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

Optimal. Leaf size=85 \[ \frac{\left (2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c}}+\frac{(c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}+\frac{e x}{c} \]

[Out]

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

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Rubi [A]  time = 0.0781866, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {773, 634, 618, 206, 628} \[ \frac{\left (2 a c e+b^2 (-e)+b c d\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c}}+\frac{(c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}+\frac{e x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*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]

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

Mathematica [A]  time = 0.0855871, size = 86, normalized size = 1.01 \[ \frac{\frac{2 \left (-2 a c e+b^2 e-b c d\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+(c d-b e) \log (a+x (b+c x))+2 c e x}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.005, size = 161, normalized size = 1.9 \begin{align*}{\frac{ex}{c}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) be}{2\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ) d}{2\,c}}-2\,{\frac{ae}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{2}e}{{c}^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{bd}{c}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55765, size = 643, normalized size = 7.56 \begin{align*} \left [\frac{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e x +{\left (b c d -{\left (b^{2} - 2 \, a c\right )} 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 ({\left (b^{2} c - 4 \, a c^{2}\right )} d -{\left (b^{3} - 4 \, a b c\right )} e\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 x + 2 \,{\left (b c d -{\left (b^{2} - 2 \, a c\right )} 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 ({\left (b^{2} c - 4 \, a c^{2}\right )} d -{\left (b^{3} - 4 \, a b c\right )} e\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(x*(e*x+d)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.49794, size = 423, normalized size = 4.98 \begin{align*} \left (- \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{2 c^{2}}\right ) \log{\left (x + \frac{- a b e - 4 a c^{2} \left (- \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{2 c^{2}}\right ) + 2 a c d + b^{2} c \left (- \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{2 c^{2}}\right )}{2 a c e - b^{2} e + b c d} \right )} + \left (\frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{2 c^{2}}\right ) \log{\left (x + \frac{- a b e - 4 a c^{2} \left (\frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{2 c^{2}}\right ) + 2 a c d + b^{2} c \left (\frac{\sqrt{- 4 a c + b^{2}} \left (2 a c e - b^{2} e + b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac{b e - c d}{2 c^{2}}\right )}{2 a c e - b^{2} e + b c d} \right )} + \frac{e x}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16992, size = 119, normalized size = 1.4 \begin{align*} \frac{x e}{c} + \frac{{\left (c d - b e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} - \frac{{\left (b c d - b^{2} e + 2 \, a c e\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(x*(e*x+d)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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