\(\int \frac {1}{x^2 (a+b x+c x^2)^{3/2} (d-f x^2)} \, dx\) [108]

Optimal result

Integrand size = 28, antiderivative size = 454 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) d x \sqrt {a+b x+c x^2}}-\frac {2 f \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) d \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x+c x^2}}{a^2 \left (b^2-4 a c\right ) d x}+\frac {3 b \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{5/2} d}+\frac {f^2 \text {arctanh}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^{3/2} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2}}+\frac {f^2 \text {arctanh}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^{3/2} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2}} \]

[Out]

3/2*b*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(5/2)/d+1/2*f^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))/d^(3/2)/(c*d+a*f-b*d^(1/2)*f
^(1/2))^(3/2)+1/2*f^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))/d^(3/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(3/2)+2*(b*c*x-2*a*c+b^2)/a/(-4*a*c+b^2)/d/x/
(c*x^2+b*x+a)^(1/2)-2*f*(b*(b^2*f-c*(3*a*f+c*d))-c*(2*a*c*f-b^2*f+2*c^2*d)*x)/(-4*a*c+b^2)/d/(b^2*d*f-(a*f+c*d
)^2)/(c*x^2+b*x+a)^(1/2)-(-8*a*c+3*b^2)*(c*x^2+b*x+a)^(1/2)/a^2/(-4*a*c+b^2)/d/x

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6857, 754, 820, 738, 212, 989, 1047} \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=\frac {3 b \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{5/2} d}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x+c x^2}}{a^2 d x \left (b^2-4 a c\right )}+\frac {f^2 \text {arctanh}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^{3/2} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {f^2 \text {arctanh}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^{3/2} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}}-\frac {2 f \left (b \left (b^2 f-c (3 a f+c d)\right )-c x \left (2 a c f+b^2 (-f)+2 c^2 d\right )\right )}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a d x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[In]

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

[Out]

(2*(b^2 - 2*a*c + b*c*x))/(a*(b^2 - 4*a*c)*d*x*Sqrt[a + b*x + c*x^2]) - (2*f*(b*(b^2*f - c*(c*d + 3*a*f)) - c*
(2*c^2*d - b^2*f + 2*a*c*f)*x))/((b^2 - 4*a*c)*d*(b^2*d*f - (c*d + a*f)^2)*Sqrt[a + b*x + c*x^2]) - ((3*b^2 -
8*a*c)*Sqrt[a + b*x + c*x^2])/(a^2*(b^2 - 4*a*c)*d*x) + (3*b*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x
^2])])/(2*a^(5/2)*d) + (f^2*ArcTanh[(b*Sqrt[d] - 2*a*Sqrt[f] + (2*c*Sqrt[d] - b*Sqrt[f])*x)/(2*Sqrt[c*d - b*Sq
rt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c*x^2])])/(2*d^(3/2)*(c*d - b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)) + (f^2*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^(3/2)*(c*d + b*Sqrt[d]*Sqrt[f] + a*f)^(3/2))

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 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 754

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

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[S
implify[m + 2*p + 3], 0]

Rule 989

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b^3*f + b*c*(c*d
- 3*a*f) + c*(2*c^2*d + b^2*f - c*(2*a*f))*x)*(a + b*x + c*x^2)^(p + 1)*((d + f*x^2)^(q + 1)/((b^2 - 4*a*c)*(b
^2*d*f + (c*d - a*f)^2)*(p + 1))), x] - Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x
+ c*x^2)^(p + 1)*(d + f*x^2)^q*Simp[2*c*(b^2*d*f + (c*d - a*f)^2)*(p + 1) - (2*c^2*d + b^2*f - c*(2*a*f))*(a*
f*(p + 1) - c*d*(p + 2)) + (2*f*(b^3*f + b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + b^2*f - c*(2*a*f))*(b*f*(
p + 1)))*x + c*f*(2*c^2*d + b^2*f - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &
&  !IGtQ[q, 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 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 {1}{d x^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {f}{d \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )}\right ) \, dx \\ & = \frac {\int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2}} \, dx}{d}+\frac {f \int \frac {1}{\left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx}{d} \\ & = \frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) d x \sqrt {a+b x+c x^2}}-\frac {2 f \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) d \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {\frac {1}{2} \left (-3 b^2+8 a c\right )-b c x}{x^2 \sqrt {a+b x+c x^2}} \, dx}{a \left (b^2-4 a c\right ) d}-\frac {(2 f) \int \frac {\frac {1}{2} \left (b^2-4 a c\right ) f (c d+a f)-\frac {1}{2} b \left (b^2-4 a c\right ) f^2 x}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{\left (b^2-4 a c\right ) d \left (b^2 d f-(c d+a f)^2\right )} \\ & = \frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) d x \sqrt {a+b x+c x^2}}-\frac {2 f \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) d \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x+c x^2}}{a^2 \left (b^2-4 a c\right ) d x}-\frac {(3 b) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{2 a^2 d}-\frac {f^{5/2} \int \frac {1}{\left (-\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 d^{3/2} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )}+\frac {f^{5/2} \int \frac {1}{\left (\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 d^{3/2} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )} \\ & = \frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) d x \sqrt {a+b x+c x^2}}-\frac {2 f \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) d \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x+c x^2}}{a^2 \left (b^2-4 a c\right ) d x}+\frac {(3 b) \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 )}{a^2 d}+\frac {f^{5/2} \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^{3/2} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )}-\frac {f^{5/2} \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^{3/2} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )} \\ & = \frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) d x \sqrt {a+b x+c x^2}}-\frac {2 f \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) d \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x+c x^2}}{a^2 \left (b^2-4 a c\right ) d x}+\frac {3 b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{5/2} d}+\frac {f^2 \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^{3/2} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2}}+\frac {f^2 \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^{3/2} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 2.10 (sec) , antiderivative size = 620, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=\frac {-2 \sqrt {a} \left (4 a^4 c f^2+3 b^2 d \left (-c^2 d+b^2 f\right ) x (b+c x)+a^3 f \left (-b^2 f+4 b c f x+4 c^2 \left (2 d+f x^2\right )\right )+a^2 \left (18 b c^2 d f x-b^3 f^2 x-b^2 c f \left (6 d+f x^2\right )+4 c^3 d \left (d+3 f x^2\right )\right )+a d \left (b^4 f+10 b c^3 d x-16 b^3 c f x+8 c^4 d x^2-b^2 c^2 \left (d+14 f x^2\right )\right )\right )-6 b \left (b^2-4 a c\right ) \left (-c^2 d^2-2 a c d f+f \left (b^2 d-a^2 f\right )\right ) x \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )+a^{5/2} \left (b^2-4 a c\right ) f^2 x \sqrt {a+x (b+c x)} \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {b c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+2 a b f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 c^{3/2} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a \sqrt {c} f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-b f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{2 a^{5/2} \left (-b^2+4 a c\right ) d \left (c^2 d^2+2 a c d f+f \left (-b^2 d+a^2 f\right )\right ) x \sqrt {a+x (b+c x)}} \]

[In]

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

[Out]

(-2*Sqrt[a]*(4*a^4*c*f^2 + 3*b^2*d*(-(c^2*d) + b^2*f)*x*(b + c*x) + a^3*f*(-(b^2*f) + 4*b*c*f*x + 4*c^2*(2*d +
f*x^2)) + a^2*(18*b*c^2*d*f*x - b^3*f^2*x - b^2*c*f*(6*d + f*x^2) + 4*c^3*d*(d + 3*f*x^2)) + a*d*(b^4*f + 10*
b*c^3*d*x - 16*b^3*c*f*x + 8*c^4*d*x^2 - b^2*c^2*(d + 14*f*x^2))) - 6*b*(b^2 - 4*a*c)*(-(c^2*d^2) - 2*a*c*d*f
+ f*(b^2*d - a^2*f))*x*Sqrt[a + x*(b + c*x)]*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]] + a^(5/2)*(b
^2 - 4*a*c)*f^2*x*Sqrt[a + x*(b + c*x)]*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*c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 2*a*b*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*
x^2] - #1] - 2*c^(3/2)*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*a*Sqrt[c]*f*Log[-(Sqrt[c]*x) +
Sqrt[a + b*x + c*x^2] - #1]*#1 - b*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) & ])/(2*a^(5/2)*(-b^2 + 4*a*c)*d*(c^2*d^2 + 2*a*c*d*f + f*(-(b^2*d) + a^2*f))*x*Sqrt[a
+ x*(b + c*x)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1064\) vs. \(2(382)=764\).

Time = 0.95 (sec) , antiderivative size = 1065, normalized size of antiderivative = 2.35

method	result	size
default	\(\text {Expression too large to display}\)	\(1065\)
risch	\(\text {Expression too large to display}\)	\(3015\)

[In]

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

[Out]

1/d*(-1/a/x/(c*x^2+b*x+a)^(1/2)-3/2*b/a*(1/a/(c*x^2+b*x+a)^(1/2)-b/a*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)
-1/a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x))-4*c/a*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2))-1/2
*f/d/(d*f)^(1/2)*(1/(b*(d*f)^(1/2)+f*a+c*d)*f/((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)-(2*c*(d*f)^(1/2)+b*f)/(b*(d*f)^(1/2)+f*a+c*d)*(2*c*(x-(d*f)^(1/2)/f)+(2*c*(d
*f)^(1/2)+b*f)/f)/(4*c*(b*(d*f)^(1/2)+f*a+c*d)/f-(2*c*(d*f)^(1/2)+b*f)^2/f^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)-1/(b*(d*f)^(1/2)+f*a+c*d)*f/((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)))+1/2*f/d/(d*f)^(1/2)*(f/(-b*(d*f)^(1/2)+f*a+c*d)/((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)-(-2*c*(d*f)^(1/2)+b*f)/(-b*(d*f)^(1/2)+f*
a+c*d)*(2*c*(x+(d*f)^(1/2)/f)+1/f*(-2*c*(d*f)^(1/2)+b*f))/(4*c/f*(-b*(d*f)^(1/2)+f*a+c*d)-1/f^2*(-2*c*(d*f)^(1
/2)+b*f)^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)-f/(-b*(d*f)^(1/2)+f*a+c*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 [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=- \int \frac {1}{- a d x^{2} \sqrt {a + b x + c x^{2}} + a f x^{4} \sqrt {a + b x + c x^{2}} - b d x^{3} \sqrt {a + b x + c x^{2}} + b f x^{5} \sqrt {a + b x + c x^{2}} - c d x^{4} \sqrt {a + b x + c x^{2}} + c f x^{6} \sqrt {a + b x + c x^{2}}}\, dx \]

[In]

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

[Out]

-Integral(1/(-a*d*x**2*sqrt(a + b*x + c*x**2) + a*f*x**4*sqrt(a + b*x + c*x**2) - b*d*x**3*sqrt(a + b*x + c*x*
*2) + b*f*x**5*sqrt(a + b*x + c*x**2) - c*d*x**4*sqrt(a + b*x + c*x**2) + c*f*x**6*sqrt(a + b*x + c*x**2)), x)

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=\int { -\frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (f x^{2} - d\right )} x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueDone

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx=\int \frac {1}{x^2\,\left (d-f\,x^2\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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