3.1.67 \(\int \frac {1}{\sqrt {a+c x^2} (d+e x+f x^2)} \, dx\) [67]

Optimal. Leaf size=266 \[ -\frac {\sqrt {2} f \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {\sqrt {2} f \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}} \]

[Out]

-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^(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)+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
^(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 = 0.10, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {999, 739, 212} \begin {gather*} \frac {\sqrt {2} f \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 {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 {2} f \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 {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 )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*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[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])])) + (S
qrt[2]*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[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])])

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 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 999

Int[1/(((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/q), Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[2*(c/q), Int[1/((b + q + 2*c*x)*Sqrt[
d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.31, size = 131, normalized size = 0.49 \begin {gather*} -2 \sqrt {c} \text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {\log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(588\) vs. \(2(232)=464\).
time = 0.12, size = 589, normalized size = 2.21

method	result	size
default	\(\frac {\sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}-\frac {\sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}\)	\(589\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(-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))-1/(-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))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-%e^2>0)', see `assume?`
for more det

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 4761 vs. \(2 (236) = 472\).
time = 1.01, size = 4761, normalized size = 17.90 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(1/(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/(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}{\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/((a + c*x^2)^(1/2)*(d + e*x + f*x^2)),x)

[Out]

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

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