3.1535 \(\int \frac{(b+2 c x) (d+e x)}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=55 \[ -\frac{2 e \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}-\frac{d+e x}{a+b x+c x^2} \]

[Out]

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

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Rubi [A]  time = 0.029854, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {768, 618, 206} \[ -\frac{2 e \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}-\frac{d+e x}{a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0287885, size = 58, normalized size = 1.05 \[ \frac{2 e \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}-\frac{d+e x}{a+x (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 58, normalized size = 1.1 \begin{align*}{\frac{-ex-d}{c{x}^{2}+bx+a}}+2\,{\frac{e}{\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*x-d)/(c*x^2+b*x+a)+2*e/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))

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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 = 1.33568, size = 599, normalized size = 10.89 \begin{align*} \left [-\frac{{\left (b^{2} - 4 \, a c\right )} e x -{\left (c e x^{2} + b e x + a e\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 (b^{2} - 4 \, a c\right )} d}{a b^{2} - 4 \, a^{2} c +{\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} +{\left (b^{3} - 4 \, a b c\right )} x}, -\frac{{\left (b^{2} - 4 \, a c\right )} e x + 2 \,{\left (c e x^{2} + b e x + a e\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 (b^{2} - 4 \, a c\right )} d}{a b^{2} - 4 \, a^{2} c +{\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} +{\left (b^{3} - 4 \, a b c\right )} x}\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="fricas")

[Out]

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

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Sympy [B]  time = 1.38738, size = 156, normalized size = 2.84 \begin{align*} - e \sqrt{- \frac{1}{4 a c - b^{2}}} \log{\left (x + \frac{- 4 a c e \sqrt{- \frac{1}{4 a c - b^{2}}} + b^{2} e \sqrt{- \frac{1}{4 a c - b^{2}}} + b e}{2 c e} \right )} + e \sqrt{- \frac{1}{4 a c - b^{2}}} \log{\left (x + \frac{4 a c e \sqrt{- \frac{1}{4 a c - b^{2}}} - b^{2} e \sqrt{- \frac{1}{4 a c - b^{2}}} + b e}{2 c e} \right )} - \frac{d + e x}{a + b x + c x^{2}} \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]

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

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Giac [A]  time = 1.15497, size = 77, normalized size = 1.4 \begin{align*} \frac{2 \, \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right ) e}{\sqrt{-b^{2} + 4 \, a c}} - \frac{x e + d}{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)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

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