3.1.68 \(\int \frac {1}{x \sqrt {a+c x^2} (d+e x+f x^2)} \, dx\) [68]

Optimal. Leaf size=330 \[ \frac {f \left (e+\sqrt {e^2-4 d f}\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}-\frac {f \left (e-\sqrt {e^2-4 d f}\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[Out]

-arctanh((c*x^2+a)^(1/2)/a^(1/2))/d/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))*(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*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))*(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 = 0.54, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6860, 272, 65, 214, 1048, 739, 212} \begin {gather*} \frac {f \left (\sqrt {e^2-4 d f}+e\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 {f \left (e-\sqrt {e^2-4 d 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 {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 {1}{x \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac {1}{d x \sqrt {a+c x^2}}+\frac {-e-f x}{d \sqrt {a+c x^2} \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {1}{x \sqrt {a+c x^2}} \, dx}{d}+\frac {\int \frac {-e-f x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{d}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d}-\frac {\left (f \left (1-\frac {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}-\frac {\left (f \left (1+\frac {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}\\ &=\frac {\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 (f \left (1-\frac {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}+\frac {\left (f \left (1+\frac {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}\\ &=\frac {f \left (1+\frac {e}{\sqrt {e^2-4 d f}}\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 {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {f \left (1-\frac {e}{\sqrt {e^2-4 d f}}\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 {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 )}{\sqrt {a} 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 = 0.34, size = 236, normalized size = 0.72 \begin {gather*} \frac {\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 \sqrt {c} e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-a \sqrt {c} e-4 c d \text {$\#$1}+2 a f \text {$\#$1}+3 \sqrt {c} e \text {$\#$1}^2-2 f \text {$\#$1}^3}\&\right ]}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/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*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*Sqrt[c]*e*Log[-(S
qrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - 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. \(680\) vs. \(2(287)=574\).
time = 0.13, size = 681, normalized size = 2.06

method	result	size
default	\(-\frac {2 f \sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}-\frac {2 f \sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}+\frac {4 f \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {a}}\)	\(681\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*f/(e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^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)*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^(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))/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(1/x/(f*x^2+e*x+d)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(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. 6674 vs. \(2 (291) = 582\).
time = 184.04, size = 13360, normalized size = 40.48 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(2)*a*d*sqrt(-(2*c*d^2*f^2 - 2*a*d*f^3 + c*e^4 - (4*c*d*f - a*f^2)*e^2 + (4*c^2*d^5*f - 8*a*c*d^4*f^
2 + 4*a^2*d^3*f^3 - a*c*d^2*e^4 - (c^2*d^4 - 6*a*c*d^3*f + a^2*d^2*f^2)*e^2)*sqrt(-(c^2*e^6 - 2*(2*c^2*d*f - a
*c*f^2)*e^4 + (4*c^2*d^2*f^2 - 4*a*c*d*f^3 + a^2*f^4)*e^2)/(4*c^4*d^9*f - 16*a*c^3*d^8*f^2 + 24*a^2*c^2*d^7*f^
3 - 16*a^3*c*d^6*f^4 + 4*a^4*d^5*f^5 - a^2*c^2*d^4*e^6 - 2*(a*c^3*d^6 - 4*a^2*c^2*d^5*f + a^3*c*d^4*f^2)*e^4 -
(c^4*d^8 - 12*a*c^3*d^7*f + 22*a^2*c^2*d^6*f^2 - 12*a^3*c*d^5*f^3 + a^4*d^4*f^4)*e^2)))/(4*c^2*d^5*f - 8*a*c*
d^4*f^2 + 4*a^2*d^3*f^3 - a*c*d^2*e^4 - (c^2*d^4 - 6*a*c*d^3*f + a^2*d^2*f^2)*e^2))*log((4*c^2*d*f^2*x*e^3 - 2
*a*c*f^2*e^4 - 4*(2*c^2*d^2*f^3 - a*c*d*f^4)*x*e + sqrt(2)*(c^2*d*e^6 - (6*c^2*d^2*f - a*c*d*f^2)*e^4 + 4*(2*c
^2*d^3*f^2 - a*c*d^2*f^3)*e^2 + (8*c^3*d^7*f^2 - 24*a*c^2*d^6*f^3 + 24*a^2*c*d^5*f^4 - 8*a^3*d^4*f^5 + a*c^2*d
^3*e^6 + (c^3*d^5 - 8*a*c^2*d^4*f + 3*a^2*c*d^3*f^2)*e^4 - 2*(3*c^3*d^6*f - 11*a*c^2*d^5*f^2 + 9*a^2*c*d^4*f^3
- a^3*d^3*f^4)*e^2)*sqrt(-(c^2*e^6 - 2*(2*c^2*d*f - a*c*f^2)*e^4 + (4*c^2*d^2*f^2 - 4*a*c*d*f^3 + a^2*f^4)*e^
2)/(4*c^4*d^9*f - 16*a*c^3*d^8*f^2 + 24*a^2*c^2*d^7*f^3 - 16*a^3*c*d^6*f^4 + 4*a^4*d^5*f^5 - a^2*c^2*d^4*e^6 -
2*(a*c^3*d^6 - 4*a^2*c^2*d^5*f + a^3*c*d^4*f^2)*e^4 - (c^4*d^8 - 12*a*c^3*d^7*f + 22*a^2*c^2*d^6*f^2 - 12*a^3
*c*d^5*f^3 + a^4*d^4*f^4)*e^2)))*sqrt(c*x^2 + a)*sqrt(-(2*c*d^2*f^2 - 2*a*d*f^3 + c*e^4 - (4*c*d*f - a*f^2)*e^
2 + (4*c^2*d^5*f - 8*a*c*d^4*f^2 + 4*a^2*d^3*f^3 - a*c*d^2*e^4 - (c^2*d^4 - 6*a*c*d^3*f + a^2*d^2*f^2)*e^2)*sq
rt(-(c^2*e^6 - 2*(2*c^2*d*f - a*c*f^2)*e^4 + (4*c^2*d^2*f^2 - 4*a*c*d*f^3 + a^2*f^4)*e^2)/(4*c^4*d^9*f - 16*a*
c^3*d^8*f^2 + 24*a^2*c^2*d^7*f^3 - 16*a^3*c*d^6*f^4 + 4*a^4*d^5*f^5 - a^2*c^2*d^4*e^6 - 2*(a*c^3*d^6 - 4*a^2*c
^2*d^5*f + a^3*c*d^4*f^2)*e^4 - (c^4*d^8 - 12*a*c^3*d^7*f + 22*a^2*c^2*d^6*f^2 - 12*a^3*c*d^5*f^3 + a^4*d^4*f^
4)*e^2)))/(4*c^2*d^5*f - 8*a*c*d^4*f^2 + 4*a^2*d^3*f^3 - a*c*d^2*e^4 - (c^2*d^4 - 6*a*c*d^3*f + a^2*d^2*f^2)*e
^2)) + 2*(2*a*c*d*f^3 - a^2*f^4)*e^2 - 2*(4*a*c^2*d^5*f^3 - 8*a^2*c*d^4*f^4 + 4*a^3*d^3*f^5 - a^2*c*d^2*f^2*e^
4 - (a*c^2*d^4*f^2 - 6*a^2*c*d^3*f^3 + a^3*d^2*f^4)*e^2)*sqrt(-(c^2*e^6 - 2*(2*c^2*d*f - a*c*f^2)*e^4 + (4*c^2
*d^2*f^2 - 4*a*c*d*f^3 + a^2*f^4)*e^2)/(4*c^4*d^9*f - 16*a*c^3*d^8*f^2 + 24*a^2*c^2*d^7*f^3 - 16*a^3*c*d^6*f^4
+ 4*a^4*d^5*f^5 - a^2*c^2*d^4*e^6 - 2*(a*c^3*d^6 - 4*a^2*c^2*d^5*f + a^3*c*d^4*f^2)*e^4 - (c^4*d^8 - 12*a*c^3
*d^7*f + 22*a^2*c^2*d^6*f^2 - 12*a^3*c*d^5*f^3 + a^4*d^4*f^4)*e^2)))/x) - sqrt(2)*a*d*sqrt(-(2*c*d^2*f^2 - 2*a
*d*f^3 + c*e^4 - (4*c*d*f - a*f^2)*e^2 + (4*c^2*d^5*f - 8*a*c*d^4*f^2 + 4*a^2*d^3*f^3 - a*c*d^2*e^4 - (c^2*d^4
- 6*a*c*d^3*f + a^2*d^2*f^2)*e^2)*sqrt(-(c^2*e^6 - 2*(2*c^2*d*f - a*c*f^2)*e^4 + (4*c^2*d^2*f^2 - 4*a*c*d*f^3
+ a^2*f^4)*e^2)/(4*c^4*d^9*f - 16*a*c^3*d^8*f^2 + 24*a^2*c^2*d^7*f^3 - 16*a^3*c*d^6*f^4 + 4*a^4*d^5*f^5 - a^2
*c^2*d^4*e^6 - 2*(a*c^3*d^6 - 4*a^2*c^2*d^5*f + a^3*c*d^4*f^2)*e^4 - (c^4*d^8 - 12*a*c^3*d^7*f + 22*a^2*c^2*d^
6*f^2 - 12*a^3*c*d^5*f^3 + a^4*d^4*f^4)*e^2)))/(4*c^2*d^5*f - 8*a*c*d^4*f^2 + 4*a^2*d^3*f^3 - a*c*d^2*e^4 - (c
^2*d^4 - 6*a*c*d^3*f + a^2*d^2*f^2)*e^2))*log((4*c^2*d*f^2*x*e^3 - 2*a*c*f^2*e^4 - 4*(2*c^2*d^2*f^3 - a*c*d*f^
4)*x*e - sqrt(2)*(c^2*d*e^6 - (6*c^2*d^2*f - a*c*d*f^2)*e^4 + 4*(2*c^2*d^3*f^2 - a*c*d^2*f^3)*e^2 + (8*c^3*d^7
*f^2 - 24*a*c^2*d^6*f^3 + 24*a^2*c*d^5*f^4 - 8*a^3*d^4*f^5 + a*c^2*d^3*e^6 + (c^3*d^5 - 8*a*c^2*d^4*f + 3*a^2*
c*d^3*f^2)*e^4 - 2*(3*c^3*d^6*f - 11*a*c^2*d^5*f^2 + 9*a^2*c*d^4*f^3 - a^3*d^3*f^4)*e^2)*sqrt(-(c^2*e^6 - 2*(2
*c^2*d*f - a*c*f^2)*e^4 + (4*c^2*d^2*f^2 - 4*a*c*d*f^3 + a^2*f^4)*e^2)/(4*c^4*d^9*f - 16*a*c^3*d^8*f^2 + 24*a^
2*c^2*d^7*f^3 - 16*a^3*c*d^6*f^4 + 4*a^4*d^5*f^5 - a^2*c^2*d^4*e^6 - 2*(a*c^3*d^6 - 4*a^2*c^2*d^5*f + a^3*c*d^
4*f^2)*e^4 - (c^4*d^8 - 12*a*c^3*d^7*f + 22*a^2*c^2*d^6*f^2 - 12*a^3*c*d^5*f^3 + a^4*d^4*f^4)*e^2)))*sqrt(c*x^
2 + a)*sqrt(-(2*c*d^2*f^2 - 2*a*d*f^3 + c*e^4 - (4*c*d*f - a*f^2)*e^2 + (4*c^2*d^5*f - 8*a*c*d^4*f^2 + 4*a^2*d
^3*f^3 - a*c*d^2*e^4 - (c^2*d^4 - 6*a*c*d^3*f + a^2*d^2*f^2)*e^2)*sqrt(-(c^2*e^6 - 2*(2*c^2*d*f - a*c*f^2)*e^4
+ (4*c^2*d^2*f^2 - 4*a*c*d*f^3 + a^2*f^4)*e^2)/(4*c^4*d^9*f - 16*a*c^3*d^8*f^2 + 24*a^2*c^2*d^7*f^3 - 16*a^3*
c*d^6*f^4 + 4*a^4*d^5*f^5 - a^2*c^2*d^4*e^6 - 2*(a*c^3*d^6 - 4*a^2*c^2*d^5*f + a^3*c*d^4*f^2)*e^4 - (c^4*d^8 -
12*a*c^3*d^7*f + 22*a^2*c^2*d^6*f^2 - 12*a^3*c*d^5*f^3 + a^4*d^4*f^4)*e^2)))/(4*c^2*d^5*f - 8*a*c*d^4*f^2 + 4
*a^2*d^3*f^3 - a*c*d^2*e^4 - (c^2*d^4 - 6*a*c*d^3*f + a^2*d^2*f^2)*e^2)) + 2*(2*a*c*d*f^3 - a^2*f^4)*e^2 - 2*(
4*a*c^2*d^5*f^3 - 8*a^2*c*d^4*f^4 + 4*a^3*d^3*f^5 - a^2*c*d^2*f^2*e^4 - (a*c^2*d^4*f^2 - 6*a^2*c*d^3*f^3 + a^3
*d^2*f^4)*e^2)*sqrt(-(c^2*e^6 - 2*(2*c^2*d*f - a*c*f^2)*e^4 + (4*c^2*d^2*f^2 - 4*a*c*d*f^3 + a^2*f^4)*e^2)/(4*
c^4*d^9*f - 16*a*c^3*d^8*f^2 + 24*a^2*c^2*d^7*f^3 - 16*a^3*c*d^6*f^4 + 4*a^4*d^5*f^5 - a^2*c^2*d^4*e^6 - 2*(a*
c^3*d^6 - 4*a^2*c^2*d^5*f + a^3*c*d^4*f^2)*e^4 - (c^4*d^8 - 12*a*c^3*d^7*f + 22*a^2*c^2*d^6*f^2 - 12*a^3*c*d^5
*f^3 + a^4*d^4*f^4)*e^2)))/x) + sqrt(2)*a*d*sqr...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \sqrt {a + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*sqrt(a + c*x**2)*(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(1/x/(f*x^2+e*x+d)/(c*x^2+a)^(1/2),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 {1}{x\,\sqrt {c\,x^2+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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