3.121 \(\int \frac{x^3}{(a+b x+c x^2)^{3/2} (d+e x+f x^2)} \, dx\)

Optimal. Leaf size=779 \[ \frac{2 \left (c x \left (\left (e^2-d f\right ) (a b f-2 a c e+b c d)-d e \left (-c (2 a f+b e)+b^2 f+2 c^2 d\right )\right )-\left (a d f-a e^2+b d e\right ) \left (-c (2 a f+b e)+b^2 f+2 c^2 d\right )+c d e (a b f-2 a c e+b c d)\right )}{f^2 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}+\frac{2 e (b+2 c x)}{f^2 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 (2 a+b x)}{f \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{\left (\left (e-\sqrt{e^2-4 d f}\right ) \left (a \left (e^2-d f\right )-b d e+c d^2\right )+2 d f (b d-a e)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{\left (\left (\sqrt{e^2-4 d f}+e\right ) \left (a \left (e^2-d f\right )-b d e+c d^2\right )+2 d f (b d-a e)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 14.1698, antiderivative size = 779, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {6728, 613, 636, 1016, 1032, 724, 206} \[ \frac{2 \left (c x \left (\left (e^2-d f\right ) (a b f-2 a c e+b c d)-d e \left (-c (2 a f+b e)+b^2 f+2 c^2 d\right )\right )-\left (a d f-a e^2+b d e\right ) \left (-c (2 a f+b e)+b^2 f+2 c^2 d\right )+c d e (a b f-2 a c e+b c d)\right )}{f^2 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}+\frac{2 e (b+2 c x)}{f^2 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 (2 a+b x)}{f \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{\left (\left (e-\sqrt{e^2-4 d f}\right ) \left (a \left (e^2-d f\right )-b d e+c d^2\right )+2 d f (b d-a e)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{\left (\left (\sqrt{e^2-4 d f}+e\right ) \left (a \left (e^2-d f\right )-b d e+c d^2\right )+2 d f (b d-a e)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rule 1016

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Sy
mbol] :> Simp[((a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1)*(g*c*(2*a*c*e - b*(c*d + a*f)) + (g*b - a*h
)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) - h*(b*c*d - 2*a*c*e + a*b*f)
)*x))/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), x] + Dist[1/((b^2 - 4*a*c)*((c*d - a*
f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Simp[(b*h - 2*g*c)
*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1) + (b^2*(g*f) - b*(h*c*d + g*c*e + a*h*f) + 2*(g*c*(c*d - a*
f) - a*(-(h*c*e))))*(a*f*(p + 1) - c*d*(p + 2)) - e*((g*c)*(2*a*c*e - b*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d +
b^2*f - c*(b*e + 2*a*f)))*(p + q + 2) - (2*f*((g*c)*(2*a*c*e - b*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d + b^2*f -
c*(b*e + 2*a*f)))*(p + q + 2) - (b^2*g*f - b*(h*c*d + g*c*e + a*h*f) + 2*(g*c*(c*d - a*f) - a*(-(h*c*e))))*(b
*f*(p + 1) - c*e*(2*p + q + 4)))*x - c*f*(b^2*(g*f) - b*(h*c*d + g*c*e + a*h*f) + 2*(g*c*(c*d - a*f) + a*h*c*e
))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2
- 4*d*f, 0] && LtQ[p, -1] && NeQ[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 0] &&  !( !IntegerQ[p] && ILtQ[q, -1
])

Rule 1032

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbo
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + e*x + f*x^2])
, x], x] - Dist[(2*c*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b,
c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && PosQ[b^2 - 4*a*c]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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])

Rubi steps

\begin{align*} \int \frac{x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=\int \left (-\frac{e}{f^2 \left (a+b x+c x^2\right )^{3/2}}+\frac{x}{f \left (a+b x+c x^2\right )^{3/2}}+\frac{d e+\left (e^2-d f\right ) x}{f^2 \left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{d e+\left (e^2-d f\right ) x}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx}{f^2}-\frac{e \int \frac{1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{f^2}+\frac{\int \frac{x}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{f}\\ &=\frac{2 (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt{a+b x+c x^2}}+\frac{2 e (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt{a+b x+c x^2}}+\frac{2 \left (c d e (b c d-2 a c e+a b f)-\left (b d e-a e^2+a d f\right ) \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )+c \left ((b c d-2 a c e+a b f) \left (e^2-d f\right )-d e \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{a+b x+c x^2}}-\frac{2 \int \frac{\frac{1}{2} \left (b^2-4 a c\right ) d (b d-a e) f^2-\frac{1}{2} \left (b^2-4 a c\right ) f^2 \left (c d^2-b d e+a e^2-a d f\right ) x}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{\left (b^2-4 a c\right ) f^2 \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}\\ &=\frac{2 (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt{a+b x+c x^2}}+\frac{2 e (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt{a+b x+c x^2}}+\frac{2 \left (c d e (b c d-2 a c e+a b f)-\left (b d e-a e^2+a d f\right ) \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )+c \left ((b c d-2 a c e+a b f) \left (e^2-d f\right )-d e \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{a+b x+c x^2}}-\frac{\left (2 d (b d-a e) f+\left (e-\sqrt{e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right ) \int \frac{1}{\left (e-\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+b x+c x^2}} \, dx}{\sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}+\frac{\left (2 d (b d-a e) f+\left (e+\sqrt{e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right ) \int \frac{1}{\left (e+\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+b x+c x^2}} \, dx}{\sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}\\ &=\frac{2 (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt{a+b x+c x^2}}+\frac{2 e (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt{a+b x+c x^2}}+\frac{2 \left (c d e (b c d-2 a c e+a b f)-\left (b d e-a e^2+a d f\right ) \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )+c \left ((b c d-2 a c e+a b f) \left (e^2-d f\right )-d e \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{a+b x+c x^2}}+\frac{\left (2 \left (2 d (b d-a e) f+\left (e-\sqrt{e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e-\sqrt{e^2-4 d f}\right )+4 c \left (e-\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{\sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}-\frac{\left (2 \left (2 d (b d-a e) f+\left (e+\sqrt{e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e+\sqrt{e^2-4 d f}\right )+4 c \left (e+\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{\sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}\\ &=\frac{2 (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt{a+b x+c x^2}}+\frac{2 e (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt{a+b x+c x^2}}+\frac{2 \left (c d e (b c d-2 a c e+a b f)-\left (b d e-a e^2+a d f\right ) \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )+c \left ((b c d-2 a c e+a b f) \left (e^2-d f\right )-d e \left (2 c^2 d+b^2 f-c (b e+2 a f)\right )\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{a+b x+c x^2}}+\frac{\left (2 d (b d-a e) f+\left (e-\sqrt{e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt{e^2-4 d f}}}-\frac{\left (2 d (b d-a e) f+\left (e+\sqrt{e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} \sqrt{e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt{c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt{e^2-4 d f}}}\\ \end{align*}

Mathematica [A]  time = 2.91359, size = 1066, normalized size = 1.37 \[ \frac{\frac{4 \left (2 f a^3+(-2 c d-b e+2 c e x+b f x) a^2+b (b (d-e x)-3 c d x) a+b^3 d x\right )}{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)}}+\frac{\sqrt{2} \left (c \left (\sqrt{e^2-4 d f}-e\right ) d^2+b \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right ) d+a \left (-e^3+\sqrt{e^2-4 d f} e^2+3 d f e-d f \sqrt{e^2-4 d f}\right )\right ) \log \left (-e-2 f x+\sqrt{e^2-4 d f}\right )}{\sqrt{e^2-4 d f} \sqrt{c \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right )+f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )}}+\frac{\sqrt{2} \left (c \left (e+\sqrt{e^2-4 d f}\right ) d^2-b \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right ) d+a \left (e^3+\sqrt{e^2-4 d f} e^2-3 d f e-d f \sqrt{e^2-4 d f}\right )\right ) \log \left (e+2 f x+\sqrt{e^2-4 d f}\right )}{\sqrt{e^2-4 d f} \sqrt{c \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right )+f \left (2 a f-b \left (e+\sqrt{e^2-4 d f}\right )\right )}}-\frac{\sqrt{2} \left (c \left (e+\sqrt{e^2-4 d f}\right ) d^2-b \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right ) d+a \left (e^3+\sqrt{e^2-4 d f} e^2-3 d f e-d f \sqrt{e^2-4 d f}\right )\right ) \log \left (-4 a f+2 c e x+2 c \sqrt{e^2-4 d f} x+b \left (e-2 f x+\sqrt{e^2-4 d f}\right )-2 \sqrt{2} \sqrt{c \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right )+f \left (2 a f-b \left (e+\sqrt{e^2-4 d f}\right )\right )} \sqrt{a+x (b+c x)}\right )}{\sqrt{e^2-4 d f} \sqrt{c \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right )+f \left (2 a f-b \left (e+\sqrt{e^2-4 d f}\right )\right )}}-\frac{\sqrt{2} \left (c \left (\sqrt{e^2-4 d f}-e\right ) d^2+b \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right ) d+a \left (-e^3+\sqrt{e^2-4 d f} e^2+3 d f e-d f \sqrt{e^2-4 d f}\right )\right ) \log \left (b \left (-e+2 f x+\sqrt{e^2-4 d f}\right )+2 \left (2 a f-c e x+c \sqrt{e^2-4 d f} x+\sqrt{2} \sqrt{f \left (-e b+\sqrt{e^2-4 d f} b+2 a f\right )+c \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right )} \sqrt{a+x (b+c x)}\right )\right )}{\sqrt{e^2-4 d f} \sqrt{c \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right )+f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )}}}{2 \left (c^2 d^2-b c e d+f \left (f a^2-b e a+b^2 d\right )+a c \left (e^2-2 d f\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*(2*a^3*f + b^3*d*x + a^2*(-2*c*d - b*e + 2*c*e*x + b*f*x) + a*b*(-3*c*d*x + b*(d - e*x))))/((b^2 - 4*a*c)*
Sqrt[a + x*(b + c*x)]) + (Sqrt[2]*(c*d^2*(-e + Sqrt[e^2 - 4*d*f]) + b*d*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]) +
a*(-e^3 + 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sqrt[e^2 - 4*d*f]))*Log[-e + Sqrt[e^2 - 4*d*f] - 2*f*x])/(Sqrt
[e^2 - 4*d*f]*Sqrt[c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f + b*(-e + Sqrt[e^2 - 4*d*f]))]) + (Sqrt[2]
*(c*d^2*(e + Sqrt[e^2 - 4*d*f]) - b*d*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]) + a*(e^3 - 3*d*e*f + e^2*Sqrt[e^2 -
4*d*f] - d*f*Sqrt[e^2 - 4*d*f]))*Log[e + Sqrt[e^2 - 4*d*f] + 2*f*x])/(Sqrt[e^2 - 4*d*f]*Sqrt[c*(e^2 - 2*d*f +
e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e + Sqrt[e^2 - 4*d*f]))]) - (Sqrt[2]*(c*d^2*(e + Sqrt[e^2 - 4*d*f]) - b*d
*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]) + a*(e^3 - 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sqrt[e^2 - 4*d*f]))*Log[
-4*a*f + 2*c*e*x + 2*c*Sqrt[e^2 - 4*d*f]*x + b*(e + Sqrt[e^2 - 4*d*f] - 2*f*x) - 2*Sqrt[2]*Sqrt[c*(e^2 - 2*d*f
+ e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e + Sqrt[e^2 - 4*d*f]))]*Sqrt[a + x*(b + c*x)]])/(Sqrt[e^2 - 4*d*f]*Sq
rt[c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e + Sqrt[e^2 - 4*d*f]))]) - (Sqrt[2]*(c*d^2*(-e + Sqr
t[e^2 - 4*d*f]) + b*d*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]) + a*(-e^3 + 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sq
rt[e^2 - 4*d*f]))*Log[b*(-e + Sqrt[e^2 - 4*d*f] + 2*f*x) + 2*(2*a*f - c*e*x + c*Sqrt[e^2 - 4*d*f]*x + Sqrt[2]*
Sqrt[f*(-(b*e) + 2*a*f + b*Sqrt[e^2 - 4*d*f]) + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + x*(b + c*x)])]
)/(Sqrt[e^2 - 4*d*f]*Sqrt[c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f + b*(-e + Sqrt[e^2 - 4*d*f]))]))/(2
*(c^2*d^2 - b*c*d*e + f*(b^2*d - a*b*e + a^2*f) + a*c*(e^2 - 2*d*f)))

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Maple [B]  time = 0.359, size = 14651, normalized size = 18.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + b x + c x^{2}\right )^{\frac{3}{2}} \left (d + e x + f x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(c*x**2+b*x+a)**(3/2)/(f*x**2+e*x+d),x)

[Out]

Integral(x**3/((a + b*x + c*x**2)**(3/2)*(d + e*x + f*x**2)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out