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

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

[Out]

-(b^3*B*d*f+3*A*b^2*f*(-a*f+c*d)-3*b*B*(-a*f+c*d)^2-A*c*(3*a^2*f^2-3*a*c*d*f+c^2*d^2))*x/f^3-1/2*(A*b*f*(-6*a*
c*f-b^2*f+3*c^2*d)-B*(c^3*d^2-3*a*c^2*d*f+3*a*b^2*f^2-3*c*f*(-a^2*f+b^2*d)))*x^2/f^3+1/3*(b^3*B*f+3*A*b^2*c*f-
A*c^2*(-3*a*f+c*d)-3*b*B*c*(-2*a*f+c*d))*x^3/f^2+1/4*c*(3*A*b*c*f-B*(-3*a*c*f-3*b^2*f+c^2*d))*x^4/f^2+1/5*c^2*
(A*c+3*B*b)*x^5/f+1/6*B*c^3*x^6/f+1/2*(A*b*f*(3*c^2*d^2-6*a*c*d*f-f*(-3*a^2*f+b^2*d))-B*(-a*f+c*d)*(c^2*d^2-2*
a*c*d*f-f*(-a^2*f+3*b^2*d)))*ln(f*x^2+d)/f^4+(b^3*B*d^2*f+3*A*b^2*d*f*(-a*f+c*d)-3*b*B*d*(-a*f+c*d)^2-A*(-a*f+
c*d)^3)*arctan(x*f^(1/2)/d^(1/2))/f^(7/2)/d^(1/2)

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Rubi [A]
time = 0.38, antiderivative size = 441, normalized size of antiderivative = 1.00, 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 = {1026, 649, 211, 266} \begin {gather*} \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^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 {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 {\text {ArcTan}\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 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 {c^2 x^5 (A c+3 b B)}{5 f}+\frac {B c^3 x^6}{6 f} \end {gather*}

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 211

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

Rule 266

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]

Rule 649

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 1026

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])

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*}

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Mathematica [A]
time = 0.33, size = 422, normalized size = 0.96 \begin {gather*} \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 {f x \left (10 b^3 f \left (-6 B d+3 A f x+2 B f x^2\right )+15 b^2 f \left (3 B x \left (-2 c d+2 a f+c f x^2\right )+4 A \left (-3 c d+3 a f+c f x^2\right )\right )+3 b \left (15 A c f x \left (-2 c d+4 a f+c f x^2\right )+4 B \left (15 a^2 f^2+10 a c f \left (-3 d+f x^2\right )+c^2 \left (15 d^2-5 d f x^2+3 f^2 x^4\right )\right )\right )+c \left (5 B x \left (18 a^2 f^2+9 a c f \left (-2 d+f x^2\right )+c^2 \left (6 d^2-3 d f x^2+2 f^2 x^4\right )\right )+4 A \left (45 a^2 f^2+15 a c f \left (-3 d+f x^2\right )+c^2 \left (15 d^2-5 d f x^2+3 f^2 x^4\right )\right )\right )\right )-30 \left (A b f \left (-3 c^2 d^2+b^2 d f+6 a c d f-3 a^2 f^2\right )+B (c d-a f) \left (c^2 d^2-3 b^2 d f-2 a c d f+a^2 f^2\right )\right ) \log \left (d+f x^2\right )}{60 f^4} \end {gather*}

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 = 0.20, size = 592, normalized size = 1.34

method	result	size
default	\(\frac {A \,b^{2} c \,f^{2} x^{3}+A a \,c^{2} f^{2} x^{3}+\frac {3}{4} A b \,c^{2} f^{2} x^{4}+\frac {3}{4} B a \,c^{2} f^{2} x^{4}+\frac {3}{4} B \,b^{2} c \,f^{2} x^{4}-\frac {1}{4} B \,c^{3} d f \,x^{4}-\frac {1}{3} A \,c^{3} d f \,x^{3}+\frac {3}{2} B \,a^{2} c \,f^{2} x^{2}+\frac {3}{2} B a \,b^{2} f^{2} x^{2}+3 A \,a^{2} c \,f^{2} x +3 A a \,b^{2} f^{2} x +3 B \,a^{2} b \,f^{2} x -b^{3} B d f x +3 B b \,c^{2} d^{2} x +\frac {3}{5} B b \,c^{2} f^{2} x^{5}+\frac {1}{2} B \,c^{3} d^{2} x^{2}+A \,c^{3} d^{2} x +\frac {1}{3} B \,b^{3} f^{2} x^{3}+\frac {1}{2} A \,b^{3} f^{2} x^{2}+\frac {1}{5} A \,c^{3} f^{2} x^{5}+\frac {1}{6} B \,c^{3} x^{6} f^{2}-3 A \,b^{2} c d f x -\frac {3}{2} B \,b^{2} c d f \,x^{2}-3 A a \,c^{2} d f x +2 B a b c \,f^{2} x^{3}-B b \,c^{2} d f \,x^{3}+3 A a b c \,f^{2} x^{2}-\frac {3}{2} A b \,c^{2} d f \,x^{2}-\frac {3}{2} B a \,c^{2} d f \,x^{2}-6 B a b c d f x}{f^{3}}+\frac {\frac {\left (3 A \,a^{2} b \,f^{3}-6 A a b c d \,f^{2}-A \,b^{3} d \,f^{2}+3 A b \,c^{2} d^{2} f +B \,a^{3} f^{3}-3 B \,a^{2} c d \,f^{2}-3 B a \,b^{2} d \,f^{2}+3 B a \,c^{2} d^{2} f +3 B \,b^{2} c \,d^{2} f -B \,c^{3} d^{3}\right ) \ln \left (f \,x^{2}+d \right )}{2 f}+\frac {\left (A \,a^{3} f^{3}-3 A \,a^{2} c d \,f^{2}-3 A a \,b^{2} d \,f^{2}+3 A a \,c^{2} d^{2} f +3 A \,b^{2} c \,d^{2} f -A \,c^{3} d^{3}-3 B \,a^{2} b d \,f^{2}+6 B a b c \,d^{2} f +b^{3} B \,d^{2} f -3 B b \,c^{2} d^{3}\right ) \arctan \left (\frac {f x}{\sqrt {d f}}\right )}{\sqrt {d f}}}{f^{3}}\)	\(592\)
risch	\(\text {Expression too large to display}\)	\(5733\)

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,method=_RETURNVERBOSE)

[Out]

1/f^3*(A*b^2*c*f^2*x^3+A*a*c^2*f^2*x^3+3/4*A*b*c^2*f^2*x^4+3/4*B*a*c^2*f^2*x^4+3/4*B*b^2*c*f^2*x^4-1/4*B*c^3*d
*f*x^4-1/3*A*c^3*d*f*x^3+3/2*B*a^2*c*f^2*x^2+3/2*B*a*b^2*f^2*x^2+3*A*a^2*c*f^2*x+3*A*a*b^2*f^2*x+3*B*a^2*b*f^2
*x-b^3*B*d*f*x+3*B*b*c^2*d^2*x+3/5*B*b*c^2*f^2*x^5+1/2*B*c^3*d^2*x^2+A*c^3*d^2*x+1/3*B*b^3*f^2*x^3+1/2*A*b^3*f
^2*x^2+1/5*A*c^3*f^2*x^5+1/6*B*c^3*x^6*f^2-3*A*b^2*c*d*f*x-3/2*B*b^2*c*d*f*x^2-3*A*a*c^2*d*f*x+2*B*a*b*c*f^2*x
^3-B*b*c^2*d*f*x^3+3*A*a*b*c*f^2*x^2-3/2*A*b*c^2*d*f*x^2-3/2*B*a*c^2*d*f*x^2-6*B*a*b*c*d*f*x)+1/f^3*(1/2*(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)/f*ln(f*x^2+d)+(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)/(d*f)^(1/2)*arctan(f*x/(d*f)^(1/2
)))

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Maxima [A]
time = 0.52, size = 471, normalized size = 1.07 \begin {gather*} \frac {{\left (A a^{3} f^{3} - {\left (3 \, B b c^{2} + A c^{3}\right )} d^{3} + {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d^{2} f - 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} d f^{2}\right )} \arctan \left (\frac {f x}{\sqrt {d f}}\right )}{\sqrt {d f} f^{3}} + \frac {10 \, B c^{3} f^{2} x^{6} + 12 \, {\left (3 \, B b c^{2} + A c^{3}\right )} f^{2} x^{5} - 15 \, {\left (B c^{3} d f - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} f^{2}\right )} x^{4} - 20 \, {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} d f - {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} f^{2}\right )} x^{3} + 30 \, {\left (B c^{3} d^{2} - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d f + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} f^{2}\right )} x^{2} + 60 \, {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} d^{2} - {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d f + 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} f^{2}\right )} x}{60 \, f^{3}} - \frac {{\left (B c^{3} d^{3} - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d^{2} f + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} d f^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} f^{3}\right )} \log \left (f x^{2} + d\right )}{2 \, f^{4}} \end {gather*}

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]

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

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Fricas [A]
time = 0.33, size = 1014, normalized size = 2.30 \begin {gather*} \left [\frac {10 \, B c^{3} d f^{3} x^{6} + 12 \, {\left (3 \, B b c^{2} + A c^{3}\right )} d f^{3} x^{5} - 15 \, {\left (B c^{3} d^{2} f^{2} - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d f^{3}\right )} x^{4} - 20 \, {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} d^{2} f^{2} - {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d f^{3}\right )} x^{3} + 30 \, {\left (B c^{3} d^{3} f - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d^{2} f^{2} + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} d f^{3}\right )} x^{2} - 30 \, {\left (A a^{3} f^{3} - {\left (3 \, B b c^{2} + A c^{3}\right )} d^{3} + {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d^{2} f - 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} d f^{2}\right )} \sqrt {-d f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-d f} x - d}{f x^{2} + d}\right ) + 60 \, {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} d^{3} f - {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d^{2} f^{2} + 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} d f^{3}\right )} x - 30 \, {\left (B c^{3} d^{4} - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d^{3} f + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} d^{2} f^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d f^{3}\right )} \log \left (f x^{2} + d\right )}{60 \, d f^{4}}, \frac {10 \, B c^{3} d f^{3} x^{6} + 12 \, {\left (3 \, B b c^{2} + A c^{3}\right )} d f^{3} x^{5} - 15 \, {\left (B c^{3} d^{2} f^{2} - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d f^{3}\right )} x^{4} - 20 \, {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} d^{2} f^{2} - {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d f^{3}\right )} x^{3} + 30 \, {\left (B c^{3} d^{3} f - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d^{2} f^{2} + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} d f^{3}\right )} x^{2} + 60 \, {\left (A a^{3} f^{3} - {\left (3 \, B b c^{2} + A c^{3}\right )} d^{3} + {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d^{2} f - 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} d f^{2}\right )} \sqrt {d f} \arctan \left (\frac {\sqrt {d f} x}{d}\right ) + 60 \, {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} d^{3} f - {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} d^{2} f^{2} + 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} d f^{3}\right )} x - 30 \, {\left (B c^{3} d^{4} - 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} d^{3} f + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} d^{2} f^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d f^{3}\right )} \log \left (f x^{2} + d\right )}{60 \, d f^{4}}\right ] \end {gather*}

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 [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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Giac [A]
time = 4.02, size = 623, normalized size = 1.41 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 3.79, size = 552, normalized size = 1.25 \begin {gather*} x^2\,\left (\frac {3\,B\,c\,a^2+3\,B\,a\,b^2+6\,A\,c\,a\,b+A\,b^3}{2\,f}-\frac {d\,\left (\frac {3\,B\,b^2\,c+3\,A\,b\,c^2+3\,B\,a\,c^2}{f}-\frac {B\,c^3\,d}{f^2}\right )}{2\,f}\right )+x\,\left (\frac {3\,B\,a^2\,b+3\,A\,c\,a^2+3\,A\,a\,b^2}{f}-\frac {d\,\left (\frac {B\,b^3+3\,A\,b^2\,c+6\,B\,a\,b\,c+3\,A\,a\,c^2}{f}-\frac {d\,\left (A\,c^3+3\,B\,b\,c^2\right )}{f^2}\right )}{f}\right )+x^3\,\left (\frac {B\,b^3+3\,A\,b^2\,c+6\,B\,a\,b\,c+3\,A\,a\,c^2}{3\,f}-\frac {d\,\left (A\,c^3+3\,B\,b\,c^2\right )}{3\,f^2}\right )+x^4\,\left (\frac {3\,B\,b^2\,c+3\,A\,b\,c^2+3\,B\,a\,c^2}{4\,f}-\frac {B\,c^3\,d}{4\,f^2}\right )+\frac {x^5\,\left (A\,c^3+3\,B\,b\,c^2\right )}{5\,f}+\frac {B\,c^3\,x^6}{6\,f}+\frac {\ln \left (f\,x^2+d\right )\,\left (4\,B\,a^3\,d\,f^7+12\,A\,a^2\,b\,d\,f^7-12\,B\,a^2\,c\,d^2\,f^6-12\,B\,a\,b^2\,d^2\,f^6-24\,A\,a\,b\,c\,d^2\,f^6+12\,B\,a\,c^2\,d^3\,f^5-4\,A\,b^3\,d^2\,f^6+12\,B\,b^2\,c\,d^3\,f^5+12\,A\,b\,c^2\,d^3\,f^5-4\,B\,c^3\,d^4\,f^4\right )}{8\,d\,f^8}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {d}}\right )\,\left (A\,a^3\,f^3-3\,B\,a^2\,b\,d\,f^2-3\,A\,a^2\,c\,d\,f^2-3\,A\,a\,b^2\,d\,f^2+6\,B\,a\,b\,c\,d^2\,f+3\,A\,a\,c^2\,d^2\,f+B\,b^3\,d^2\,f+3\,A\,b^2\,c\,d^2\,f-3\,B\,b\,c^2\,d^3-A\,c^3\,d^3\right )}{\sqrt {d}\,f^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*((A*b^3 + 3*B*a*b^2 + 3*B*a^2*c + 6*A*a*b*c)/(2*f) - (d*((3*A*b*c^2 + 3*B*a*c^2 + 3*B*b^2*c)/f - (B*c^3*d)
/f^2))/(2*f)) + x*((3*A*a*b^2 + 3*A*a^2*c + 3*B*a^2*b)/f - (d*((B*b^3 + 3*A*a*c^2 + 3*A*b^2*c + 6*B*a*b*c)/f -
(d*(A*c^3 + 3*B*b*c^2))/f^2))/f) + x^3*((B*b^3 + 3*A*a*c^2 + 3*A*b^2*c + 6*B*a*b*c)/(3*f) - (d*(A*c^3 + 3*B*b
*c^2))/(3*f^2)) + x^4*((3*A*b*c^2 + 3*B*a*c^2 + 3*B*b^2*c)/(4*f) - (B*c^3*d)/(4*f^2)) + (x^5*(A*c^3 + 3*B*b*c^
2))/(5*f) + (B*c^3*x^6)/(6*f) + (log(d + f*x^2)*(4*B*a^3*d*f^7 - 4*A*b^3*d^2*f^6 - 4*B*c^3*d^4*f^4 - 12*B*a*b^
2*d^2*f^6 + 12*A*b*c^2*d^3*f^5 + 12*B*a*c^2*d^3*f^5 - 12*B*a^2*c*d^2*f^6 + 12*B*b^2*c*d^3*f^5 + 12*A*a^2*b*d*f
^7 - 24*A*a*b*c*d^2*f^6))/(8*d*f^8) + (atan((f^(1/2)*x)/d^(1/2))*(A*a^3*f^3 - A*c^3*d^3 - 3*B*b*c^2*d^3 + B*b^
3*d^2*f - 3*A*a*b^2*d*f^2 + 3*A*a*c^2*d^2*f - 3*A*a^2*c*d*f^2 - 3*B*a^2*b*d*f^2 + 3*A*b^2*c*d^2*f + 6*B*a*b*c*
d^2*f))/(d^(1/2)*f^(7/2))

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