3.14.50 \(\int \frac {(b+2 c x) (d+e x)^2}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac {e (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {2 e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {768, 638, 618, 206} \begin {gather*} -\frac {e (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {2 e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(d + e*x)^2/(2*(a + b*x + c*x^2)^2) - (e*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)) +
(2*e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}+e \int \frac {d+e x}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}-\frac {e (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(e (2 c d-b e)) \int \frac {1}{a+b x+c x^2} \, dx}{b^2-4 a c}\\ &=-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}-\frac {e (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {(2 e (2 c d-b e)) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{b^2-4 a c}\\ &=-\frac {(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}-\frac {e (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {2 e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 143, normalized size = 1.28 \begin {gather*} \frac {1}{2} \left (\frac {e \left (4 c (c d x-2 a e)+b^2 e+2 b c (d-e x)\right )}{c \left (4 a c-b^2\right ) (a+x (b+c x))}-\frac {4 e (b e-2 c d) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+\frac {e^2 (a+b x)-c d (d+2 e x)}{c (a+x (b+c x))^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*(b^2*e + 4*c*(-2*a*e + c*d*x) + 2*b*c*(d - e*x)))/(c*(-b^2 + 4*a*c)*(a + x*(b + c*x))) + (e^2*(a + b*x) -
c*d*(d + 2*e*x))/(c*(a + x*(b + c*x))^2) - (4*e*(-2*c*d + b*e)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 +
4*a*c)^(3/2))/2

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 1004, normalized size = 8.96 \begin {gather*} \left [-\frac {2 \, {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{2}\right )} x^{3} + {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{2} + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} d e - 4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e^{2} + {\left (6 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e - {\left (b^{4} + 4 \, a b^{2} c - 32 \, a^{2} c^{2}\right )} e^{2}\right )} x^{2} - 2 \, {\left (2 \, a^{2} c d e - a^{2} b e^{2} + {\left (2 \, c^{3} d e - b c^{2} e^{2}\right )} x^{4} + 2 \, {\left (2 \, b c^{2} d e - b^{2} c e^{2}\right )} x^{3} + {\left (2 \, {\left (b^{2} c + 2 \, a c^{2}\right )} d e - {\left (b^{3} + 2 \, a b c\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, a b c d e - a b^{2} e^{2}\right )} x\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, {\left (2 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e - 3 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} e^{2}\right )} x}{2 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}}, -\frac {2 \, {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{2}\right )} x^{3} + {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{2} + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} d e - 4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e^{2} + {\left (6 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e - {\left (b^{4} + 4 \, a b^{2} c - 32 \, a^{2} c^{2}\right )} e^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c d e - a^{2} b e^{2} + {\left (2 \, c^{3} d e - b c^{2} e^{2}\right )} x^{4} + 2 \, {\left (2 \, b c^{2} d e - b^{2} c e^{2}\right )} x^{3} + {\left (2 \, {\left (b^{2} c + 2 \, a c^{2}\right )} d e - {\left (b^{3} + 2 \, a b c\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, a b c d e - a b^{2} e^{2}\right )} x\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (2 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e - 3 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} e^{2}\right )} x}{2 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(2*(b^2*c^2 - 4*a*c^3)*d*e - (b^3*c - 4*a*b*c^2)*e^2)*x^3 + (b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^2 + 2*(a
*b^3 - 4*a^2*b*c)*d*e - 4*(a^2*b^2 - 4*a^3*c)*e^2 + (6*(b^3*c - 4*a*b*c^2)*d*e - (b^4 + 4*a*b^2*c - 32*a^2*c^2
)*e^2)*x^2 - 2*(2*a^2*c*d*e - a^2*b*e^2 + (2*c^3*d*e - b*c^2*e^2)*x^4 + 2*(2*b*c^2*d*e - b^2*c*e^2)*x^3 + (2*(
b^2*c + 2*a*c^2)*d*e - (b^3 + 2*a*b*c)*e^2)*x^2 + 2*(2*a*b*c*d*e - a*b^2*e^2)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*
x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 2*(2*(b^4 - 5*a*b^2*c + 4*a^
2*c^2)*d*e - 3*(a*b^3 - 4*a^2*b*c)*e^2)*x)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a
^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a
^2*b^3*c + 16*a^3*b*c^2)*x), -1/2*(2*(2*(b^2*c^2 - 4*a*c^3)*d*e - (b^3*c - 4*a*b*c^2)*e^2)*x^3 + (b^4 - 8*a*b^
2*c + 16*a^2*c^2)*d^2 + 2*(a*b^3 - 4*a^2*b*c)*d*e - 4*(a^2*b^2 - 4*a^3*c)*e^2 + (6*(b^3*c - 4*a*b*c^2)*d*e - (
b^4 + 4*a*b^2*c - 32*a^2*c^2)*e^2)*x^2 - 4*(2*a^2*c*d*e - a^2*b*e^2 + (2*c^3*d*e - b*c^2*e^2)*x^4 + 2*(2*b*c^2
*d*e - b^2*c*e^2)*x^3 + (2*(b^2*c + 2*a*c^2)*d*e - (b^3 + 2*a*b*c)*e^2)*x^2 + 2*(2*a*b*c*d*e - a*b^2*e^2)*x)*s
qrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*d
*e - 3*(a*b^3 - 4*a^2*b*c)*e^2)*x)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*
x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c
+ 16*a^3*b*c^2)*x)]

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giac [A]  time = 0.17, size = 183, normalized size = 1.63 \begin {gather*} -\frac {2 \, {\left (2 \, c d e - b e^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {4 \, c^{2} d x^{3} e - 2 \, b c x^{3} e^{2} + 6 \, b c d x^{2} e - b^{2} x^{2} e^{2} - 8 \, a c x^{2} e^{2} + 4 \, b^{2} d x e - 4 \, a c d x e + b^{2} d^{2} - 4 \, a c d^{2} - 6 \, a b x e^{2} + 2 \, a b d e - 4 \, a^{2} e^{2}}{2 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (b^{2} - 4 \, a c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*c*d*e - b*e^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c)*sqrt(-b^2 + 4*a*c)) - 1/2*(4*c^2*d*
x^3*e - 2*b*c*x^3*e^2 + 6*b*c*d*x^2*e - b^2*x^2*e^2 - 8*a*c*x^2*e^2 + 4*b^2*d*x*e - 4*a*c*d*x*e + b^2*d^2 - 4*
a*c*d^2 - 6*a*b*x*e^2 + 2*a*b*d*e - 4*a^2*e^2)/((c*x^2 + b*x + a)^2*(b^2 - 4*a*c))

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maple [B]  time = 0.05, size = 229, normalized size = 2.04 \begin {gather*} -\frac {2 b \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {4 c d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {-\frac {\left (b e -2 c d \right ) c e \,x^{3}}{4 a c -b^{2}}-\frac {\left (8 a c e +e \,b^{2}-6 b c d \right ) e \,x^{2}}{2 \left (4 a c -b^{2}\right )}-\frac {\left (3 a b e +2 a c d -2 b^{2} d \right ) e x}{4 a c -b^{2}}-\frac {4 a^{2} e^{2}-2 a b d e +4 a c \,d^{2}-b^{2} d^{2}}{2 \left (4 a c -b^{2}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(b*e-2*c*d)*c*e/(4*a*c-b^2)*x^3-1/2*e*(8*a*c*e+b^2*e-6*b*c*d)/(4*a*c-b^2)*x^2-e*(3*a*b*e+2*a*c*d-2*b^2*d)/(4
*a*c-b^2)*x-1/2*(4*a^2*e^2-2*a*b*d*e+4*a*c*d^2-b^2*d^2)/(4*a*c-b^2))/(c*x^2+b*x+a)^2-2*e^2/(4*a*c-b^2)^(3/2)*a
rctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b+4*e/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c*d

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 2.00, size = 284, normalized size = 2.54 \begin {gather*} \frac {2\,e\,\mathrm {atan}\left (\frac {\left (4\,a\,c-b^2\right )\,\left (\frac {e\,\left (b^3-4\,a\,b\,c\right )\,\left (b\,e-2\,c\,d\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {2\,c\,e\,x\,\left (b\,e-2\,c\,d\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{b\,e^2-2\,c\,d\,e}\right )\,\left (b\,e-2\,c\,d\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {\frac {4\,a^2\,e^2-2\,a\,b\,d\,e+4\,c\,a\,d^2-b^2\,d^2}{2\,\left (4\,a\,c-b^2\right )}-\frac {e\,x^3\,\left (2\,c^2\,d-b\,c\,e\right )}{4\,a\,c-b^2}+\frac {e\,x\,\left (-2\,d\,b^2+3\,a\,e\,b+2\,a\,c\,d\right )}{4\,a\,c-b^2}+\frac {e\,x^2\,\left (e\,b^2-6\,c\,d\,b+8\,a\,c\,e\right )}{2\,\left (4\,a\,c-b^2\right )}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*e*atan(((4*a*c - b^2)*((e*(b^3 - 4*a*b*c)*(b*e - 2*c*d))/(4*a*c - b^2)^(5/2) - (2*c*e*x*(b*e - 2*c*d))/(4*a
*c - b^2)^(3/2)))/(b*e^2 - 2*c*d*e))*(b*e - 2*c*d))/(4*a*c - b^2)^(3/2) - ((4*a^2*e^2 - b^2*d^2 + 4*a*c*d^2 -
2*a*b*d*e)/(2*(4*a*c - b^2)) - (e*x^3*(2*c^2*d - b*c*e))/(4*a*c - b^2) + (e*x*(3*a*b*e - 2*b^2*d + 2*a*c*d))/(
4*a*c - b^2) + (e*x^2*(b^2*e + 8*a*c*e - 6*b*c*d))/(2*(4*a*c - b^2)))/(x^2*(2*a*c + b^2) + a^2 + c^2*x^4 + 2*a
*b*x + 2*b*c*x^3)

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sympy [B]  time = 12.58, size = 530, normalized size = 4.73 \begin {gather*} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) \log {\left (x + \frac {- 16 a^{2} c^{2} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + 8 a b^{2} c e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) - b^{4} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{2} e^{2} - 2 b c d e}{2 b c e^{2} - 4 c^{2} d e} \right )} - e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) \log {\left (x + \frac {16 a^{2} c^{2} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) - 8 a b^{2} c e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{4} e \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{2} e^{2} - 2 b c d e}{2 b c e^{2} - 4 c^{2} d e} \right )} + \frac {- 4 a^{2} e^{2} + 2 a b d e - 4 a c d^{2} + b^{2} d^{2} + x^{3} \left (- 2 b c e^{2} + 4 c^{2} d e\right ) + x^{2} \left (- 8 a c e^{2} - b^{2} e^{2} + 6 b c d e\right ) + x \left (- 6 a b e^{2} - 4 a c d e + 4 b^{2} d e\right )}{8 a^{3} c - 2 a^{2} b^{2} + x^{4} \left (8 a c^{3} - 2 b^{2} c^{2}\right ) + x^{3} \left (16 a b c^{2} - 4 b^{3} c\right ) + x^{2} \left (16 a^{2} c^{2} + 4 a b^{2} c - 2 b^{4}\right ) + x \left (16 a^{2} b c - 4 a b^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d)*log(x + (-16*a**2*c**2*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) +
8*a*b**2*c*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) - b**4*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**
2*e**2 - 2*b*c*d*e)/(2*b*c*e**2 - 4*c**2*d*e)) - e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d)*log(x + (16*a**2*c
**2*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) - 8*a*b**2*c*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**4*
e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**2*e**2 - 2*b*c*d*e)/(2*b*c*e**2 - 4*c**2*d*e)) + (-4*a**2*e**2
+ 2*a*b*d*e - 4*a*c*d**2 + b**2*d**2 + x**3*(-2*b*c*e**2 + 4*c**2*d*e) + x**2*(-8*a*c*e**2 - b**2*e**2 + 6*b*
c*d*e) + x*(-6*a*b*e**2 - 4*a*c*d*e + 4*b**2*d*e))/(8*a**3*c - 2*a**2*b**2 + x**4*(8*a*c**3 - 2*b**2*c**2) + x
**3*(16*a*b*c**2 - 4*b**3*c) + x**2*(16*a**2*c**2 + 4*a*b**2*c - 2*b**4) + x*(16*a**2*b*c - 4*a*b**3))

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