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

Optimal. Leaf size=394 \[ -\frac{\tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{a^{3/2} d}-\frac{2 f \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\right )}{d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac{2 \left (-2 a c+b^2+b c x\right )}{a d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{f^{3/2} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 d \left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right )^{3/2}}+\frac{f^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 d \left (a f+b \sqrt{d} \sqrt{f}+c d\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 1.18181, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6725, 740, 12, 724, 206, 1018, 1033} \[ -\frac{\tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{a^{3/2} d}-\frac{2 f \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\right )}{d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac{2 \left (-2 a c+b^2+b c x\right )}{a d \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{f^{3/2} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 d \left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right )^{3/2}}+\frac{f^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 d \left (a f+b \sqrt{d} \sqrt{f}+c d\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

Rule 6725

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]

Rule 740

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 1018

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[((a + b*x + c*x^2)^(p + 1)*(d + f*x^2)^(q + 1)*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*(
2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(2*a*f)) - h*(b*c*d + a*b*f))*x))/((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[(b*h - 2*g*c)*((c*d - a*f)^2 - (b*d)*(-(b*f)))*(p + 1) + (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*
(c*d - a*f)))*(a*f*(p + 1) - c*d*(p + 2)) - (2*f*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*
(2*a*f)))*(p + q + 2) - (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(b*f*(p + 1)))*x - c*f*(b^2*(g*f
) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, 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
])

Rule 1033

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)]

Rubi steps

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

Mathematica [A]  time = 0.999934, size = 436, normalized size = 1.11 \[ \frac{-\frac{\tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{a^{3/2}}-\frac{2 f \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\right )}{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \left (b^2 d f-(a f+c d)^2\right )}+\frac{f^{3/2} \left (\left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right )^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+b \sqrt{d}+b \sqrt{f} x+2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )+\left (a f+b \sqrt{d} \sqrt{f}+c d\right )^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}-b \sqrt{d}+b \sqrt{f} x-2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )\right )}{2 \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d} \left ((a f+c d)^2-b^2 d f\right )}+\frac{2 \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a+x (b+c x)}}}{d} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.26, size = 1518, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (f x^{2} - d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2