\(\int \frac {\sqrt {a+b x+c x^2}}{x^3 (d-f x^2)} \, dx\) [83]

Optimal result

Integrand size = 28, antiderivative size = 353 \[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=-\frac {(2 a+b x) \sqrt {a+b x+c x^2}}{4 a d x^2}+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{3/2} d}-\frac {\sqrt {a} f \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}-\frac {\sqrt {f} \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \text {arctanh}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^2}+\frac {\sqrt {f} \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \text {arctanh}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^2} \]

[Out]

1/8*(-4*a*c+b^2)*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(3/2)/d-f*arctanh(1/2*(b*x+2*a)/a^(1/2)/
(c*x^2+b*x+a)^(1/2))*a^(1/2)/d^2-1/4*(b*x+2*a)*(c*x^2+b*x+a)^(1/2)/a/d/x^2-1/2*arctanh(1/2*(b*d^(1/2)-2*a*f^(1
/2)+x*(2*c*d^(1/2)-b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1/2))^(1/2))*f^(1/2)*(c*d+a*f-b*d^(1/
2)*f^(1/2))^(1/2)/d^2+1/2*arctanh(1/2*(b*d^(1/2)+2*a*f^(1/2)+x*(2*c*d^(1/2)+b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c
*d+a*f+b*d^(1/2)*f^(1/2))^(1/2))*f^(1/2)*(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2)/d^2

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6857, 734, 738, 212, 748, 857, 635, 1035, 1092, 1047} \[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{3/2} d}-\frac {\sqrt {a} f \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}-\frac {\sqrt {f} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d} \text {arctanh}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^2}+\frac {\sqrt {f} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d} \text {arctanh}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^2}-\frac {(2 a+b x) \sqrt {a+b x+c x^2}}{4 a d x^2} \]

[In]

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

[Out]

-1/4*((2*a + b*x)*Sqrt[a + b*x + c*x^2])/(a*d*x^2) + ((b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*
x + c*x^2])])/(8*a^(3/2)*d) - (Sqrt[a]*f*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/d^2 - (Sqrt[f
]*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(b*Sqrt[d] - 2*a*Sqrt[f] + (2*c*Sqrt[d] - b*Sqrt[f])*x)/(2*Sqrt[
c*d - b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c*x^2])])/(2*d^2) + (Sqrt[f]*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]
*ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*Sqrt[d] + b*Sqrt[f])*x)/(2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a
+ b*x + c*x^2])])/(2*d^2)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 734

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

Rule 738

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 748

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 857

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 1035

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[h*(a + b*x + c*x^2)^p*((d + 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 + f*x^2)^q*Simp[h*p*(b*d) + a*(-2*g*f)*(p + q + 1) + (2*h*p*(c*d - a*f) + b*(-2*g*f)*(p + q
+ 1))*x + (h*p*((-b)*f) + c*(-2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}, x] && NeQ
[b^2 - 4*a*c, 0] && GtQ[p, 0] && NeQ[p + q + 1, 0]

Rule 1047

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

Rule 1092

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

Rule 6857

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.60 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=\frac {-\frac {d (2 a+b x) \sqrt {a+x (b+c x)}}{a x^2}+\frac {\left (b^2 d-4 a (c d+2 a f)\right ) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{a^{3/2}}-2 f \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {b^2 d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a^2 f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 b \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+a f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{4 d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.36

method	result	size
risch	\(-\frac {\left (b x +2 a \right ) \sqrt {c \,x^{2}+b x +a}}{4 a d \,x^{2}}-\frac {-\frac {\left (-8 a^{2} f -4 a c d +b^{2} d \right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{d \sqrt {a}}-\frac {4 a \left (b \sqrt {d f}+f a +c d \right ) \ln \left (\frac {\frac {2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{d \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}}+\frac {4 a \left (b \sqrt {d f}-f a -c d \right ) \ln \left (\frac {\frac {-2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{d \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}}}{8 a d}\)	\(479\)
default	\(\text {Expression too large to display}\)	\(1143\)

[In]

int((c*x^2+b*x+a)^(1/2)/x^3/(-f*x^2+d),x,method=_RETURNVERBOSE)

[Out]

-1/4*(b*x+2*a)*(c*x^2+b*x+a)^(1/2)/a/d/x^2-1/8/a/d*(-(-8*a^2*f-4*a*c*d+b^2*d)/d/a^(1/2)*ln((2*a+b*x+2*a^(1/2)*
(c*x^2+b*x+a)^(1/2))/x)-4*a*(b*(d*f)^(1/2)+f*a+c*d)/d/((b*(d*f)^(1/2)+f*a+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+f
*a+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+f*a+c*d)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c
+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+f*a+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))+4*a*(b*(d*f)^(
1/2)-f*a-c*d)/d/(1/f*(-b*(d*f)^(1/2)+f*a+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+f*a+c*d)+1/f*(-2*c*(d*f)^(1/2)+b*
f)*(x+(d*f)^(1/2)/f)+2*(1/f*(-b*(d*f)^(1/2)+f*a+c*d))^(1/2)*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*
(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+f*a+c*d))^(1/2))/(x+(d*f)^(1/2)/f)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 738 vs. \(2 (275) = 550\).

Time = 81.78 (sec) , antiderivative size = 1485, normalized size of antiderivative = 4.21 \[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=\text {Too large to display} \]

[In]

integrate((c*x^2+b*x+a)^(1/2)/x^3/(-f*x^2+d),x, algorithm="fricas")

[Out]

[1/16*(4*a^2*d^2*x^2*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4)*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b
^2*f^3/d^7)*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4) + 2*b*c*f^2*x + b^2*f^2 + (b*d^3*f*x + 2*a*d^3*f
)*sqrt(b^2*f^3/d^7))/x) - 4*a^2*d^2*x^2*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4)*log(-(2*sqrt(c*x^2 +
b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 - (b
*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) - 4*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4)
*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4) + 2*b*c
*f^2*x + b^2*f^2 - (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) + 4*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7)
- c*d*f - a*f^2)/d^4)*log(-(2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f
- a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 + (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) + (8*a^2*f - (b^2 - 4*a
*c)*d)*sqrt(a)*x^2*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^
2) - 4*(a*b*d*x + 2*a^2*d)*sqrt(c*x^2 + b*x + a))/(a^2*d^2*x^2), 1/8*(2*a^2*d^2*x^2*sqrt((d^4*sqrt(b^2*f^3/d^7
) + c*d*f + a*f^2)/d^4)*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt((d^4*sqrt(b^2*f^3/d^7) + c*d*f
+ a*f^2)/d^4) + 2*b*c*f^2*x + b^2*f^2 + (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) - 2*a^2*d^2*x^2*sqrt((d
^4*sqrt(b^2*f^3/d^7) + c*d*f + a*f^2)/d^4)*log(-(2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/d^7)*sqrt((d^4*sqrt(
b^2*f^3/d^7) + c*d*f + a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 - (b*d^3*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) - 2
*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4)*log((2*sqrt(c*x^2 + b*x + a)*d^5*sqrt(b^2*f^3/
d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4) + 2*b*c*f^2*x + b^2*f^2 - (b*d^3*f*x + 2*a*d^3*f)*sqrt
(b^2*f^3/d^7))/x) + 2*a^2*d^2*x^2*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4)*log(-(2*sqrt(c*x^2 + b*x
+ a)*d^5*sqrt(b^2*f^3/d^7)*sqrt(-(d^4*sqrt(b^2*f^3/d^7) - c*d*f - a*f^2)/d^4) - 2*b*c*f^2*x - b^2*f^2 + (b*d^3
*f*x + 2*a*d^3*f)*sqrt(b^2*f^3/d^7))/x) + (8*a^2*f - (b^2 - 4*a*c)*d)*sqrt(-a)*x^2*arctan(1/2*sqrt(c*x^2 + b*x
+ a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 2*(a*b*d*x + 2*a^2*d)*sqrt(c*x^2 + b*x + a))/(a^2*d^2*x^
2)]

Sympy [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=- \int \frac {\sqrt {a + b x + c x^{2}}}{- d x^{3} + f x^{5}}\, dx \]

[In]

integrate((c*x**2+b*x+a)**(1/2)/x**3/(-f*x**2+d),x)

[Out]

-Integral(sqrt(a + b*x + c*x**2)/(-d*x**3 + f*x**5), x)

Maxima [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=\int { -\frac {\sqrt {c x^{2} + b x + a}}{{\left (f x^{2} - d\right )} x^{3}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(1/2)/x^3/(-f*x^2+d),x, algorithm="maxima")

[Out]

-integrate(sqrt(c*x^2 + b*x + a)/((f*x^2 - d)*x^3), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=\text {Exception raised: AttributeError} \]

[In]

integrate((c*x^2+b*x+a)^(1/2)/x^3/(-f*x^2+d),x, algorithm="giac")

[Out]

Exception raised: AttributeError >> type

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{x^3 \left (d-f x^2\right )} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{x^3\,\left (d-f\,x^2\right )} \,d x \]

[In]

int((a + b*x + c*x^2)^(1/2)/(x^3*(d - f*x^2)),x)

[Out]

int((a + b*x + c*x^2)^(1/2)/(x^3*(d - f*x^2)), x)