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

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

________________________________________________________________________________________

Rubi [A]  time = 14.17, antiderivative size = 779, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.62, size = 1066, normalized size = 1.37 \begin {gather*} \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 )} \end {gather*}

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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IntegrateAlgebraic [C]  time = 1.69, size = 729, normalized size = 0.94 \begin {gather*} \frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+\text {$\#$1}^2 b e+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e-4 \text {$\#$1} b \sqrt {c} d+a^2 f-a b e+b^2 d\&,\frac {-\text {$\#$1}^2 c d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 b d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 a d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-\text {$\#$1}^2 a e^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a^2 d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a^2 e^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^2 d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a c d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b \sqrt {c} d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 a b d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} a \sqrt {c} d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{-2 \text {$\#$1}^3 f+3 \text {$\#$1}^2 \sqrt {c} e+2 \text {$\#$1} a f-\text {$\#$1} b e-4 \text {$\#$1} c d-a \sqrt {c} e+2 b \sqrt {c} d}\&\right ]}{a^2 f^2-a b e f-2 a c d f+a c e^2+b^2 d f-b c d e+c^2 d^2}+\frac {2 \left (2 a^3 f-a^2 b e+a^2 b f x-2 a^2 c d+2 a^2 c e x+a b^2 d-a b^2 e x-3 a b c d x+b^3 d x\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a^2 f^2-a b e f-2 a c d f+a c e^2+b^2 d f-b c d e+c^2 d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*b^2*d - 2*a^2*c*d - a^2*b*e + 2*a^3*f + b^3*d*x - 3*a*b*c*d*x - a*b^2*e*x + 2*a^2*c*e*x + a^2*b*f*x))/((
b^2 - 4*a*c)*(c^2*d^2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b*e*f + a^2*f^2)*Sqrt[a + b*x + c*x^2]) +
RootSum[b^2*d - a*b*e + a^2*f - 4*b*Sqrt[c]*d*#1 + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 + b*e*#1^2 - 2*a*f*#1^2 - 2*S
qrt[c]*e*#1^3 + f*#1^4 & , (b^2*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + a*c*d^2*Log[-(Sqrt[c]*x)
+ Sqrt[a + b*x + c*x^2] - #1] - 2*a*b*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + a^2*e^2*Log[-(Sqrt[
c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^2*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b*Sqrt[c]*d^2
*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*a*Sqrt[c]*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2]
- #1]*#1 - c*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + b*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x +
c*x^2] - #1]*#1^2 - a*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + a*d*f*Log[-(Sqrt[c]*x) + Sqrt
[a + b*x + c*x^2] - #1]*#1^2)/(2*b*Sqrt[c]*d - a*Sqrt[c]*e - 4*c*d*#1 - b*e*#1 + 2*a*f*#1 + 3*Sqrt[c]*e*#1^2 -
2*f*#1^3) & ]/(c^2*d^2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b*e*f + a^2*f^2)

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fricas [F(-1)]  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(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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giac [F(-1)]  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(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.04, size = 14651, normalized size = 18.81 \begin {gather*} \text {output too large to display} \end {gather*}

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` f
or more details)Is 4*d*f-e^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3}{{\left (c\,x^2+b\,x+a\right )}^{3/2}\,\left (f\,x^2+e\,x+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  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(x**3/(c*x**2+b*x+a)**(3/2)/(f*x**2+e*x+d),x)

[Out]

Timed out

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