3.1.54 \(\int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx\)
Optimal. Leaf size=298 \[ -\frac {\sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f}}+\frac {\sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f} \]
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Rubi [A] time = 0.38, antiderivative size = 298, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used =
{991, 217, 206, 1034, 725} \begin {gather*} -\frac {\sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\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} f \sqrt {e^2-4 d f}}+\frac {\sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\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} f \sqrt {e^2-4 d f}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + c*x^2]/(d + e*x + f*x^2),x]
[Out]
(Sqrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/f - (Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*ArcT
anh[(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]*f*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[(
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]*f*Sqrt[e^2 - 4*d*f])
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])
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
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 991
Int[Sqrt[(a_) + (c_.)*(x_)^2]/((d_) + (e_.)*(x_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sqrt[a + c*x^2]
, x], x] - Dist[1/f, Int[(c*d - a*f + c*e*x)/(Sqrt[a + c*x^2]*(d + e*x + f*x^2)), x], x] /; FreeQ[{a, c, d, e,
f}, x] && NeQ[e^2 - 4*d*f, 0]
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]
Rubi steps
\begin {align*} \int \frac {\sqrt {a+c x^2}}{d+e x+f x^2} \, dx &=-\frac {\int \frac {c d-a f+c e x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{f}+\frac {c \int \frac {1}{\sqrt {a+c x^2}} \, dx}{f}\\ &=\frac {c \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{f}+\frac {\left (2 f (c d-a f)-c e \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}{f \sqrt {e^2-4 d f}}+\frac {\left (2 a f^2+c \left (e^2-2 d f-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}{f \sqrt {e^2-4 d f}}\\ &=\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f}-\frac {\left (2 f (c d-a f)-c e \left (e+\sqrt {e^2-4 d f}\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 )}{f \sqrt {e^2-4 d f}}-\frac {\left (2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\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 )}{f \sqrt {e^2-4 d f}}\\ &=\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f}-\frac {\sqrt {2 a f^2+c \left (e^2-2 d f-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} f \sqrt {e^2-4 d f}}+\frac {\sqrt {2 a f^2+c \left (e^2-2 d f+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} f \sqrt {e^2-4 d f}}\\ \end {align*}
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Mathematica [A] time = 0.36, size = 282, normalized size = 0.95 \begin {gather*} \frac {-\sqrt {4 a f^2-2 c \left (e \sqrt {e^2-4 d f}+2 d f-e^2\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 {4 a f^2+2 c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\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 )+2 \sqrt {c} \sqrt {e^2-4 d f} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 f \sqrt {e^2-4 d f}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + c*x^2]/(d + e*x + f*x^2),x]
[Out]
(2*Sqrt[c]*Sqrt[e^2 - 4*d*f]*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] - Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + 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[4*a*f^2 + 2*c*(e^2 - 2*d*f + 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])])/(2*f*Sqrt[e^
2 - 4*d*f])
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IntegrateAlgebraic [C] time = 0.45, size = 267, normalized size = 0.90 \begin {gather*} \frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e+a^2 f\&,\frac {\text {$\#$1}^2 (-c) e \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} c^{3/2} d \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+a c e \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} a \sqrt {c} f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 \text {$\#$1}^3 f-3 \text {$\#$1}^2 \sqrt {c} e-2 \text {$\#$1} a f+4 \text {$\#$1} c d+a \sqrt {c} e}\&\right ]}{f}-\frac {\sqrt {c} \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{f} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[Sqrt[a + c*x^2]/(d + e*x + f*x^2),x]
[Out]
-((Sqrt[c]*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/f) + 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*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(3/2)*d*Log[-(Sqrt[c]*
x) + Sqrt[a + c*x^2] - #1]*#1 - 2*a*Sqrt[c]*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - c*e*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) & ]/f
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fricas [F(-1)] 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)/(f*x^2+e*x+d),x, algorithm="fricas")
[Out]
Timed out
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giac [F(-1)] 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)/(f*x^2+e*x+d),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 0.01, size = 3249, normalized size = 10.90 \begin {gather*} \text {output too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*x^2+a)^(1/2)/(f*x^2+e*x+d),x)
[Out]
-1/2/(-4*d*f+e^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/(-4*d*f+e^2)^(1/2)*c^(1/2)/f*l
n((-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))*e+1/2*c^(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/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))*c*e+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))*
a-1/(-4*d*f+e^2)^(1/2)/f*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))*c*d+1/2/(-4*d*f+e^2)^(1/2)/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))*c*e^2+1/2/(-4*d*f+e^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)/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^(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))*e+1/2/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))*c*e-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))*a+
1/(-4*d*f+e^2)^(1/2)/f*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))*c*d-1/2/(-4*d*f+e^2)^(1/2)/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))*c*e^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((c*x^2+a)^(1/2)/(f*x^2+e*x+d),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?` f
or more details)Is 4*d*f-e^2 positive, negative or zero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^2+a}}{f\,x^2+e\,x+d} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + c*x^2)^(1/2)/(d + e*x + f*x^2),x)
[Out]
int((a + c*x^2)^(1/2)/(d + e*x + f*x^2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + c x^{2}}}{d + e x + f x^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x**2+a)**(1/2)/(f*x**2+e*x+d),x)
[Out]
Integral(sqrt(a + c*x**2)/(d + e*x + f*x**2), x)
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