3.3 \(\int \frac{(A+B x) (a+b x+c x^2)^3}{d+f x^2} \, dx\)

Optimal. Leaf size=441 \[ -\frac{x^2 \left (A b f \left (-6 a c f+b^2 (-f)+3 c^2 d\right )-B \left (-3 c f \left (b^2 d-a^2 f\right )+3 a b^2 f^2-3 a c^2 d f+c^3 d^2\right )\right )}{2 f^3}+\frac{\log \left (d+f x^2\right ) \left (A b f \left (-f \left (b^2 d-3 a^2 f\right )-6 a c d f+3 c^2 d^2\right )-B (c d-a f) \left (-f \left (3 b^2 d-a^2 f\right )-2 a c d f+c^2 d^2\right )\right )}{2 f^4}-\frac{x \left (-A c \left (3 a^2 f^2-3 a c d f+c^2 d^2\right )+3 A b^2 f (c d-a f)-3 b B (c d-a f)^2+b^3 B d f\right )}{f^3}+\frac{c x^4 \left (3 A b c f-B \left (-3 a c f-3 b^2 f+c^2 d\right )\right )}{4 f^2}+\frac{x^3 \left (-A c^2 (c d-3 a f)-3 b B c (c d-2 a f)+3 A b^2 c f+b^3 B f\right )}{3 f^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right ) \left (3 A b^2 d f (c d-a f)-A (c d-a f)^3-3 b B d (c d-a f)^2+b^3 B d^2 f\right )}{\sqrt{d} f^{7/2}}+\frac{c^2 x^5 (A c+3 b B)}{5 f}+\frac{B c^3 x^6}{6 f} \]

[Out]

-(((b^3*B*d*f + 3*A*b^2*f*(c*d - a*f) - 3*b*B*(c*d - a*f)^2 - A*c*(c^2*d^2 - 3*a*c*d*f + 3*a^2*f^2))*x)/f^3) -
((A*b*f*(3*c^2*d - b^2*f - 6*a*c*f) - B*(c^3*d^2 - 3*a*c^2*d*f + 3*a*b^2*f^2 - 3*c*f*(b^2*d - a^2*f)))*x^2)/(
2*f^3) + ((b^3*B*f + 3*A*b^2*c*f - A*c^2*(c*d - 3*a*f) - 3*b*B*c*(c*d - 2*a*f))*x^3)/(3*f^2) + (c*(3*A*b*c*f -
B*(c^2*d - 3*b^2*f - 3*a*c*f))*x^4)/(4*f^2) + (c^2*(3*b*B + A*c)*x^5)/(5*f) + (B*c^3*x^6)/(6*f) + ((b^3*B*d^2
*f + 3*A*b^2*d*f*(c*d - a*f) - 3*b*B*d*(c*d - a*f)^2 - A*(c*d - a*f)^3)*ArcTan[(Sqrt[f]*x)/Sqrt[d]])/(Sqrt[d]*
f^(7/2)) + ((A*b*f*(3*c^2*d^2 - 6*a*c*d*f - f*(b^2*d - 3*a^2*f)) - B*(c*d - a*f)*(c^2*d^2 - 2*a*c*d*f - f*(3*b
^2*d - a^2*f)))*Log[d + f*x^2])/(2*f^4)

________________________________________________________________________________________

Rubi [A]  time = 0.621031, antiderivative size = 441, 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} \[ -\frac{x^2 \left (A b f \left (-6 a c f+b^2 (-f)+3 c^2 d\right )-B \left (-3 c f \left (b^2 d-a^2 f\right )+3 a b^2 f^2-3 a c^2 d f+c^3 d^2\right )\right )}{2 f^3}+\frac{\log \left (d+f x^2\right ) \left (A b f \left (-f \left (b^2 d-3 a^2 f\right )-6 a c d f+3 c^2 d^2\right )-B (c d-a f) \left (-f \left (3 b^2 d-a^2 f\right )-2 a c d f+c^2 d^2\right )\right )}{2 f^4}-\frac{x \left (-A c \left (3 a^2 f^2-3 a c d f+c^2 d^2\right )+3 A b^2 f (c d-a f)-3 b B (c d-a f)^2+b^3 B d f\right )}{f^3}+\frac{c x^4 \left (3 A b c f-B \left (-3 a c f-3 b^2 f+c^2 d\right )\right )}{4 f^2}+\frac{x^3 \left (-A c^2 (c d-3 a f)-3 b B c (c d-2 a f)+3 A b^2 c f+b^3 B f\right )}{3 f^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right ) \left (3 A b^2 d f (c d-a f)-A (c d-a f)^3-3 b B d (c d-a f)^2+b^3 B d^2 f\right )}{\sqrt{d} f^{7/2}}+\frac{c^2 x^5 (A c+3 b B)}{5 f}+\frac{B c^3 x^6}{6 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^3)/(d + f*x^2),x]

[Out]

-(((b^3*B*d*f + 3*A*b^2*f*(c*d - a*f) - 3*b*B*(c*d - a*f)^2 - A*c*(c^2*d^2 - 3*a*c*d*f + 3*a^2*f^2))*x)/f^3) -
((A*b*f*(3*c^2*d - b^2*f - 6*a*c*f) - B*(c^3*d^2 - 3*a*c^2*d*f + 3*a*b^2*f^2 - 3*c*f*(b^2*d - a^2*f)))*x^2)/(
2*f^3) + ((b^3*B*f + 3*A*b^2*c*f - A*c^2*(c*d - 3*a*f) - 3*b*B*c*(c*d - 2*a*f))*x^3)/(3*f^2) + (c*(3*A*b*c*f -
B*(c^2*d - 3*b^2*f - 3*a*c*f))*x^4)/(4*f^2) + (c^2*(3*b*B + A*c)*x^5)/(5*f) + (B*c^3*x^6)/(6*f) + ((b^3*B*d^2
*f + 3*A*b^2*d*f*(c*d - a*f) - 3*b*B*d*(c*d - a*f)^2 - A*(c*d - a*f)^3)*ArcTan[(Sqrt[f]*x)/Sqrt[d]])/(Sqrt[d]*
f^(7/2)) + ((A*b*f*(3*c^2*d^2 - 6*a*c*d*f - f*(b^2*d - 3*a^2*f)) - B*(c*d - a*f)*(c^2*d^2 - 2*a*c*d*f - f*(3*b
^2*d - a^2*f)))*Log[d + f*x^2])/(2*f^4)

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 )^3}{d+f x^2} \, dx &=\int \left (-\frac{b^3 B d f+3 A b^2 f (c d-a f)-3 b B (c d-a f)^2-A c \left (c^2 d^2-3 a c d f+3 a^2 f^2\right )}{f^3}-\frac{\left (A b f \left (3 c^2 d-b^2 f-6 a c f\right )-B \left (c^3 d^2-3 a c^2 d f+3 a b^2 f^2-3 c f \left (b^2 d-a^2 f\right )\right )\right ) x}{f^3}+\frac{\left (b^3 B f+3 A b^2 c f-A c^2 (c d-3 a f)-3 b B c (c d-2 a f)\right ) x^2}{f^2}+\frac{c \left (3 A b c f-B \left (c^2 d-3 b^2 f-3 a c f\right )\right ) x^3}{f^2}+\frac{c^2 (3 b B+A c) x^4}{f}+\frac{B c^3 x^5}{f}-\frac{-b^3 B d^2 f-3 A b^2 d f (c d-a f)+3 b B d (c d-a f)^2+A (c d-a f)^3-\left (A b f \left (3 c^2 d^2-6 a c d f-f \left (b^2 d-3 a^2 f\right )\right )-B (c d-a f) \left (c^2 d^2-2 a c d f-f \left (3 b^2 d-a^2 f\right )\right )\right ) x}{f^3 \left (d+f x^2\right )}\right ) \, dx\\ &=-\frac{\left (b^3 B d f+3 A b^2 f (c d-a f)-3 b B (c d-a f)^2-A c \left (c^2 d^2-3 a c d f+3 a^2 f^2\right )\right ) x}{f^3}-\frac{\left (A b f \left (3 c^2 d-b^2 f-6 a c f\right )-B \left (c^3 d^2-3 a c^2 d f+3 a b^2 f^2-3 c f \left (b^2 d-a^2 f\right )\right )\right ) x^2}{2 f^3}+\frac{\left (b^3 B f+3 A b^2 c f-A c^2 (c d-3 a f)-3 b B c (c d-2 a f)\right ) x^3}{3 f^2}+\frac{c \left (3 A b c f-B \left (c^2 d-3 b^2 f-3 a c f\right )\right ) x^4}{4 f^2}+\frac{c^2 (3 b B+A c) x^5}{5 f}+\frac{B c^3 x^6}{6 f}-\frac{\int \frac{-b^3 B d^2 f-3 A b^2 d f (c d-a f)+3 b B d (c d-a f)^2+A (c d-a f)^3-\left (A b f \left (3 c^2 d^2-6 a c d f-f \left (b^2 d-3 a^2 f\right )\right )-B (c d-a f) \left (c^2 d^2-2 a c d f-f \left (3 b^2 d-a^2 f\right )\right )\right ) x}{d+f x^2} \, dx}{f^3}\\ &=-\frac{\left (b^3 B d f+3 A b^2 f (c d-a f)-3 b B (c d-a f)^2-A c \left (c^2 d^2-3 a c d f+3 a^2 f^2\right )\right ) x}{f^3}-\frac{\left (A b f \left (3 c^2 d-b^2 f-6 a c f\right )-B \left (c^3 d^2-3 a c^2 d f+3 a b^2 f^2-3 c f \left (b^2 d-a^2 f\right )\right )\right ) x^2}{2 f^3}+\frac{\left (b^3 B f+3 A b^2 c f-A c^2 (c d-3 a f)-3 b B c (c d-2 a f)\right ) x^3}{3 f^2}+\frac{c \left (3 A b c f-B \left (c^2 d-3 b^2 f-3 a c f\right )\right ) x^4}{4 f^2}+\frac{c^2 (3 b B+A c) x^5}{5 f}+\frac{B c^3 x^6}{6 f}+\frac{\left (b^3 B d^2 f+3 A b^2 d f (c d-a f)-3 b B d (c d-a f)^2-A (c d-a f)^3\right ) \int \frac{1}{d+f x^2} \, dx}{f^3}+\frac{\left (A b f \left (3 c^2 d^2-6 a c d f-f \left (b^2 d-3 a^2 f\right )\right )-B (c d-a f) \left (c^2 d^2-2 a c d f-f \left (3 b^2 d-a^2 f\right )\right )\right ) \int \frac{x}{d+f x^2} \, dx}{f^3}\\ &=-\frac{\left (b^3 B d f+3 A b^2 f (c d-a f)-3 b B (c d-a f)^2-A c \left (c^2 d^2-3 a c d f+3 a^2 f^2\right )\right ) x}{f^3}-\frac{\left (A b f \left (3 c^2 d-b^2 f-6 a c f\right )-B \left (c^3 d^2-3 a c^2 d f+3 a b^2 f^2-3 c f \left (b^2 d-a^2 f\right )\right )\right ) x^2}{2 f^3}+\frac{\left (b^3 B f+3 A b^2 c f-A c^2 (c d-3 a f)-3 b B c (c d-2 a f)\right ) x^3}{3 f^2}+\frac{c \left (3 A b c f-B \left (c^2 d-3 b^2 f-3 a c f\right )\right ) x^4}{4 f^2}+\frac{c^2 (3 b B+A c) x^5}{5 f}+\frac{B c^3 x^6}{6 f}+\frac{\left (b^3 B d^2 f+3 A b^2 d f (c d-a f)-3 b B d (c d-a f)^2-A (c d-a f)^3\right ) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right )}{\sqrt{d} f^{7/2}}+\frac{\left (A b f \left (3 c^2 d^2-6 a c d f-f \left (b^2 d-3 a^2 f\right )\right )-B (c d-a f) \left (c^2 d^2-2 a c d f-f \left (3 b^2 d-a^2 f\right )\right )\right ) \log \left (d+f x^2\right )}{2 f^4}\\ \end{align*}

Mathematica [A]  time = 0.492801, size = 422, normalized size = 0.96 \[ \frac{f x \left (3 b \left (4 B \left (15 a^2 f^2+10 a c f \left (f x^2-3 d\right )+c^2 \left (15 d^2-5 d f x^2+3 f^2 x^4\right )\right )+15 A c f x \left (4 a f-2 c d+c f x^2\right )\right )+c \left (4 A \left (45 a^2 f^2+15 a c f \left (f x^2-3 d\right )+c^2 \left (15 d^2-5 d f x^2+3 f^2 x^4\right )\right )+5 B x \left (18 a^2 f^2+9 a c f \left (f x^2-2 d\right )+c^2 \left (6 d^2-3 d f x^2+2 f^2 x^4\right )\right )\right )+15 b^2 f \left (4 A \left (3 a f-3 c d+c f x^2\right )+3 B x \left (2 a f-2 c d+c f x^2\right )\right )+10 b^3 f \left (3 A f x-6 B d+2 B f x^2\right )\right )-30 \log \left (d+f x^2\right ) \left (A b f \left (-3 a^2 f^2+6 a c d f+b^2 d f-3 c^2 d^2\right )+B (c d-a f) \left (a^2 f^2-2 a c d f-3 b^2 d f+c^2 d^2\right )\right )}{60 f^4}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{d}}\right ) \left (3 A b^2 d f (c d-a f)-A (c d-a f)^3-3 b B d (c d-a f)^2+b^3 B d^2 f\right )}{\sqrt{d} f^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^3)/(d + f*x^2),x]

[Out]

((b^3*B*d^2*f + 3*A*b^2*d*f*(c*d - a*f) - 3*b*B*d*(c*d - a*f)^2 - A*(c*d - a*f)^3)*ArcTan[(Sqrt[f]*x)/Sqrt[d]]
)/(Sqrt[d]*f^(7/2)) + (f*x*(10*b^3*f*(-6*B*d + 3*A*f*x + 2*B*f*x^2) + 15*b^2*f*(3*B*x*(-2*c*d + 2*a*f + c*f*x^
2) + 4*A*(-3*c*d + 3*a*f + c*f*x^2)) + 3*b*(15*A*c*f*x*(-2*c*d + 4*a*f + c*f*x^2) + 4*B*(15*a^2*f^2 + 10*a*c*f
*(-3*d + f*x^2) + c^2*(15*d^2 - 5*d*f*x^2 + 3*f^2*x^4))) + c*(5*B*x*(18*a^2*f^2 + 9*a*c*f*(-2*d + f*x^2) + c^2
*(6*d^2 - 3*d*f*x^2 + 2*f^2*x^4)) + 4*A*(45*a^2*f^2 + 15*a*c*f*(-3*d + f*x^2) + c^2*(15*d^2 - 5*d*f*x^2 + 3*f^
2*x^4)))) - 30*(A*b*f*(-3*c^2*d^2 + b^2*d*f + 6*a*c*d*f - 3*a^2*f^2) + B*(c*d - a*f)*(c^2*d^2 - 3*b^2*d*f - 2*
a*c*d*f + a^2*f^2))*Log[d + f*x^2])/(60*f^4)

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Maple [A]  time = 0.053, size = 822, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^3/(f*x^2+d),x)

[Out]

-3/2/f^2*ln(f*x^2+d)*B*a^2*c*d+6/f^2/(d*f)^(1/2)*arctan(x*f/(d*f)^(1/2))*B*a*b*c*d^2+3/f^2/(d*f)^(1/2)*arctan(
x*f/(d*f)^(1/2))*A*a*c^2*d^2-3/f/(d*f)^(1/2)*arctan(x*f/(d*f)^(1/2))*B*a^2*b*d-3/f^3/(d*f)^(1/2)*arctan(x*f/(d
*f)^(1/2))*B*b*c^2*d^3-6/f^2*B*a*b*c*d*x+3/f^2/(d*f)^(1/2)*arctan(x*f/(d*f)^(1/2))*A*b^2*c*d^2-3/f/(d*f)^(1/2)
*arctan(x*f/(d*f)^(1/2))*A*a^2*c*d-3/f/(d*f)^(1/2)*arctan(x*f/(d*f)^(1/2))*A*a*b^2*d+1/3/f*B*x^3*b^3+1/5/f*A*x
^5*c^3+3/f*a*b^2*A*x+3/2/f*B*x^2*a^2*c+3/f*b*a^2*B*x-1/f^2*b^3*B*d*x+3/2/f*B*x^2*a*b^2+1/f^3*A*c^3*d^2*x+1/2/f
^3*B*x^2*c^3*d^2+3/2/f*ln(f*x^2+d)*A*a^2*b+3/4/f*B*x^4*b^2*c-1/4/f^2*B*x^4*c^3*d-1/3/f^2*A*x^3*c^3*d+3/4/f*A*x
^4*b*c^2+3/4/f*B*x^4*a*c^2+3/5/f*B*x^5*b*c^2+3/f*a^2*c*A*x+1/(d*f)^(1/2)*arctan(x*f/(d*f)^(1/2))*A*a^3+1/2/f*l
n(f*x^2+d)*B*a^3+1/2/f*A*x^2*b^3+1/f*A*x^3*b^2*c+1/f*A*x^3*a*c^2+1/f^2/(d*f)^(1/2)*arctan(x*f/(d*f)^(1/2))*b^3
*B*d^2-3/2/f^2*ln(f*x^2+d)*B*a*b^2*d+3/2/f^3*ln(f*x^2+d)*B*a*c^2*d^2+3/2/f^3*ln(f*x^2+d)*B*b^2*c*d^2-1/2/f^2*l
n(f*x^2+d)*A*b^3*d-1/2/f^4*ln(f*x^2+d)*B*c^3*d^3-3/2/f^2*B*x^2*b^2*c*d-3/f^2*ln(f*x^2+d)*A*a*b*c*d-1/f^3/(d*f)
^(1/2)*arctan(x*f/(d*f)^(1/2))*A*c^3*d^3-3/f^2*A*a*c^2*d*x-3/f^2*A*b^2*c*d*x+3/2/f^3*ln(f*x^2+d)*A*b*c^2*d^2+1
/6*B*c^3*x^6/f+3/f*A*x^2*a*b*c-3/2/f^2*A*x^2*b*c^2*d-3/2/f^2*B*x^2*a*c^2*d-1/f^2*B*x^3*b*c^2*d+3/f^3*B*b*c^2*d
^2*x+2/f*B*x^3*a*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)^3/(f*x^2+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.21747, size = 2164, normalized size = 4.91 \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)^3/(f*x^2+d),x, algorithm="fricas")

[Out]

[1/60*(10*B*c^3*d*f^3*x^6 + 12*(3*B*b*c^2 + A*c^3)*d*f^3*x^5 - 15*(B*c^3*d^2*f^2 - 3*(B*b^2*c + (B*a + A*b)*c^
2)*d*f^3)*x^4 - 20*((3*B*b*c^2 + A*c^3)*d^2*f^2 - (B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*d*f^3)*x^3 + 30*
(B*c^3*d^3*f - 3*(B*b^2*c + (B*a + A*b)*c^2)*d^2*f^2 + (3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*d*f^3)*x^2
- 30*(A*a^3*f^3 - (3*B*b*c^2 + A*c^3)*d^3 + (B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*d^2*f - 3*(B*a^2*b + A
*a*b^2 + A*a^2*c)*d*f^2)*sqrt(-d*f)*log((f*x^2 - 2*sqrt(-d*f)*x - d)/(f*x^2 + d)) + 60*((3*B*b*c^2 + A*c^3)*d^
3*f - (B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*d^2*f^2 + 3*(B*a^2*b + A*a*b^2 + A*a^2*c)*d*f^3)*x - 30*(B*c
^3*d^4 - 3*(B*b^2*c + (B*a + A*b)*c^2)*d^3*f + (3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*d^2*f^2 - (B*a^3 +
3*A*a^2*b)*d*f^3)*log(f*x^2 + d))/(d*f^4), 1/60*(10*B*c^3*d*f^3*x^6 + 12*(3*B*b*c^2 + A*c^3)*d*f^3*x^5 - 15*(B
*c^3*d^2*f^2 - 3*(B*b^2*c + (B*a + A*b)*c^2)*d*f^3)*x^4 - 20*((3*B*b*c^2 + A*c^3)*d^2*f^2 - (B*b^3 + 3*A*a*c^2
+ 3*(2*B*a*b + A*b^2)*c)*d*f^3)*x^3 + 30*(B*c^3*d^3*f - 3*(B*b^2*c + (B*a + A*b)*c^2)*d^2*f^2 + (3*B*a*b^2 +
A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*d*f^3)*x^2 + 60*(A*a^3*f^3 - (3*B*b*c^2 + A*c^3)*d^3 + (B*b^3 + 3*A*a*c^2 + 3*(
2*B*a*b + A*b^2)*c)*d^2*f - 3*(B*a^2*b + A*a*b^2 + A*a^2*c)*d*f^2)*sqrt(d*f)*arctan(sqrt(d*f)*x/d) + 60*((3*B*
b*c^2 + A*c^3)*d^3*f - (B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*d^2*f^2 + 3*(B*a^2*b + A*a*b^2 + A*a^2*c)*d
*f^3)*x - 30*(B*c^3*d^4 - 3*(B*b^2*c + (B*a + A*b)*c^2)*d^3*f + (3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*d^
2*f^2 - (B*a^3 + 3*A*a^2*b)*d*f^3)*log(f*x^2 + d))/(d*f^4)]

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Sympy [B]  time = 25.993, size = 1940, normalized size = 4.4 \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)**3/(f*x**2+d),x)

[Out]

B*c**3*x**6/(6*f) + ((3*A*a**2*b*f**3 - 6*A*a*b*c*d*f**2 - A*b**3*d*f**2 + 3*A*b*c**2*d**2*f + B*a**3*f**3 - 3
*B*a**2*c*d*f**2 - 3*B*a*b**2*d*f**2 + 3*B*a*c**2*d**2*f + 3*B*b**2*c*d**2*f - B*c**3*d**3)/(2*f**4) - sqrt(-d
*f**9)*(A*a**3*f**3 - 3*A*a**2*c*d*f**2 - 3*A*a*b**2*d*f**2 + 3*A*a*c**2*d**2*f + 3*A*b**2*c*d**2*f - A*c**3*d
**3 - 3*B*a**2*b*d*f**2 + 6*B*a*b*c*d**2*f + B*b**3*d**2*f - 3*B*b*c**2*d**3)/(2*d*f**8))*log(x + (-3*A*a**2*b
*d*f**3 + 6*A*a*b*c*d**2*f**2 + A*b**3*d**2*f**2 - 3*A*b*c**2*d**3*f - B*a**3*d*f**3 + 3*B*a**2*c*d**2*f**2 +
3*B*a*b**2*d**2*f**2 - 3*B*a*c**2*d**3*f - 3*B*b**2*c*d**3*f + B*c**3*d**4 + 2*d*f**4*((3*A*a**2*b*f**3 - 6*A*
a*b*c*d*f**2 - A*b**3*d*f**2 + 3*A*b*c**2*d**2*f + B*a**3*f**3 - 3*B*a**2*c*d*f**2 - 3*B*a*b**2*d*f**2 + 3*B*a
*c**2*d**2*f + 3*B*b**2*c*d**2*f - B*c**3*d**3)/(2*f**4) - sqrt(-d*f**9)*(A*a**3*f**3 - 3*A*a**2*c*d*f**2 - 3*
A*a*b**2*d*f**2 + 3*A*a*c**2*d**2*f + 3*A*b**2*c*d**2*f - A*c**3*d**3 - 3*B*a**2*b*d*f**2 + 6*B*a*b*c*d**2*f +
B*b**3*d**2*f - 3*B*b*c**2*d**3)/(2*d*f**8)))/(A*a**3*f**4 - 3*A*a**2*c*d*f**3 - 3*A*a*b**2*d*f**3 + 3*A*a*c*
*2*d**2*f**2 + 3*A*b**2*c*d**2*f**2 - A*c**3*d**3*f - 3*B*a**2*b*d*f**3 + 6*B*a*b*c*d**2*f**2 + B*b**3*d**2*f*
*2 - 3*B*b*c**2*d**3*f)) + ((3*A*a**2*b*f**3 - 6*A*a*b*c*d*f**2 - A*b**3*d*f**2 + 3*A*b*c**2*d**2*f + B*a**3*f
**3 - 3*B*a**2*c*d*f**2 - 3*B*a*b**2*d*f**2 + 3*B*a*c**2*d**2*f + 3*B*b**2*c*d**2*f - B*c**3*d**3)/(2*f**4) +
sqrt(-d*f**9)*(A*a**3*f**3 - 3*A*a**2*c*d*f**2 - 3*A*a*b**2*d*f**2 + 3*A*a*c**2*d**2*f + 3*A*b**2*c*d**2*f - A
*c**3*d**3 - 3*B*a**2*b*d*f**2 + 6*B*a*b*c*d**2*f + B*b**3*d**2*f - 3*B*b*c**2*d**3)/(2*d*f**8))*log(x + (-3*A
*a**2*b*d*f**3 + 6*A*a*b*c*d**2*f**2 + A*b**3*d**2*f**2 - 3*A*b*c**2*d**3*f - B*a**3*d*f**3 + 3*B*a**2*c*d**2*
f**2 + 3*B*a*b**2*d**2*f**2 - 3*B*a*c**2*d**3*f - 3*B*b**2*c*d**3*f + B*c**3*d**4 + 2*d*f**4*((3*A*a**2*b*f**3
- 6*A*a*b*c*d*f**2 - A*b**3*d*f**2 + 3*A*b*c**2*d**2*f + B*a**3*f**3 - 3*B*a**2*c*d*f**2 - 3*B*a*b**2*d*f**2
+ 3*B*a*c**2*d**2*f + 3*B*b**2*c*d**2*f - B*c**3*d**3)/(2*f**4) + sqrt(-d*f**9)*(A*a**3*f**3 - 3*A*a**2*c*d*f*
*2 - 3*A*a*b**2*d*f**2 + 3*A*a*c**2*d**2*f + 3*A*b**2*c*d**2*f - A*c**3*d**3 - 3*B*a**2*b*d*f**2 + 6*B*a*b*c*d
**2*f + B*b**3*d**2*f - 3*B*b*c**2*d**3)/(2*d*f**8)))/(A*a**3*f**4 - 3*A*a**2*c*d*f**3 - 3*A*a*b**2*d*f**3 + 3
*A*a*c**2*d**2*f**2 + 3*A*b**2*c*d**2*f**2 - A*c**3*d**3*f - 3*B*a**2*b*d*f**3 + 6*B*a*b*c*d**2*f**2 + B*b**3*
d**2*f**2 - 3*B*b*c**2*d**3*f)) + x**5*(A*c**3 + 3*B*b*c**2)/(5*f) + x**4*(3*A*b*c**2*f + 3*B*a*c**2*f + 3*B*b
**2*c*f - B*c**3*d)/(4*f**2) + x**3*(3*A*a*c**2*f + 3*A*b**2*c*f - A*c**3*d + 6*B*a*b*c*f + B*b**3*f - 3*B*b*c
**2*d)/(3*f**2) + x**2*(6*A*a*b*c*f**2 + A*b**3*f**2 - 3*A*b*c**2*d*f + 3*B*a**2*c*f**2 + 3*B*a*b**2*f**2 - 3*
B*a*c**2*d*f - 3*B*b**2*c*d*f + B*c**3*d**2)/(2*f**3) + x*(3*A*a**2*c*f**2 + 3*A*a*b**2*f**2 - 3*A*a*c**2*d*f
- 3*A*b**2*c*d*f + A*c**3*d**2 + 3*B*a**2*b*f**2 - 6*B*a*b*c*d*f - B*b**3*d*f + 3*B*b*c**2*d**2)/f**3

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Giac [A]  time = 1.33529, size = 841, normalized size = 1.91 \begin{align*} -\frac{{\left (3 \, B b c^{2} d^{3} + A c^{3} d^{3} - B b^{3} d^{2} f - 6 \, B a b c d^{2} f - 3 \, A b^{2} c d^{2} f - 3 \, A a c^{2} d^{2} f + 3 \, B a^{2} b d f^{2} + 3 \, A a b^{2} d f^{2} + 3 \, A a^{2} c d f^{2} - A a^{3} f^{3}\right )} \arctan \left (\frac{f x}{\sqrt{d f}}\right )}{\sqrt{d f} f^{3}} - \frac{{\left (B c^{3} d^{3} - 3 \, B b^{2} c d^{2} f - 3 \, B a c^{2} d^{2} f - 3 \, A b c^{2} d^{2} f + 3 \, B a b^{2} d f^{2} + A b^{3} d f^{2} + 3 \, B a^{2} c d f^{2} + 6 \, A a b c d f^{2} - B a^{3} f^{3} - 3 \, A a^{2} b f^{3}\right )} \log \left (f x^{2} + d\right )}{2 \, f^{4}} + \frac{10 \, B c^{3} f^{5} x^{6} + 36 \, B b c^{2} f^{5} x^{5} + 12 \, A c^{3} f^{5} x^{5} - 15 \, B c^{3} d f^{4} x^{4} + 45 \, B b^{2} c f^{5} x^{4} + 45 \, B a c^{2} f^{5} x^{4} + 45 \, A b c^{2} f^{5} x^{4} - 60 \, B b c^{2} d f^{4} x^{3} - 20 \, A c^{3} d f^{4} x^{3} + 20 \, B b^{3} f^{5} x^{3} + 120 \, B a b c f^{5} x^{3} + 60 \, A b^{2} c f^{5} x^{3} + 60 \, A a c^{2} f^{5} x^{3} + 30 \, B c^{3} d^{2} f^{3} x^{2} - 90 \, B b^{2} c d f^{4} x^{2} - 90 \, B a c^{2} d f^{4} x^{2} - 90 \, A b c^{2} d f^{4} x^{2} + 90 \, B a b^{2} f^{5} x^{2} + 30 \, A b^{3} f^{5} x^{2} + 90 \, B a^{2} c f^{5} x^{2} + 180 \, A a b c f^{5} x^{2} + 180 \, B b c^{2} d^{2} f^{3} x + 60 \, A c^{3} d^{2} f^{3} x - 60 \, B b^{3} d f^{4} x - 360 \, B a b c d f^{4} x - 180 \, A b^{2} c d f^{4} x - 180 \, A a c^{2} d f^{4} x + 180 \, B a^{2} b f^{5} x + 180 \, A a b^{2} f^{5} x + 180 \, A a^{2} c f^{5} x}{60 \, f^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/(f*x^2+d),x, algorithm="giac")

[Out]

-(3*B*b*c^2*d^3 + A*c^3*d^3 - B*b^3*d^2*f - 6*B*a*b*c*d^2*f - 3*A*b^2*c*d^2*f - 3*A*a*c^2*d^2*f + 3*B*a^2*b*d*
f^2 + 3*A*a*b^2*d*f^2 + 3*A*a^2*c*d*f^2 - A*a^3*f^3)*arctan(f*x/sqrt(d*f))/(sqrt(d*f)*f^3) - 1/2*(B*c^3*d^3 -
3*B*b^2*c*d^2*f - 3*B*a*c^2*d^2*f - 3*A*b*c^2*d^2*f + 3*B*a*b^2*d*f^2 + A*b^3*d*f^2 + 3*B*a^2*c*d*f^2 + 6*A*a*
b*c*d*f^2 - B*a^3*f^3 - 3*A*a^2*b*f^3)*log(f*x^2 + d)/f^4 + 1/60*(10*B*c^3*f^5*x^6 + 36*B*b*c^2*f^5*x^5 + 12*A
*c^3*f^5*x^5 - 15*B*c^3*d*f^4*x^4 + 45*B*b^2*c*f^5*x^4 + 45*B*a*c^2*f^5*x^4 + 45*A*b*c^2*f^5*x^4 - 60*B*b*c^2*
d*f^4*x^3 - 20*A*c^3*d*f^4*x^3 + 20*B*b^3*f^5*x^3 + 120*B*a*b*c*f^5*x^3 + 60*A*b^2*c*f^5*x^3 + 60*A*a*c^2*f^5*
x^3 + 30*B*c^3*d^2*f^3*x^2 - 90*B*b^2*c*d*f^4*x^2 - 90*B*a*c^2*d*f^4*x^2 - 90*A*b*c^2*d*f^4*x^2 + 90*B*a*b^2*f
^5*x^2 + 30*A*b^3*f^5*x^2 + 90*B*a^2*c*f^5*x^2 + 180*A*a*b*c*f^5*x^2 + 180*B*b*c^2*d^2*f^3*x + 60*A*c^3*d^2*f^
3*x - 60*B*b^3*d*f^4*x - 360*B*a*b*c*d*f^4*x - 180*A*b^2*c*d*f^4*x - 180*A*a*c^2*d*f^4*x + 180*B*a^2*b*f^5*x +
180*A*a*b^2*f^5*x + 180*A*a^2*c*f^5*x)/f^6