3.1537 \(\int \frac{b+2 c x}{(d+e x) (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=236 \[ \frac{e \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac{\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac{e^2 (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac{e^2 (2 c d-b e) \log (d+e x)}{\left (a e^2-b d e+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.358326, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {822, 800, 634, 618, 206, 628} \[ \frac{e \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac{\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac{e^2 (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac{e^2 (2 c d-b e) \log (d+e x)}{\left (a e^2-b d e+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 822

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 634

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

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 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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.278336, size = 173, normalized size = 0.73 \[ \frac{\frac{2 e \left (2 c e (a e+b d)-b^2 e^2-2 c^2 d^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+\frac{2 \left (e (a e-b d)+c d^2\right ) (b e-c d+c e x)}{a+x (b+c x)}-e^2 (b e-2 c d) \log (a+x (b+c x))+2 e^2 (b e-2 c d) \log (d+e x)}{2 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.026, size = 668, normalized size = 2.8 \begin{align*}{\frac{{e}^{3}\ln \left ( ex+d \right ) b}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-2\,{\frac{{e}^{2}\ln \left ( ex+d \right ) cd}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+{\frac{axc{e}^{3}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{bxcd{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}+{\frac{e{d}^{2}x{c}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}+{\frac{ab{e}^{3}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{acd{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{{b}^{2}d{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}+2\,{\frac{b{d}^{2}ce}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{{c}^{2}{d}^{3}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) }}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ){e}^{3}b}{2\, \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) d{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+2\,{\frac{c{e}^{3}a}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{e}^{3}{b}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+2\,{\frac{{e}^{2}dcb}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-2\,{\frac{{c}^{2}e{d}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 13.5486, size = 3758, normalized size = 15.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.17916, size = 482, normalized size = 2.04 \begin{align*} \frac{{\left (2 \, c d e^{2} - b e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} - \frac{{\left (2 \, c d e^{3} - b e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}} - \frac{{\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + b^{2} e^{3} - 2 \, a c e^{3}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{c^{2} d^{3} - 2 \, b c d^{2} e + b^{2} d e^{2} + a c d e^{2} - a b e^{3} -{\left (c^{2} d^{2} e - b c d e^{2} + a c e^{3}\right )} x}{{\left (c d^{2} - b d e + a e^{2}\right )}^{2}{\left (c x^{2} + b x + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c*d*e^2 - b*e^3)*log(c*x^2 + b*x + a)/(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^
3 + a^2*e^4) - (2*c*d*e^3 - b*e^4)*log(abs(x*e + d))/(c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3 + 2*a*c*d^2*e^3
- 2*a*b*d*e^4 + a^2*e^5) - (2*c^2*d^2*e - 2*b*c*d*e^2 + b^2*e^3 - 2*a*c*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*
a*c))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*sqrt(-b^2 + 4*a*c)) - (c^
2*d^3 - 2*b*c*d^2*e + b^2*d*e^2 + a*c*d*e^2 - a*b*e^3 - (c^2*d^2*e - b*c*d*e^2 + a*c*e^3)*x)/((c*d^2 - b*d*e +
a*e^2)^2*(c*x^2 + b*x + a))