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

Optimal. Leaf size=736 \[ -\frac {f \left (a \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )-b d \left (\sqrt {e^2-4 d f}+e\right )+2 c d^2\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (a \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )-b d \left (e-\sqrt {e^2-4 d f}\right )+2 c d^2\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {f \left (2 a f-b \left (\sqrt {e^2-4 d f}+e\right )\right )+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\sqrt {a} e \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}-\frac {b e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {\sqrt {a+b x+c x^2}}{d x}-\frac {b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {a} d}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{d} \]

________________________________________________________________________________________

Rubi [A]  time = 3.48, antiderivative size = 736, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6728, 732, 843, 621, 206, 724, 734, 1019, 1076, 1032} \begin {gather*} -\frac {f \left (a \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )-b d \left (\sqrt {e^2-4 d f}+e\right )+2 c d^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 (a \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )-b d \left (e-\sqrt {e^2-4 d f}\right )+2 c d^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 {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 {a} e \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}-\frac {b e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {\sqrt {a+b x+c x^2}}{d x}-\frac {b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {a} d}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && 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 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1019

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Sy
mbol] :> Simp[(h*(a + b*x + c*x^2)^p*(d + e*x + f*x^2)^(q + 1))/(2*f*(p + q + 1)), x] - Dist[1/(2*f*(p + q + 1
)), Int[(a + b*x + c*x^2)^(p - 1)*(d + e*x + f*x^2)^q*Simp[h*p*(b*d - a*e) + a*(h*e - 2*g*f)*(p + q + 1) + (2*
h*p*(c*d - a*f) + b*(h*e - 2*g*f)*(p + q + 1))*x + (h*p*(c*e - b*f) + c*(h*e - 2*g*f)*(p + q + 1))*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] && GtQ[p, 0] && Ne
Q[p + q + 1, 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 1076

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x
_)^2]), x_Symbol] :> Dist[C/c, Int[1/Sqrt[d + e*x + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + (B*c - b*C)*x)
/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b^2 - 4*a*c
, 0] && NeQ[e^2 - 4*d*f, 0]

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

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Mathematica [A]  time = 1.68, size = 520, normalized size = 0.71 \begin {gather*} \frac {(2 a e-b d) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{2 \sqrt {a} d^2}-\frac {4 d f \sqrt {e^2-4 d f} \sqrt {a+x (b+c x)}+\sqrt {2} x \left (e \sqrt {e^2-4 d f}+2 d f-e^2\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 )} \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 {2} x \left (e \sqrt {e^2-4 d f}-2 d f+e^2\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 )} \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 )}{4 d^2 f x \sqrt {e^2-4 d f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 0.88, size = 589, normalized size = 0.80 \begin {gather*} \frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+\text {$\#$1}^2 b e+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e-4 \text {$\#$1} b \sqrt {c} d+a^2 f-a b e+b^2 d\&,\frac {\text {$\#$1}^2 b d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-\text {$\#$1}^2 a e f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a^2 e f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^2 d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} c^{3/2} d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-b c d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b \sqrt {c} d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} a \sqrt {c} d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a b e^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} a \sqrt {c} e^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{-2 \text {$\#$1}^3 f+3 \text {$\#$1}^2 \sqrt {c} e+2 \text {$\#$1} a f-\text {$\#$1} b e-4 \text {$\#$1} c d-a \sqrt {c} e+2 b \sqrt {c} d}\&\right ]}{d^2}+\frac {(2 a e-b d) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} d^2}-\frac {\sqrt {a+b x+c x^2}}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x + c*x^2]/(d*x)) + ((-(b*d) + 2*a*e)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(S
qrt[a]*d^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]) + b^2*d*e*
Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*b*e^2*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*b*Sqrt[c]*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*a*Sqrt[c]*e^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*L
og[-(Sqrt[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) & ]/d^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((c*x^2+b*x+a)^(1/2)/x^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((c*x^2+b*x+a)^(1/2)/x^2/(f*x^2+e*x+d),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.02, size = 6765, normalized size = 9.19 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c x^{2} + b x + a}}{{\left (f x^{2} + e x + d\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out

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