3.119 \(\int \frac{1}{x^2 \sqrt{a+b x+c x^2} (d+e x+f x^2)} \, dx\)

Optimal. Leaf size=543 \[ \frac{b \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2} d}-\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{f \left (2 a f-b \left (e-\sqrt{e^2-4 d f}\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+b x+c x^2}}{a d x} \]

[Out]

-(Sqrt[a + b*x + c*x^2]/(a*d*x)) + (b*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(2*a^(3/2)*d) +
(e*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(Sqrt[a]*d^2) - (f*(e^2 - 2*d*f + 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]*d^2*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]))]) + (f*(e^2 - 2*d*f
- 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]*d^2*Sqrt[e^2 - 4*d*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]])

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Rubi [A]  time = 4.59176, antiderivative size = 543, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6728, 730, 724, 206, 1032} \[ \frac{b \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2} d}-\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{f \left (2 a f-b \left (e-\sqrt{e^2-4 d f}\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+b x+c x^2}}{a d x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]

[Out]

-(Sqrt[a + b*x + c*x^2]/(a*d*x)) + (b*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(2*a^(3/2)*d) +
(e*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(Sqrt[a]*d^2) - (f*(e^2 - 2*d*f + 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]*d^2*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]))]) + (f*(e^2 - 2*d*f
- 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]*d^2*Sqrt[e^2 - 4*d*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 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 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 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 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]

Rubi steps

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

Mathematica [A]  time = 1.59828, size = 533, normalized size = 0.98 \[ -\frac{-\frac{b d \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{a^{3/2}}+\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (\sqrt{e^2-4 d f}+e-2 f x\right )-2 c x \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{e^2-4 d f} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{2} f \left (\frac{e^2-2 d f}{\sqrt{e^2-4 d f}}+e\right ) \tanh ^{-1}\left (\frac{4 a f+b \left (\sqrt{e^2-4 d f}-e+2 f x\right )+2 c x \left (\sqrt{e^2-4 d f}-e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{2 d \sqrt{a+x (b+c x)}}{a x}-\frac{2 e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{\sqrt{a}}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]

[Out]

-((2*d*Sqrt[a + x*(b + c*x)])/(a*x) - (b*d*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/a^(3/2) - (
2*e*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/Sqrt[a] + (Sqrt[2]*f*(-e^2 + 2*d*f + e*Sqrt[e^2 -
4*d*f])*ArcTanh[(4*a*f - 2*c*(e + 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]*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]*f*(
e + (e^2 - 2*d*f)/Sqrt[e^2 - 4*d*f])*ArcTanh[(4*a*f + 2*c*(-e + 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]))]*S
qrt[a + x*(b + c*x)])])/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*d^2)

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Maple [B]  time = 0.344, size = 983, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x)

[Out]

-4*f^2/(-e+(-4*d*f+e^2)^(1/2))^2/(-4*d*f+e^2)^(1/2)*2^(1/2)/(((-4*d*f+e^2)^(1/2)*b*f-(-4*d*f+e^2)^(1/2)*c*e+2*
a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)*ln((((-4*d*f+e^2)^(1/2)*b*f-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f
+c*e^2)/f^2+(c*(-4*d*f+e^2)^(1/2)+b*f-c*e)/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*(((-4*d*f+e^2)^(1/2
)*b*f-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c+
4*(c*(-4*d*f+e^2)^(1/2)+b*f-c*e)/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+2*((-4*d*f+e^2)^(1/2)*b*f-(-4*d*f+e^2)^(1
/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2))/(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f))+16*f^2*e/(-e+(-4*d*f+e^2)^
(1/2))^2/(e+(-4*d*f+e^2)^(1/2))^2/a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+4*f^2/(e+(-4*d*f+e^2)^
(1/2))^2/(-4*d*f+e^2)^(1/2)*2^(1/2)/((-(-4*d*f+e^2)^(1/2)*b*f+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e
^2)/f^2)^(1/2)*ln(((-(-4*d*f+e^2)^(1/2)*b*f+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2+1/f*(-c*(-
4*d*f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*((-(-4*d*f+e^2)^(1/2)*b*f+(-4*d*f+e^2)^
(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c+4/f*(-c*(-4*d*f+e^2)
^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+2*(-(-4*d*f+e^2)^(1/2)*b*f+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e
*f-2*c*d*f+c*e^2)/f^2)^(1/2))/(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))+4*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1
/2))/a/x*(c*x^2+b*x+a)^(1/2)-2*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1/2))*b/a^(3/2)*ln((2*a+b*x+2*a^(1/2
)*(c*x^2+b*x+a)^(1/2))/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{2} + b x + a}{\left (f x^{2} + e x + d\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^2 + b*x + a)*(f*x^2 + e*x + d)*x^2), x)

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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(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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(1/x**2/(c*x**2+b*x+a)**(1/2)/(f*x**2+e*x+d),x)

[Out]

Timed out

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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(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="giac")

[Out]

Timed out