3.1530 \(\int \frac{b+2 c x}{(d+e x)^2 (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=210 \[ \frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac{\log (d+e x) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{e \sqrt{b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{2 c d-b e}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]

[Out]

(2*c*d - b*e)/((c*d^2 - b*d*e + a*e^2)*(d + e*x)) + (Sqrt[b^2 - 4*a*c]*e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/Sqr
t[b^2 - 4*a*c]])/(c*d^2 - b*d*e + a*e^2)^2 - ((2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*Log[d + e*x])/(c*d^2 -
b*d*e + a*e^2)^2 + ((2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*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 = 0.299542, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {800, 634, 618, 206, 628} \[ \frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac{\log (d+e x) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{e \sqrt{b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{2 c d-b e}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)^2 \left (a+b x+c x^2\right )} \, dx &=\int \left (\frac{e (-2 c d+b e)}{\left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac{e \left (-2 c^2 d^2-b^2 e^2+2 c e (b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac{-2 b^2 c d e+4 a c^2 d e+b^3 e^2+b c \left (c d^2-3 a e^2\right )+c \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) x}{\left (c d^2-b d e+a e^2\right )^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}+\frac{\int \frac{-2 b^2 c d e+4 a c^2 d e+b^3 e^2+b c \left (c d^2-3 a e^2\right )+c \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) x}{a+b x+c x^2} \, dx}{\left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}-\frac{\left (\left (b^2-4 a c\right ) e (2 c d-b e)\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 (2 c^2 d^2+b^2 e^2-2 c e (b d+a 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{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}+\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^2}+\frac{\left (\left (b^2-4 a c\right ) e (2 c d-b e)\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{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac{\sqrt{b^2-4 a c} e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (c d^2-b d e+a e^2\right )^2}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}+\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \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 = 0.323495, size = 177, normalized size = 0.84 \[ \frac{\log (d+e x) \left (4 c e (a e+b d)-2 b^2 e^2-4 c^2 d^2\right )+\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log (a+x (b+c x))-2 e \sqrt{4 a c-b^2} (b e-2 c d) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )+\frac{2 (2 c d-b e) \left (e (a e-b d)+c d^2\right )}{d+e x}}{2 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 560, normalized size = 2.7 \begin{align*} 2\,{\frac{\ln \left ( ex+d \right ) ac{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( ex+d \right ){b}^{2}{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+2\,{\frac{\ln \left ( ex+d \right ) bcde}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( ex+d \right ){c}^{2}{d}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-{\frac{be}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) \left ( ex+d \right ) }}+2\,{\frac{cd}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) \left ( ex+d \right ) }}-{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) a{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{2}{e}^{2}}{2\, \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) bde}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+{\frac{{c}^{2}\ln \left ( c{x}^{2}+bx+a \right ){d}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-4\,{\frac{c{e}^{2}ab}{ \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 ) }+8\,{\frac{a{c}^{2}de}{ \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{{b}^{3}{e}^{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{{b}^{2}cde}{ \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)^2/(c*x^2+b*x+a),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 11.6136, size = 1634, normalized size = 7.78 \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)^2/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

[1/2*(4*c^2*d^3 - 6*b*c*d^2*e - 2*a*b*e^3 + 2*(b^2 + 2*a*c)*d*e^2 - (2*c*d^2*e - b*d*e^2 + (2*c*d*e^2 - b*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*c^2*d^3 - 2*b*c*d^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^2*d^2*e - 2*b*c*d*e^2 + (b^2 - 2*a*c)*e^3)*x)*log(c
*x^2 + b*x + a) - 2*(2*c^2*d^3 - 2*b*c*d^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^2*d^2*e - 2*b*c*d*e^2 + (b^2 - 2*a*c
)*e^3)*x)*log(e*x + d))/(c^2*d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4 + (b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*
e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4 + a^2*e^5 + (b^2 + 2*a*c)*d^2*e^3)*x), 1/2*(4*c^2*d^3 - 6*b*c*d^2*e - 2*a*b*e^
3 + 2*(b^2 + 2*a*c)*d*e^2 + 2*(2*c*d^2*e - b*d*e^2 + (2*c*d*e^2 - b*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*c^2*d^3 - 2*b*c*d^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^2*d^2*e - 2*b*c
*d*e^2 + (b^2 - 2*a*c)*e^3)*x)*log(c*x^2 + b*x + a) - 2*(2*c^2*d^3 - 2*b*c*d^2*e + (b^2 - 2*a*c)*d*e^2 + (2*c^
2*d^2*e - 2*b*c*d*e^2 + (b^2 - 2*a*c)*e^3)*x)*log(e*x + d))/(c^2*d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4
+ (b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4 + a^2*e^5 + (b^2 + 2*a*c)*d^2*e^3)*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)**2/(c*x**2+b*x+a),x)

[Out]

Timed out

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Giac [A]  time = 1.15526, size = 494, normalized size = 2.35 \begin{align*} \frac{{\left (2 \, b^{2} c d e^{3} - 8 \, a c^{2} d e^{3} - b^{3} e^{4} + 4 \, a b c e^{4}\right )} \arctan \left (-\frac{{\left (2 \, c d - \frac{2 \, c d^{2}}{x e + d} - b e + \frac{2 \, b d e}{x e + d} - \frac{2 \, a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{-b^{2} + 4 \, a c}}\right ) e^{\left (-2\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{{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} \log \left (-c + \frac{2 \, c d}{x e + d} - \frac{c d^{2}}{{\left (x e + d\right )}^{2}} - \frac{b e}{x e + d} + \frac{b d e}{{\left (x e + d\right )}^{2}} - \frac{a e^{2}}{{\left (x e + d\right )}^{2}}\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{\frac{2 \, c d e^{2}}{x e + d} - \frac{b e^{3}}{x e + d}}{c d^{2} e^{2} - b d e^{3} + a e^{4}} \end{align*}

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),x, algorithm="giac")

[Out]

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