3.2365 \(\int \frac{(A+B x) (d+e x)^2}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=205 \[ \frac{\log \left (a+b x+c x^2\right ) \left (B \left (-c e (a e+2 b d)+b^2 e^2+c^2 d^2\right )+A c e (2 c d-b e)\right )}{2 c^3}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \left (b c \left (-3 a B e^2+2 A c d e+B c d^2\right )-2 c^2 \left (-a A e^2-2 a B d e+A c d^2\right )-b^2 c e (A e+2 B d)+b^3 B e^2\right )}{c^3 \sqrt{b^2-4 a c}}+\frac{e x (A c e-b B e+2 B c d)}{c^2}+\frac{B e^2 x^2}{2 c} \]

[Out]

(e*(2*B*c*d - b*B*e + A*c*e)*x)/c^2 + (B*e^2*x^2)/(2*c) + ((b^3*B*e^2 - b^2*c*e*(2*B*d + A*e) - 2*c^2*(A*c*d^2
- 2*a*B*d*e - a*A*e^2) + b*c*(B*c*d^2 + 2*A*c*d*e - 3*a*B*e^2))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*
Sqrt[b^2 - 4*a*c]) + ((A*c*e*(2*c*d - b*e) + B*(c^2*d^2 + b^2*e^2 - c*e*(2*b*d + a*e)))*Log[a + b*x + c*x^2])/
(2*c^3)

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Rubi [A]  time = 0.370986, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {800, 634, 618, 206, 628} \[ \frac{\log \left (a+b x+c x^2\right ) \left (B \left (-c e (a e+2 b d)+b^2 e^2+c^2 d^2\right )+A c e (2 c d-b e)\right )}{2 c^3}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \left (b c \left (-3 a B e^2+2 A c d e+B c d^2\right )-2 c^2 \left (-a A e^2-2 a B d e+A c d^2\right )-b^2 c e (A e+2 B d)+b^3 B e^2\right )}{c^3 \sqrt{b^2-4 a c}}+\frac{e x (A c e-b B e+2 B c d)}{c^2}+\frac{B e^2 x^2}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(2*B*c*d - b*B*e + A*c*e)*x)/c^2 + (B*e^2*x^2)/(2*c) + ((b^3*B*e^2 - b^2*c*e*(2*B*d + A*e) - 2*c^2*(A*c*d^2
- 2*a*B*d*e - a*A*e^2) + b*c*(B*c*d^2 + 2*A*c*d*e - 3*a*B*e^2))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*
Sqrt[b^2 - 4*a*c]) + ((A*c*e*(2*c*d - b*e) + B*(c^2*d^2 + b^2*e^2 - c*e*(2*b*d + a*e)))*Log[a + b*x + c*x^2])/
(2*c^3)

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{(A+B x) (d+e x)^2}{a+b x+c x^2} \, dx &=\int \left (\frac{e (2 B c d-b B e+A c e)}{c^2}+\frac{B e^2 x}{c}+\frac{-a B e (2 c d-b e)+A c \left (c d^2-a e^2\right )+\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) x}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac{e (2 B c d-b B e+A c e) x}{c^2}+\frac{B e^2 x^2}{2 c}+\frac{\int \frac{-a B e (2 c d-b e)+A c \left (c d^2-a e^2\right )+\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) x}{a+b x+c x^2} \, dx}{c^2}\\ &=\frac{e (2 B c d-b B e+A c e) x}{c^2}+\frac{B e^2 x^2}{2 c}-\frac{\left (b^3 B e^2-b^2 c e (2 B d+A e)-2 c^2 \left (A c d^2-2 a B d e-a A e^2\right )+b c \left (B c d^2+2 A c d e-3 a B e^2\right )\right ) \int \frac{1}{a+b x+c x^2} \, dx}{2 c^3}+\frac{\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) \int \frac{b+2 c x}{a+b x+c x^2} \, dx}{2 c^3}\\ &=\frac{e (2 B c d-b B e+A c e) x}{c^2}+\frac{B e^2 x^2}{2 c}+\frac{\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^3}+\frac{\left (b^3 B e^2-b^2 c e (2 B d+A e)-2 c^2 \left (A c d^2-2 a B d e-a A e^2\right )+b c \left (B c d^2+2 A c d e-3 a B 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^3}\\ &=\frac{e (2 B c d-b B e+A c e) x}{c^2}+\frac{B e^2 x^2}{2 c}+\frac{\left (b^3 B e^2-b^2 c e (2 B d+A e)-2 c^2 \left (A c d^2-2 a B d e-a A e^2\right )+b c \left (B c d^2+2 A c d e-3 a B e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^3 \sqrt{b^2-4 a c}}+\frac{\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.184288, size = 203, normalized size = 0.99 \[ \frac{\log (a+x (b+c x)) \left (B \left (-c e (a e+2 b d)+b^2 e^2+c^2 d^2\right )+A c e (2 c d-b e)\right )-\frac{2 \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right ) \left (b c \left (-3 a B e^2+2 A c d e+B c d^2\right )+2 c^2 \left (a A e^2+2 a B d e-A c d^2\right )-b^2 c e (A e+2 B d)+b^3 B e^2\right )}{\sqrt{4 a c-b^2}}+2 c e x (A c e-b B e+2 B c d)+B c^2 e^2 x^2}{2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*e*(2*B*c*d - b*B*e + A*c*e)*x + B*c^2*e^2*x^2 - (2*(b^3*B*e^2 - b^2*c*e*(2*B*d + A*e) + 2*c^2*(-(A*c*d^2)
+ 2*a*B*d*e + a*A*e^2) + b*c*(B*c*d^2 + 2*A*c*d*e - 3*a*B*e^2))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[
-b^2 + 4*a*c] + (A*c*e*(2*c*d - b*e) + B*(c^2*d^2 + b^2*e^2 - c*e*(2*b*d + a*e)))*Log[a + x*(b + c*x)])/(2*c^3
)

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Maple [B]  time = 0.004, size = 543, normalized size = 2.7 \begin{align*}{\frac{B{e}^{2}{x}^{2}}{2\,c}}+{\frac{A{e}^{2}x}{c}}-{\frac{B{e}^{2}bx}{{c}^{2}}}+2\,{\frac{Bdex}{c}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) Ab{e}^{2}}{2\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ) Ade}{c}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) aB{e}^{2}}{2\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ) B{e}^{2}{b}^{2}}{2\,{c}^{3}}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) Bbde}{{c}^{2}}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ) B{d}^{2}}{2\,c}}-2\,{\frac{aA{e}^{2}}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{A{d}^{2}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+3\,{\frac{B{e}^{2}ab}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-4\,{\frac{aBde}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{A{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{Abde}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{3}B{e}^{2}}{{c}^{3}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+2\,{\frac{B{b}^{2}de}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{bB{d}^{2}}{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((B*x+A)*(e*x+d)^2/(c*x^2+b*x+a),x)

[Out]

1/2*B*e^2*x^2/c+e^2/c*A*x-e^2/c^2*b*B*x+2*e/c*B*d*x-1/2/c^2*ln(c*x^2+b*x+a)*A*b*e^2+1/c*ln(c*x^2+b*x+a)*A*d*e-
1/2/c^2*ln(c*x^2+b*x+a)*a*B*e^2+1/2/c^3*ln(c*x^2+b*x+a)*B*e^2*b^2-1/c^2*ln(c*x^2+b*x+a)*B*b*d*e+1/2/c*ln(c*x^2
+b*x+a)*B*d^2-2/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*A*e^2+2/(4*a*c-b^2)^(1/2)*arctan((2*
c*x+b)/(4*a*c-b^2)^(1/2))*A*d^2+3/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*e^2*a*b-4/c/(4*a
*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*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))*A*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))*A*b*d*e-1/c^3/(4*a*c-b^2)^(1/2)*ar
ctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^3*B*e^2+2/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*b^2*
d*e-1/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*b*d^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((B*x+A)*(e*x+d)^2/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68313, size = 1420, normalized size = 6.93 \begin{align*} \left [\frac{{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} e^{2} x^{2} +{\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 2 \,{\left (B b^{2} c -{\left (2 \, B a + A b\right )} c^{2}\right )} d e +{\left (B b^{3} + 2 \, A a c^{2} -{\left (3 \, B a b + A b^{2}\right )} 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 ) + 2 \,{\left (2 \,{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d e -{\left (B b^{3} c + 4 \, A a c^{3} -{\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x +{\left ({\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d^{2} - 2 \,{\left (B b^{3} c + 4 \, A a c^{3} -{\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} d e +{\left (B b^{4} + 4 \,{\left (B a^{2} + A a b\right )} c^{2} -{\left (5 \, B a b^{2} + A b^{3}\right )} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac{{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} e^{2} x^{2} + 2 \,{\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 2 \,{\left (B b^{2} c -{\left (2 \, B a + A b\right )} c^{2}\right )} d e +{\left (B b^{3} + 2 \, A a c^{2} -{\left (3 \, B a b + A b^{2}\right )} 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 ) + 2 \,{\left (2 \,{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d e -{\left (B b^{3} c + 4 \, A a c^{3} -{\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x +{\left ({\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d^{2} - 2 \,{\left (B b^{3} c + 4 \, A a c^{3} -{\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} d e +{\left (B b^{4} + 4 \,{\left (B a^{2} + A a b\right )} c^{2} -{\left (5 \, B a b^{2} + A b^{3}\right )} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((B*b^2*c^2 - 4*B*a*c^3)*e^2*x^2 + ((B*b*c^2 - 2*A*c^3)*d^2 - 2*(B*b^2*c - (2*B*a + A*b)*c^2)*d*e + (B*b^
3 + 2*A*a*c^2 - (3*B*a*b + A*b^2)*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*(2*(B*b^2*c^2 - 4*B*a*c^3)*d*e - (B*b^3*c + 4*A*a*c^3 - (4*B*a*b
+ A*b^2)*c^2)*e^2)*x + ((B*b^2*c^2 - 4*B*a*c^3)*d^2 - 2*(B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*d*e + (B
*b^4 + 4*(B*a^2 + A*a*b)*c^2 - (5*B*a*b^2 + A*b^3)*c)*e^2)*log(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4), 1/2*((B*
b^2*c^2 - 4*B*a*c^3)*e^2*x^2 + 2*((B*b*c^2 - 2*A*c^3)*d^2 - 2*(B*b^2*c - (2*B*a + A*b)*c^2)*d*e + (B*b^3 + 2*A
*a*c^2 - (3*B*a*b + A*b^2)*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*(2*(B*b^2*c^2 - 4*B*a*c^3)*d*e - (B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*e^2)*x + ((B*b^2*c^2 - 4*B*a*
c^3)*d^2 - 2*(B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*d*e + (B*b^4 + 4*(B*a^2 + A*a*b)*c^2 - (5*B*a*b^2 +
A*b^3)*c)*e^2)*log(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4)]

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Sympy [B]  time = 10.094, size = 1532, normalized size = 7.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*e**2*x**2/(2*c) + (-sqrt(-4*a*c + b**2)*(-2*A*a*c**2*e**2 + A*b**2*c*e**2 - 2*A*b*c**2*d*e + 2*A*c**3*d**2 +
3*B*a*b*c*e**2 - 4*B*a*c**2*d*e - B*b**3*e**2 + 2*B*b**2*c*d*e - B*b*c**2*d**2)/(2*c**3*(4*a*c - b**2)) - (A*
b*c*e**2 - 2*A*c**2*d*e + B*a*c*e**2 - B*b**2*e**2 + 2*B*b*c*d*e - B*c**2*d**2)/(2*c**3))*log(x + (A*a*b*c*e**
2 - 4*A*a*c**2*d*e + A*b*c**2*d**2 + 2*B*a**2*c*e**2 - B*a*b**2*e**2 + 2*B*a*b*c*d*e - 2*B*a*c**2*d**2 + 4*a*c
**3*(-sqrt(-4*a*c + b**2)*(-2*A*a*c**2*e**2 + A*b**2*c*e**2 - 2*A*b*c**2*d*e + 2*A*c**3*d**2 + 3*B*a*b*c*e**2
- 4*B*a*c**2*d*e - B*b**3*e**2 + 2*B*b**2*c*d*e - B*b*c**2*d**2)/(2*c**3*(4*a*c - b**2)) - (A*b*c*e**2 - 2*A*c
**2*d*e + B*a*c*e**2 - B*b**2*e**2 + 2*B*b*c*d*e - B*c**2*d**2)/(2*c**3)) - b**2*c**2*(-sqrt(-4*a*c + b**2)*(-
2*A*a*c**2*e**2 + A*b**2*c*e**2 - 2*A*b*c**2*d*e + 2*A*c**3*d**2 + 3*B*a*b*c*e**2 - 4*B*a*c**2*d*e - B*b**3*e*
*2 + 2*B*b**2*c*d*e - B*b*c**2*d**2)/(2*c**3*(4*a*c - b**2)) - (A*b*c*e**2 - 2*A*c**2*d*e + B*a*c*e**2 - B*b**
2*e**2 + 2*B*b*c*d*e - B*c**2*d**2)/(2*c**3)))/(-2*A*a*c**2*e**2 + A*b**2*c*e**2 - 2*A*b*c**2*d*e + 2*A*c**3*d
**2 + 3*B*a*b*c*e**2 - 4*B*a*c**2*d*e - B*b**3*e**2 + 2*B*b**2*c*d*e - B*b*c**2*d**2)) + (sqrt(-4*a*c + b**2)*
(-2*A*a*c**2*e**2 + A*b**2*c*e**2 - 2*A*b*c**2*d*e + 2*A*c**3*d**2 + 3*B*a*b*c*e**2 - 4*B*a*c**2*d*e - B*b**3*
e**2 + 2*B*b**2*c*d*e - B*b*c**2*d**2)/(2*c**3*(4*a*c - b**2)) - (A*b*c*e**2 - 2*A*c**2*d*e + B*a*c*e**2 - B*b
**2*e**2 + 2*B*b*c*d*e - B*c**2*d**2)/(2*c**3))*log(x + (A*a*b*c*e**2 - 4*A*a*c**2*d*e + A*b*c**2*d**2 + 2*B*a
**2*c*e**2 - B*a*b**2*e**2 + 2*B*a*b*c*d*e - 2*B*a*c**2*d**2 + 4*a*c**3*(sqrt(-4*a*c + b**2)*(-2*A*a*c**2*e**2
+ A*b**2*c*e**2 - 2*A*b*c**2*d*e + 2*A*c**3*d**2 + 3*B*a*b*c*e**2 - 4*B*a*c**2*d*e - B*b**3*e**2 + 2*B*b**2*c
*d*e - B*b*c**2*d**2)/(2*c**3*(4*a*c - b**2)) - (A*b*c*e**2 - 2*A*c**2*d*e + B*a*c*e**2 - B*b**2*e**2 + 2*B*b*
c*d*e - B*c**2*d**2)/(2*c**3)) - b**2*c**2*(sqrt(-4*a*c + b**2)*(-2*A*a*c**2*e**2 + A*b**2*c*e**2 - 2*A*b*c**2
*d*e + 2*A*c**3*d**2 + 3*B*a*b*c*e**2 - 4*B*a*c**2*d*e - B*b**3*e**2 + 2*B*b**2*c*d*e - B*b*c**2*d**2)/(2*c**3
*(4*a*c - b**2)) - (A*b*c*e**2 - 2*A*c**2*d*e + B*a*c*e**2 - B*b**2*e**2 + 2*B*b*c*d*e - B*c**2*d**2)/(2*c**3)
))/(-2*A*a*c**2*e**2 + A*b**2*c*e**2 - 2*A*b*c**2*d*e + 2*A*c**3*d**2 + 3*B*a*b*c*e**2 - 4*B*a*c**2*d*e - B*b*
*3*e**2 + 2*B*b**2*c*d*e - B*b*c**2*d**2)) - x*(-A*c*e**2 + B*b*e**2 - 2*B*c*d*e)/c**2

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Giac [A]  time = 1.12342, size = 296, normalized size = 1.44 \begin{align*} \frac{B c x^{2} e^{2} + 4 \, B c d x e - 2 \, B b x e^{2} + 2 \, A c x e^{2}}{2 \, c^{2}} + \frac{{\left (B c^{2} d^{2} - 2 \, B b c d e + 2 \, A c^{2} d e + B b^{2} e^{2} - B a c e^{2} - A b c e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} - \frac{{\left (B b c^{2} d^{2} - 2 \, A c^{3} d^{2} - 2 \, B b^{2} c d e + 4 \, B a c^{2} d e + 2 \, A b c^{2} d e + B b^{3} e^{2} - 3 \, B a b c e^{2} - A b^{2} c e^{2} + 2 \, A a c^{2} e^{2}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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