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

Optimal. Leaf size=228 \[ -\frac{\log \left (d+f x^2\right ) \left (2 A b f (c d-a f)-B \left (-f \left (b^2 d-a^2 f\right )-2 a c d f+c^2 d^2\right )\right )}{2 f^3}+\frac{x^2 \left (2 A b c f-B \left (-2 a c f+b^2 (-f)+c^2 d\right )\right )}{2 f^2}+\frac{x \left (-A c (c d-2 a f)-b B (2 c d-2 a f)+A b^2 f\right )}{f^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right ) \left (-A (c d-a f)^2-2 b B d (c d-a f)+A b^2 d f\right )}{\sqrt{d} f^{5/2}}+\frac{c x^3 (A c+2 b B)}{3 f}+\frac{B c^2 x^4}{4 f} \]

[Out]

((A*b^2*f - A*c*(c*d - 2*a*f) - b*B*(2*c*d - 2*a*f))*x)/f^2 + ((2*A*b*c*f - B*(c^2*d - b^2*f - 2*a*c*f))*x^2)/
(2*f^2) + (c*(2*b*B + A*c)*x^3)/(3*f) + (B*c^2*x^4)/(4*f) - ((A*b^2*d*f - 2*b*B*d*(c*d - a*f) - A*(c*d - a*f)^
2)*ArcTan[(Sqrt[f]*x)/Sqrt[d]])/(Sqrt[d]*f^(5/2)) - ((2*A*b*f*(c*d - a*f) - B*(c^2*d^2 - 2*a*c*d*f - f*(b^2*d
- a^2*f)))*Log[d + f*x^2])/(2*f^3)

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Rubi [A]  time = 0.329642, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1012, 635, 205, 260} \[ -\frac{\log \left (d+f x^2\right ) \left (2 A b f (c d-a f)-B \left (-f \left (b^2 d-a^2 f\right )-2 a c d f+c^2 d^2\right )\right )}{2 f^3}+\frac{x^2 \left (2 A b c f-B \left (-2 a c f+b^2 (-f)+c^2 d\right )\right )}{2 f^2}+\frac{x \left (-A c (c d-2 a f)-b B (2 c d-2 a f)+A b^2 f\right )}{f^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right ) \left (-A (c d-a f)^2-2 b B d (c d-a f)+A b^2 d f\right )}{\sqrt{d} f^{5/2}}+\frac{c x^3 (A c+2 b B)}{3 f}+\frac{B c^2 x^4}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b^2*f - A*c*(c*d - 2*a*f) - b*B*(2*c*d - 2*a*f))*x)/f^2 + ((2*A*b*c*f - B*(c^2*d - b^2*f - 2*a*c*f))*x^2)/
(2*f^2) + (c*(2*b*B + A*c)*x^3)/(3*f) + (B*c^2*x^4)/(4*f) - ((A*b^2*d*f - 2*b*B*d*(c*d - a*f) - A*(c*d - a*f)^
2)*ArcTan[(Sqrt[f]*x)/Sqrt[d]])/(Sqrt[d]*f^(5/2)) - ((2*A*b*f*(c*d - a*f) - B*(c^2*d^2 - 2*a*c*d*f - f*(b^2*d
- a^2*f)))*Log[d + f*x^2])/(2*f^3)

Rule 1012

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Int[
ExpandIntegrand[(a + c*x^2)^p*(d + e*x + f*x^2)^q*(g + h*x), x], x] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[
e^2 - 4*d*f, 0] && IntegersQ[p, q] && (GtQ[p, 0] || GtQ[q, 0])

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^2}{d+f x^2} \, dx &=\int \left (\frac{A b^2 f-A c (c d-2 a f)-b B (2 c d-2 a f)}{f^2}+\frac{\left (2 A b c f-B \left (c^2 d-b^2 f-2 a c f\right )\right ) x}{f^2}+\frac{c (2 b B+A c) x^2}{f}+\frac{B c^2 x^3}{f}+\frac{-A b^2 d f+2 b B d (c d-a f)+A (c d-a f)^2-\left (2 A b f (c d-a f)-B \left (c^2 d^2-2 a c d f-f \left (b^2 d-a^2 f\right )\right )\right ) x}{f^2 \left (d+f x^2\right )}\right ) \, dx\\ &=\frac{\left (A b^2 f-A c (c d-2 a f)-b B (2 c d-2 a f)\right ) x}{f^2}+\frac{\left (2 A b c f-B \left (c^2 d-b^2 f-2 a c f\right )\right ) x^2}{2 f^2}+\frac{c (2 b B+A c) x^3}{3 f}+\frac{B c^2 x^4}{4 f}+\frac{\int \frac{-A b^2 d f+2 b B d (c d-a f)+A (c d-a f)^2-\left (2 A b f (c d-a f)-B \left (c^2 d^2-2 a c d f-f \left (b^2 d-a^2 f\right )\right )\right ) x}{d+f x^2} \, dx}{f^2}\\ &=\frac{\left (A b^2 f-A c (c d-2 a f)-b B (2 c d-2 a f)\right ) x}{f^2}+\frac{\left (2 A b c f-B \left (c^2 d-b^2 f-2 a c f\right )\right ) x^2}{2 f^2}+\frac{c (2 b B+A c) x^3}{3 f}+\frac{B c^2 x^4}{4 f}-\frac{\left (A b^2 d f-2 b B d (c d-a f)-A (c d-a f)^2\right ) \int \frac{1}{d+f x^2} \, dx}{f^2}-\frac{\left (2 A b f (c d-a f)-B \left (c^2 d^2-2 a c d f-f \left (b^2 d-a^2 f\right )\right )\right ) \int \frac{x}{d+f x^2} \, dx}{f^2}\\ &=\frac{\left (A b^2 f-A c (c d-2 a f)-b B (2 c d-2 a f)\right ) x}{f^2}+\frac{\left (2 A b c f-B \left (c^2 d-b^2 f-2 a c f\right )\right ) x^2}{2 f^2}+\frac{c (2 b B+A c) x^3}{3 f}+\frac{B c^2 x^4}{4 f}-\frac{\left (A b^2 d f-2 b B d (c d-a f)-A (c d-a f)^2\right ) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right )}{\sqrt{d} f^{5/2}}-\frac{\left (2 A b f (c d-a f)-B \left (c^2 d^2-2 a c d f-f \left (b^2 d-a^2 f\right )\right )\right ) \log \left (d+f x^2\right )}{2 f^3}\\ \end{align*}

Mathematica [A]  time = 0.222411, size = 204, normalized size = 0.89 \[ \frac{6 \log \left (d+f x^2\right ) \left (B \left (a^2 f^2-2 a c d f+b^2 (-d) f+c^2 d^2\right )+2 A b f (a f-c d)\right )+f x \left (4 A c \left (6 a f-3 c d+c f x^2\right )+4 b B \left (6 a f-6 c d+2 c f x^2\right )+3 B c x \left (4 a f-2 c d+c f x^2\right )+6 b^2 f (2 A+B x)+12 A b c f x\right )}{12 f^3}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right ) \left (A (c d-a f)^2+2 b B d (c d-a f)-A b^2 d f\right )}{\sqrt{d} f^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-(A*b^2*d*f) + 2*b*B*d*(c*d - a*f) + A*(c*d - a*f)^2)*ArcTan[(Sqrt[f]*x)/Sqrt[d]])/(Sqrt[d]*f^(5/2)) + (f*x*
(12*A*b*c*f*x + 6*b^2*f*(2*A + B*x) + 3*B*c*x*(-2*c*d + 4*a*f + c*f*x^2) + 4*A*c*(-3*c*d + 6*a*f + c*f*x^2) +
4*b*B*(-6*c*d + 6*a*f + 2*c*f*x^2)) + 6*(2*A*b*f*(-(c*d) + a*f) + B*(c^2*d^2 - b^2*d*f - 2*a*c*d*f + a^2*f^2))
*Log[d + f*x^2])/(12*f^3)

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Maple [A]  time = 0.052, size = 373, normalized size = 1.6 \begin{align*}{\frac{B{c}^{2}{x}^{4}}{4\,f}}+{\frac{A{x}^{3}{c}^{2}}{3\,f}}+{\frac{2\,B{x}^{3}bc}{3\,f}}+{\frac{Abc{x}^{2}}{f}}+{\frac{B{x}^{2}ac}{f}}+{\frac{B{x}^{2}{b}^{2}}{2\,f}}-{\frac{B{c}^{2}{x}^{2}d}{2\,{f}^{2}}}+2\,{\frac{aAcx}{f}}+{\frac{A{b}^{2}x}{f}}-{\frac{A{c}^{2}dx}{{f}^{2}}}+2\,{\frac{abBx}{f}}-2\,{\frac{Bbcdx}{{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+d \right ) Aab}{f}}-{\frac{\ln \left ( f{x}^{2}+d \right ) Abcd}{{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+d \right ) B{a}^{2}}{2\,f}}-{\frac{\ln \left ( f{x}^{2}+d \right ) Bacd}{{f}^{2}}}-{\frac{\ln \left ( f{x}^{2}+d \right ) B{b}^{2}d}{2\,{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+d \right ) B{c}^{2}{d}^{2}}{2\,{f}^{3}}}+{A{a}^{2}\arctan \left ({fx{\frac{1}{\sqrt{df}}}} \right ){\frac{1}{\sqrt{df}}}}-2\,{\frac{aAcd}{f\sqrt{df}}\arctan \left ({\frac{fx}{\sqrt{df}}} \right ) }-{\frac{A{b}^{2}d}{f}\arctan \left ({fx{\frac{1}{\sqrt{df}}}} \right ){\frac{1}{\sqrt{df}}}}+{\frac{A{c}^{2}{d}^{2}}{{f}^{2}}\arctan \left ({fx{\frac{1}{\sqrt{df}}}} \right ){\frac{1}{\sqrt{df}}}}-2\,{\frac{abBd}{f\sqrt{df}}\arctan \left ({\frac{fx}{\sqrt{df}}} \right ) }+2\,{\frac{Bbc{d}^{2}}{{f}^{2}\sqrt{df}}\arctan \left ({\frac{fx}{\sqrt{df}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9131, size = 1088, normalized size = 4.77 \begin{align*} \left [\frac{3 \, B c^{2} d f^{2} x^{4} + 4 \,{\left (2 \, B b c + A c^{2}\right )} d f^{2} x^{3} - 6 \,{\left (B c^{2} d^{2} f -{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d f^{2}\right )} x^{2} - 6 \,{\left (A a^{2} f^{2} +{\left (2 \, B b c + A c^{2}\right )} d^{2} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d f\right )} \sqrt{-d f} \log \left (\frac{f x^{2} - 2 \, \sqrt{-d f} x - d}{f x^{2} + d}\right ) - 12 \,{\left ({\left (2 \, B b c + A c^{2}\right )} d^{2} f -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d f^{2}\right )} x + 6 \,{\left (B c^{2} d^{3} -{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d^{2} f +{\left (B a^{2} + 2 \, A a b\right )} d f^{2}\right )} \log \left (f x^{2} + d\right )}{12 \, d f^{3}}, \frac{3 \, B c^{2} d f^{2} x^{4} + 4 \,{\left (2 \, B b c + A c^{2}\right )} d f^{2} x^{3} - 6 \,{\left (B c^{2} d^{2} f -{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d f^{2}\right )} x^{2} + 12 \,{\left (A a^{2} f^{2} +{\left (2 \, B b c + A c^{2}\right )} d^{2} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d f\right )} \sqrt{d f} \arctan \left (\frac{\sqrt{d f} x}{d}\right ) - 12 \,{\left ({\left (2 \, B b c + A c^{2}\right )} d^{2} f -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d f^{2}\right )} x + 6 \,{\left (B c^{2} d^{3} -{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d^{2} f +{\left (B a^{2} + 2 \, A a b\right )} d f^{2}\right )} \log \left (f x^{2} + d\right )}{12 \, d f^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*B*c^2*d*f^2*x^4 + 4*(2*B*b*c + A*c^2)*d*f^2*x^3 - 6*(B*c^2*d^2*f - (B*b^2 + 2*(B*a + A*b)*c)*d*f^2)*x
^2 - 6*(A*a^2*f^2 + (2*B*b*c + A*c^2)*d^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d*f)*sqrt(-d*f)*log((f*x^2 - 2*sqrt(-d
*f)*x - d)/(f*x^2 + d)) - 12*((2*B*b*c + A*c^2)*d^2*f - (2*B*a*b + A*b^2 + 2*A*a*c)*d*f^2)*x + 6*(B*c^2*d^3 -
(B*b^2 + 2*(B*a + A*b)*c)*d^2*f + (B*a^2 + 2*A*a*b)*d*f^2)*log(f*x^2 + d))/(d*f^3), 1/12*(3*B*c^2*d*f^2*x^4 +
4*(2*B*b*c + A*c^2)*d*f^2*x^3 - 6*(B*c^2*d^2*f - (B*b^2 + 2*(B*a + A*b)*c)*d*f^2)*x^2 + 12*(A*a^2*f^2 + (2*B*b
*c + A*c^2)*d^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d*f)*sqrt(d*f)*arctan(sqrt(d*f)*x/d) - 12*((2*B*b*c + A*c^2)*d^2
*f - (2*B*a*b + A*b^2 + 2*A*a*c)*d*f^2)*x + 6*(B*c^2*d^3 - (B*b^2 + 2*(B*a + A*b)*c)*d^2*f + (B*a^2 + 2*A*a*b)
*d*f^2)*log(f*x^2 + d))/(d*f^3)]

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Sympy [B]  time = 8.99137, size = 928, normalized size = 4.07 \begin{align*} \frac{B c^{2} x^{4}}{4 f} + \left (\frac{2 A a b f^{2} - 2 A b c d f + B a^{2} f^{2} - 2 B a c d f - B b^{2} d f + B c^{2} d^{2}}{2 f^{3}} - \frac{\sqrt{- d f^{7}} \left (A a^{2} f^{2} - 2 A a c d f - A b^{2} d f + A c^{2} d^{2} - 2 B a b d f + 2 B b c d^{2}\right )}{2 d f^{6}}\right ) \log{\left (x + \frac{- 2 A a b d f^{2} + 2 A b c d^{2} f - B a^{2} d f^{2} + 2 B a c d^{2} f + B b^{2} d^{2} f - B c^{2} d^{3} + 2 d f^{3} \left (\frac{2 A a b f^{2} - 2 A b c d f + B a^{2} f^{2} - 2 B a c d f - B b^{2} d f + B c^{2} d^{2}}{2 f^{3}} - \frac{\sqrt{- d f^{7}} \left (A a^{2} f^{2} - 2 A a c d f - A b^{2} d f + A c^{2} d^{2} - 2 B a b d f + 2 B b c d^{2}\right )}{2 d f^{6}}\right )}{A a^{2} f^{3} - 2 A a c d f^{2} - A b^{2} d f^{2} + A c^{2} d^{2} f - 2 B a b d f^{2} + 2 B b c d^{2} f} \right )} + \left (\frac{2 A a b f^{2} - 2 A b c d f + B a^{2} f^{2} - 2 B a c d f - B b^{2} d f + B c^{2} d^{2}}{2 f^{3}} + \frac{\sqrt{- d f^{7}} \left (A a^{2} f^{2} - 2 A a c d f - A b^{2} d f + A c^{2} d^{2} - 2 B a b d f + 2 B b c d^{2}\right )}{2 d f^{6}}\right ) \log{\left (x + \frac{- 2 A a b d f^{2} + 2 A b c d^{2} f - B a^{2} d f^{2} + 2 B a c d^{2} f + B b^{2} d^{2} f - B c^{2} d^{3} + 2 d f^{3} \left (\frac{2 A a b f^{2} - 2 A b c d f + B a^{2} f^{2} - 2 B a c d f - B b^{2} d f + B c^{2} d^{2}}{2 f^{3}} + \frac{\sqrt{- d f^{7}} \left (A a^{2} f^{2} - 2 A a c d f - A b^{2} d f + A c^{2} d^{2} - 2 B a b d f + 2 B b c d^{2}\right )}{2 d f^{6}}\right )}{A a^{2} f^{3} - 2 A a c d f^{2} - A b^{2} d f^{2} + A c^{2} d^{2} f - 2 B a b d f^{2} + 2 B b c d^{2} f} \right )} + \frac{x^{3} \left (A c^{2} + 2 B b c\right )}{3 f} + \frac{x^{2} \left (2 A b c f + 2 B a c f + B b^{2} f - B c^{2} d\right )}{2 f^{2}} + \frac{x \left (2 A a c f + A b^{2} f - A c^{2} d + 2 B a b f - 2 B b c d\right )}{f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*c**2*x**4/(4*f) + ((2*A*a*b*f**2 - 2*A*b*c*d*f + B*a**2*f**2 - 2*B*a*c*d*f - B*b**2*d*f + B*c**2*d**2)/(2*f*
*3) - sqrt(-d*f**7)*(A*a**2*f**2 - 2*A*a*c*d*f - A*b**2*d*f + A*c**2*d**2 - 2*B*a*b*d*f + 2*B*b*c*d**2)/(2*d*f
**6))*log(x + (-2*A*a*b*d*f**2 + 2*A*b*c*d**2*f - B*a**2*d*f**2 + 2*B*a*c*d**2*f + B*b**2*d**2*f - B*c**2*d**3
+ 2*d*f**3*((2*A*a*b*f**2 - 2*A*b*c*d*f + B*a**2*f**2 - 2*B*a*c*d*f - B*b**2*d*f + B*c**2*d**2)/(2*f**3) - sq
rt(-d*f**7)*(A*a**2*f**2 - 2*A*a*c*d*f - A*b**2*d*f + A*c**2*d**2 - 2*B*a*b*d*f + 2*B*b*c*d**2)/(2*d*f**6)))/(
A*a**2*f**3 - 2*A*a*c*d*f**2 - A*b**2*d*f**2 + A*c**2*d**2*f - 2*B*a*b*d*f**2 + 2*B*b*c*d**2*f)) + ((2*A*a*b*f
**2 - 2*A*b*c*d*f + B*a**2*f**2 - 2*B*a*c*d*f - B*b**2*d*f + B*c**2*d**2)/(2*f**3) + sqrt(-d*f**7)*(A*a**2*f**
2 - 2*A*a*c*d*f - A*b**2*d*f + A*c**2*d**2 - 2*B*a*b*d*f + 2*B*b*c*d**2)/(2*d*f**6))*log(x + (-2*A*a*b*d*f**2
+ 2*A*b*c*d**2*f - B*a**2*d*f**2 + 2*B*a*c*d**2*f + B*b**2*d**2*f - B*c**2*d**3 + 2*d*f**3*((2*A*a*b*f**2 - 2*
A*b*c*d*f + B*a**2*f**2 - 2*B*a*c*d*f - B*b**2*d*f + B*c**2*d**2)/(2*f**3) + sqrt(-d*f**7)*(A*a**2*f**2 - 2*A*
a*c*d*f - A*b**2*d*f + A*c**2*d**2 - 2*B*a*b*d*f + 2*B*b*c*d**2)/(2*d*f**6)))/(A*a**2*f**3 - 2*A*a*c*d*f**2 -
A*b**2*d*f**2 + A*c**2*d**2*f - 2*B*a*b*d*f**2 + 2*B*b*c*d**2*f)) + x**3*(A*c**2 + 2*B*b*c)/(3*f) + x**2*(2*A*
b*c*f + 2*B*a*c*f + B*b**2*f - B*c**2*d)/(2*f**2) + x*(2*A*a*c*f + A*b**2*f - A*c**2*d + 2*B*a*b*f - 2*B*b*c*d
)/f**2

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Giac [A]  time = 1.4019, size = 355, normalized size = 1.56 \begin{align*} \frac{{\left (2 \, B b c d^{2} + A c^{2} d^{2} - 2 \, B a b d f - A b^{2} d f - 2 \, A a c d f + A a^{2} f^{2}\right )} \arctan \left (\frac{f x}{\sqrt{d f}}\right )}{\sqrt{d f} f^{2}} + \frac{{\left (B c^{2} d^{2} - B b^{2} d f - 2 \, B a c d f - 2 \, A b c d f + B a^{2} f^{2} + 2 \, A a b f^{2}\right )} \log \left (f x^{2} + d\right )}{2 \, f^{3}} + \frac{3 \, B c^{2} f^{3} x^{4} + 8 \, B b c f^{3} x^{3} + 4 \, A c^{2} f^{3} x^{3} - 6 \, B c^{2} d f^{2} x^{2} + 6 \, B b^{2} f^{3} x^{2} + 12 \, B a c f^{3} x^{2} + 12 \, A b c f^{3} x^{2} - 24 \, B b c d f^{2} x - 12 \, A c^{2} d f^{2} x + 24 \, B a b f^{3} x + 12 \, A b^{2} f^{3} x + 24 \, A a c f^{3} x}{12 \, f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*b*c*d^2 + A*c^2*d^2 - 2*B*a*b*d*f - A*b^2*d*f - 2*A*a*c*d*f + A*a^2*f^2)*arctan(f*x/sqrt(d*f))/(sqrt(d*f)
*f^2) + 1/2*(B*c^2*d^2 - B*b^2*d*f - 2*B*a*c*d*f - 2*A*b*c*d*f + B*a^2*f^2 + 2*A*a*b*f^2)*log(f*x^2 + d)/f^3 +
1/12*(3*B*c^2*f^3*x^4 + 8*B*b*c*f^3*x^3 + 4*A*c^2*f^3*x^3 - 6*B*c^2*d*f^2*x^2 + 6*B*b^2*f^3*x^2 + 12*B*a*c*f^
3*x^2 + 12*A*b*c*f^3*x^2 - 24*B*b*c*d*f^2*x - 12*A*c^2*d*f^2*x + 24*B*a*b*f^3*x + 12*A*b^2*f^3*x + 24*A*a*c*f^
3*x)/f^4