3.2291 \(\int \frac {\sqrt {d+e x}}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=198 \[ \frac {\sqrt {2} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{\sqrt {c} \sqrt {b^2-4 a c}}-\frac {\sqrt {2} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{\sqrt {c} \sqrt {b^2-4 a c}} \]

[Out]

-arctanh(2^(1/2)*c^(1/2)*(e*x+d)^(1/2)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))^(1/2))*2^(1/2)*(2*c*d-e*(b-(-4*a*c+b^2
)^(1/2)))^(1/2)/c^(1/2)/(-4*a*c+b^2)^(1/2)+arctanh(2^(1/2)*c^(1/2)*(e*x+d)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2
)))^(1/2))*2^(1/2)*(2*c*d-e*(b+(-4*a*c+b^2)^(1/2)))^(1/2)/c^(1/2)/(-4*a*c+b^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {699, 1130, 208} \[ \frac {\sqrt {2} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{\sqrt {c} \sqrt {b^2-4 a c}}-\frac {\sqrt {2} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{\sqrt {c} \sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - S
qrt[b^2 - 4*a*c])*e]])/(Sqrt[c]*Sqrt[b^2 - 4*a*c])) + (Sqrt[2]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*ArcTanh
[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[c]*Sqrt[b^2 - 4*a*c])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 699

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

Rule 1130

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2*(b/q + 1))/2, Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2*(b/q - 1))/2, Int[(d*x)^(m - 2)/(b
/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.46, size = 175, normalized size = 0.88 \[ \frac {\sqrt {2} \left (\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )-\sqrt {e \sqrt {b^2-4 a c}-b e+2 c d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d+e x}}{\sqrt {e \sqrt {b^2-4 a c}-b e+2 c d}}\right )\right )}{\sqrt {c} \sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.04, size = 715, normalized size = 3.61 \[ -\frac {1}{2} \, \sqrt {2} \sqrt {\frac {2 \, c d - b e + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (\sqrt {2} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt {\frac {2 \, c d - b e + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} + 2 \, \sqrt {e x + d} e\right ) + \frac {1}{2} \, \sqrt {2} \sqrt {\frac {2 \, c d - b e + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (-\sqrt {2} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt {\frac {2 \, c d - b e + {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} + 2 \, \sqrt {e x + d} e\right ) + \frac {1}{2} \, \sqrt {2} \sqrt {\frac {2 \, c d - b e - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (\sqrt {2} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt {\frac {2 \, c d - b e - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} + 2 \, \sqrt {e x + d} e\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {\frac {2 \, c d - b e - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} \log \left (-\sqrt {2} {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}} \sqrt {\frac {2 \, c d - b e - {\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {\frac {e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}}{b^{2} c - 4 \, a c^{2}}} + 2 \, \sqrt {e x + d} e\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*sqrt((2*c*d - b*e + (b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3)))/(b^2*c - 4*a*c^2))*log(sqrt(
2)*(b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3))*sqrt((2*c*d - b*e + (b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4
*a*c^3)))/(b^2*c - 4*a*c^2)) + 2*sqrt(e*x + d)*e) + 1/2*sqrt(2)*sqrt((2*c*d - b*e + (b^2*c - 4*a*c^2)*sqrt(e^2
/(b^2*c^2 - 4*a*c^3)))/(b^2*c - 4*a*c^2))*log(-sqrt(2)*(b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3))*sqrt((2
*c*d - b*e + (b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3)))/(b^2*c - 4*a*c^2)) + 2*sqrt(e*x + d)*e) + 1/2*sq
rt(2)*sqrt((2*c*d - b*e - (b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3)))/(b^2*c - 4*a*c^2))*log(sqrt(2)*(b^2
*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3))*sqrt((2*c*d - b*e - (b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3)
))/(b^2*c - 4*a*c^2)) + 2*sqrt(e*x + d)*e) - 1/2*sqrt(2)*sqrt((2*c*d - b*e - (b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c
^2 - 4*a*c^3)))/(b^2*c - 4*a*c^2))*log(-sqrt(2)*(b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3))*sqrt((2*c*d -
b*e - (b^2*c - 4*a*c^2)*sqrt(e^2/(b^2*c^2 - 4*a*c^3)))/(b^2*c - 4*a*c^2)) + 2*sqrt(e*x + d)*e)

________________________________________________________________________________________

giac [A]  time = 0.29, size = 223, normalized size = 1.13 \[ -\frac {\sqrt {-4 \, c^{2} d + 2 \, {\left (b c - \sqrt {b^{2} - 4 \, a c} c\right )} e} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {x e + d}}{\sqrt {-\frac {2 \, c d - b e + \sqrt {{\left (2 \, c d - b e\right )}^{2} - 4 \, {\left (c d^{2} - b d e + a e^{2}\right )} c}}{c}}}\right )}{\sqrt {b^{2} - 4 \, a c} {\left | c \right |}} + \frac {\sqrt {-4 \, c^{2} d + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} e} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {x e + d}}{\sqrt {-\frac {2 \, c d - b e - \sqrt {{\left (2 \, c d - b e\right )}^{2} - 4 \, {\left (c d^{2} - b d e + a e^{2}\right )} c}}{c}}}\right )}{\sqrt {b^{2} - 4 \, a c} {\left | c \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-4*c^2*d + 2*(b*c - sqrt(b^2 - 4*a*c)*c)*e)*arctan(2*sqrt(1/2)*sqrt(x*e + d)/sqrt(-(2*c*d - b*e + sqrt((
2*c*d - b*e)^2 - 4*(c*d^2 - b*d*e + a*e^2)*c))/c))/(sqrt(b^2 - 4*a*c)*abs(c)) + sqrt(-4*c^2*d + 2*(b*c + sqrt(
b^2 - 4*a*c)*c)*e)*arctan(2*sqrt(1/2)*sqrt(x*e + d)/sqrt(-(2*c*d - b*e - sqrt((2*c*d - b*e)^2 - 4*(c*d^2 - b*d
*e + a*e^2)*c))/c))/(sqrt(b^2 - 4*a*c)*abs(c))

________________________________________________________________________________________

maple [B]  time = 0.15, size = 545, normalized size = 2.75 \[ \frac {\sqrt {2}\, b \,e^{2} \arctanh \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\, \sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}+\frac {\sqrt {2}\, b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\, \sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}-\frac {2 \sqrt {2}\, c d e \arctanh \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\, \sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}-\frac {2 \sqrt {2}\, c d e \arctan \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\, \sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}-\frac {\sqrt {2}\, e \arctanh \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}+\frac {\sqrt {2}\, e \arctan \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^2/(-(4*a*c-b^2)*e^2)^(1/2)*2^(1/2)/((-b*e+2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*2^(
1/2)/((-b*e+2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*c)*b-2*c*e/(-(4*a*c-b^2)*e^2)^(1/2)*2^(1/2)/((-b*e+2*c*d+
(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*2^(1/2)/((-b*e+2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/
2)*c)*d-e*2^(1/2)/((-b*e+2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*2^(1/2)/((-b*e+2*c*d+(
-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*c)+e^2/(-(4*a*c-b^2)*e^2)^(1/2)*2^(1/2)/((b*e-2*c*d+(-(4*a*c-b^2)*e^2)^(1/2)
)*c)^(1/2)*arctan((e*x+d)^(1/2)*2^(1/2)/((b*e-2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*c)*b-2*c*e/(-(4*a*c-b^2
)*e^2)^(1/2)*2^(1/2)/((b*e-2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*2^(1/2)/((b*e-2*c*d+(
-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*c)*d+e*2^(1/2)/((b*e-2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)
^(1/2)*2^(1/2)/((b*e-2*c*d+(-(4*a*c-b^2)*e^2)^(1/2))*c)^(1/2)*c)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e x + d}}{c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.18, size = 709, normalized size = 3.58 \[ -2\,\mathrm {atanh}\left (\frac {2\,\left (\sqrt {d+e\,x}\,\left (-8\,b^2\,c\,e^4+16\,b\,c^2\,d\,e^3-16\,c^3\,d^2\,e^2+16\,a\,c^2\,e^4\right )+\frac {\sqrt {d+e\,x}\,\left (8\,b^3\,c^2\,e^3-16\,d\,b^2\,c^3\,e^2-32\,a\,b\,c^3\,e^3+64\,a\,d\,c^4\,e^2\right )\,\left (b^3\,e+e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+8\,a\,c^2\,d-2\,b^2\,c\,d-4\,a\,b\,c\,e\right )}{2\,\left (16\,a^2\,c^3-8\,a\,b^2\,c^2+b^4\,c\right )}\right )\,\sqrt {-\frac {b^3\,e+e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+8\,a\,c^2\,d-2\,b^2\,c\,d-4\,a\,b\,c\,e}{2\,\left (16\,a^2\,c^3-8\,a\,b^2\,c^2+b^4\,c\right )}}}{16\,c^2\,d^2\,e^3-16\,b\,c\,d\,e^4+16\,a\,c\,e^5}\right )\,\sqrt {-\frac {b^3\,e+e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+8\,a\,c^2\,d-2\,b^2\,c\,d-4\,a\,b\,c\,e}{2\,\left (16\,a^2\,c^3-8\,a\,b^2\,c^2+b^4\,c\right )}}-2\,\mathrm {atanh}\left (\frac {2\,\left (\sqrt {d+e\,x}\,\left (-8\,b^2\,c\,e^4+16\,b\,c^2\,d\,e^3-16\,c^3\,d^2\,e^2+16\,a\,c^2\,e^4\right )-\frac {\sqrt {d+e\,x}\,\left (8\,b^3\,c^2\,e^3-16\,d\,b^2\,c^3\,e^2-32\,a\,b\,c^3\,e^3+64\,a\,d\,c^4\,e^2\right )\,\left (e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,e-8\,a\,c^2\,d+2\,b^2\,c\,d+4\,a\,b\,c\,e\right )}{2\,\left (16\,a^2\,c^3-8\,a\,b^2\,c^2+b^4\,c\right )}\right )\,\sqrt {\frac {e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,e-8\,a\,c^2\,d+2\,b^2\,c\,d+4\,a\,b\,c\,e}{2\,\left (16\,a^2\,c^3-8\,a\,b^2\,c^2+b^4\,c\right )}}}{16\,c^2\,d^2\,e^3-16\,b\,c\,d\,e^4+16\,a\,c\,e^5}\right )\,\sqrt {\frac {e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-b^3\,e-8\,a\,c^2\,d+2\,b^2\,c\,d+4\,a\,b\,c\,e}{2\,\left (16\,a^2\,c^3-8\,a\,b^2\,c^2+b^4\,c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*atanh((2*((d + e*x)^(1/2)*(16*a*c^2*e^4 - 8*b^2*c*e^4 - 16*c^3*d^2*e^2 + 16*b*c^2*d*e^3) + ((d + e*x)^(1/2
)*(8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2 - 32*a*b*c^3*e^3 + 64*a*c^4*d*e^2)*(b^3*e + e*(-(4*a*c - b^2)^3)^(1/2) + 8
*a*c^2*d - 2*b^2*c*d - 4*a*b*c*e))/(2*(b^4*c + 16*a^2*c^3 - 8*a*b^2*c^2)))*(-(b^3*e + e*(-(4*a*c - b^2)^3)^(1/
2) + 8*a*c^2*d - 2*b^2*c*d - 4*a*b*c*e)/(2*(b^4*c + 16*a^2*c^3 - 8*a*b^2*c^2)))^(1/2))/(16*c^2*d^2*e^3 + 16*a*
c*e^5 - 16*b*c*d*e^4))*(-(b^3*e + e*(-(4*a*c - b^2)^3)^(1/2) + 8*a*c^2*d - 2*b^2*c*d - 4*a*b*c*e)/(2*(b^4*c +
16*a^2*c^3 - 8*a*b^2*c^2)))^(1/2) - 2*atanh((2*((d + e*x)^(1/2)*(16*a*c^2*e^4 - 8*b^2*c*e^4 - 16*c^3*d^2*e^2 +
16*b*c^2*d*e^3) - ((d + e*x)^(1/2)*(8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2 - 32*a*b*c^3*e^3 + 64*a*c^4*d*e^2)*(e*(-
(4*a*c - b^2)^3)^(1/2) - b^3*e - 8*a*c^2*d + 2*b^2*c*d + 4*a*b*c*e))/(2*(b^4*c + 16*a^2*c^3 - 8*a*b^2*c^2)))*(
(e*(-(4*a*c - b^2)^3)^(1/2) - b^3*e - 8*a*c^2*d + 2*b^2*c*d + 4*a*b*c*e)/(2*(b^4*c + 16*a^2*c^3 - 8*a*b^2*c^2)
))^(1/2))/(16*c^2*d^2*e^3 + 16*a*c*e^5 - 16*b*c*d*e^4))*((e*(-(4*a*c - b^2)^3)^(1/2) - b^3*e - 8*a*c^2*d + 2*b
^2*c*d + 4*a*b*c*e)/(2*(b^4*c + 16*a^2*c^3 - 8*a*b^2*c^2)))^(1/2)

________________________________________________________________________________________

sympy [A]  time = 49.08, size = 155, normalized size = 0.78 \[ 2 e \operatorname {RootSum} {\left (t^{4} \left (256 a^{2} c^{3} e^{4} - 128 a b^{2} c^{2} e^{4} + 16 b^{4} c e^{4}\right ) + t^{2} \left (- 16 a b c e^{3} + 32 a c^{2} d e^{2} + 4 b^{3} e^{3} - 8 b^{2} c d e^{2}\right ) + a e^{2} - b d e + c d^{2}, \left (t \mapsto t \log {\left (64 t^{3} a c^{2} e^{2} - 16 t^{3} b^{2} c e^{2} - 2 t b e + 4 t c d + \sqrt {d + e x} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e*RootSum(_t**4*(256*a**2*c**3*e**4 - 128*a*b**2*c**2*e**4 + 16*b**4*c*e**4) + _t**2*(-16*a*b*c*e**3 + 32*a*
c**2*d*e**2 + 4*b**3*e**3 - 8*b**2*c*d*e**2) + a*e**2 - b*d*e + c*d**2, Lambda(_t, _t*log(64*_t**3*a*c**2*e**2
- 16*_t**3*b**2*c*e**2 - 2*_t*b*e + 4*_t*c*d + sqrt(d + e*x))))

________________________________________________________________________________________