### 3.267 $$\int \frac{\sqrt{d+e x} (A+B x+C x^2)}{\sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=557 $\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} (4 b C e-5 B c e+2 c C d) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{15 c^3 e^2 \sqrt{d+e x} \sqrt{a+b x+c x^2}}+\frac{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-c e (9 a C e+10 b B e+3 b C d)+c^2 \left (-\left (2 C d^2-5 e (3 A e+B d)\right )\right )+8 b^2 C e^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 c^3 e^2 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{2 \sqrt{d+e x} \sqrt{a+b x+c x^2} (4 b C e-5 B c e+2 c C d)}{15 c^2 e}+\frac{2 C (d+e x)^{3/2} \sqrt{a+b x+c x^2}}{5 c e}$

[Out]

(-2*(2*c*C*d - 5*B*c*e + 4*b*C*e)*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])/(15*c^2*e) + (2*C*(d + e*x)^(3/2)*Sqrt[
a + b*x + c*x^2])/(5*c*e) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(8*b^2*C*e^2 - c*e*(3*b*C*d + 10*b*B*e + 9*a*C*e) - c^2
*(2*C*d^2 - 5*e*(B*d + 3*A*e)))*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sq
rt[(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)])/(15*c^3*e^2*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2]) +
(2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*c*C*d - 5*B*c*e + 4*b*C*e)*(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[Sqrt[(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)])/
(15*c^3*e^2*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.888835, antiderivative size = 557, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 34, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {1653, 832, 843, 718, 424, 419} $\frac{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-c e (9 a C e+10 b B e+3 b C d)+c^2 \left (-\left (2 C d^2-5 e (3 A e+B d)\right )\right )+8 b^2 C e^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 c^3 e^2 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} (4 b C e-5 B c e+2 c C d) F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 c^3 e^2 \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{d+e x} \sqrt{a+b x+c x^2} (4 b C e-5 B c e+2 c C d)}{15 c^2 e}+\frac{2 C (d+e x)^{3/2} \sqrt{a+b x+c x^2}}{5 c e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(2*c*C*d - 5*B*c*e + 4*b*C*e)*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])/(15*c^2*e) + (2*C*(d + e*x)^(3/2)*Sqrt[
a + b*x + c*x^2])/(5*c*e) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(8*b^2*C*e^2 - c*e*(3*b*C*d + 10*b*B*e + 9*a*C*e) - c^2
*(2*C*d^2 - 5*e*(B*d + 3*A*e)))*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sq
rt[(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)])/(15*c^3*e^2*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2]) +
(2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*c*C*d - 5*B*c*e + 4*b*C*e)*(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[Sqrt[(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)])/
(15*c^3*e^2*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

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]

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 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 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)])

Rubi steps

\begin{align*} \int \frac{\sqrt{d+e x} \left (A+B x+C x^2\right )}{\sqrt{a+b x+c x^2}} \, dx &=\frac{2 C (d+e x)^{3/2} \sqrt{a+b x+c x^2}}{5 c e}+\frac{2 \int \frac{\sqrt{d+e x} \left (-\frac{1}{2} e (b C d-5 A c e+3 a C e)-\frac{1}{2} e (2 c C d-5 B c e+4 b C e) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{5 c e^2}\\ &=-\frac{2 (2 c C d-5 B c e+4 b C e) \sqrt{d+e x} \sqrt{a+b x+c x^2}}{15 c^2 e}+\frac{2 C (d+e x)^{3/2} \sqrt{a+b x+c x^2}}{5 c e}+\frac{4 \int \frac{\frac{1}{4} e \left (4 b^2 C d e+4 a b C e^2-b c d (C d+5 B e)+c e (15 A c d-7 a C d-5 a B e)\right )+\frac{1}{4} e \left (8 b^2 C e^2-c e (3 b C d+10 b B e+9 a C e)-c^2 \left (2 C d^2-5 e (B d+3 A e)\right )\right ) x}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{15 c^2 e^2}\\ &=-\frac{2 (2 c C d-5 B c e+4 b C e) \sqrt{d+e x} \sqrt{a+b x+c x^2}}{15 c^2 e}+\frac{2 C (d+e x)^{3/2} \sqrt{a+b x+c x^2}}{5 c e}+\frac{\left ((2 c C d-5 B c e+4 b C e) \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}{15 c^2 e^2}+\frac{\left (8 b^2 C e^2-c e (3 b C d+10 b B e+9 a C e)-c^2 \left (2 C d^2-5 e (B d+3 A e)\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x+c x^2}} \, dx}{15 c^2 e^2}\\ &=-\frac{2 (2 c C d-5 B c e+4 b C e) \sqrt{d+e x} \sqrt{a+b x+c x^2}}{15 c^2 e}+\frac{2 C (d+e x)^{3/2} \sqrt{a+b x+c x^2}}{5 c e}+\frac{\left (\sqrt{2} \sqrt{b^2-4 a c} \left (8 b^2 C e^2-c e (3 b C d+10 b B e+9 a C e)-c^2 \left (2 C d^2-5 e (B d+3 A e)\right )\right ) \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 )}{15 c^3 e^2 \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 (2 \sqrt{2} \sqrt{b^2-4 a c} (2 c C d-5 B c e+4 b C e) \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 )}{15 c^3 e^2 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ &=-\frac{2 (2 c C d-5 B c e+4 b C e) \sqrt{d+e x} \sqrt{a+b x+c x^2}}{15 c^2 e}+\frac{2 C (d+e x)^{3/2} \sqrt{a+b x+c x^2}}{5 c e}+\frac{\sqrt{2} \sqrt{b^2-4 a c} \left (8 b^2 C e^2-c e (3 b C d+10 b B e+9 a C e)-c^2 \left (2 C d^2-5 e (B d+3 A e)\right )\right ) \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 )}{15 c^3 e^2 \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{2 \sqrt{2} \sqrt{b^2-4 a c} (2 c C d-5 B c e+4 b C e) \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 )}{15 c^3 e^2 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 11.7188, size = 992, normalized size = 1.78 $\frac{\left (\frac{2 (c C d+5 B c e-4 b C e)}{15 c^2 e}+\frac{2 C x}{5 c}\right ) \sqrt{d+e x} \left (c x^2+b x+a\right )}{\sqrt{a+x (b+c x)}}-\frac{2 (d+e x)^{3/2} \sqrt{c x^2+b x+a} \left (\left (\left (2 C d^2-5 e (B d+3 A e)\right ) c^2+e (3 b C d+10 b B e+9 a C e) c-8 b^2 C e^2\right ) \left (c \left (\frac{d}{d+e x}-1\right )^2+\frac{e \left (-\frac{d b}{d+e x}+b+\frac{a e}{d+e x}\right )}{d+e x}\right )+\frac{i \sqrt{1-\frac{2 \left (c d^2+e (a e-b d)\right )}{\left (2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt{\frac{2 \left (c d^2+e (a e-b d)\right )}{\left (-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}+1} \left (\left (2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) \left (\left (5 e (B d+3 A e)-2 C d^2\right ) c^2-e (3 b C d+10 b B e+9 a C e) c+8 b^2 C e^2\right ) 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 )+\left (8 b^3 C e^3-b^2 \left (11 c d C+8 \sqrt{\left (b^2-4 a c\right ) e^2} C+10 B c e\right ) e^2+b c \left (15 A c e^2-17 a C e^2+5 B \left (3 c d e+2 \sqrt{\left (b^2-4 a c\right ) e^2} e\right )+3 C d \sqrt{\left (b^2-4 a c\right ) e^2}\right ) e+c \left (-15 A c \left (2 c d+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) e^2+a \left (14 c d C+9 \sqrt{\left (b^2-4 a c\right ) e^2} C+10 B c e\right ) e^2+c d \sqrt{\left (b^2-4 a c\right ) e^2} (2 C d-5 B e)\right )\right ) \text{EllipticF}\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 )\right )}{2 \sqrt{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}}} \sqrt{d+e x}}\right )}{15 c^3 e^3 \sqrt{a+x (b+c x)} \sqrt{\frac{(d+e x)^2 \left (c \left (\frac{d}{d+e x}-1\right )^2+\frac{e \left (-\frac{d b}{d+e x}+b+\frac{a e}{d+e x}\right )}{d+e x}\right )}{e^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(((2*(c*C*d + 5*B*c*e - 4*b*C*e))/(15*c^2*e) + (2*C*x)/(5*c))*Sqrt[d + e*x]*(a + b*x + c*x^2))/Sqrt[a + x*(b +
c*x)] - (2*(d + e*x)^(3/2)*Sqrt[a + b*x + c*x^2]*((-8*b^2*C*e^2 + c*e*(3*b*C*d + 10*b*B*e + 9*a*C*e) + c^2*(2
*C*d^2 - 5*e*(B*d + 3*A*e)))*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d)/(d + e*x) + (a*e)/(d + e*x)))/(d + e*x))
+ ((I/2)*Sqrt[1 - (2*(c*d^2 + e*(-(b*d) + a*e)))/((2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*Sqrt[1 +
(2*(c*d^2 + e*(-(b*d) + a*e)))/((-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*((2*c*d - b*e + Sqrt[(b^
2 - 4*a*c)*e^2])*(8*b^2*C*e^2 - c*e*(3*b*C*d + 10*b*B*e + 9*a*C*e) + c^2*(-2*C*d^2 + 5*e*(B*d + 3*A*e)))*Ellip
ticE[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]))] + (8*b^3*C*e^3 - b^2*e^
2*(11*c*C*d + 10*B*c*e + 8*C*Sqrt[(b^2 - 4*a*c)*e^2]) + c*(c*d*Sqrt[(b^2 - 4*a*c)*e^2]*(2*C*d - 5*B*e) - 15*A*
c*e^2*(2*c*d + Sqrt[(b^2 - 4*a*c)*e^2]) + a*e^2*(14*c*C*d + 10*B*c*e + 9*C*Sqrt[(b^2 - 4*a*c)*e^2])) + b*c*e*(
15*A*c*e^2 - 17*a*C*e^2 + 3*C*d*Sqrt[(b^2 - 4*a*c)*e^2] + 5*B*(3*c*d*e + 2*e*Sqrt[(b^2 - 4*a*c)*e^2])))*Ellipt
icF[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[2]*Sqrt[(c*d^2 +
e*(-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*Sqrt[d + e*x])))/(15*c^3*e^3*Sqrt[a + x*(b + c*x)
]*Sqrt[((d + e*x)^2*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d)/(d + e*x) + (a*e)/(d + e*x)))/(d + e*x)))/e^2])

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Maple [B]  time = 0.361, size = 8161, normalized size = 14.7 \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C x^{2} + B x + A\right )} \sqrt{e x + d}}{\sqrt{c x^{2} + b x + a}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C x^{2} + B x + A\right )} \sqrt{e x + d}}{\sqrt{c x^{2} + b x + a}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d + e x} \left (A + B x + C x^{2}\right )}{\sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out