3.55 \(\int \frac{A+B x+C x^2}{(d+e x)^2 (a+c x^2)^2} \, dx\)

Optimal. Leaf size=374 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+a \left (a^2 C e^4-6 a c d e^2 (C d-B e)+c^2 d^3 (C d-2 B e)\right )\right )}{2 a^{3/2} \sqrt{c} \left (a e^2+c d^2\right )^3}-\frac{a \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )-x \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac{e \log \left (a+c x^2\right ) \left (a e^2 (2 C d-B e)-c d \left (2 C d^2-e (3 B d-4 A e)\right )\right )}{2 \left (a e^2+c d^2\right )^3}-\frac{e \left (A e^2-B d e+C d^2\right )}{(d+e x) \left (a e^2+c d^2\right )^2}-\frac{e \log (d+e x) \left (a e^2 (2 C d-B e)-c d \left (2 C d^2-e (3 B d-4 A e)\right )\right )}{\left (a e^2+c d^2\right )^3} \]

[Out]

-((e*(C*d^2 - B*d*e + A*e^2))/((c*d^2 + a*e^2)^2*(d + e*x))) - (a*(B*c*d^2 - 2*A*c*d*e + 2*a*C*d*e - a*B*e^2)
- (A*c*(c*d^2 - a*e^2) + a*(a*C*e^2 - c*d*(C*d - 2*B*e)))*x)/(2*a*(c*d^2 + a*e^2)^2*(a + c*x^2)) + ((A*c*(c^2*
d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4) + a*(a^2*C*e^4 + c^2*d^3*(C*d - 2*B*e) - 6*a*c*d*e^2*(C*d - B*e)))*ArcTan[(Sq
rt[c]*x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]*(c*d^2 + a*e^2)^3) - (e*(a*e^2*(2*C*d - B*e) - c*d*(2*C*d^2 - e*(3*B*d -
4*A*e)))*Log[d + e*x])/(c*d^2 + a*e^2)^3 + (e*(a*e^2*(2*C*d - B*e) - c*d*(2*C*d^2 - e*(3*B*d - 4*A*e)))*Log[a
+ c*x^2])/(2*(c*d^2 + a*e^2)^3)

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Rubi [A]  time = 0.950404, antiderivative size = 371, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {1647, 1629, 635, 205, 260} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+a \left (a^2 C e^4-6 a c d e^2 (C d-B e)+c^2 d^3 (C d-2 B e)\right )\right )}{2 a^{3/2} \sqrt{c} \left (a e^2+c d^2\right )^3}-\frac{a \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )-x \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}-\frac{e \log \left (a+c x^2\right ) \left (-a e^2 (2 C d-B e)-c d e (3 B d-4 A e)+2 c C d^3\right )}{2 \left (a e^2+c d^2\right )^3}-\frac{e \left (A e^2-B d e+C d^2\right )}{(d+e x) \left (a e^2+c d^2\right )^2}+\frac{e \log (d+e x) \left (-a e^2 (2 C d-B e)-c d e (3 B d-4 A e)+2 c C d^3\right )}{\left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/((d + e*x)^2*(a + c*x^2)^2),x]

[Out]

-((e*(C*d^2 - B*d*e + A*e^2))/((c*d^2 + a*e^2)^2*(d + e*x))) - (a*(B*c*d^2 - 2*A*c*d*e + 2*a*C*d*e - a*B*e^2)
- (A*c*(c*d^2 - a*e^2) + a*(a*C*e^2 - c*d*(C*d - 2*B*e)))*x)/(2*a*(c*d^2 + a*e^2)^2*(a + c*x^2)) + ((A*c*(c^2*
d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4) + a*(a^2*C*e^4 + c^2*d^3*(C*d - 2*B*e) - 6*a*c*d*e^2*(C*d - B*e)))*ArcTan[(Sq
rt[c]*x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[c]*(c*d^2 + a*e^2)^3) + (e*(2*c*C*d^3 - c*d*e*(3*B*d - 4*A*e) - a*e^2*(2*C*
d - B*e))*Log[d + e*x])/(c*d^2 + a*e^2)^3 - (e*(2*c*C*d^3 - c*d*e*(3*B*d - 4*A*e) - a*e^2*(2*C*d - B*e))*Log[a
+ c*x^2])/(2*(c*d^2 + a*e^2)^3)

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

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]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx &=-\frac{a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac{\int \frac{-\frac{c \left (A \left (c^2 d^4+5 a c d^2 e^2+2 a^2 e^4\right )-a d^2 \left (a C e^2-c d (C d-2 B e)\right )\right )}{\left (c d^2+a e^2\right )^2}-\frac{2 c e (A c d-a C d+a B e) x}{c d^2+a e^2}-\frac{c e^2 \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x^2}{\left (c d^2+a e^2\right )^2}}{(d+e x)^2 \left (a+c x^2\right )} \, dx}{2 a c}\\ &=-\frac{a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac{\int \left (-\frac{2 a c e^2 \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)^2}+\frac{2 a c e^2 \left (-2 c C d^3+c d e (3 B d-4 A e)+a e^2 (2 C d-B e)\right )}{\left (c d^2+a e^2\right )^3 (d+e x)}+\frac{c \left (-A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\right )+2 a c e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) x\right )}{\left (c d^2+a e^2\right )^3 \left (a+c x^2\right )}\right ) \, dx}{2 a c}\\ &=-\frac{e \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac{a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac{\int \frac{-A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\right )+2 a c e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) x}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=-\frac{e \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac{a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac{\left (c e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right )\right ) \int \frac{x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}+\frac{\left (A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\right )\right ) \int \frac{1}{a+c x^2} \, dx}{2 a \left (c d^2+a e^2\right )^3}\\ &=-\frac{e \left (C d^2-B d e+A e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac{a \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )-\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) x}{2 a \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{\left (A c \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+a \left (a^2 C e^4+c^2 d^3 (C d-2 B e)-6 a c d e^2 (C d-B e)\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{c} \left (c d^2+a e^2\right )^3}+\frac{e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}-\frac{e \left (2 c C d^3-c d e (3 B d-4 A e)-a e^2 (2 C d-B e)\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.467226, size = 320, normalized size = 0.86 \[ \frac{\frac{\left (a e^2+c d^2\right ) \left (a^2 e (B e-2 C d+C e x)-a c \left (A e (e x-2 d)+B d (d-2 e x)+C d^2 x\right )+A c^2 d^2 x\right )}{a \left (a+c x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+a \left (a^2 C e^4+6 a c d e^2 (B e-C d)+c^2 d^3 (C d-2 B e)\right )\right )}{a^{3/2} \sqrt{c}}-e \log \left (a+c x^2\right ) \left (a e^2 (B e-2 C d)+c d e (4 A e-3 B d)+2 c C d^3\right )-\frac{2 e \left (a e^2+c d^2\right ) \left (e (A e-B d)+C d^2\right )}{d+e x}+2 e \log (d+e x) \left (a e^2 (B e-2 C d)+c d e (4 A e-3 B d)+2 c C d^3\right )}{2 \left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/((d + e*x)^2*(a + c*x^2)^2),x]

[Out]

((-2*e*(c*d^2 + a*e^2)*(C*d^2 + e*(-(B*d) + A*e)))/(d + e*x) + ((c*d^2 + a*e^2)*(A*c^2*d^2*x + a^2*e*(-2*C*d +
B*e + C*e*x) - a*c*(C*d^2*x + B*d*(d - 2*e*x) + A*e*(-2*d + e*x))))/(a*(a + c*x^2)) + ((A*c*(c^2*d^4 + 6*a*c*
d^2*e^2 - 3*a^2*e^4) + a*(a^2*C*e^4 + c^2*d^3*(C*d - 2*B*e) + 6*a*c*d*e^2*(-(C*d) + B*e)))*ArcTan[(Sqrt[c]*x)/
Sqrt[a]])/(a^(3/2)*Sqrt[c]) + 2*e*(2*c*C*d^3 + c*d*e*(-3*B*d + 4*A*e) + a*e^2*(-2*C*d + B*e))*Log[d + e*x] - e
*(2*c*C*d^3 + c*d*e*(-3*B*d + 4*A*e) + a*e^2*(-2*C*d + B*e))*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)

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Maple [B]  time = 0.067, size = 1036, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)/(e*x+d)^2/(c*x^2+a)^2,x)

[Out]

e^4/(a*e^2+c*d^2)^3*ln(e*x+d)*B*a+e^2/(a*e^2+c*d^2)^2/(e*x+d)*B*d-e/(a*e^2+c*d^2)^2/(e*x+d)*C*d^2+1/2/(a*e^2+c
*d^2)^3/(c*x^2+a)*B*a^2*e^4-1/2/(a*e^2+c*d^2)^3/(c*x^2+a)*B*c^2*d^4-1/2/(a*e^2+c*d^2)^3*a*ln(c*x^2+a)*B*e^4+1/
(a*e^2+c*d^2)^3/(c*x^2+a)*B*c^2*d^3*e*x+1/(a*e^2+c*d^2)^3/(c*x^2+a)*A*a*c*d*e^3-1/(a*e^2+c*d^2)^3/(c*x^2+a)*C*
a*c*d^3*e-1/2/(a*e^2+c*d^2)^3/(c*x^2+a)*A*a*c*e^4*x-e^3/(a*e^2+c*d^2)^2/(e*x+d)*A+3/(a*e^2+c*d^2)^3/(a*c)^(1/2
)*arctan(x*c/(a*c)^(1/2))*A*c^2*d^2*e^2+1/2/(a*e^2+c*d^2)^3/a/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*c^3*d^4-3/
(a*e^2+c*d^2)^3*a/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*C*c*d^2*e^2+3/(a*e^2+c*d^2)^3*a/(a*c)^(1/2)*arctan(x*c/(
a*c)^(1/2))*B*d*e^3*c+1/(a*e^2+c*d^2)^3/(c*x^2+a)*B*a*c*d*e^3*x+1/2/(a*e^2+c*d^2)^3/(c*x^2+a)*x/a*A*c^3*d^4-1/
(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*c^2*d^3*e-3/2/(a*e^2+c*d^2)^3*a/(a*c)^(1/2)*arctan(x*c/(
a*c)^(1/2))*A*e^4*c+1/(a*e^2+c*d^2)^3/(c*x^2+a)*A*c^2*d^3*e-1/(a*e^2+c*d^2)^3/(c*x^2+a)*C*a^2*d*e^3+1/2/(a*e^2
+c*d^2)^3/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*C*c^2*d^4-1/(a*e^2+c*d^2)^3*c*ln(c*x^2+a)*C*d^3*e+1/2/(a*e^2+c*d
^2)^3/(c*x^2+a)*a^2*C*e^4*x-2*e^3/(a*e^2+c*d^2)^3*ln(e*x+d)*C*a*d+2*e/(a*e^2+c*d^2)^3*ln(e*x+d)*C*c*d^3+4*e^3/
(a*e^2+c*d^2)^3*ln(e*x+d)*A*c*d+1/2/(a*e^2+c*d^2)^3*a^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*C*e^4+3/2/(a*e^2+c
*d^2)^3*c*ln(c*x^2+a)*B*d^2*e^2+1/(a*e^2+c*d^2)^3*a*ln(c*x^2+a)*C*d*e^3-2/(a*e^2+c*d^2)^3*c*ln(c*x^2+a)*A*d*e^
3-1/2/(a*e^2+c*d^2)^3/(c*x^2+a)*C*c^2*d^4*x-3*e^2/(a*e^2+c*d^2)^3*ln(e*x+d)*B*c*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((C*x^2+B*x+A)/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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((C*x^2+B*x+A)/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

Timed out

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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((C*x**2+B*x+A)/(e*x+d)**2/(c*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20344, size = 821, normalized size = 2.2 \begin{align*} \frac{{\left (C a c^{2} d^{4} e^{2} + A c^{3} d^{4} e^{2} - 2 \, B a c^{2} d^{3} e^{3} - 6 \, C a^{2} c d^{2} e^{4} + 6 \, A a c^{2} d^{2} e^{4} + 6 \, B a^{2} c d e^{5} + C a^{3} e^{6} - 3 \, A a^{2} c e^{6}\right )} \arctan \left (\frac{{\left (c d - \frac{c d^{2}}{x e + d} - \frac{a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{a c}}\right ) e^{\left (-2\right )}}{2 \,{\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt{a c}} - \frac{{\left (2 \, C c d^{3} e - 3 \, B c d^{2} e^{2} - 2 \, C a d e^{3} + 4 \, A c d e^{3} + B a e^{4}\right )} \log \left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \,{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac{\frac{C d^{2} e^{5}}{x e + d} - \frac{B d e^{6}}{x e + d} + \frac{A e^{7}}{x e + d}}{c^{2} d^{4} e^{4} + 2 \, a c d^{2} e^{6} + a^{2} e^{8}} - \frac{\frac{C a c^{2} d^{3} e - A c^{3} d^{3} e - 3 \, B a c^{2} d^{2} e^{2} - 3 \, C a^{2} c d e^{3} + 3 \, A a c^{2} d e^{3} + B a^{2} c e^{4}}{c d^{2} + a e^{2}} - \frac{{\left (C a c^{2} d^{4} e^{2} - A c^{3} d^{4} e^{2} - 4 \, B a c^{2} d^{3} e^{3} - 6 \, C a^{2} c d^{2} e^{4} + 6 \, A a c^{2} d^{2} e^{4} + 4 \, B a^{2} c d e^{5} + C a^{3} e^{6} - A a^{2} c e^{6}\right )} e^{\left (-1\right )}}{{\left (c d^{2} + a e^{2}\right )}{\left (x e + d\right )}}}{2 \,{\left (c d^{2} + a e^{2}\right )}^{2} a{\left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)^2/(c*x^2+a)^2,x, algorithm="giac")

[Out]

1/2*(C*a*c^2*d^4*e^2 + A*c^3*d^4*e^2 - 2*B*a*c^2*d^3*e^3 - 6*C*a^2*c*d^2*e^4 + 6*A*a*c^2*d^2*e^4 + 6*B*a^2*c*d
*e^5 + C*a^3*e^6 - 3*A*a^2*c*e^6)*arctan((c*d - c*d^2/(x*e + d) - a*e^2/(x*e + d))*e^(-1)/sqrt(a*c))*e^(-2)/((
a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3*c*d^2*e^4 + a^4*e^6)*sqrt(a*c)) - 1/2*(2*C*c*d^3*e - 3*B*c*d^2*e^2 - 2*C
*a*d*e^3 + 4*A*c*d*e^3 + B*a*e^4)*log(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2)/(c^3*d^6 +
3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6) - (C*d^2*e^5/(x*e + d) - B*d*e^6/(x*e + d) + A*e^7/(x*e + d))/(c^
2*d^4*e^4 + 2*a*c*d^2*e^6 + a^2*e^8) - 1/2*((C*a*c^2*d^3*e - A*c^3*d^3*e - 3*B*a*c^2*d^2*e^2 - 3*C*a^2*c*d*e^3
+ 3*A*a*c^2*d*e^3 + B*a^2*c*e^4)/(c*d^2 + a*e^2) - (C*a*c^2*d^4*e^2 - A*c^3*d^4*e^2 - 4*B*a*c^2*d^3*e^3 - 6*C
*a^2*c*d^2*e^4 + 6*A*a*c^2*d^2*e^4 + 4*B*a^2*c*d*e^5 + C*a^3*e^6 - A*a^2*c*e^6)*e^(-1)/((c*d^2 + a*e^2)*(x*e +
d)))/((c*d^2 + a*e^2)^2*a*(c - 2*c*d/(x*e + d) + c*d^2/(x*e + d)^2 + a*e^2/(x*e + d)^2))