\(\int \frac {1}{x (a+c x^2)^{3/2} (d+e x+f x^2)} \, dx\) [75]

Optimal result

Integrand size = 27, antiderivative size = 526 \[ \int \frac {1}{x \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\frac {1}{a d \sqrt {a+c x^2}}-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}-\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d} \]

[Out]

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

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 526, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6860, 272, 53, 65, 214, 1031, 1048, 739, 212} \[ \int \frac {1}{x \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )}+\frac {1}{a d \sqrt {a+c x^2}} \]

[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 53

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 65

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 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 214

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

Rule 272

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 739

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 1031

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)/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 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), 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 1048

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 6860

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*} \text {integral}& = \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 {\text {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 {\text {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 {\text {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 ) \text {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 ) \text {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] (verified)

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

Time = 0.82 (sec) , antiderivative size = 542, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=-\frac {c (-c d+a f+c e x)}{a \left (c^2 d^2+a^2 f^2+a c \left (e^2-2 d f\right )\right ) \sqrt {a+c x^2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {\text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a c e^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-a c d f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a^2 f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 c^{3/2} e^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-4 c^{3/2} d e f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a \sqrt {c} e f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-c e^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+c d f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{d \left (c^2 d^2+a^2 f^2+a c \left (e^2-2 d f\right )\right )} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1564\) vs. \(2(480)=960\).

Time = 0.68 (sec) , antiderivative size = 1565, normalized size of antiderivative = 2.98

method	result	size
default	\(\text {Expression too large to display}\)	\(1565\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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