3.2184 \(\int \frac{(d+e x)^2}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c}}+\frac{e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}+\frac{e^2 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.12332, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {701, 634, 618, 206, 628} \[ -\frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c}}+\frac{e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}+\frac{e^2 x}{c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + b*x + c*x^2),x]

[Out]

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

Rule 701

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

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

Mathematica [A]  time = 0.0674587, size = 101, normalized size = 1. \[ \frac{\frac{2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+e (2 c d-b e) \log (a+x (b+c x))+2 c e^2 x}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.153, size = 207, normalized size = 2.1 \begin{align*}{\frac{{e}^{2}x}{c}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) b{e}^{2}}{2\,{c}^{2}}}+{\frac{d\ln \left ( c{x}^{2}+bx+a \right ) e}{c}}-2\,{\frac{a{e}^{2}}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{{d}^{2}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{2}{e}^{2}}{{c}^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-2\,{\frac{bde}{c\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((e*x+d)^2/(c*x^2+b*x+a),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.15431, size = 713, normalized size = 7.06 \begin{align*} \left [\frac{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x -{\left (2 \, c^{2} d^{2} - 2 \, b c d e +{\left (b^{2} - 2 \, a c\right )} e^{2}\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 )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x - 2 \,{\left (2 \, c^{2} d^{2} - 2 \, b c d e +{\left (b^{2} - 2 \, a c\right )} e^{2}\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 )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.7585, size = 588, normalized size = 5.82 \begin{align*} \left (- \frac{e \left (b e - 2 c d\right )}{2 c^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log{\left (x + \frac{- a b e^{2} - 4 a c^{2} \left (- \frac{e \left (b e - 2 c d\right )}{2 c^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + 4 a c d e + b^{2} c \left (- \frac{e \left (b e - 2 c d\right )}{2 c^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) - b c d^{2}}{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}} \right )} + \left (- \frac{e \left (b e - 2 c d\right )}{2 c^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log{\left (x + \frac{- a b e^{2} - 4 a c^{2} \left (- \frac{e \left (b e - 2 c d\right )}{2 c^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + 4 a c d e + b^{2} c \left (- \frac{e \left (b e - 2 c d\right )}{2 c^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) - b c d^{2}}{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}} \right )} + \frac{e^{2} x}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10346, size = 135, normalized size = 1.34 \begin{align*} \frac{x e^{2}}{c} + \frac{{\left (2 \, c d e - b e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} + \frac{{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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