3.1.55 \(\int \frac {\sqrt {a+c x^2}}{x (d+e x+f x^2)} \, dx\) [55]

Optimal. Leaf size=358 \[ \frac {\left (2 a e f+(c d-a f) \left (e-\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}-\frac {\left (2 a e f+(c d-a f) \left (e+\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.83, antiderivative size = 358, normalized size of antiderivative = 1.00, 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 = {6860, 272, 52, 65, 214, 1034, 1048, 739, 212} \begin {gather*} \frac {\left (\left (e-\sqrt {e^2-4 d f}\right ) (c d-a f)+2 a e f\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} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {\left (\left (\sqrt {e^2-4 d f}+e\right ) (c d-a f)+2 a e f\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} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 1034

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

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.37, size = 299, normalized size = 0.84 \begin {gather*} \frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )+\text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {-a c d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 a \sqrt {c} e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+c d \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a]*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]] + 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*d*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]) + a^2*f*Log[-(Sqrt
[c]*x) + Sqrt[a + c*x^2] - #1] + 2*a*Sqrt[c]*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + c*d*Log[-(Sqrt[c]
*x) + Sqrt[a + c*x^2] - #1]*#1^2 - a*f*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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1315\) vs. \(2(315)=630\).
time = 0.13, size = 1316, normalized size = 3.68

method	result	size
default	\(\frac {2 f \left (\frac {\sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}-\frac {\sqrt {c}\, \left (e +\sqrt {-4 d f +e^{2}}\right ) \ln \left (\frac {-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right )}{2 f}+c \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{2 f^{2}}}\right )}{2 f}-\frac {\left (\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}\right ) \sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{2} \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}\right )}{\left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}}+\frac {2 f \left (\frac {\sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}-\frac {\sqrt {c}\, \left (e -\sqrt {-4 d f +e^{2}}\right ) \ln \left (\frac {-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right )}{2 f}+c \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{\sqrt {c}}+\sqrt {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{2 f^{2}}}\right )}{2 f}-\frac {\left (-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}\right ) \sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{2} \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}}-\frac {4 f \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right )}\)	\(1316\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^2 + a)/((f*x^2 + x*e + d)*x), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1127 vs. \(2 (322) = 644\).
time = 12.36, size = 2266, normalized size = 6.33 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(sqrt(2)*d*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 + (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4
*d^3*f - d^2*e^2))*log((2*a*c*d*x*e + sqrt(2)*(4*d^4*f - d^3*e^2)*sqrt(c*x^2 + a)*sqrt(-a^2*e^2/(4*d^5*f - d^4
*e^2))*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 + (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d
^2*e^2)) - a^2*e^2 - (4*a*d^3*f - a*d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/x) - sqrt(2)*d*sqrt(-(2*c*d^2
- 2*a*d*f + a*e^2 + (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2))*log((2*a*c*d
*x*e - sqrt(2)*(4*d^4*f - d^3*e^2)*sqrt(c*x^2 + a)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2))*sqrt(-(2*c*d^2 - 2*a*d*f
+ a*e^2 + (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2)) - a^2*e^2 - (4*a*d^3*f
- a*d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/x) - sqrt(2)*d*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 - (4*d^3*f -
d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2))*log((2*a*c*d*x*e + sqrt(2)*(4*d^4*f - d^3*e
^2)*sqrt(c*x^2 + a)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2))*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 - (4*d^3*f - d^2*e^2)*
sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2)) - a^2*e^2 + (4*a*d^3*f - a*d^2*e^2)*sqrt(-a^2*e^2/(4*
d^5*f - d^4*e^2)))/x) + sqrt(2)*d*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 - (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*
f - d^4*e^2)))/(4*d^3*f - d^2*e^2))*log((2*a*c*d*x*e - sqrt(2)*(4*d^4*f - d^3*e^2)*sqrt(c*x^2 + a)*sqrt(-a^2*e
^2/(4*d^5*f - d^4*e^2))*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 - (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^
2)))/(4*d^3*f - d^2*e^2)) - a^2*e^2 + (4*a*d^3*f - a*d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/x) - 2*sqrt(
a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2))/d, -1/4*(sqrt(2)*d*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 +
(4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2))*log((2*a*c*d*x*e + sqrt(2)*(4*d^4
*f - d^3*e^2)*sqrt(c*x^2 + a)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2))*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 + (4*d^3*f -
d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2)) - a^2*e^2 - (4*a*d^3*f - a*d^2*e^2)*sqrt(-a
^2*e^2/(4*d^5*f - d^4*e^2)))/x) - sqrt(2)*d*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 + (4*d^3*f - d^2*e^2)*sqrt(-a^2*e
^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2))*log((2*a*c*d*x*e - sqrt(2)*(4*d^4*f - d^3*e^2)*sqrt(c*x^2 + a)*s
qrt(-a^2*e^2/(4*d^5*f - d^4*e^2))*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 + (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*
f - d^4*e^2)))/(4*d^3*f - d^2*e^2)) - a^2*e^2 - (4*a*d^3*f - a*d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/x)
- sqrt(2)*d*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 - (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3
*f - d^2*e^2))*log((2*a*c*d*x*e + sqrt(2)*(4*d^4*f - d^3*e^2)*sqrt(c*x^2 + a)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2
))*sqrt(-(2*c*d^2 - 2*a*d*f + a*e^2 - (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e
^2)) - a^2*e^2 + (4*a*d^3*f - a*d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/x) + sqrt(2)*d*sqrt(-(2*c*d^2 - 2
*a*d*f + a*e^2 - (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2))*log((2*a*c*d*x*e
- sqrt(2)*(4*d^4*f - d^3*e^2)*sqrt(c*x^2 + a)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2))*sqrt(-(2*c*d^2 - 2*a*d*f + a
*e^2 - (4*d^3*f - d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/(4*d^3*f - d^2*e^2)) - a^2*e^2 + (4*a*d^3*f - a
*d^2*e^2)*sqrt(-a^2*e^2/(4*d^5*f - d^4*e^2)))/x) - 4*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)))/d]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + c x^{2}}}{x \left (d + e x + f x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^2+a}}{x\,\left (f\,x^2+e\,x+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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