3.15 \(\int \frac{(a+b x^2) (c+d x^2)^2}{(e+f x^2)^4} \, dx\)

Optimal. Leaf size=240 \[ -\frac{x \left (a f \left (-15 c^2 f^2+4 c d e f+3 d^2 e^2\right )+b e \left (-3 c^2 f^2-4 c d e f+15 d^2 e^2\right )\right )}{48 e^3 f^3 \left (e+f x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a f \left (5 c^2 f^2+2 c d e f+d^2 e^2\right )+b e \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{16 e^{7/2} f^{7/2}}-\frac{x \left (c+d x^2\right ) (d e (a f+5 b e)-c f (5 a f+b e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac{x \left (c+d x^2\right )^2 (b e-a f)}{6 e f \left (e+f x^2\right )^3} \]

[Out]

-((b*e - a*f)*x*(c + d*x^2)^2)/(6*e*f*(e + f*x^2)^3) - ((d*e*(5*b*e + a*f) - c*f*(b*e + 5*a*f))*x*(c + d*x^2))
/(24*e^2*f^2*(e + f*x^2)^2) - ((a*f*(3*d^2*e^2 + 4*c*d*e*f - 15*c^2*f^2) + b*e*(15*d^2*e^2 - 4*c*d*e*f - 3*c^2
*f^2))*x)/(48*e^3*f^3*(e + f*x^2)) + ((b*e*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*c^
2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(7/2))

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Rubi [A]  time = 0.280437, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {526, 385, 205} \[ -\frac{x \left (a f \left (-15 c^2 f^2+4 c d e f+3 d^2 e^2\right )+b e \left (-3 c^2 f^2-4 c d e f+15 d^2 e^2\right )\right )}{48 e^3 f^3 \left (e+f x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a f \left (5 c^2 f^2+2 c d e f+d^2 e^2\right )+b e \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{16 e^{7/2} f^{7/2}}-\frac{x \left (c+d x^2\right ) (d e (a f+5 b e)-c f (5 a f+b e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac{x \left (c+d x^2\right )^2 (b e-a f)}{6 e f \left (e+f x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e - a*f)*x*(c + d*x^2)^2)/(6*e*f*(e + f*x^2)^3) - ((d*e*(5*b*e + a*f) - c*f*(b*e + 5*a*f))*x*(c + d*x^2))
/(24*e^2*f^2*(e + f*x^2)^2) - ((a*f*(3*d^2*e^2 + 4*c*d*e*f - 15*c^2*f^2) + b*e*(15*d^2*e^2 - 4*c*d*e*f - 3*c^2
*f^2))*x)/(48*e^3*f^3*(e + f*x^2)) + ((b*e*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*c^
2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(7/2))

Rule 526

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*b*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n
)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))*x
^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx &=-\frac{(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac{\int \frac{\left (c+d x^2\right ) \left (-c (b e+5 a f)-d (5 b e+a f) x^2\right )}{\left (e+f x^2\right )^3} \, dx}{6 e f}\\ &=-\frac{(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac{(d e (5 b e+a f)-c f (b e+5 a f)) x \left (c+d x^2\right )}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac{\int \frac{c (d e (5 b e+a f)+3 c f (b e+5 a f))+d (b e (15 d e+c f)+a f (3 d e+5 c f)) x^2}{\left (e+f x^2\right )^2} \, dx}{24 e^2 f^2}\\ &=-\frac{(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac{(d e (5 b e+a f)-c f (b e+5 a f)) x \left (c+d x^2\right )}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac{\left (a f \left (3 d^2 e^2+4 c d e f-15 c^2 f^2\right )+b e \left (15 d^2 e^2-4 c d e f-3 c^2 f^2\right )\right ) x}{48 e^3 f^3 \left (e+f x^2\right )}+\frac{\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \int \frac{1}{e+f x^2} \, dx}{16 e^3 f^3}\\ &=-\frac{(b e-a f) x \left (c+d x^2\right )^2}{6 e f \left (e+f x^2\right )^3}-\frac{(d e (5 b e+a f)-c f (b e+5 a f)) x \left (c+d x^2\right )}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac{\left (a f \left (3 d^2 e^2+4 c d e f-15 c^2 f^2\right )+b e \left (15 d^2 e^2-4 c d e f-3 c^2 f^2\right )\right ) x}{48 e^3 f^3 \left (e+f x^2\right )}+\frac{\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{16 e^{7/2} f^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.160394, size = 242, normalized size = 1.01 \[ \frac{x \left (a f \left (5 c^2 f^2+2 c d e f+d^2 e^2\right )+b e \left (c^2 f^2+2 c d e f-11 d^2 e^2\right )\right )}{16 e^3 f^3 \left (e+f x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a f \left (5 c^2 f^2+2 c d e f+d^2 e^2\right )+b e \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{16 e^{7/2} f^{7/2}}+\frac{x (d e-c f) (b e (13 d e-c f)-a f (5 c f+7 d e))}{24 e^2 f^3 \left (e+f x^2\right )^2}-\frac{x (b e-a f) (d e-c f)^2}{6 e f^3 \left (e+f x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e - a*f)*(d*e - c*f)^2*x)/(6*e*f^3*(e + f*x^2)^3) + ((d*e - c*f)*(b*e*(13*d*e - c*f) - a*f*(7*d*e + 5*c*f
))*x)/(24*e^2*f^3*(e + f*x^2)^2) + ((b*e*(-11*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*c^
2*f^2))*x)/(16*e^3*f^3*(e + f*x^2)) + ((b*e*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*c
^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(7/2))

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Maple [A]  time = 0.01, size = 360, normalized size = 1.5 \begin{align*}{\frac{1}{ \left ( f{x}^{2}+e \right ) ^{3}} \left ({\frac{ \left ( 5\,a{c}^{2}{f}^{3}+2\,acde{f}^{2}+a{d}^{2}{e}^{2}f+b{c}^{2}e{f}^{2}+2\,bcd{e}^{2}f-11\,b{d}^{2}{e}^{3} \right ){x}^{5}}{16\,{e}^{3}f}}+{\frac{ \left ( 5\,a{c}^{2}{f}^{3}+2\,acde{f}^{2}-a{d}^{2}{e}^{2}f+b{c}^{2}e{f}^{2}-2\,bcd{e}^{2}f-5\,b{d}^{2}{e}^{3} \right ){x}^{3}}{6\,{e}^{2}{f}^{2}}}+{\frac{ \left ( 11\,a{c}^{2}{f}^{3}-2\,acde{f}^{2}-a{d}^{2}{e}^{2}f-b{c}^{2}e{f}^{2}-2\,bcd{e}^{2}f-5\,b{d}^{2}{e}^{3} \right ) x}{16\,{f}^{3}e}} \right ) }+{\frac{5\,a{c}^{2}}{16\,{e}^{3}}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{acd}{8\,{e}^{2}f}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{a{d}^{2}}{16\,e{f}^{2}}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{b{c}^{2}}{16\,{e}^{2}f}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{bcd}{8\,e{f}^{2}}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{5\,b{d}^{2}}{16\,{f}^{3}}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x^2+c)^2/(f*x^2+e)^4,x)

[Out]

(1/16*(5*a*c^2*f^3+2*a*c*d*e*f^2+a*d^2*e^2*f+b*c^2*e*f^2+2*b*c*d*e^2*f-11*b*d^2*e^3)/e^3/f*x^5+1/6*(5*a*c^2*f^
3+2*a*c*d*e*f^2-a*d^2*e^2*f+b*c^2*e*f^2-2*b*c*d*e^2*f-5*b*d^2*e^3)/e^2/f^2*x^3+1/16*(11*a*c^2*f^3-2*a*c*d*e*f^
2-a*d^2*e^2*f-b*c^2*e*f^2-2*b*c*d*e^2*f-5*b*d^2*e^3)/f^3/e*x)/(f*x^2+e)^3+5/16/e^3/(e*f)^(1/2)*arctan(x*f/(e*f
)^(1/2))*a*c^2+1/8/e^2/f/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*a*c*d+1/16/e/f^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/
2))*a*d^2+1/16/e^2/f/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*c^2+1/8/e/f^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b
*c*d+5/16/f^3/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*d^2

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60621, size = 2117, normalized size = 8.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(6*(11*b*d^2*e^4*f^3 - 5*a*c^2*e*f^6 - (2*b*c*d + a*d^2)*e^3*f^4 - (b*c^2 + 2*a*c*d)*e^2*f^5)*x^5 + 16*
(5*b*d^2*e^5*f^2 - 5*a*c^2*e^2*f^5 + (2*b*c*d + a*d^2)*e^4*f^3 - (b*c^2 + 2*a*c*d)*e^3*f^4)*x^3 + 3*(5*b*d^2*e
^6 + 5*a*c^2*e^3*f^3 + (2*b*c*d + a*d^2)*e^5*f + (b*c^2 + 2*a*c*d)*e^4*f^2 + (5*b*d^2*e^3*f^3 + 5*a*c^2*f^6 +
(2*b*c*d + a*d^2)*e^2*f^4 + (b*c^2 + 2*a*c*d)*e*f^5)*x^6 + 3*(5*b*d^2*e^4*f^2 + 5*a*c^2*e*f^5 + (2*b*c*d + a*d
^2)*e^3*f^3 + (b*c^2 + 2*a*c*d)*e^2*f^4)*x^4 + 3*(5*b*d^2*e^5*f + 5*a*c^2*e^2*f^4 + (2*b*c*d + a*d^2)*e^4*f^2
+ (b*c^2 + 2*a*c*d)*e^3*f^3)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x - e)/(f*x^2 + e)) + 6*(5*b*d^2*e^6*f
- 11*a*c^2*e^3*f^4 + (2*b*c*d + a*d^2)*e^5*f^2 + (b*c^2 + 2*a*c*d)*e^4*f^3)*x)/(e^4*f^7*x^6 + 3*e^5*f^6*x^4 +
3*e^6*f^5*x^2 + e^7*f^4), -1/48*(3*(11*b*d^2*e^4*f^3 - 5*a*c^2*e*f^6 - (2*b*c*d + a*d^2)*e^3*f^4 - (b*c^2 + 2*
a*c*d)*e^2*f^5)*x^5 + 8*(5*b*d^2*e^5*f^2 - 5*a*c^2*e^2*f^5 + (2*b*c*d + a*d^2)*e^4*f^3 - (b*c^2 + 2*a*c*d)*e^3
*f^4)*x^3 - 3*(5*b*d^2*e^6 + 5*a*c^2*e^3*f^3 + (2*b*c*d + a*d^2)*e^5*f + (b*c^2 + 2*a*c*d)*e^4*f^2 + (5*b*d^2*
e^3*f^3 + 5*a*c^2*f^6 + (2*b*c*d + a*d^2)*e^2*f^4 + (b*c^2 + 2*a*c*d)*e*f^5)*x^6 + 3*(5*b*d^2*e^4*f^2 + 5*a*c^
2*e*f^5 + (2*b*c*d + a*d^2)*e^3*f^3 + (b*c^2 + 2*a*c*d)*e^2*f^4)*x^4 + 3*(5*b*d^2*e^5*f + 5*a*c^2*e^2*f^4 + (2
*b*c*d + a*d^2)*e^4*f^2 + (b*c^2 + 2*a*c*d)*e^3*f^3)*x^2)*sqrt(e*f)*arctan(sqrt(e*f)*x/e) + 3*(5*b*d^2*e^6*f -
11*a*c^2*e^3*f^4 + (2*b*c*d + a*d^2)*e^5*f^2 + (b*c^2 + 2*a*c*d)*e^4*f^3)*x)/(e^4*f^7*x^6 + 3*e^5*f^6*x^4 + 3
*e^6*f^5*x^2 + e^7*f^4)]

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Sympy [B]  time = 96.292, size = 486, normalized size = 2.02 \begin{align*} - \frac{\sqrt{- \frac{1}{e^{7} f^{7}}} \left (5 a c^{2} f^{3} + 2 a c d e f^{2} + a d^{2} e^{2} f + b c^{2} e f^{2} + 2 b c d e^{2} f + 5 b d^{2} e^{3}\right ) \log{\left (- e^{4} f^{3} \sqrt{- \frac{1}{e^{7} f^{7}}} + x \right )}}{32} + \frac{\sqrt{- \frac{1}{e^{7} f^{7}}} \left (5 a c^{2} f^{3} + 2 a c d e f^{2} + a d^{2} e^{2} f + b c^{2} e f^{2} + 2 b c d e^{2} f + 5 b d^{2} e^{3}\right ) \log{\left (e^{4} f^{3} \sqrt{- \frac{1}{e^{7} f^{7}}} + x \right )}}{32} + \frac{x^{5} \left (15 a c^{2} f^{5} + 6 a c d e f^{4} + 3 a d^{2} e^{2} f^{3} + 3 b c^{2} e f^{4} + 6 b c d e^{2} f^{3} - 33 b d^{2} e^{3} f^{2}\right ) + x^{3} \left (40 a c^{2} e f^{4} + 16 a c d e^{2} f^{3} - 8 a d^{2} e^{3} f^{2} + 8 b c^{2} e^{2} f^{3} - 16 b c d e^{3} f^{2} - 40 b d^{2} e^{4} f\right ) + x \left (33 a c^{2} e^{2} f^{3} - 6 a c d e^{3} f^{2} - 3 a d^{2} e^{4} f - 3 b c^{2} e^{3} f^{2} - 6 b c d e^{4} f - 15 b d^{2} e^{5}\right )}{48 e^{6} f^{3} + 144 e^{5} f^{4} x^{2} + 144 e^{4} f^{5} x^{4} + 48 e^{3} f^{6} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(d*x**2+c)**2/(f*x**2+e)**4,x)

[Out]

-sqrt(-1/(e**7*f**7))*(5*a*c**2*f**3 + 2*a*c*d*e*f**2 + a*d**2*e**2*f + b*c**2*e*f**2 + 2*b*c*d*e**2*f + 5*b*d
**2*e**3)*log(-e**4*f**3*sqrt(-1/(e**7*f**7)) + x)/32 + sqrt(-1/(e**7*f**7))*(5*a*c**2*f**3 + 2*a*c*d*e*f**2 +
a*d**2*e**2*f + b*c**2*e*f**2 + 2*b*c*d*e**2*f + 5*b*d**2*e**3)*log(e**4*f**3*sqrt(-1/(e**7*f**7)) + x)/32 +
(x**5*(15*a*c**2*f**5 + 6*a*c*d*e*f**4 + 3*a*d**2*e**2*f**3 + 3*b*c**2*e*f**4 + 6*b*c*d*e**2*f**3 - 33*b*d**2*
e**3*f**2) + x**3*(40*a*c**2*e*f**4 + 16*a*c*d*e**2*f**3 - 8*a*d**2*e**3*f**2 + 8*b*c**2*e**2*f**3 - 16*b*c*d*
e**3*f**2 - 40*b*d**2*e**4*f) + x*(33*a*c**2*e**2*f**3 - 6*a*c*d*e**3*f**2 - 3*a*d**2*e**4*f - 3*b*c**2*e**3*f
**2 - 6*b*c*d*e**4*f - 15*b*d**2*e**5))/(48*e**6*f**3 + 144*e**5*f**4*x**2 + 144*e**4*f**5*x**4 + 48*e**3*f**6
*x**6)

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Giac [A]  time = 1.15767, size = 420, normalized size = 1.75 \begin{align*} \frac{{\left (5 \, a c^{2} f^{3} + b c^{2} f^{2} e + 2 \, a c d f^{2} e + 2 \, b c d f e^{2} + a d^{2} f e^{2} + 5 \, b d^{2} e^{3}\right )} \arctan \left (\sqrt{f} x e^{\left (-\frac{1}{2}\right )}\right ) e^{\left (-\frac{7}{2}\right )}}{16 \, f^{\frac{7}{2}}} + \frac{{\left (15 \, a c^{2} f^{5} x^{5} + 3 \, b c^{2} f^{4} x^{5} e + 6 \, a c d f^{4} x^{5} e + 6 \, b c d f^{3} x^{5} e^{2} + 3 \, a d^{2} f^{3} x^{5} e^{2} - 33 \, b d^{2} f^{2} x^{5} e^{3} + 40 \, a c^{2} f^{4} x^{3} e + 8 \, b c^{2} f^{3} x^{3} e^{2} + 16 \, a c d f^{3} x^{3} e^{2} - 16 \, b c d f^{2} x^{3} e^{3} - 8 \, a d^{2} f^{2} x^{3} e^{3} - 40 \, b d^{2} f x^{3} e^{4} + 33 \, a c^{2} f^{3} x e^{2} - 3 \, b c^{2} f^{2} x e^{3} - 6 \, a c d f^{2} x e^{3} - 6 \, b c d f x e^{4} - 3 \, a d^{2} f x e^{4} - 15 \, b d^{2} x e^{5}\right )} e^{\left (-3\right )}}{48 \,{\left (f x^{2} + e\right )}^{3} f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*a*c^2*f^3 + b*c^2*f^2*e + 2*a*c*d*f^2*e + 2*b*c*d*f*e^2 + a*d^2*f*e^2 + 5*b*d^2*e^3)*arctan(sqrt(f)*x*
e^(-1/2))*e^(-7/2)/f^(7/2) + 1/48*(15*a*c^2*f^5*x^5 + 3*b*c^2*f^4*x^5*e + 6*a*c*d*f^4*x^5*e + 6*b*c*d*f^3*x^5*
e^2 + 3*a*d^2*f^3*x^5*e^2 - 33*b*d^2*f^2*x^5*e^3 + 40*a*c^2*f^4*x^3*e + 8*b*c^2*f^3*x^3*e^2 + 16*a*c*d*f^3*x^3
*e^2 - 16*b*c*d*f^2*x^3*e^3 - 8*a*d^2*f^2*x^3*e^3 - 40*b*d^2*f*x^3*e^4 + 33*a*c^2*f^3*x*e^2 - 3*b*c^2*f^2*x*e^
3 - 6*a*c*d*f^2*x*e^3 - 6*b*c*d*f*x*e^4 - 3*a*d^2*f*x*e^4 - 15*b*d^2*x*e^5)*e^(-3)/((f*x^2 + e)^3*f^3)