3.1526 \(\int \frac{(b+2 c x) (d+e x)^2}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=114 \[ \frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac{e \sqrt{b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2}+e x \left (4 d-\frac{b e}{c}\right )+e^2 x^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.133303, antiderivative size = 114, 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} \[ \frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac{e \sqrt{b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^2}+e x \left (4 d-\frac{b e}{c}\right )+e^2 x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

e*(4*d - (b*e)/c)*x + e^2*x^2 - (Sqrt[b^2 - 4*a*c]*e*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/c^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^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}{a+b x+c x^2} \, dx &=\int \left (e \left (4 d-\frac{b e}{c}\right )+2 e^2 x+\frac{b c d^2-4 a c d e+a b e^2+\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) x}{c \left (a+b x+c x^2\right )}\right ) \, dx\\ &=e \left (4 d-\frac{b e}{c}\right ) x+e^2 x^2+\frac{\int \frac{b c d^2-4 a c d e+a b e^2+\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}{c}\\ &=e \left (4 d-\frac{b e}{c}\right ) x+e^2 x^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 c^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 c^2}\\ &=e \left (4 d-\frac{b e}{c}\right ) x+e^2 x^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 c^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 )}{c^2}\\ &=e \left (4 d-\frac{b e}{c}\right ) x+e^2 x^2-\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 )}{c^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 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0991354, size = 111, normalized size = 0.97 \[ \frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log (a+x (b+c x))+2 e \sqrt{4 a c-b^2} (b e-2 c d) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )+2 c e x (-b e+4 c d+c e x)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.003, size = 264, normalized size = 2.3 \begin{align*}{e}^{2}{x}^{2}-{\frac{b{e}^{2}x}{c}}+4\,edx-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) a{e}^{2}}{c}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{2}{e}^{2}}{2\,{c}^{2}}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) bde}{c}}+\ln \left ( c{x}^{2}+bx+a \right ){d}^{2}+4\,{\frac{ab{e}^{2}}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-8\,{\frac{ade}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{3}{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{{b}^{2}de}{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((2*c*x+b)*(e*x+d)^2/(c*x^2+b*x+a),x)

[Out]

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

________________________________________________________________________________________

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

________________________________________________________________________________________

Fricas [A]  time = 1.49549, size = 632, normalized size = 5.54 \begin{align*} \left [\frac{2 \, c^{2} e^{2} x^{2} -{\left (2 \, c d e - b 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 ) + 2 \,{\left (4 \, c^{2} d e - b c e^{2}\right )} x +{\left (2 \, c^{2} d^{2} - 2 \, b c d e +{\left (b^{2} - 2 \, a c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}}, \frac{2 \, c^{2} e^{2} x^{2} - 2 \,{\left (2 \, c d e - b 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 ) + 2 \,{\left (4 \, c^{2} d e - b c e^{2}\right )} x +{\left (2 \, c^{2} d^{2} - 2 \, b c d e +{\left (b^{2} - 2 \, a c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}}\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),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 1.99729, size = 337, normalized size = 2.96 \begin{align*} e^{2} x^{2} + \left (- \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right ) \log{\left (x + \frac{a e^{2} - c d^{2} + c \left (- \frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right )}{b e^{2} - 2 c d e} \right )} + \left (\frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right ) \log{\left (x + \frac{a e^{2} - c d^{2} + c \left (\frac{e \sqrt{- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right )}{b e^{2} - 2 c d e} \right )} - \frac{x \left (b e^{2} - 4 c d e\right )}{c} \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]

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

________________________________________________________________________________________

Giac [A]  time = 1.15763, size = 194, normalized size = 1.7 \begin{align*} \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 x^{2} + b x + a\right )}{2 \, c^{2}} + \frac{{\left (2 \, b^{2} c d e - 8 \, a c^{2} d e - b^{3} e^{2} + 4 \, a b 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}} + \frac{c^{2} x^{2} e^{2} + 4 \, c^{2} d x e - b c x e^{2}}{c^{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),x, algorithm="giac")

[Out]

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