3.1650 \(\int \frac{(b+2 c x) (d+e x)^{5/2}}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=494 \[ -\frac{20 \sqrt{2} e \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{10 e \sqrt{d+e x} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{5 \sqrt{2} e \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{2 (d+e x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(d + e*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) - (10*e*Sqrt[d + e*x]*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^
2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (5*Sqrt[2]*e*(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]]/Sqrt[2]], (-2*Sqrt[b^2 -
4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c*Sqrt[b^2 - 4*a*c]*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b
^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2]) - (20*Sqrt[2]*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)])/(3*
c*Sqrt[b^2 - 4*a*c]*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.414939, antiderivative size = 494, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {768, 738, 843, 718, 424, 419} \[ -\frac{20 \sqrt{2} e \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{10 e \sqrt{d+e x} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{5 \sqrt{2} e \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{2 (d+e x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) - (10*e*Sqrt[d + e*x]*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^
2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (5*Sqrt[2]*e*(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]]/Sqrt[2]], (-2*Sqrt[b^2 -
4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c*Sqrt[b^2 - 4*a*c]*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b
^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2]) - (20*Sqrt[2]*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)])/(3*
c*Sqrt[b^2 - 4*a*c]*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

Rule 768

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))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

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

Mathematica [C]  time = 10.4069, size = 1150, normalized size = 2.33 \[ \frac{\sqrt{d+e x} \left (\frac{2 \left (10 d e x c^2-14 a e^2 c+5 b d e c-5 b e^2 x c+b^2 e^2\right )}{3 c \left (4 a c-b^2\right ) \left (c x^2+b x+a\right )}-\frac{2 \left (c d^2+2 c e x d-a e^2-b e^2 x\right )}{3 c \left (c x^2+b x+a\right )^2}\right ) \left (c x^2+b x+a\right )^3}{(a+x (b+c x))^{5/2}}+\frac{5 (d+e x)^{3/2} \left (-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}}} \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{2} (b e-2 c d) \left (2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) \sqrt{\frac{-\frac{2 a e^2}{d+e x}+b \left (\frac{2 d}{d+e x}-1\right ) e-2 c d \left (\frac{d}{d+e x}-1\right )+\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}}} \sqrt{\frac{\frac{2 a e^2}{d+e x}+2 c d \left (\frac{d}{d+e x}-1\right )+b \left (e-\frac{2 d e}{d+e x}\right )+\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}}} 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{-\frac{2 a e^2}{d+e x}+b \left (\frac{2 d}{d+e x}-1\right ) e-2 c d \left (\frac{d}{d+e x}-1\right )+\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}}} \sqrt{\frac{\frac{2 a e^2}{d+e x}+2 c d \left (\frac{d}{d+e x}-1\right )+b \left (e-\frac{2 d e}{d+e x}\right )+\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}}} \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 )}{\sqrt{d+e x}}\right ) \left (c x^2+b x+a\right )^{5/2}}{6 c \left (b^2-4 a c\right ) \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))^{5/2} \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[((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[d + e*x]*(a + b*x + c*x^2)^3*((-2*(c*d^2 - a*e^2 + 2*c*d*e*x - b*e^2*x))/(3*c*(a + b*x + c*x^2)^2) + (2*
(5*b*c*d*e + b^2*e^2 - 14*a*c*e^2 + 10*c^2*d*e*x - 5*b*c*e^2*x))/(3*c*(-b^2 + 4*a*c)*(a + b*x + c*x^2))))/(a +
x*(b + c*x))^(5/2) + (5*(d + e*x)^(3/2)*(a + b*x + c*x^2)^(5/2)*(-4*(-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])]*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d)/(d + e*x) + (a*e)/(d +
e*x)))/(d + e*x)) + (I*Sqrt[2]*(-2*c*d + b*e)*(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])*Sqrt[(Sqrt[(b^2 - 4*a*c
)*e^2] - (2*a*e^2)/(d + e*x) - 2*c*d*(-1 + d/(d + e*x)) + b*e*(-1 + (2*d)/(d + e*x)))/(2*c*d - b*e + Sqrt[(b^2
- 4*a*c)*e^2])]*Sqrt[(Sqrt[(b^2 - 4*a*c)*e^2] + (2*a*e^2)/(d + e*x) + 2*c*d*(-1 + d/(d + e*x)) + b*(e - (2*d*
e)/(d + e*x)))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*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])])/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] - (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[(Sqrt[(b^2 - 4*a*c)*e^2] - (2*a*e^2)/(d + e*x) - 2*c*d*(-1 +
d/(d + e*x)) + b*e*(-1 + (2*d)/(d + e*x)))/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*Sqrt[(Sqrt[(b^2 - 4*a*c)*e
^2] + (2*a*e^2)/(d + e*x) + 2*c*d*(-1 + d/(d + e*x)) + b*(e - (2*d*e)/(d + e*x)))/(-2*c*d + b*e + Sqrt[(b^2 -
4*a*c)*e^2])]*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]))])/Sqr
t[d + e*x]))/(6*c*(b^2 - 4*a*c)*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))^(5/2)*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 = 0.106, size = 5517, normalized size = 11.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(5/2),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 (2 \, c x + b\right )}{\left (e x + d\right )}^{\frac{5}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((2*c*x + b)*(e*x + d)^(5/2)/(c*x^2 + b*x + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (2 \, c e^{2} x^{3} + b d^{2} +{\left (4 \, c d e + b e^{2}\right )} x^{2} + 2 \,{\left (c d^{2} + b d e\right )} x\right )} \sqrt{c x^{2} + b x + a} \sqrt{e x + d}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral((2*c*e^2*x^3 + b*d^2 + (4*c*d*e + b*e^2)*x^2 + 2*(c*d^2 + b*d*e)*x)*sqrt(c*x^2 + b*x + a)*sqrt(e*x +
d)/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*
x^2), x)

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Sympy [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((2*c*x+b)*(e*x+d)**(5/2)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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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((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

Timed out