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

Optimal. Leaf size=184 \[ -\frac{\log \left (d+e x+f x^2\right ) \left (A f (c e-b f)-B \left (a f^2-b e f-c d f+c e^2\right )\right )}{2 f^3}-\frac{\tanh ^{-1}\left (\frac{e+2 f x}{\sqrt{e^2-4 d f}}\right ) \left (A f \left (2 a f^2-b e f-2 c d f+c e^2\right )+B \left (f \left (-a e f-2 b d f+b e^2\right )-c \left (e^3-3 d e f\right )\right )\right )}{f^3 \sqrt{e^2-4 d f}}-\frac{x (-A c f-b B f+B c e)}{f^2}+\frac{B c x^2}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.345593, antiderivative size = 182, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {1628, 634, 618, 206, 628} \[ -\frac{\log \left (d+e x+f x^2\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )}{2 f^3}-\frac{\tanh ^{-1}\left (\frac{e+2 f x}{\sqrt{e^2-4 d f}}\right ) \left (A f \left (2 a f^2-b e f-2 c d f+c e^2\right )+B f \left (-a e f-2 b d f+b e^2\right )-B c \left (e^3-3 d e f\right )\right )}{f^3 \sqrt{e^2-4 d f}}-\frac{x (-A c f-b B f+B c e)}{f^2}+\frac{B c x^2}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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{(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx &=\int \left (-\frac{B c e-b B f-A c f}{f^2}+\frac{B c x}{f}+\frac{-A f (c d-a f)+B d (c e-b f)-\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) x}{f^2 \left (d+e x+f x^2\right )}\right ) \, dx\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}+\frac{\int \frac{-A f (c d-a f)+B d (c e-b f)-\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) x}{d+e x+f x^2} \, dx}{f^2}\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}+\frac{\left (-B f (b e-a f)-A f (c e-b f)+B c \left (e^2-d f\right )\right ) \int \frac{e+2 f x}{d+e x+f x^2} \, dx}{2 f^3}+\frac{\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \int \frac{1}{d+e x+f x^2} \, dx}{2 f^3}\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}-\frac{\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3}-\frac{\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{e^2-4 d f-x^2} \, dx,x,e+2 f x\right )}{f^3}\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}-\frac{\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \tanh ^{-1}\left (\frac{e+2 f x}{\sqrt{e^2-4 d f}}\right )}{f^3 \sqrt{e^2-4 d f}}-\frac{\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3}\\ \end{align*}

Mathematica [A]  time = 0.202927, size = 175, normalized size = 0.95 \[ \frac{-\frac{2 \tan ^{-1}\left (\frac{e+2 f x}{\sqrt{4 d f-e^2}}\right ) \left (A f \left (-2 a f^2+b e f+2 c d f-c e^2\right )+B f \left (a e f+2 b d f-b e^2\right )+B c \left (e^3-3 d e f\right )\right )}{\sqrt{4 d f-e^2}}+\log (d+x (e+f x)) \left (B f (a f-b e)+A f (b f-c e)+B c \left (e^2-d f\right )\right )+2 f x (A c f+b B f-B c e)+B c f^2 x^2}{2 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.175, size = 510, normalized size = 2.8 \begin{align*}{\frac{Bc{x}^{2}}{2\,f}}+{\frac{Acx}{f}}+{\frac{Bbx}{f}}-{\frac{Bcex}{{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Ab}{2\,f}}-{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Ace}{2\,{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Ba}{2\,f}}-{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Bbe}{2\,{f}^{2}}}-{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Bcd}{2\,{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Bc{e}^{2}}{2\,{f}^{3}}}+2\,{\frac{Aa}{\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }-2\,{\frac{Acd}{f\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }-2\,{\frac{Bbd}{f\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }+3\,{\frac{Bcde}{{f}^{2}\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }-{\frac{Abe}{f}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}+{\frac{Ac{e}^{2}}{{f}^{2}}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}-{\frac{aBe}{f}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}+{\frac{bB{e}^{2}}{{f}^{2}}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}-{\frac{Bc{e}^{3}}{{f}^{3}}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*c*x^2/f+1/f*A*c*x+1/f*B*b*x-1/f^2*B*c*e*x+1/2/f*ln(f*x^2+e*x+d)*A*b-1/2/f^2*ln(f*x^2+e*x+d)*A*c*e+1/2/f*
ln(f*x^2+e*x+d)*B*a-1/2/f^2*ln(f*x^2+e*x+d)*B*b*e-1/2/f^2*ln(f*x^2+e*x+d)*B*c*d+1/2/f^3*ln(f*x^2+e*x+d)*B*c*e^
2+2/(4*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2))*A*a-2/f/(4*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^
2)^(1/2))*A*c*d-2/f/(4*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2))*B*b*d+3/f^2/(4*d*f-e^2)^(1/2)*arctan
((2*f*x+e)/(4*d*f-e^2)^(1/2))*B*c*d*e-1/f/(4*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2))*e*A*b+1/f^2/(4
*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2))*A*c*e^2-1/f/(4*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)
^(1/2))*e*B*a+1/f^2/(4*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2))*b*B*e^2-1/f^3/(4*d*f-e^2)^(1/2)*arct
an((2*f*x+e)/(4*d*f-e^2)^(1/2))*e^3*B*c

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79901, size = 1280, normalized size = 6.96 \begin{align*} \left [\frac{{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} -{\left (B c e^{3} - 2 \, A a f^{3} +{\left (2 \,{\left (B b + A c\right )} d +{\left (B a + A b\right )} e\right )} f^{2} -{\left (3 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt{e^{2} - 4 \, d f} \log \left (\frac{2 \, f^{2} x^{2} + 2 \, e f x + e^{2} - 2 \, d f - \sqrt{e^{2} - 4 \, d f}{\left (2 \, f x + e\right )}}{f x^{2} + e x + d}\right ) - 2 \,{\left (B c e^{3} f + 4 \,{\left (B b + A c\right )} d f^{3} -{\left (4 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x +{\left (B c e^{4} - 4 \,{\left (B a + A b\right )} d f^{3} +{\left (4 \, B c d^{2} + 4 \,{\left (B b + A c\right )} d e +{\left (B a + A b\right )} e^{2}\right )} f^{2} -{\left (5 \, B c d e^{2} +{\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \,{\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}, \frac{{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} + 2 \,{\left (B c e^{3} - 2 \, A a f^{3} +{\left (2 \,{\left (B b + A c\right )} d +{\left (B a + A b\right )} e\right )} f^{2} -{\left (3 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt{-e^{2} + 4 \, d f} \arctan \left (-\frac{\sqrt{-e^{2} + 4 \, d f}{\left (2 \, f x + e\right )}}{e^{2} - 4 \, d f}\right ) - 2 \,{\left (B c e^{3} f + 4 \,{\left (B b + A c\right )} d f^{3} -{\left (4 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x +{\left (B c e^{4} - 4 \,{\left (B a + A b\right )} d f^{3} +{\left (4 \, B c d^{2} + 4 \,{\left (B b + A c\right )} d e +{\left (B a + A b\right )} e^{2}\right )} f^{2} -{\left (5 \, B c d e^{2} +{\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \,{\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 25.2751, size = 1260, normalized size = 6.85 \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)*(c*x**2+b*x+a)/(f*x**2+e*x+d),x)

[Out]

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

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Giac [A]  time = 1.14854, size = 258, normalized size = 1.4 \begin{align*} \frac{B c f x^{2} + 2 \, B b f x + 2 \, A c f x - 2 \, B c x e}{2 \, f^{2}} - \frac{{\left (B c d f - B a f^{2} - A b f^{2} + B b f e + A c f e - B c e^{2}\right )} \log \left (f x^{2} + x e + d\right )}{2 \, f^{3}} - \frac{{\left (2 \, B b d f^{2} + 2 \, A c d f^{2} - 2 \, A a f^{3} - 3 \, B c d f e + B a f^{2} e + A b f^{2} e - B b f e^{2} - A c f e^{2} + B c e^{3}\right )} \arctan \left (\frac{2 \, f x + e}{\sqrt{4 \, d f - e^{2}}}\right )}{\sqrt{4 \, d f - e^{2}} f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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