3.1.53 \(\int \frac {x \sqrt {a+c x^2}}{d+e x+f x^2} \, dx\) [53]

Optimal. Leaf size=395 \[ \frac {\sqrt {a+c x^2}}{f}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f^2}-\frac {\left (2 c d e f-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {\left (2 c d e f-\left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}} \]

[Out]

-e*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))*c^(1/2)/f^2+(c*x^2+a)^(1/2)/f-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*c*d*e*f-(a*f^2+c*(-d*f
+e^2))*(e-(-4*d*f+e^2)^(1/2)))/f^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*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*c*d*e*f-(a*f^2+c*(-d*f+e^2))*(e+(-4*d*f+e^2)^(1/2)))/f^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 = 0.64, antiderivative size = 395, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1034, 1094, 223, 212, 1048, 739} \begin {gather*} -\frac {\left (2 c d e f-\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 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^2 \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 (2 c d e f-\left (\sqrt {e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\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^2 \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 {c} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f^2}+\frac {\sqrt {a+c x^2}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a + c*x^2]/f - (Sqrt[c]*e*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/f^2 - ((2*c*d*e*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]*f^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])]) + ((2*c*d*e*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]*f^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 223

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 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 1094

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (f_.)*(x_)^2]), x_Sym
bol] :> Dist[C/c, Int[1/Sqrt[d + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + (B*c - b*C)*x)/((a + b*x + c*x^2)
*Sqrt[d + f*x^2]), x], x] /; FreeQ[{a, b, c, d, f, A, B, C}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x \sqrt {a+c x^2}}{d+e x+f x^2} \, dx &=\frac {\sqrt {a+c x^2}}{f}+\frac {\int \frac {-(c d-a f) x-c e x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{f}\\ &=\frac {\sqrt {a+c x^2}}{f}+\frac {\int \frac {c d e+\left (c e^2+f (-c d+a f)\right ) x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{f^2}-\frac {(c e) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{f^2}\\ &=\frac {\sqrt {a+c x^2}}{f}-\frac {(c e) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{f^2}+\frac {\left (2 c d e f-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \int \frac {1}{\left (e-\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{f^2 \sqrt {e^2-4 d f}}-\frac {\left (2 c d e f-\left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \int \frac {1}{\left (e+\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{f^2 \sqrt {e^2-4 d f}}\\ &=\frac {\sqrt {a+c x^2}}{f}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f^2}-\frac {\left (2 c d e f-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\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 )}{f^2 \sqrt {e^2-4 d f}}+\frac {\left (2 c d e f-\left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\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 )}{f^2 \sqrt {e^2-4 d f}}\\ &=\frac {\sqrt {a+c x^2}}{f}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f^2}-\frac {\left (2 c d e f-\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} f^2 \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 c d e f-\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} f^2 \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 = 0.37, size = 379, normalized size = 0.96 \begin {gather*} \frac {f \sqrt {a+c x^2}+\sqrt {c} e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )-\text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a c e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-a c d f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 c^{3/2} d e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-c e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+c d f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f*Sqrt[a + c*x^2] + Sqrt[c]*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]] - 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*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - a*c*d*f*
Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + a^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(3/2)*d*e*Lo
g[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - c*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + c*d*f*Log[-(
Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - a*f^2*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^2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1244\) vs. \(2(350)=700\).
time = 0.14, size = 1245, normalized size = 3.15

method	result	size
default	\(\frac {\left (e +\sqrt {-4 d f +e^{2}}\right ) \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 )}{2 \sqrt {-4 d f +e^{2}}\, f}+\frac {\left (-e +\sqrt {-4 d f +e^{2}}\right ) \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 )}{2 \sqrt {-4 d f +e^{2}}\, f}\)	\(1245\)
risch	\(\text {Expression too large to display}\)	\(3576\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)/f*(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)))+1/2*(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)/f*(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)))))

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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(x*(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?`
for more det

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

[Out]

Timed out

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

[Out]

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

[Out]

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

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