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

Optimal. Leaf size=146 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (c d (2 a B e+a C d+A c d)+a e^2 (A c-3 a C)\right )}{2 a^{3/2} c^{5/2}}-\frac {(d+e x)^2 (a B-x (A c-a C))}{2 a c \left (a+c x^2\right )}-\frac {e^2 x (A c-3 a C)}{2 a c^2}+\frac {e \log \left (a+c x^2\right ) (B e+2 C d)}{2 c^2} \]

[Out]

-1/2*(A*c-3*C*a)*e^2*x/a/c^2-1/2*(a*B-(A*c-C*a)*x)*(e*x+d)^2/a/c/(c*x^2+a)+1/2*(a*(A*c-3*C*a)*e^2+c*d*(A*c*d+2
*B*a*e+C*a*d))*arctan(x*c^(1/2)/a^(1/2))/a^(3/2)/c^(5/2)+1/2*e*(B*e+2*C*d)*ln(c*x^2+a)/c^2

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Rubi [A]  time = 0.25, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1645, 774, 635, 205, 260} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (c d (2 a B e+a C d+A c d)+a e^2 (A c-3 a C)\right )}{2 a^{3/2} c^{5/2}}-\frac {(d+e x)^2 (a B-x (A c-a C))}{2 a c \left (a+c x^2\right )}-\frac {e^2 x (A c-3 a C)}{2 a c^2}+\frac {e \log \left (a+c x^2\right ) (B e+2 C d)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*c - 3*a*C)*e^2*x)/(2*a*c^2) - ((a*B - (A*c - a*C)*x)*(d + e*x)^2)/(2*a*c*(a + c*x^2)) + ((a*(A*c - 3*a*C)
*e^2 + c*d*(A*c*d + a*C*d + 2*a*B*e))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*c^(5/2)) + (e*(2*C*d + B*e)*Log[
a + c*x^2])/(2*c^2)

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]

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 774

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

Rule 1645

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 175, normalized size = 1.20 \[ \frac {\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c \left (a e^2+c d^2\right )+a \left (c d (2 B e+C d)-3 a C e^2\right )\right )}{a^{3/2}}+\frac {\sqrt {c} \left (a^2 e (B e+2 C d+C e x)-a c \left (A e (2 d+e x)+B d (d+2 e x)+C d^2 x\right )+A c^2 d^2 x\right )}{a \left (a+c x^2\right )}+\sqrt {c} e \log \left (a+c x^2\right ) (B e+2 C d)+2 \sqrt {c} C e^2 x}{2 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*C*e^2*x + (Sqrt[c]*(A*c^2*d^2*x + a^2*e*(2*C*d + B*e + C*e*x) - a*c*(C*d^2*x + A*e*(2*d + e*x) + B*
d*(d + 2*e*x))))/(a*(a + c*x^2)) + ((A*c*(c*d^2 + a*e^2) + a*(-3*a*C*e^2 + c*d*(C*d + 2*B*e)))*ArcTan[(Sqrt[c]
*x)/Sqrt[a]])/a^(3/2) + Sqrt[c]*e*(2*C*d + B*e)*Log[a + c*x^2])/(2*c^(5/2))

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fricas [B]  time = 0.97, size = 631, normalized size = 4.32 \[ \left [\frac {4 \, C a^{2} c^{2} e^{2} x^{3} - 2 \, B a^{2} c^{2} d^{2} + 2 \, B a^{3} c e^{2} + 4 \, {\left (C a^{3} c - A a^{2} c^{2}\right )} d e - {\left (2 \, B a^{2} c d e + {\left (C a^{2} c + A a c^{2}\right )} d^{2} - {\left (3 \, C a^{3} - A a^{2} c\right )} e^{2} + {\left (2 \, B a c^{2} d e + {\left (C a c^{2} + A c^{3}\right )} d^{2} - {\left (3 \, C a^{2} c - A a c^{2}\right )} e^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (2 \, B a^{2} c^{2} d e + {\left (C a^{2} c^{2} - A a c^{3}\right )} d^{2} - {\left (3 \, C a^{3} c - A a^{2} c^{2}\right )} e^{2}\right )} x + 2 \, {\left (2 \, C a^{3} c d e + B a^{3} c e^{2} + {\left (2 \, C a^{2} c^{2} d e + B a^{2} c^{2} e^{2}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}, \frac {2 \, C a^{2} c^{2} e^{2} x^{3} - B a^{2} c^{2} d^{2} + B a^{3} c e^{2} + 2 \, {\left (C a^{3} c - A a^{2} c^{2}\right )} d e + {\left (2 \, B a^{2} c d e + {\left (C a^{2} c + A a c^{2}\right )} d^{2} - {\left (3 \, C a^{3} - A a^{2} c\right )} e^{2} + {\left (2 \, B a c^{2} d e + {\left (C a c^{2} + A c^{3}\right )} d^{2} - {\left (3 \, C a^{2} c - A a c^{2}\right )} e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (2 \, B a^{2} c^{2} d e + {\left (C a^{2} c^{2} - A a c^{3}\right )} d^{2} - {\left (3 \, C a^{3} c - A a^{2} c^{2}\right )} e^{2}\right )} x + {\left (2 \, C a^{3} c d e + B a^{3} c e^{2} + {\left (2 \, C a^{2} c^{2} d e + B a^{2} c^{2} e^{2}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*C*a^2*c^2*e^2*x^3 - 2*B*a^2*c^2*d^2 + 2*B*a^3*c*e^2 + 4*(C*a^3*c - A*a^2*c^2)*d*e - (2*B*a^2*c*d*e + (
C*a^2*c + A*a*c^2)*d^2 - (3*C*a^3 - A*a^2*c)*e^2 + (2*B*a*c^2*d*e + (C*a*c^2 + A*c^3)*d^2 - (3*C*a^2*c - A*a*c
^2)*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*(2*B*a^2*c^2*d*e + (C*a^2*c^2 - A*a
*c^3)*d^2 - (3*C*a^3*c - A*a^2*c^2)*e^2)*x + 2*(2*C*a^3*c*d*e + B*a^3*c*e^2 + (2*C*a^2*c^2*d*e + B*a^2*c^2*e^2
)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3), 1/2*(2*C*a^2*c^2*e^2*x^3 - B*a^2*c^2*d^2 + B*a^3*c*e^2 + 2*(C*
a^3*c - A*a^2*c^2)*d*e + (2*B*a^2*c*d*e + (C*a^2*c + A*a*c^2)*d^2 - (3*C*a^3 - A*a^2*c)*e^2 + (2*B*a*c^2*d*e +
(C*a*c^2 + A*c^3)*d^2 - (3*C*a^2*c - A*a*c^2)*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (2*B*a^2*c^2*d*e +
(C*a^2*c^2 - A*a*c^3)*d^2 - (3*C*a^3*c - A*a^2*c^2)*e^2)*x + (2*C*a^3*c*d*e + B*a^3*c*e^2 + (2*C*a^2*c^2*d*e +
B*a^2*c^2*e^2)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3)]

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giac [A]  time = 0.17, size = 184, normalized size = 1.26 \[ \frac {C x e^{2}}{c^{2}} + \frac {{\left (2 \, C d e + B e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (C a c d^{2} + A c^{2} d^{2} + 2 \, B a c d e - 3 \, C a^{2} e^{2} + A a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} - \frac {B a c d^{2} - 2 \, C a^{2} d e + 2 \, A a c d e - B a^{2} e^{2} + {\left (C a c d^{2} - A c^{2} d^{2} + 2 \, B a c d e - C a^{2} e^{2} + A a c e^{2}\right )} x}{2 \, {\left (c x^{2} + a\right )} a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 323, normalized size = 2.21 \[ \frac {A \,d^{2} x}{2 \left (c \,x^{2}+a \right ) a}+\frac {A \,d^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}-\frac {A \,e^{2} x}{2 \left (c \,x^{2}+a \right ) c}+\frac {A \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c}-\frac {B d e x}{\left (c \,x^{2}+a \right ) c}+\frac {B d e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {C a \,e^{2} x}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {3 C a \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c^{2}}-\frac {C \,d^{2} x}{2 \left (c \,x^{2}+a \right ) c}+\frac {C \,d^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c}-\frac {A d e}{\left (c \,x^{2}+a \right ) c}+\frac {B a \,e^{2}}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {B \,d^{2}}{2 \left (c \,x^{2}+a \right ) c}+\frac {B \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {C a d e}{\left (c \,x^{2}+a \right ) c^{2}}+\frac {C d e \ln \left (c \,x^{2}+a \right )}{c^{2}}+\frac {C \,e^{2} x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C*e^2/c^2*x-1/2/c/(c*x^2+a)*A*e^2*x+1/2/(c*x^2+a)/a*x*A*d^2-1/c/(c*x^2+a)*B*d*e*x+1/2/c^2/(c*x^2+a)*a*C*e^2*x-
1/2/c/(c*x^2+a)*C*d^2*x-1/c/(c*x^2+a)*A*d*e+1/2/c^2/(c*x^2+a)*B*a*e^2-1/2/c/(c*x^2+a)*B*d^2+1/c^2/(c*x^2+a)*C*
a*d*e+1/2/c^2*ln(c*x^2+a)*B*e^2+1/c^2*ln(c*x^2+a)*C*d*e+1/2/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*e^2+1/2/
a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*d^2+1/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*d*e-3/2/c^2*a/(a*c)^
(1/2)*arctan(1/(a*c)^(1/2)*c*x)*C*e^2+1/2/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*C*d^2

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maxima [A]  time = 0.96, size = 188, normalized size = 1.29 \[ \frac {C e^{2} x}{c^{2}} - \frac {B a c d^{2} - B a^{2} e^{2} - 2 \, {\left (C a^{2} - A a c\right )} d e + {\left (2 \, B a c d e + {\left (C a c - A c^{2}\right )} d^{2} - {\left (C a^{2} - A a c\right )} e^{2}\right )} x}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}} + \frac {{\left (2 \, C d e + B e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (2 \, B a c d e + {\left (C a c + A c^{2}\right )} d^{2} - {\left (3 \, C a^{2} - A a c\right )} e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.23, size = 195, normalized size = 1.34 \[ \frac {C\,e^2\,x}{c^2}-\frac {\frac {x\,\left (-C\,a^2\,e^2+C\,a\,c\,d^2+2\,B\,a\,c\,d\,e+A\,a\,c\,e^2-A\,c^2\,d^2\right )}{2\,a}-\frac {B\,a\,e^2}{2}+\frac {B\,c\,d^2}{2}+A\,c\,d\,e-C\,a\,d\,e}{c^3\,x^2+a\,c^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-3\,C\,a^2\,e^2+C\,a\,c\,d^2+2\,B\,a\,c\,d\,e+A\,a\,c\,e^2+A\,c^2\,d^2\right )}{2\,a^{3/2}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (16\,B\,a^3\,c^3\,e^2+32\,C\,d\,a^3\,c^3\,e\right )}{32\,a^3\,c^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*e^2*x)/c^2 - ((x*(A*a*c*e^2 - C*a^2*e^2 - A*c^2*d^2 + C*a*c*d^2 + 2*B*a*c*d*e))/(2*a) - (B*a*e^2)/2 + (B*c*
d^2)/2 + A*c*d*e - C*a*d*e)/(a*c^2 + c^3*x^2) + (atan((c^(1/2)*x)/a^(1/2))*(A*c^2*d^2 - 3*C*a^2*e^2 + A*a*c*e^
2 + C*a*c*d^2 + 2*B*a*c*d*e))/(2*a^(3/2)*c^(5/2)) + (log(a + c*x^2)*(16*B*a^3*c^3*e^2 + 32*C*a^3*c^3*d*e))/(32
*a^3*c^5)

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sympy [B]  time = 18.40, size = 593, normalized size = 4.06 \[ \frac {C e^{2} x}{c^{2}} + \left (\frac {e \left (B e + 2 C d\right )}{2 c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (- A a c e^{2} - A c^{2} d^{2} - 2 B a c d e + 3 C a^{2} e^{2} - C a c d^{2}\right )}{4 a^{3} c^{5}}\right ) \log {\left (x + \frac {2 B a^{2} e^{2} + 4 C a^{2} d e - 4 a^{2} c^{2} \left (\frac {e \left (B e + 2 C d\right )}{2 c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (- A a c e^{2} - A c^{2} d^{2} - 2 B a c d e + 3 C a^{2} e^{2} - C a c d^{2}\right )}{4 a^{3} c^{5}}\right )}{- A a c e^{2} - A c^{2} d^{2} - 2 B a c d e + 3 C a^{2} e^{2} - C a c d^{2}} \right )} + \left (\frac {e \left (B e + 2 C d\right )}{2 c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (- A a c e^{2} - A c^{2} d^{2} - 2 B a c d e + 3 C a^{2} e^{2} - C a c d^{2}\right )}{4 a^{3} c^{5}}\right ) \log {\left (x + \frac {2 B a^{2} e^{2} + 4 C a^{2} d e - 4 a^{2} c^{2} \left (\frac {e \left (B e + 2 C d\right )}{2 c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (- A a c e^{2} - A c^{2} d^{2} - 2 B a c d e + 3 C a^{2} e^{2} - C a c d^{2}\right )}{4 a^{3} c^{5}}\right )}{- A a c e^{2} - A c^{2} d^{2} - 2 B a c d e + 3 C a^{2} e^{2} - C a c d^{2}} \right )} + \frac {- 2 A a c d e + B a^{2} e^{2} - B a c d^{2} + 2 C a^{2} d e + x \left (- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e + C a^{2} e^{2} - C a c d^{2}\right )}{2 a^{2} c^{2} + 2 a c^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C*e**2*x/c**2 + (e*(B*e + 2*C*d)/(2*c**2) - sqrt(-a**3*c**5)*(-A*a*c*e**2 - A*c**2*d**2 - 2*B*a*c*d*e + 3*C*a*
*2*e**2 - C*a*c*d**2)/(4*a**3*c**5))*log(x + (2*B*a**2*e**2 + 4*C*a**2*d*e - 4*a**2*c**2*(e*(B*e + 2*C*d)/(2*c
**2) - sqrt(-a**3*c**5)*(-A*a*c*e**2 - A*c**2*d**2 - 2*B*a*c*d*e + 3*C*a**2*e**2 - C*a*c*d**2)/(4*a**3*c**5)))
/(-A*a*c*e**2 - A*c**2*d**2 - 2*B*a*c*d*e + 3*C*a**2*e**2 - C*a*c*d**2)) + (e*(B*e + 2*C*d)/(2*c**2) + sqrt(-a
**3*c**5)*(-A*a*c*e**2 - A*c**2*d**2 - 2*B*a*c*d*e + 3*C*a**2*e**2 - C*a*c*d**2)/(4*a**3*c**5))*log(x + (2*B*a
**2*e**2 + 4*C*a**2*d*e - 4*a**2*c**2*(e*(B*e + 2*C*d)/(2*c**2) + sqrt(-a**3*c**5)*(-A*a*c*e**2 - A*c**2*d**2
- 2*B*a*c*d*e + 3*C*a**2*e**2 - C*a*c*d**2)/(4*a**3*c**5)))/(-A*a*c*e**2 - A*c**2*d**2 - 2*B*a*c*d*e + 3*C*a**
2*e**2 - C*a*c*d**2)) + (-2*A*a*c*d*e + B*a**2*e**2 - B*a*c*d**2 + 2*C*a**2*d*e + x*(-A*a*c*e**2 + A*c**2*d**2
- 2*B*a*c*d*e + C*a**2*e**2 - C*a*c*d**2))/(2*a**2*c**2 + 2*a*c**3*x**2)

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