3.1534 \(\int \frac{(b+2 c x) (d+e x)^2}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=87 \[ -\frac{2 e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c}}-\frac{(d+e x)^2}{a+b x+c x^2}+\frac{e^2 \log \left (a+b x+c x^2\right )}{c} \]

[Out]

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

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Rubi [A]  time = 0.0575549, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {768, 634, 618, 206, 628} \[ -\frac{2 e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c}}-\frac{(d+e x)^2}{a+b x+c x^2}+\frac{e^2 \log \left (a+b x+c x^2\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 0.124553, size = 98, normalized size = 1.13 \[ \frac{-\frac{2 e (b e-2 c d) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+\frac{e^2 (a+b x)-c d (d+2 e x)}{a+x (b+c x)}+e^2 \log (a+x (b+c x))}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 141, normalized size = 1.6 \begin{align*}{\frac{1}{c{x}^{2}+bx+a} \left ({\frac{e \left ( be-2\,cd \right ) x}{c}}+{\frac{a{e}^{2}-c{d}^{2}}{c}} \right ) }+{\frac{{e}^{2}\ln \left ( c{x}^{2}+bx+a \right ) }{c}}+4\,{\frac{de}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-2\,{\frac{b{e}^{2}}{\sqrt{4\,ac-{b}^{2}}c}\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)^2,x)

[Out]

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

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.30802, size = 1202, normalized size = 13.82 \begin{align*} \left [-\frac{{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} -{\left (a b^{2} - 4 \, a^{2} c\right )} e^{2} +{\left (2 \, a c d e - a b e^{2} +{\left (2 \, c^{2} d e - b c e^{2}\right )} x^{2} +{\left (2 \, b c d e - 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 ) +{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e -{\left (b^{3} - 4 \, a b c\right )} e^{2}\right )} x -{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x^{2} +{\left (b^{3} - 4 \, a b c\right )} e^{2} x +{\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{a b^{2} c - 4 \, a^{2} c^{2} +{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} +{\left (b^{3} c - 4 \, a b c^{2}\right )} x}, -\frac{{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} -{\left (a b^{2} - 4 \, a^{2} c\right )} e^{2} + 2 \,{\left (2 \, a c d e - a b e^{2} +{\left (2 \, c^{2} d e - b c e^{2}\right )} x^{2} +{\left (2 \, b c d e - 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 ) +{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e -{\left (b^{3} - 4 \, a b c\right )} e^{2}\right )} x -{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x^{2} +{\left (b^{3} - 4 \, a b c\right )} e^{2} x +{\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{a b^{2} c - 4 \, a^{2} c^{2} +{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} +{\left (b^{3} c - 4 \, a b c^{2}\right )} x}\right ] \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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 4.18444, size = 340, normalized size = 3.91 \begin{align*} \left (\frac{e^{2}}{c} - \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{c \left (4 a c - b^{2}\right )}\right ) \log{\left (x + \frac{- 4 a c \left (\frac{e^{2}}{c} - \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{c \left (4 a c - b^{2}\right )}\right ) + 4 a e^{2} + b^{2} \left (\frac{e^{2}}{c} - \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{c \left (4 a c - b^{2}\right )}\right ) - 2 b d e}{2 b e^{2} - 4 c d e} \right )} + \left (\frac{e^{2}}{c} + \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{c \left (4 a c - b^{2}\right )}\right ) \log{\left (x + \frac{- 4 a c \left (\frac{e^{2}}{c} + \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{c \left (4 a c - b^{2}\right )}\right ) + 4 a e^{2} + b^{2} \left (\frac{e^{2}}{c} + \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{c \left (4 a c - b^{2}\right )}\right ) - 2 b d e}{2 b e^{2} - 4 c d e} \right )} + \frac{a e^{2} - c d^{2} + x \left (b e^{2} - 2 c d e\right )}{a c + b c x + c^{2} x^{2}} \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)**2,x)

[Out]

(e**2/c - e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(c*(4*a*c - b**2)))*log(x + (-4*a*c*(e**2/c - e*sqrt(-4*a*c + b*
*2)*(b*e - 2*c*d)/(c*(4*a*c - b**2))) + 4*a*e**2 + b**2*(e**2/c - e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(c*(4*a*
c - b**2))) - 2*b*d*e)/(2*b*e**2 - 4*c*d*e)) + (e**2/c + e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(c*(4*a*c - b**2)
))*log(x + (-4*a*c*(e**2/c + e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(c*(4*a*c - b**2))) + 4*a*e**2 + b**2*(e**2/c
+ e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(c*(4*a*c - b**2))) - 2*b*d*e)/(2*b*e**2 - 4*c*d*e)) + (a*e**2 - c*d**2
+ x*(b*e**2 - 2*c*d*e))/(a*c + b*c*x + c**2*x**2)

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Giac [A]  time = 1.19068, size = 154, normalized size = 1.77 \begin{align*} \frac{e^{2} \log \left (c x^{2} + b x + a\right )}{c} + \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 )}{\sqrt{-b^{2} + 4 \, a c} c} - \frac{\frac{{\left (2 \, c d e - b e^{2}\right )} x}{c} + \frac{c d^{2} - a e^{2}}{c}}{c x^{2} + b x + a} \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)^2,x, algorithm="giac")

[Out]

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