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

Optimal. Leaf size=567 \[ \frac{4 \sqrt{2} e \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a g^2-b f g+c f^2\right ) \sqrt{\frac{c (f+g x)}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}} (2 b e g-5 c d g+c e f) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 g \sqrt{b^2-4 a c}}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}\right )}{15 c^3 g^2 \sqrt{f+g x} \sqrt{a+b x+c x^2}}+\frac{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{f+g x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-c e g (9 a e g+20 b d g+3 b e f)+8 b^2 e^2 g^2+c^2 \left (-\left (-15 d^2 g^2-10 d e f g+2 e^2 f^2\right )\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} g}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}\right )}{15 c^3 g^2 \sqrt{a+b x+c x^2} \sqrt{\frac{c (f+g x)}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 e \sqrt{f+g x} \sqrt{a+b x+c x^2} (-4 b e g+7 c d g+c e f)}{15 c^2 g}+\frac{2 e (d+e x) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{5 c} \]

[Out]

(2*e*(c*e*f + 7*c*d*g - 4*b*e*g)*Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2])/(15*c^2*g) + (2*e*(d + e*x)*Sqrt[f + g*x
]*Sqrt[a + b*x + c*x^2])/(5*c) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(8*b^2*e^2*g^2 - c*e*g*(3*b*e*f + 20*b*d*g + 9*a*e
*g) - c^2*(2*e^2*f^2 - 10*d*e*f*g - 15*d^2*g^2))*Sqrt[f + g*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*El
lipticE[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]*g)/(2*c
*f - (b + Sqrt[b^2 - 4*a*c])*g)])/(15*c^3*g^2*Sqrt[(c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[a +
b*x + c*x^2]) + (4*Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*(c*e*f - 5*c*d*g + 2*b*e*g)*(c*f^2 - b*f*g + a*g^2)*Sqrt[(c*(f
+ g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[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]*g)/(2*c*f - (b + Sqrt[b^
2 - 4*a*c])*g)])/(15*c^3*g^2*Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.93054, antiderivative size = 567, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {941, 1653, 843, 718, 424, 419} \[ \frac{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{f+g x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-c e g (9 a e g+20 b d g+3 b e f)+8 b^2 e^2 g^2+c^2 \left (-\left (-15 d^2 g^2-10 d e f g+2 e^2 f^2\right )\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} g}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}\right )}{15 c^3 g^2 \sqrt{a+b x+c x^2} \sqrt{\frac{c (f+g x)}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{4 \sqrt{2} e \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a g^2-b f g+c f^2\right ) \sqrt{\frac{c (f+g x)}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}} (2 b e g-5 c d g+c e f) F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} g}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}\right )}{15 c^3 g^2 \sqrt{f+g x} \sqrt{a+b x+c x^2}}+\frac{2 e \sqrt{f+g x} \sqrt{a+b x+c x^2} (-4 b e g+7 c d g+c e f)}{15 c^2 g}+\frac{2 e (d+e x) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(c*e*f + 7*c*d*g - 4*b*e*g)*Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2])/(15*c^2*g) + (2*e*(d + e*x)*Sqrt[f + g*x
]*Sqrt[a + b*x + c*x^2])/(5*c) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(8*b^2*e^2*g^2 - c*e*g*(3*b*e*f + 20*b*d*g + 9*a*e
*g) - c^2*(2*e^2*f^2 - 10*d*e*f*g - 15*d^2*g^2))*Sqrt[f + g*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*El
lipticE[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]*g)/(2*c
*f - (b + Sqrt[b^2 - 4*a*c])*g)])/(15*c^3*g^2*Sqrt[(c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[a +
b*x + c*x^2]) + (4*Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*(c*e*f - 5*c*d*g + 2*b*e*g)*(c*f^2 - b*f*g + a*g^2)*Sqrt[(c*(f
+ g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[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]*g)/(2*c*f - (b + Sqrt[b^
2 - 4*a*c])*g)])/(15*c^3*g^2*Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2])

Rule 941

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

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 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{(d+e x)^2 \sqrt{f+g x}}{\sqrt{a+b x+c x^2}} \, dx &=\frac{2 e (d+e x) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{5 c}-\frac{\int \frac{-5 c d^2 f+e (b d f+2 a e f+a d g)-(c d (8 e f+5 d g)-e (3 b e f+2 b d g+3 a e g)) x-e (c e f+7 c d g-4 b e g) x^2}{\sqrt{f+g x} \sqrt{a+b x+c x^2}} \, dx}{5 c}\\ &=\frac{2 e (c e f+7 c d g-4 b e g) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{15 c^2 g}+\frac{2 e (d+e x) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{5 c}-\frac{2 \int \frac{-\frac{1}{2} g \left (4 b^2 e^2 f g+b e \left (4 a e g^2-c f (e f+10 d g)\right )+c g \left (15 c d^2 f-a e (7 e f+10 d g)\right )\right )-\frac{1}{2} g \left (8 b^2 e^2 g^2-c e g (3 b e f+20 b d g+9 a e g)-c^2 \left (2 e^2 f^2-10 d e f g-15 d^2 g^2\right )\right ) x}{\sqrt{f+g x} \sqrt{a+b x+c x^2}} \, dx}{15 c^2 g^2}\\ &=\frac{2 e (c e f+7 c d g-4 b e g) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{15 c^2 g}+\frac{2 e (d+e x) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{5 c}+\frac{\left (2 e (c e f-5 c d g+2 b e g) \left (c f^2-b f g+a g^2\right )\right ) \int \frac{1}{\sqrt{f+g x} \sqrt{a+b x+c x^2}} \, dx}{15 c^2 g^2}+\frac{\left (8 b^2 e^2 g^2-c e g (3 b e f+20 b d g+9 a e g)-c^2 \left (2 e^2 f^2-10 d e f g-15 d^2 g^2\right )\right ) \int \frac{\sqrt{f+g x}}{\sqrt{a+b x+c x^2}} \, dx}{15 c^2 g^2}\\ &=\frac{2 e (c e f+7 c d g-4 b e g) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{15 c^2 g}+\frac{2 e (d+e x) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{5 c}+\frac{\left (\sqrt{2} \sqrt{b^2-4 a c} \left (8 b^2 e^2 g^2-c e g (3 b e f+20 b d g+9 a e g)-c^2 \left (2 e^2 f^2-10 d e f g-15 d^2 g^2\right )\right ) \sqrt{f+g 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} g x^2}{2 c f-b g-\sqrt{b^2-4 a c} g}}}{\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 g^2 \sqrt{\frac{c (f+g x)}{2 c f-b g-\sqrt{b^2-4 a c} g}} \sqrt{a+b x+c x^2}}+\frac{\left (4 \sqrt{2} \sqrt{b^2-4 a c} e (c e f-5 c d g+2 b e g) \left (c f^2-b f g+a g^2\right ) \sqrt{\frac{c (f+g x)}{2 c f-b g-\sqrt{b^2-4 a c} g}} \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} g x^2}{2 c f-b g-\sqrt{b^2-4 a c} g}}} \, 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 g^2 \sqrt{f+g x} \sqrt{a+b x+c x^2}}\\ &=\frac{2 e (c e f+7 c d g-4 b e g) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{15 c^2 g}+\frac{2 e (d+e x) \sqrt{f+g x} \sqrt{a+b x+c x^2}}{5 c}+\frac{\sqrt{2} \sqrt{b^2-4 a c} \left (8 b^2 e^2 g^2-c e g (3 b e f+20 b d g+9 a e g)-c^2 \left (2 e^2 f^2-10 d e f g-15 d^2 g^2\right )\right ) \sqrt{f+g 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} g}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}\right )}{15 c^3 g^2 \sqrt{\frac{c (f+g x)}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}} \sqrt{a+b x+c x^2}}+\frac{4 \sqrt{2} \sqrt{b^2-4 a c} e (c e f-5 c d g+2 b e g) \left (c f^2-b f g+a g^2\right ) \sqrt{\frac{c (f+g x)}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}} \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} g}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}\right )}{15 c^3 g^2 \sqrt{f+g x} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 11.5826, size = 1002, normalized size = 1.77 \[ \frac{\left (\frac{2 e^2 x}{5 c}-\frac{2 e (-c e f-10 c d g+4 b e g)}{15 c^2 g}\right ) \sqrt{f+g x} \left (c x^2+b x+a\right )}{\sqrt{a+x (b+c x)}}-\frac{2 (f+g x)^{3/2} \sqrt{c x^2+b x+a} \left (\left (\left (2 e^2 f^2-10 d e g f-15 d^2 g^2\right ) c^2+e g (3 b e f+20 b d g+9 a e g) c-8 b^2 e^2 g^2\right ) \left (c \left (\frac{f}{f+g x}-1\right )^2+\frac{g \left (-\frac{f b}{f+g x}+b+\frac{a g}{f+g x}\right )}{f+g x}\right )+\frac{i \sqrt{1-\frac{2 \left (c f^2+g (a g-b f)\right )}{\left (2 c f-b g+\sqrt{\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}} \sqrt{\frac{2 \left (c f^2+g (a g-b f)\right )}{\left (-2 c f+b g+\sqrt{\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}+1} \left (\left (2 c f-b g+\sqrt{\left (b^2-4 a c\right ) g^2}\right ) \left (\left (-2 e^2 f^2+10 d e g f+15 d^2 g^2\right ) c^2-e g (3 b e f+20 b d g+9 a e g) c+8 b^2 e^2 g^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{c f^2-b g f+a g^2}{-2 c f+b g+\sqrt{\left (b^2-4 a c\right ) g^2}}}}{\sqrt{f+g x}}\right )|-\frac{-2 c f+b g+\sqrt{\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt{\left (b^2-4 a c\right ) g^2}}\right )+\left (-30 d^2 f g^2 c^3-\left (-15 b d (2 e f+d g) g^2-2 a e (7 e f+10 d g) g^2+\sqrt{\left (b^2-4 a c\right ) g^2} \left (-2 e^2 f^2+10 d e g f+15 d^2 g^2\right )\right ) c^2+e g \left (-g (11 e f+20 d g) b^2-17 a e g^2 b+\sqrt{\left (b^2-4 a c\right ) g^2} (3 e f+20 d g) b+9 a e g \sqrt{\left (b^2-4 a c\right ) g^2}\right ) c+8 b^2 e^2 g^2 \left (b g-\sqrt{\left (b^2-4 a c\right ) g^2}\right )\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{c f^2-b g f+a g^2}{-2 c f+b g+\sqrt{\left (b^2-4 a c\right ) g^2}}}}{\sqrt{f+g x}}\right ),-\frac{-2 c f+b g+\sqrt{\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt{\left (b^2-4 a c\right ) g^2}}\right )\right )}{2 \sqrt{2} \sqrt{\frac{c f^2+g (a g-b f)}{-2 c f+b g+\sqrt{\left (b^2-4 a c\right ) g^2}}} \sqrt{f+g x}}\right )}{15 c^3 g^3 \sqrt{a+x (b+c x)} \sqrt{\frac{(f+g x)^2 \left (c \left (\frac{f}{f+g x}-1\right )^2+\frac{g \left (-\frac{f b}{f+g x}+b+\frac{a g}{f+g x}\right )}{f+g x}\right )}{g^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*e*(-(c*e*f) - 10*c*d*g + 4*b*e*g))/(15*c^2*g) + (2*e^2*x)/(5*c))*Sqrt[f + g*x]*(a + b*x + c*x^2))/Sqrt[a
+ x*(b + c*x)] - (2*(f + g*x)^(3/2)*Sqrt[a + b*x + c*x^2]*((-8*b^2*e^2*g^2 + c*e*g*(3*b*e*f + 20*b*d*g + 9*a*
e*g) + c^2*(2*e^2*f^2 - 10*d*e*f*g - 15*d^2*g^2))*(c*(-1 + f/(f + g*x))^2 + (g*(b - (b*f)/(f + g*x) + (a*g)/(f
+ g*x)))/(f + g*x)) + ((I/2)*Sqrt[1 - (2*(c*f^2 + g*(-(b*f) + a*g)))/((2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2])
*(f + g*x))]*Sqrt[1 + (2*(c*f^2 + g*(-(b*f) + a*g)))/((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(f + g*x))]*((2
*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(8*b^2*e^2*g^2 - c*e*g*(3*b*e*f + 20*b*d*g + 9*a*e*g) + c^2*(-2*e^2*f^2
+ 10*d*e*f*g + 15*d^2*g^2))*EllipticE[I*ArcSinh[(Sqrt[2]*Sqrt[(c*f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^
2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c
)*g^2]))] + (-30*c^3*d^2*f*g^2 + 8*b^2*e^2*g^2*(b*g - Sqrt[(b^2 - 4*a*c)*g^2]) + c*e*g*(-17*a*b*e*g^2 + 9*a*e*
g*Sqrt[(b^2 - 4*a*c)*g^2] + b*Sqrt[(b^2 - 4*a*c)*g^2]*(3*e*f + 20*d*g) - b^2*g*(11*e*f + 20*d*g)) - c^2*(-15*b
*d*g^2*(2*e*f + d*g) - 2*a*e*g^2*(7*e*f + 10*d*g) + Sqrt[(b^2 - 4*a*c)*g^2]*(-2*e^2*f^2 + 10*d*e*f*g + 15*d^2*
g^2)))*EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[(c*f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sq
rt[f + g*x]], -((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2]))]))/(Sqrt[2]*
Sqrt[(c*f^2 + g*(-(b*f) + a*g))/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])]*Sqrt[f + g*x])))/(15*c^3*g^3*Sqrt[a
+ x*(b + c*x)]*Sqrt[((f + g*x)^2*(c*(-1 + f/(f + g*x))^2 + (g*(b - (b*f)/(f + g*x) + (a*g)/(f + g*x)))/(f + g*
x)))/g^2])

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Maple [B]  time = 0.348, size = 8248, normalized size = 14.6 \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)^2*(g*x+f)^(1/2)/(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 (e x + d\right )}^{2} \sqrt{g x + f}}{\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)^2*(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^2*sqrt(g*x + f)/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 (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt{g x + f}}{\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)^2*(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral((e^2*x^2 + 2*d*e*x + d^2)*sqrt(g*x + f)/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{\left (d + e x\right )^{2} \sqrt{f + g x}}{\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)**2*(g*x+f)**(1/2)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d + e*x)**2*sqrt(f + g*x)/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)^2*(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out