3.1639 \(\int \frac {(b+2 c x) \sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=441 \[ -\frac {4 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e \sqrt {d+e x} \sqrt {a+b x+c x^2}}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {4}{3} \sqrt {d+e x} \sqrt {a+b x+c x^2} \]

[Out]

4/3*(e*x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)+1/3*(-b*e+2*c*d)*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2
)^(1/2))^(1/2)*2^(1/2),(-2*e*(-4*a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*2^(1/2)*(-4*a*c+b^2)^
(1/2)*(e*x+d)^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/c/e/(c*x^2+b*x+a)^(1/2)/(c*(e*x+d)/(2*c*d-e*(b+(-4*a
*c+b^2)^(1/2))))^(1/2)-4/3*(a*e^2-b*d*e+c*d^2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))
^(1/2)*2^(1/2),(-2*e*(-4*a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*2^(1/2)*(-4*a*c+b^2)^(1/2)*(-
c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*(c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2)/c/e/(e*x+d)^(1/2)/(c*x^
2+b*x+a)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 441, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {832, 843, 718, 424, 419} \[ -\frac {4 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e \sqrt {d+e x} \sqrt {a+b x+c x^2}}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {4}{3} \sqrt {d+e x} \sqrt {a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])/3 + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*c*d - b*e)*Sqrt[d + e*x]*Sqrt[-((c*(
a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqr
t[2]], (-2*Sqrt[b^2 - 4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c*e*Sqrt[(c*(d + e*x))/(2*c*d - (b +
Sqrt[b^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2]) - (4*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)*Sqrt[(c*(
d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[S
qrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*e)/(2*c*d - (b + Sqrt[b
^2 - 4*a*c])*e)])/(3*c*e*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 718

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*Rt[b^2 - 4*a*c, 2]
*(d + e*x)^m*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))])/(c*Sqrt[a + b*x + c*x^2]*((2*c*(d + e*x))/(2*c*d -
b*e - e*Rt[b^2 - 4*a*c, 2]))^m), Subst[Int[(1 + (2*e*Rt[b^2 - 4*a*c, 2]*x^2)/(2*c*d - b*e - e*Rt[b^2 - 4*a*c,
2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], 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] && EqQ[m^2, 1/4]

Rule 832

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

Rule 843

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 {(b+2 c x) \sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx &=\frac {4}{3} \sqrt {d+e x} \sqrt {a+b x+c x^2}+\frac {2 \int \frac {\frac {1}{2} c (b d-2 a e)+\frac {1}{2} c (2 c d-b e) x}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 c}\\ &=\frac {4}{3} \sqrt {d+e x} \sqrt {a+b x+c x^2}+\frac {(2 c d-b e) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{3 e}-\frac {\left (2 \left (c d^2-b d e+a e^2\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 e}\\ &=\frac {4}{3} \sqrt {d+e x} \sqrt {a+b x+c x^2}+\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} (2 c d-b e) \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {2 \sqrt {b^2-4 a c} e x^2}{2 c d-b e-\sqrt {b^2-4 a c} e}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )}{3 c e \sqrt {\frac {c (d+e x)}{2 c d-b e-\sqrt {b^2-4 a c} e}} \sqrt {a+b x+c x^2}}-\frac {\left (4 \sqrt {2} \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-b e-\sqrt {b^2-4 a c} e}} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {2 \sqrt {b^2-4 a c} e x^2}{2 c d-b e-\sqrt {b^2-4 a c} e}}} \, dx,x,\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )}{3 c e \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=\frac {4}{3} \sqrt {d+e x} \sqrt {a+b x+c x^2}+\frac {\sqrt {2} \sqrt {b^2-4 a c} (2 c d-b e) \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e \sqrt {\frac {c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {a+b x+c x^2}}-\frac {4 \sqrt {2} \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 7.73, size = 933, normalized size = 2.12 \[ \frac {\sqrt {d+e x} \left (8 (a+x (b+c x))+\frac {(d+e x) \left (-\frac {4 (b e-2 c d) \sqrt {\frac {c d^2+e (a e-b d)}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}} (a+x (b+c x)) e^2}{(d+e x)^2}-\frac {i \sqrt {2} (2 c d-b e) \left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) \sqrt {\frac {-2 a e^2+2 c d x e+\sqrt {\left (b^2-4 a c\right ) e^2} x e+b (d-e x) e+d \sqrt {\left (b^2-4 a c\right ) e^2}}{\left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt {\frac {2 a e^2-2 c d x e+\sqrt {\left (b^2-4 a c\right ) e^2} x e+b (e x-d) e+d \sqrt {\left (b^2-4 a c\right ) e^2}}{\left (-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} E\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )}{\sqrt {d+e x}}+\frac {i \sqrt {2} \left (b^2 e^2-4 a c e^2-b \sqrt {\left (b^2-4 a c\right ) e^2} e+2 c d \sqrt {\left (b^2-4 a c\right ) e^2}\right ) \sqrt {\frac {-2 a e^2+2 c d x e+\sqrt {\left (b^2-4 a c\right ) e^2} x e+b (d-e x) e+d \sqrt {\left (b^2-4 a c\right ) e^2}}{\left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt {\frac {2 a e^2-2 c d x e+\sqrt {\left (b^2-4 a c\right ) e^2} x e+b (e x-d) e+d \sqrt {\left (b^2-4 a c\right ) e^2}}{\left (-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )}{\sqrt {d+e x}}\right )}{c e^2 \sqrt {\frac {c d^2+e (a e-b d)}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}\right )}{6 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(8*(a + x*(b + c*x)) + ((d + e*x)*((-4*e^2*(-2*c*d + b*e)*Sqrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c
*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*(a + x*(b + c*x)))/(d + e*x)^2 - (I*Sqrt[2]*(2*c*d - b*e)*(2*c*d - b*e +
Sqrt[(b^2 - 4*a*c)*e^2])*Sqrt[(-2*a*e^2 + d*Sqrt[(b^2 - 4*a*c)*e^2] + 2*c*d*e*x + e*Sqrt[(b^2 - 4*a*c)*e^2]*x
+ b*e*(d - e*x))/((2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*Sqrt[(2*a*e^2 + d*Sqrt[(b^2 - 4*a*c)*e^2
] - 2*c*d*e*x + e*Sqrt[(b^2 - 4*a*c)*e^2]*x + b*e*(-d + e*x))/((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(d + e
*x))]*EllipticE[I*ArcSinh[(Sqrt[2]*Sqrt[(c*d^2 - b*d*e + a*e^2)/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])])/Sqr
t[d + e*x]], -((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2]))])/Sqrt[d + e*
x] + (I*Sqrt[2]*(b^2*e^2 - 4*a*c*e^2 + 2*c*d*Sqrt[(b^2 - 4*a*c)*e^2] - b*e*Sqrt[(b^2 - 4*a*c)*e^2])*Sqrt[(-2*a
*e^2 + d*Sqrt[(b^2 - 4*a*c)*e^2] + 2*c*d*e*x + e*Sqrt[(b^2 - 4*a*c)*e^2]*x + b*e*(d - e*x))/((2*c*d - b*e + Sq
rt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*Sqrt[(2*a*e^2 + d*Sqrt[(b^2 - 4*a*c)*e^2] - 2*c*d*e*x + e*Sqrt[(b^2 - 4*a*c
)*e^2]*x + b*e*(-d + e*x))/((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*EllipticF[I*ArcSinh[(Sqrt[2]*
Sqrt[(c*d^2 - b*d*e + a*e^2)/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])])/Sqrt[d + e*x]], -((-2*c*d + b*e + Sqrt
[(b^2 - 4*a*c)*e^2])/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2]))])/Sqrt[d + e*x]))/(c*e^2*Sqrt[(c*d^2 + e*(-(b*d)
 + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])])))/(6*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {e x + d}}{\sqrt {c x^{2} + b x + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.11, size = 1854, normalized size = 4.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(e*x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c*(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-
b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*
a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+
(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*(-4*a*c+b^2)^(1/2)*a*e^3-2^(1/2)*(-(e*x+d)/(b*
e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/
2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*
c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2)
)*(-4*a*c+b^2)^(1/2)*b*d*e^2+2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^
2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/
2)*e)*e)^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)
^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*(-4*a*c+b^2)^(1/2)*c*d^2*e+2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-
4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x
+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*
c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*a*b*e^3-
2*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a
*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^
(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a
*c+b^2)^(1/2)*e))^(1/2))*a*c*d*e^2-2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*
a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^
2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*
c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*b^2*d*e^2+3*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2
)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*
c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1
/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*b*c*d^2*e-2*2^(1/
2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)
^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(
-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)
^(1/2)*e))^(1/2))*c^2*d^3+2*x^3*c^2*e^3+2*b*c*e^3*x^2+2*x^2*c^2*d*e^2+2*a*c*e^3*x+2*x*b*c*d*e^2+2*a*c*d*e^2)/e
^2/(c*e*x^3+b*e*x^2+c*d*x^2+a*e*x+b*d*x+a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (b+2\,c\,x\right )\,\sqrt {d+e\,x}}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b + 2 c x\right ) \sqrt {d + e x}}{\sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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