2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{3 f^3 (e+f x)^{3/2} (d e-c f)^2}-\frac{2 (b e-a f)^3}{5 f^3 (e+f x)^{5/2} (d e-c f)}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{\sqrt{d} (d e-c f)^{7/2}} \]

[Out]

(-2*(b*e - a*f)^3)/(5*f^3*(d*e - c*f)*(e + f*x)^(5/2)) + (2*(b*e - a*f)^2*(2*b*d*e - 3*b*c*f + a*d*f))/(3*f^3*
(d*e - c*f)^2*(e + f*x)^(3/2)) - (2*(b*e - a*f)*(a^2*d^2*f^2 + a*b*d*f*(d*e - 3*c*f) + b^2*(d^2*e^2 - 3*c*d*e*
f + 3*c^2*f^2)))/(f^3*(d*e - c*f)^3*Sqrt[e + f*x]) + (2*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e
- c*f]])/(Sqrt[d]*(d*e - c*f)^(7/2))

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Rubi [A]  time = 0.279823, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {87, 63, 208} \[ -\frac{2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )\right )}{f^3 \sqrt{e+f x} (d e-c f)^3}+\frac{2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{3 f^3 (e+f x)^{3/2} (d e-c f)^2}-\frac{2 (b e-a f)^3}{5 f^3 (e+f x)^{5/2} (d e-c f)}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{\sqrt{d} (d e-c f)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(-2*(b*e - a*f)^3)/(5*f^3*(d*e - c*f)*(e + f*x)^(5/2)) + (2*(b*e - a*f)^2*(2*b*d*e - 3*b*c*f + a*d*f))/(3*f^3*
(d*e - c*f)^2*(e + f*x)^(3/2)) - (2*(b*e - a*f)*(a^2*d^2*f^2 + a*b*d*f*(d*e - 3*c*f) + b^2*(d^2*e^2 - 3*c*d*e*
f + 3*c^2*f^2)))/(f^3*(d*e - c*f)^3*Sqrt[e + f*x]) + (2*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e
- c*f]])/(Sqrt[d]*(d*e - c*f)^(7/2))

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rubi steps

\begin{align*} \int \frac{(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx &=\int \left (\frac{(-b e+a f)^3}{f^2 (-d e+c f) (e+f x)^{7/2}}+\frac{(-b e+a f)^2 (-2 b d e+3 b c f-a d f)}{f^2 (-d e+c f)^2 (e+f x)^{5/2}}+\frac{-3 a b^2 c^2 f^3+3 a^2 b c d f^3-a^3 d^2 f^3+b^3 e \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )}{f^2 (d e-c f)^3 (e+f x)^{3/2}}-\frac{(b c-a d)^3}{(d e-c f)^3 (c+d x) \sqrt{e+f x}}\right ) \, dx\\ &=-\frac{2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac{2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac{2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt{e+f x}}-\frac{(b c-a d)^3 \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{(d e-c f)^3}\\ &=-\frac{2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac{2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac{2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt{e+f x}}-\frac{\left (2 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{c-\frac{d e}{f}+\frac{d x^2}{f}} \, dx,x,\sqrt{e+f x}\right )}{f (d e-c f)^3}\\ &=-\frac{2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac{2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac{2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt{e+f x}}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{\sqrt{d} (d e-c f)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.13605, size = 166, normalized size = 0.73 \[ \frac{2 \left (-\frac{3 b \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{f^3}+\frac{5 b^2 d (e+f x) (-3 a d f+b c f+2 b d e)}{f^3}+\frac{3 (b c-a d)^3 \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{d (e+f x)}{d e-c f}\right )}{c f-d e}-\frac{15 b^3 d^2 (e+f x)^2}{f^3}\right )}{15 d^3 (e+f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(2*((-3*b*(3*a^2*d^2*f^2 - 3*a*b*d*f*(d*e + c*f) + b^2*(d^2*e^2 + c*d*e*f + c^2*f^2)))/f^3 + (5*b^2*d*(2*b*d*e
+ b*c*f - 3*a*d*f)*(e + f*x))/f^3 - (15*b^3*d^2*(e + f*x)^2)/f^3 + (3*(b*c - a*d)^3*Hypergeometric2F1[-5/2, 1
, -3/2, (d*(e + f*x))/(d*e - c*f)])/(-(d*e) + c*f)))/(15*d^3*(e + f*x)^(5/2))

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Maple [B]  time = 0.019, size = 649, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x)

[Out]

-2/5/(c*f-d*e)/(f*x+e)^(5/2)*a^3+6/5/f/(c*f-d*e)/(f*x+e)^(5/2)*a^2*b*e-6/5/f^2/(c*f-d*e)/(f*x+e)^(5/2)*a*b^2*e
^2+2/5/f^3/(c*f-d*e)/(f*x+e)^(5/2)*b^3*e^3+2/3/(c*f-d*e)^2/(f*x+e)^(3/2)*a^3*d-2/(c*f-d*e)^2/(f*x+e)^(3/2)*a^2
*b*c+4/f/(c*f-d*e)^2/(f*x+e)^(3/2)*a*b^2*c*e-2/f^2/(c*f-d*e)^2/(f*x+e)^(3/2)*a*b^2*d*e^2-2/f^2/(c*f-d*e)^2/(f*
x+e)^(3/2)*b^3*c*e^2+4/3/f^3/(c*f-d*e)^2/(f*x+e)^(3/2)*b^3*d*e^3-2/(c*f-d*e)^3/(f*x+e)^(1/2)*a^3*d^2+6/(c*f-d*
e)^3/(f*x+e)^(1/2)*a^2*b*c*d-6/(c*f-d*e)^3/(f*x+e)^(1/2)*a*b^2*c^2+6/f/(c*f-d*e)^3/(f*x+e)^(1/2)*b^3*c^2*e-6/f
^2/(c*f-d*e)^3/(f*x+e)^(1/2)*b^3*c*d*e^2+2/f^3/(c*f-d*e)^3/(f*x+e)^(1/2)*b^3*d^2*e^3-2/(c*f-d*e)^3/((c*f-d*e)*
d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^3*d^3+6/(c*f-d*e)^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^
(1/2)*d/((c*f-d*e)*d)^(1/2))*a^2*c*b*d^2-6/(c*f-d*e)^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d
)^(1/2))*a*b^2*c^2*d+2/(c*f-d*e)^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^3*c^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.85155, size = 4080, normalized size = 17.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x, algorithm="fricas")

[Out]

[1/15*(15*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^6*x^3 + 3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*
c*d^2 - a^3*d^3)*e*f^5*x^2 + 3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e^2*f^4*x + (b^3*c^3 - 3*a*
b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e^3*f^3)*sqrt(d^2*e - c*d*f)*log((d*f*x + 2*d*e - c*f + 2*sqrt(d^2*e - c*
d*f)*sqrt(f*x + e))/(d*x + c)) - 2*(8*b^3*d^4*e^6 + 3*a^3*c^3*d*f^6 - 2*(17*b^3*c*d^3 - 3*a*b^2*d^4)*e^5*f + (
59*b^3*c^2*d^2 - 33*a*b^2*c*d^3 + 9*a^2*b*d^4)*e^4*f^2 - (33*b^3*c^3*d - 3*a*b^2*c^2*d^2 - 33*a^2*b*c*d^3 + 23
*a^3*d^4)*e^3*f^3 + 2*(12*a*b^2*c^3*d - 24*a^2*b*c^2*d^2 + 17*a^3*c*d^3)*e^2*f^4 + 2*(3*a^2*b*c^3*d - 7*a^3*c^
2*d^2)*e*f^5 + 15*(b^3*d^4*e^4*f^2 - 4*b^3*c*d^3*e^3*f^3 + 6*b^3*c^2*d^2*e^2*f^4 - (3*b^3*c^3*d + 3*a*b^2*c^2*
d^2 - 3*a^2*b*c*d^3 + a^3*d^4)*e*f^5 + (3*a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + a^3*c*d^3)*f^6)*x^2 + 5*(4*b^3*d^4*e
^5*f - (17*b^3*c*d^3 - 3*a*b^2*d^4)*e^4*f^2 + 4*(7*b^3*c^2*d^2 - 3*a*b^2*c*d^3)*e^3*f^3 - (15*b^3*c^3*d + 3*a*
b^2*c^2*d^2 - 21*a^2*b*c*d^3 + 7*a^3*d^4)*e^2*f^4 + 4*(3*a*b^2*c^3*d - 6*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*e*f^5 +
(3*a^2*b*c^3*d - a^3*c^2*d^2)*f^6)*x)*sqrt(f*x + e))/(d^5*e^7*f^3 - 4*c*d^4*e^6*f^4 + 6*c^2*d^3*e^5*f^5 - 4*c^
3*d^2*e^4*f^6 + c^4*d*e^3*f^7 + (d^5*e^4*f^6 - 4*c*d^4*e^3*f^7 + 6*c^2*d^3*e^2*f^8 - 4*c^3*d^2*e*f^9 + c^4*d*f
^10)*x^3 + 3*(d^5*e^5*f^5 - 4*c*d^4*e^4*f^6 + 6*c^2*d^3*e^3*f^7 - 4*c^3*d^2*e^2*f^8 + c^4*d*e*f^9)*x^2 + 3*(d^
5*e^6*f^4 - 4*c*d^4*e^5*f^5 + 6*c^2*d^3*e^4*f^6 - 4*c^3*d^2*e^3*f^7 + c^4*d*e^2*f^8)*x), -2/15*(15*((b^3*c^3 -
3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^6*x^3 + 3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e*f^
5*x^2 + 3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e^2*f^4*x + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c
*d^2 - a^3*d^3)*e^3*f^3)*sqrt(-d^2*e + c*d*f)*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e)) + (8*b^
3*d^4*e^6 + 3*a^3*c^3*d*f^6 - 2*(17*b^3*c*d^3 - 3*a*b^2*d^4)*e^5*f + (59*b^3*c^2*d^2 - 33*a*b^2*c*d^3 + 9*a^2*
b*d^4)*e^4*f^2 - (33*b^3*c^3*d - 3*a*b^2*c^2*d^2 - 33*a^2*b*c*d^3 + 23*a^3*d^4)*e^3*f^3 + 2*(12*a*b^2*c^3*d -
24*a^2*b*c^2*d^2 + 17*a^3*c*d^3)*e^2*f^4 + 2*(3*a^2*b*c^3*d - 7*a^3*c^2*d^2)*e*f^5 + 15*(b^3*d^4*e^4*f^2 - 4*b
^3*c*d^3*e^3*f^3 + 6*b^3*c^2*d^2*e^2*f^4 - (3*b^3*c^3*d + 3*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 + a^3*d^4)*e*f^5 + (
3*a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + a^3*c*d^3)*f^6)*x^2 + 5*(4*b^3*d^4*e^5*f - (17*b^3*c*d^3 - 3*a*b^2*d^4)*e^4*
f^2 + 4*(7*b^3*c^2*d^2 - 3*a*b^2*c*d^3)*e^3*f^3 - (15*b^3*c^3*d + 3*a*b^2*c^2*d^2 - 21*a^2*b*c*d^3 + 7*a^3*d^4
)*e^2*f^4 + 4*(3*a*b^2*c^3*d - 6*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*e*f^5 + (3*a^2*b*c^3*d - a^3*c^2*d^2)*f^6)*x)*sq
rt(f*x + e))/(d^5*e^7*f^3 - 4*c*d^4*e^6*f^4 + 6*c^2*d^3*e^5*f^5 - 4*c^3*d^2*e^4*f^6 + c^4*d*e^3*f^7 + (d^5*e^4
*f^6 - 4*c*d^4*e^3*f^7 + 6*c^2*d^3*e^2*f^8 - 4*c^3*d^2*e*f^9 + c^4*d*f^10)*x^3 + 3*(d^5*e^5*f^5 - 4*c*d^4*e^4*
f^6 + 6*c^2*d^3*e^3*f^7 - 4*c^3*d^2*e^2*f^8 + c^4*d*e*f^9)*x^2 + 3*(d^5*e^6*f^4 - 4*c*d^4*e^5*f^5 + 6*c^2*d^3*
e^4*f^6 - 4*c^3*d^2*e^3*f^7 + c^4*d*e^2*f^8)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3/(d*x+c)/(f*x+e)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 1.7622, size = 826, normalized size = 3.64 \begin{align*} \frac{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c^{3} f^{3} - 3 \, c^{2} d f^{2} e + 3 \, c d^{2} f e^{2} - d^{3} e^{3}\right )} \sqrt{c d f - d^{2} e}} - \frac{2 \,{\left (45 \,{\left (f x + e\right )}^{2} a b^{2} c^{2} f^{3} - 45 \,{\left (f x + e\right )}^{2} a^{2} b c d f^{3} + 15 \,{\left (f x + e\right )}^{2} a^{3} d^{2} f^{3} + 15 \,{\left (f x + e\right )} a^{2} b c^{2} f^{4} - 5 \,{\left (f x + e\right )} a^{3} c d f^{4} + 3 \, a^{3} c^{2} f^{5} - 45 \,{\left (f x + e\right )}^{2} b^{3} c^{2} f^{2} e - 30 \,{\left (f x + e\right )} a b^{2} c^{2} f^{3} e - 15 \,{\left (f x + e\right )} a^{2} b c d f^{3} e + 5 \,{\left (f x + e\right )} a^{3} d^{2} f^{3} e - 9 \, a^{2} b c^{2} f^{4} e - 6 \, a^{3} c d f^{4} e + 45 \,{\left (f x + e\right )}^{2} b^{3} c d f e^{2} + 15 \,{\left (f x + e\right )} b^{3} c^{2} f^{2} e^{2} + 45 \,{\left (f x + e\right )} a b^{2} c d f^{2} e^{2} + 9 \, a b^{2} c^{2} f^{3} e^{2} + 18 \, a^{2} b c d f^{3} e^{2} + 3 \, a^{3} d^{2} f^{3} e^{2} - 15 \,{\left (f x + e\right )}^{2} b^{3} d^{2} e^{3} - 25 \,{\left (f x + e\right )} b^{3} c d f e^{3} - 15 \,{\left (f x + e\right )} a b^{2} d^{2} f e^{3} - 3 \, b^{3} c^{2} f^{2} e^{3} - 18 \, a b^{2} c d f^{2} e^{3} - 9 \, a^{2} b d^{2} f^{2} e^{3} + 10 \,{\left (f x + e\right )} b^{3} d^{2} e^{4} + 6 \, b^{3} c d f e^{4} + 9 \, a b^{2} d^{2} f e^{4} - 3 \, b^{3} d^{2} e^{5}\right )}}{15 \,{\left (c^{3} f^{6} - 3 \, c^{2} d f^{5} e + 3 \, c d^{2} f^{4} e^{2} - d^{3} f^{3} e^{3}\right )}{\left (f x + e\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x, algorithm="giac")

[Out]

2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c^3*f^3 -
3*c^2*d*f^2*e + 3*c*d^2*f*e^2 - d^3*e^3)*sqrt(c*d*f - d^2*e)) - 2/15*(45*(f*x + e)^2*a*b^2*c^2*f^3 - 45*(f*x +
e)^2*a^2*b*c*d*f^3 + 15*(f*x + e)^2*a^3*d^2*f^3 + 15*(f*x + e)*a^2*b*c^2*f^4 - 5*(f*x + e)*a^3*c*d*f^4 + 3*a^
3*c^2*f^5 - 45*(f*x + e)^2*b^3*c^2*f^2*e - 30*(f*x + e)*a*b^2*c^2*f^3*e - 15*(f*x + e)*a^2*b*c*d*f^3*e + 5*(f*
x + e)*a^3*d^2*f^3*e - 9*a^2*b*c^2*f^4*e - 6*a^3*c*d*f^4*e + 45*(f*x + e)^2*b^3*c*d*f*e^2 + 15*(f*x + e)*b^3*c
^2*f^2*e^2 + 45*(f*x + e)*a*b^2*c*d*f^2*e^2 + 9*a*b^2*c^2*f^3*e^2 + 18*a^2*b*c*d*f^3*e^2 + 3*a^3*d^2*f^3*e^2 -
15*(f*x + e)^2*b^3*d^2*e^3 - 25*(f*x + e)*b^3*c*d*f*e^3 - 15*(f*x + e)*a*b^2*d^2*f*e^3 - 3*b^3*c^2*f^2*e^3 -
18*a*b^2*c*d*f^2*e^3 - 9*a^2*b*d^2*f^2*e^3 + 10*(f*x + e)*b^3*d^2*e^4 + 6*b^3*c*d*f*e^4 + 9*a*b^2*d^2*f*e^4 -
3*b^3*d^2*e^5)/((c^3*f^6 - 3*c^2*d*f^5*e + 3*c*d^2*f^4*e^2 - d^3*f^3*e^3)*(f*x + e)^(5/2))