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

Optimal. Leaf size=526 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{a^{3/2} d}-\frac{a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )}+\frac{f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{1}{a d \sqrt{a+c x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.18307, antiderivative size = 526, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6728, 266, 51, 63, 208, 1017, 1034, 725, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{a^{3/2} d}-\frac{a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )}+\frac{f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{1}{a d \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 1017

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[((a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1)*(g*c*(2*a*c*e) + (-(a*h))*(2*c^2*d - c*(2*a*f)) + c*(g*(2*c^2*
d - c*(2*a*f)) - h*(-2*a*c*e))*x))/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), x] + Dist[1/((-4*a*c)*(a*c*e^
2 + (c*d - a*f)^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Simp[(-2*g*c)*((c*d - a*f)^2 - (-(a*e
))*(c*e))*(p + 1) + (2*(g*c*(c*d - a*f) - a*(-(h*c*e))))*(a*f*(p + 1) - c*d*(p + 2)) - e*((g*c)*(2*a*c*e) + (-
(a*h))*(2*c^2*d - c*((Plus[2])*a*f)))*(p + q + 2) - (2*f*((g*c)*(2*a*c*e) + (-(a*h))*(2*c^2*d - c*((Plus[2])*a
*f)))*(p + q + 2) - (2*(g*c*(c*d - a*f) - a*(-(h*c*e))))*(-(c*e*(2*p + q + 4))))*x - c*f*(2*(g*c*(c*d - a*f) -
a*(-(h*c*e))))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h, q}, x] && NeQ[e^2 - 4*d*f, 0] &
& LtQ[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1])

Rule 1034

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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])

Rubi steps

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

Mathematica [C]  time = 4.02099, size = 497, normalized size = 0.94 \[ \frac{-\frac{f \left (\frac{e}{\sqrt{e^2-4 d f}}+1\right ) \left (2 a f+c x \left (e-\sqrt{e^2-4 d f}\right )\right )}{a \sqrt{a+c x^2} \left (4 a f^2+c \left (e-\sqrt{e^2-4 d f}\right )^2\right )}-\frac{f \left (1-\frac{e}{\sqrt{e^2-4 d f}}\right ) \left (2 a f+c x \left (\sqrt{e^2-4 d f}+e\right )\right )}{a \sqrt{a+c x^2} \left (4 a f^2+c \left (\sqrt{e^2-4 d f}+e\right )^2\right )}+\frac{\sqrt{2} f^3 \left (\sqrt{e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac{\sqrt{2} f^3 \left (\sqrt{e^2-4 d f}-e\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x^2}{a}+1\right )}{a \sqrt{a+c x^2}}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((f*(1 + e/Sqrt[e^2 - 4*d*f])*(2*a*f + c*(e - Sqrt[e^2 - 4*d*f])*x))/(a*(4*a*f^2 + c*(e - Sqrt[e^2 - 4*d*f])
^2)*Sqrt[a + c*x^2])) - (f*(1 - e/Sqrt[e^2 - 4*d*f])*(2*a*f + c*(e + Sqrt[e^2 - 4*d*f])*x))/(a*(4*a*f^2 + c*(e
+ Sqrt[e^2 - 4*d*f])^2)*Sqrt[a + c*x^2]) + (Sqrt[2]*f^3*(e + Sqrt[e^2 - 4*d*f])*ArcTanh[(2*a*f + c*(-e + Sqrt
[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[e^2 - 4*d
*f]*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]))^(3/2)) + (Sqrt[2]*f^3*(-e + Sqrt[e^2 - 4*d*f])*ArcTanh[(
2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])
])/(Sqrt[e^2 - 4*d*f]*(2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]))^(3/2)) + Hypergeometric2F1[-1/2, 1, 1/
2, 1 + (c*x^2)/a]/(a*Sqrt[a + c*x^2]))/d

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Maple [B]  time = 0.273, size = 1945, normalized size = 3.7 \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+a)^(3/2)/(f*x^2+e*x+d),x)

[Out]

4*f^3/(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/((x-1/2*(-e+(
-4*d*f+e^2)^(1/2))/f)^2*c-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*(-(-4*d*f+e^2)^(1/2
)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)-8*f^2/(-e+(-4*d*f+e^2)^(1/2))*c^2/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c
*d*f+c*e^2)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-1/f^2*(-4*d*f+e^2)*c^2)/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e-(
-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2
)^(1/2)*x+8*f^2/(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)*c^2/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)
/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-1/f^2*(-4*d*f+e^2)*c^2)/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e-(-4*d*f+e^2)
^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*x*e
-4*f^3/(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)/(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)*2^(1/2)/((-(
-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^
2-c*(e-(-4*d*f+e^2)^(1/2))/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)*c*e+2*a*f^2-2
*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*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*
d*f+e^2)^(1/2))/f)+2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*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/(c*x^2+a)^(1/2)+4*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*
d*f+e^2)^(1/2))/a^(3/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)+4*f^3/(e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)/
((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/
f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)+8*f^2/(e+(-4*
d*f+e^2)^(1/2))*c^2/((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-1/f^2*(-4*d*f+
e^2)*c^2)/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/
2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*x+8*f^2/(e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)*
c^2/((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-1/f^2*(-4*d*f+e^2)*c^2)/((x+1/
2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*((-4*d*f+e^2)^
(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*x*e-4*f^3/(e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)/((-4*d*f+e^2)^
(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)*2^(1/2)/(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*ln((((-4*d
*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2-c*(e+(-4*d*f+e^2)^(1/2))/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)*c*e+2*a*f^2-2*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*
(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2
)^(1/2))/(x+1/2*(e+(-4*d*f+e^2)^(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} + a\right )}^{\frac{3}{2}}{\left (f x^{2} + e x + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*(a + c*x**2)**(3/2)*(d + e*x + f*x**2)), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError