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

Optimal. Leaf size=609 \[ -\frac {2 \left (c x \left (-a b e-2 a (c d-a f)+b^2 d\right )+a (a b f-2 a c e+b c d)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}-\frac {f \left (2 d (c d-a f)-\left (e-\sqrt {e^2-4 d f}\right ) (b d-a e)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {f \left (2 d (c d-a f)-\left (\sqrt {e^2-4 d f}+e\right ) (b d-a e)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}} \]

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Rubi [A]  time = 5.84, antiderivative size = 609, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1061, 1032, 724, 206} \begin {gather*} -\frac {2 \left (c x \left (-a b e-2 a (c d-a f)+b^2 d\right )+a (a b f-2 a c e+b c d)\right )}{\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 {f \left (2 d (c d-a f)-\left (e-\sqrt {e^2-4 d f}\right ) (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 {f \left (2 d (c d-a f)-\left (\sqrt {e^2-4 d f}+e\right ) (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^2/((a + b*x + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]

[Out]

(-2*(a*(b*c*d - 2*a*c*e + a*b*f) + c*(b^2*d - a*b*e - 2*a*(c*d - a*f))*x))/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*
d - a*e)*(c*e - b*f))*Sqrt[a + b*x + c*x^2]) - (f*(2*d*(c*d - a*f) - (b*d - a*e)*(e - Sqrt[e^2 - 4*d*f]))*ArcT
anh[(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]])
+ (f*(2*d*(c*d - a*f) - (b*d - a*e)*(e + Sqrt[e^2 - 4*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 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 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 1061

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((A_.) + (C_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_
Symbol] :> Simp[((a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1)*((A*c - a*C)*(2*a*c*e - b*(c*d + a*f)) +
(A*b)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) + c*(A*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) + C*(b^2*d - a*b*e - 2*a*
(c*d - a*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[(
-2*A*c - 2*a*C)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1) + (b^2*(C*d + A*f) - b*((Plus[A])*c*e + a*C*
e) + 2*(A*c*(c*d - a*f) - a*(c*C*d - a*C*f)))*(a*f*(p + 1) - c*d*(p + 2)) - e*((A*c - a*C)*(2*a*c*e - b*(c*d +
a*f)) + (A*b)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)))*(p + q + 2) - (2*f*((A*c - a*C)*(2*a*c*e - b*(c*d + a*f))
+ (A*b)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)))*(p + q + 2) - (b^2*(C*d + A*f) - b*(A*c*e + a*C*e) + 2*(A*c*(c*d
- a*f) - a*(c*C*d - a*C*f)))*(b*f*(p + 1) - c*e*(2*p + q + 4)))*x - c*f*(b^2*(C*d + A*f) - b*(A*c*e + a*C*e) +
2*(A*c*(c*d - a*f) - a*(c*C*d - a*C*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, 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]) &&  !IGtQ[q, 0]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=-\frac {2 \left (a (b c d-2 a c e+a b f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x\right )}{\left (b^2-4 a c\right ) \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 (c d-a f)-\frac {1}{2} \left (b^2-4 a c\right ) (b d-a e) f x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}\\ &=-\frac {2 \left (a (b c d-2 a c e+a b f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x\right )}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {a+b x+c x^2}}+\frac {\left (f \left (2 d (c d-a f)-(b d-a e) \left (e-\sqrt {e^2-4 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 (f \left (2 d (c d-a f)-(b d-a e) \left (e+\sqrt {e^2-4 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 \left (a (b c d-2 a c e+a b f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x\right )}{\left (b^2-4 a c\right ) \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 f \left (2 d (c d-a f)-(b d-a e) \left (e-\sqrt {e^2-4 d f}\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 f \left (2 d (c d-a f)-(b d-a e) \left (e+\sqrt {e^2-4 d f}\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 \left (a (b c d-2 a c e+a b f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x\right )}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {a+b x+c x^2}}-\frac {f \left (2 d (c d-a f)-(b d-a e) \left (e-\sqrt {e^2-4 d f}\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 {f \left (2 d (c d-a f)-(b d-a e) \left (e+\sqrt {e^2-4 d f}\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 = 6.59, size = 1097, normalized size = 1.80 \begin {gather*} \frac {16 \sqrt {2} f \sqrt {c e^2-b f e-c \sqrt {e^2-4 d f} e+2 a f^2-2 c d f+b f \sqrt {e^2-4 d f}} \left (e+\frac {2 d f-e^2}{\sqrt {e^2-4 d f}}\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )-\left (2 c \left (e-\sqrt {e^2-4 d f}\right )-2 b f\right ) x}{2 \sqrt {2} \sqrt {c e^2-b f e-c \sqrt {e^2-4 d f} e+2 a f^2-2 c d f+b f \sqrt {e^2-4 d f}} \sqrt {c x^2+b x+a}}\right ) \left (c x^2+b x+a\right )^{3/2}}{\left (4 a f^2-2 b \left (e-\sqrt {e^2-4 d f}\right ) f+c \left (e-\sqrt {e^2-4 d f}\right )^2\right ) \left (16 a f^2-8 b \left (e-\sqrt {e^2-4 d f}\right ) f+4 c \left (e-\sqrt {e^2-4 d f}\right )^2\right ) (a+x (b+c x))^{3/2}}+\frac {16 \sqrt {2} f \sqrt {c e^2-b f e+c \sqrt {e^2-4 d f} e+2 a f^2-2 c d f-b f \sqrt {e^2-4 d f}} \left (e-\frac {2 d f-e^2}{\sqrt {e^2-4 d f}}\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )-\left (2 c \left (e+\sqrt {e^2-4 d f}\right )-2 b f\right ) x}{2 \sqrt {2} \sqrt {c e^2-b f e+c \sqrt {e^2-4 d f} e+2 a f^2-2 c d f-b f \sqrt {e^2-4 d f}} \sqrt {c x^2+b x+a}}\right ) \left (c x^2+b x+a\right )^{3/2}}{\left (4 a f^2-2 b \left (e+\sqrt {e^2-4 d f}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )^2\right ) \left (16 a f^2-8 b \left (e+\sqrt {e^2-4 d f}\right ) f+4 c \left (e+\sqrt {e^2-4 d f}\right )^2\right ) (a+x (b+c x))^{3/2}}-\frac {2 \left (e-\frac {e^2-2 d f}{\sqrt {e^2-4 d f}}\right ) \left (2 f b^2-c \left (e-\sqrt {e^2-4 d f}\right ) b-4 a c f+2 c \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x\right ) \left (c x^2+b x+a\right )}{\left (b^2-4 a c\right ) f \left (4 a f^2-2 b \left (e-\sqrt {e^2-4 d f}\right ) f+c \left (e-\sqrt {e^2-4 d f}\right )^2\right ) (a+x (b+c x))^{3/2}}-\frac {2 \left (e+\frac {e^2-2 d f}{\sqrt {e^2-4 d f}}\right ) \left (2 f b^2-c \left (e+\sqrt {e^2-4 d f}\right ) b-4 a c f+2 c \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x\right ) \left (c x^2+b x+a\right )}{\left (b^2-4 a c\right ) f \left (4 a f^2-2 b \left (e+\sqrt {e^2-4 d f}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )^2\right ) (a+x (b+c x))^{3/2}}+\frac {4 (b+2 c x) \left (-\frac {c \left (c x^2+b x+a\right )}{b^2-4 a c}\right )^{3/2}}{c f (a+x (b+c x))^{3/2} \sqrt {1-\frac {(b+2 c x)^2}{b^2-4 a c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(e - (e^2 - 2*d*f)/Sqrt[e^2 - 4*d*f])*(2*b^2*f - 4*a*c*f - b*c*(e - Sqrt[e^2 - 4*d*f]) + 2*c*(b*f - c*(e -
Sqrt[e^2 - 4*d*f]))*x)*(a + b*x + c*x^2))/((b^2 - 4*a*c)*f*(4*a*f^2 - 2*b*f*(e - Sqrt[e^2 - 4*d*f]) + c*(e -
Sqrt[e^2 - 4*d*f])^2)*(a + x*(b + c*x))^(3/2)) - (2*(e + (e^2 - 2*d*f)/Sqrt[e^2 - 4*d*f])*(2*b^2*f - 4*a*c*f -
b*c*(e + Sqrt[e^2 - 4*d*f]) + 2*c*(b*f - c*(e + Sqrt[e^2 - 4*d*f]))*x)*(a + b*x + c*x^2))/((b^2 - 4*a*c)*f*(4
*a*f^2 - 2*b*f*(e + Sqrt[e^2 - 4*d*f]) + c*(e + Sqrt[e^2 - 4*d*f])^2)*(a + x*(b + c*x))^(3/2)) + (4*(b + 2*c*x
)*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^(3/2))/(c*f*(a + x*(b + c*x))^(3/2)*Sqrt[1 - (b + 2*c*x)^2/(b^2 - 4
*a*c)]) + (16*Sqrt[2]*f*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 - c*e*Sqrt[e^2 - 4*d*f] + b*f*Sqrt[e^2 - 4*d*f]
]*(e + (-e^2 + 2*d*f)/Sqrt[e^2 - 4*d*f])*(a + b*x + c*x^2)^(3/2)*ArcTanh[(4*a*f - b*(e - Sqrt[e^2 - 4*d*f]) -
(-2*b*f + 2*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*Sqrt[e^2 - 4
*d*f] + b*f*Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/((4*a*f^2 - 2*b*f*(e - Sqrt[e^2 - 4*d*f]) + c*(e - Sqr
t[e^2 - 4*d*f])^2)*(16*a*f^2 - 8*b*f*(e - Sqrt[e^2 - 4*d*f]) + 4*c*(e - Sqrt[e^2 - 4*d*f])^2)*(a + x*(b + c*x)
)^(3/2)) + (16*Sqrt[2]*f*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 + c*e*Sqrt[e^2 - 4*d*f] - b*f*Sqrt[e^2 - 4*d*f
]]*(e - (-e^2 + 2*d*f)/Sqrt[e^2 - 4*d*f])*(a + b*x + c*x^2)^(3/2)*ArcTanh[(4*a*f - b*(e + Sqrt[e^2 - 4*d*f]) -
(-2*b*f + 2*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*Sqrt[e^2 -
4*d*f] - b*f*Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/((4*a*f^2 - 2*b*f*(e + Sqrt[e^2 - 4*d*f]) + c*(e + Sq
rt[e^2 - 4*d*f])^2)*(16*a*f^2 - 8*b*f*(e + Sqrt[e^2 - 4*d*f]) + 4*c*(e + Sqrt[e^2 - 4*d*f])^2)*(a + x*(b + c*x
))^(3/2))

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IntegrateAlgebraic [C]  time = 1.40, size = 570, normalized size = 0.94 \begin {gather*} \frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+\text {$\#$1}^2 b e+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e-4 \text {$\#$1} b \sqrt {c} d+a^2 f-a b e+b^2 d\&,\frac {\text {$\#$1}^2 b d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-\text {$\#$1}^2 a e f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a^2 e f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} c^{3/2} d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b c d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 a b d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} a \sqrt {c} d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{-2 \text {$\#$1}^3 f+3 \text {$\#$1}^2 \sqrt {c} e+2 \text {$\#$1} a f-\text {$\#$1} b e-4 \text {$\#$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 (a^2 b f-2 a^2 c e+2 a^2 c f x+a b c d-a b c e x-2 a c^2 d x+b^2 c 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^2/((a + b*x + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]

[Out]

(-2*(a*b*c*d - 2*a^2*c*e + a^2*b*f + b^2*c*d*x - 2*a*c^2*d*x - a*b*c*e*x + 2*a^2*c*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*Sqrt[c]*e*#1^3 + f*#
1^4 & , (b*c*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*a*b*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x +
c*x^2] - #1] + a^2*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*c^(3/2)*d^2*Log[-(Sqrt[c]*x) + Sqrt[
a + b*x + c*x^2] - #1]*#1 + 2*a*Sqrt[c]*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + b*d*f*Log[-(Sq
rt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - a*e*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^2/(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]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2

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maple [B]  time = 0.02, size = 11341, normalized size = 18.62 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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