3.9.57 \(\int \frac {\sqrt {a+b x+c x^2}}{d+e x} \, dx\) [857]

Optimal. Leaf size=152 \[ \frac {\sqrt {a+b x+c x^2}}{e}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} e^2}+\frac {\sqrt {c d^2-b d e+a e^2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {748, 857, 635, 212, 738} \begin {gather*} \frac {\sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^2}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} e^2}+\frac {\sqrt {a+b x+c x^2}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 140, normalized size = 0.92 \begin {gather*} \frac {2 e \sqrt {a+x (b+c x)}+4 \sqrt {-c d^2+b d e-a e^2} \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )+\frac {(2 c d-b e) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{\sqrt {c}}}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*Sqrt[a + x*(b + c*x)] + 4*Sqrt[-(c*d^2) + b*d*e - a*e^2]*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + x*(b + c*
x)])/Sqrt[-(c*d^2) + e*(b*d - a*e)]] + ((2*c*d - b*e)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/Sqrt[c
])/(2*e^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(328\) vs. \(2(132)=264\).
time = 0.13, size = 329, normalized size = 2.16

method	result	size
default	\(\frac {\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}+\frac {\left (e b -2 c d \right ) \ln \left (\frac {\frac {e b -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\right )}{2 e \sqrt {c}}-\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{e}\)	\(329\)
risch	\(\frac {\sqrt {c \,x^{2}+b x +a}}{e}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b}{2 e \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {c}\, d}{e^{2}}-\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) a}{e \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}+\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) b d}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right ) c \,d^{2}}{e^{3} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\)	\(560\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+b*x+a)^(1/2)/(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(c*d^2-%e*b*d+%e^2*a>0)', see `
assume?` for

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Fricas [A]
time = 5.28, size = 990, normalized size = 6.51 \begin {gather*} \left [\frac {{\left (4 \, \sqrt {c x^{2} + b x + a} c e - {\left (2 \, c d - b e\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 2 \, \sqrt {c d^{2} - b d e + a e^{2}} c \log \left (-\frac {8 \, c^{2} d^{2} x^{2} + 8 \, b c d^{2} x + {\left (b^{2} + 4 \, a c\right )} d^{2} + 4 \, \sqrt {c d^{2} - b d e + a e^{2}} {\left (2 \, c d x + b d - {\left (b x + 2 \, a\right )} e\right )} \sqrt {c x^{2} + b x + a} + {\left (8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 8 \, a^{2}\right )} e^{2} - 2 \, {\left (4 \, b c d x^{2} + 4 \, a b d + {\left (3 \, b^{2} + 4 \, a c\right )} d x\right )} e}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right )\right )} e^{\left (-2\right )}}{4 \, c}, \frac {{\left ({\left (2 \, c d - b e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, \sqrt {c x^{2} + b x + a} c e + \sqrt {c d^{2} - b d e + a e^{2}} c \log \left (-\frac {8 \, c^{2} d^{2} x^{2} + 8 \, b c d^{2} x + {\left (b^{2} + 4 \, a c\right )} d^{2} + 4 \, \sqrt {c d^{2} - b d e + a e^{2}} {\left (2 \, c d x + b d - {\left (b x + 2 \, a\right )} e\right )} \sqrt {c x^{2} + b x + a} + {\left (8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 8 \, a^{2}\right )} e^{2} - 2 \, {\left (4 \, b c d x^{2} + 4 \, a b d + {\left (3 \, b^{2} + 4 \, a c\right )} d x\right )} e}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right )\right )} e^{\left (-2\right )}}{2 \, c}, \frac {{\left (4 \, \sqrt {-c d^{2} + b d e - a e^{2}} c \arctan \left (-\frac {\sqrt {-c d^{2} + b d e - a e^{2}} {\left (2 \, c d x + b d - {\left (b x + 2 \, a\right )} e\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} d^{2} x^{2} + b c d^{2} x + a c d^{2} + {\left (a c x^{2} + a b x + a^{2}\right )} e^{2} - {\left (b c d x^{2} + b^{2} d x + a b d\right )} e\right )}}\right ) + 4 \, \sqrt {c x^{2} + b x + a} c e - {\left (2 \, c d - b e\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right )\right )} e^{\left (-2\right )}}{4 \, c}, \frac {{\left (2 \, \sqrt {-c d^{2} + b d e - a e^{2}} c \arctan \left (-\frac {\sqrt {-c d^{2} + b d e - a e^{2}} {\left (2 \, c d x + b d - {\left (b x + 2 \, a\right )} e\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} d^{2} x^{2} + b c d^{2} x + a c d^{2} + {\left (a c x^{2} + a b x + a^{2}\right )} e^{2} - {\left (b c d x^{2} + b^{2} d x + a b d\right )} e\right )}}\right ) + {\left (2 \, c d - b e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, \sqrt {c x^{2} + b x + a} c e\right )} e^{\left (-2\right )}}{2 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*sqrt(c*x^2 + b*x + a)*c*e - (2*c*d - b*e)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x
+ a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 2*sqrt(c*d^2 - b*d*e + a*e^2)*c*log(-(8*c^2*d^2*x^2 + 8*b*c*d^2*x + (b^2 +
4*a*c)*d^2 + 4*sqrt(c*d^2 - b*d*e + a*e^2)*(2*c*d*x + b*d - (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a) + (8*a*b*x +
(b^2 + 4*a*c)*x^2 + 8*a^2)*e^2 - 2*(4*b*c*d*x^2 + 4*a*b*d + (3*b^2 + 4*a*c)*d*x)*e)/(x^2*e^2 + 2*d*x*e + d^2)
))*e^(-2)/c, 1/2*((2*c*d - b*e)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*
x + a*c)) + 2*sqrt(c*x^2 + b*x + a)*c*e + sqrt(c*d^2 - b*d*e + a*e^2)*c*log(-(8*c^2*d^2*x^2 + 8*b*c*d^2*x + (b
^2 + 4*a*c)*d^2 + 4*sqrt(c*d^2 - b*d*e + a*e^2)*(2*c*d*x + b*d - (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a) + (8*a*b
*x + (b^2 + 4*a*c)*x^2 + 8*a^2)*e^2 - 2*(4*b*c*d*x^2 + 4*a*b*d + (3*b^2 + 4*a*c)*d*x)*e)/(x^2*e^2 + 2*d*x*e +
d^2)))*e^(-2)/c, 1/4*(4*sqrt(-c*d^2 + b*d*e - a*e^2)*c*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*(2*c*d*x + b*d
- (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a)/(c^2*d^2*x^2 + b*c*d^2*x + a*c*d^2 + (a*c*x^2 + a*b*x + a^2)*e^2 - (b*
c*d*x^2 + b^2*d*x + a*b*d)*e)) + 4*sqrt(c*x^2 + b*x + a)*c*e - (2*c*d - b*e)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x
- b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c))*e^(-2)/c, 1/2*(2*sqrt(-c*d^2 + b*d*e - a*e^2)*c*
arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*(2*c*d*x + b*d - (b*x + 2*a)*e)*sqrt(c*x^2 + b*x + a)/(c^2*d^2*x^2 +
b*c*d^2*x + a*c*d^2 + (a*c*x^2 + a*b*x + a^2)*e^2 - (b*c*d*x^2 + b^2*d*x + a*b*d)*e)) + (2*c*d - b*e)*sqrt(-c)
*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*sqrt(c*x^2 + b*x + a)*c*e)
*e^(-2)/c]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x+a}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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