### 3.2451 $$\int \frac{(a+b x+c x^2)^{3/2}}{(d+e x)^{5/2}} \, dx$$

Optimal. Leaf size=499 $\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{3 c e^4 \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{8 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 e^4 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 \sqrt{a+b x+c x^2} (-3 b e+8 c d+2 c e x)}{3 e^3 \sqrt{d+e x}}-\frac{2 \left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}$

[Out]

(2*(8*c*d - 3*b*e + 2*c*e*x)*Sqrt[a + b*x + c*x^2])/(3*e^3*Sqrt[d + e*x]) - (2*(a + b*x + c*x^2)^(3/2))/(3*e*(
d + e*x)^(3/2)) - (8*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]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*
a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*e^4*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*S
qrt[a + b*x + c*x^2]) + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(16*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(4*b*d - a*e))*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[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)])/(3*c*e^4*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.452133, antiderivative size = 499, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {732, 812, 843, 718, 424, 419} $\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c e^4 \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{8 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 e^4 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 \sqrt{a+b x+c x^2} (-3 b e+8 c d+2 c e x)}{3 e^3 \sqrt{d+e x}}-\frac{2 \left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

(2*(8*c*d - 3*b*e + 2*c*e*x)*Sqrt[a + b*x + c*x^2])/(3*e^3*Sqrt[d + e*x]) - (2*(a + b*x + c*x^2)^(3/2))/(3*e*(
d + e*x)^(3/2)) - (8*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]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*
a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*e^4*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*S
qrt[a + b*x + c*x^2]) + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(16*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(4*b*d - a*e))*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[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)])/(3*c*e^4*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 812

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

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{\left (a+b x+c x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx &=-\frac{2 \left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}+\frac{\int \frac{(b+2 c x) \sqrt{a+b x+c x^2}}{(d+e x)^{3/2}} \, dx}{e}\\ &=\frac{2 (8 c d-3 b e+2 c e x) \sqrt{a+b x+c x^2}}{3 e^3 \sqrt{d+e x}}-\frac{2 \left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}-\frac{2 \int \frac{\frac{1}{2} \left (8 b c d-3 b^2 e-4 a c e\right )+4 c (2 c d-b e) x}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{3 e^3}\\ &=\frac{2 (8 c d-3 b e+2 c e x) \sqrt{a+b x+c x^2}}{3 e^3 \sqrt{d+e x}}-\frac{2 \left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}-\frac{(8 c (2 c d-b e)) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x+c x^2}} \, dx}{3 e^4}+\frac{\left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{3 e^4}\\ &=\frac{2 (8 c d-3 b e+2 c e x) \sqrt{a+b x+c x^2}}{3 e^3 \sqrt{d+e x}}-\frac{2 \left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}-\frac{\left (8 \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 e^4 \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} \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\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^4 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ &=\frac{2 (8 c d-3 b e+2 c e x) \sqrt{a+b x+c x^2}}{3 e^3 \sqrt{d+e x}}-\frac{2 \left (a+b x+c x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}-\frac{8 \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 e^4 \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} \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\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^4 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 8.97314, size = 978, normalized size = 1.96 $\frac{\sqrt{d+e x} \left (\frac{2 (a+x (b+c x)) \left (c \left (8 d^2+10 e x d+e^2 x^2\right )-e (3 b d+a e+4 b e x)\right )}{e^3 (d+e x)^2}-\frac{(d+e x) \left (-\frac{16 (b e-2 c d) \sqrt{\frac{c d^2+e (a e-b d)}{-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}}} (a+x (b+c x)) e^2}{(d+e x)^2}-\frac{4 i \sqrt{2} (2 c d-b e) \left (2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) \sqrt{\frac{-2 a e^2+2 c d x e+\sqrt{\left (b^2-4 a c\right ) e^2} x e+b (d-e x) e+d \sqrt{\left (b^2-4 a c\right ) e^2}}{\left (2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt{\frac{2 a e^2-2 c d x e+\sqrt{\left (b^2-4 a c\right ) e^2} x e+b (e x-d) e+d \sqrt{\left (b^2-4 a c\right ) e^2}}{\left (-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{c d^2-b e d+a e^2}{-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}}}}{\sqrt{d+e x}}\right )|-\frac{-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}}\right )}{\sqrt{d+e x}}+\frac{i \sqrt{2} \left (b^2 e^2-4 a c e^2-4 b \sqrt{\left (b^2-4 a c\right ) e^2} e+8 c d \sqrt{\left (b^2-4 a c\right ) e^2}\right ) \sqrt{\frac{-2 a e^2+2 c d x e+\sqrt{\left (b^2-4 a c\right ) e^2} x e+b (d-e x) e+d \sqrt{\left (b^2-4 a c\right ) e^2}}{\left (2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt{\frac{2 a e^2-2 c d x e+\sqrt{\left (b^2-4 a c\right ) e^2} x e+b (e x-d) e+d \sqrt{\left (b^2-4 a c\right ) e^2}}{\left (-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{c d^2-b e d+a e^2}{-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}}}}{\sqrt{d+e x}}\right ),-\frac{-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt{\left (b^2-4 a c\right ) e^2}}\right )}{\sqrt{d+e x}}\right )}{e^5 \sqrt{\frac{c d^2+e (a e-b d)}{-2 c d+b e+\sqrt{\left (b^2-4 a c\right ) e^2}}}}\right )}{3 \sqrt{a+x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

(Sqrt[d + e*x]*((2*(a + x*(b + c*x))*(-(e*(3*b*d + a*e + 4*b*e*x)) + c*(8*d^2 + 10*d*e*x + e^2*x^2)))/(e^3*(d
+ e*x)^2) - ((d + e*x)*((-16*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 - ((4*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*Sqr
t[(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*Arc
Sinh[(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 + 8*c*d*Sqrt[(b^2 - 4*a*c)*e^2] - 4*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))]*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]))/(e^5*Sqrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*
e + Sqrt[(b^2 - 4*a*c)*e^2])])))/(3*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.395, size = 5874, normalized size = 11.8 \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(e*x+d)^(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 (c x^{2} + b x + a\right )}^{\frac{3}{2}}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(e*x + d)^(5/2), 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 )}^{\frac{3}{2}} \sqrt{e x + d}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x + c x^{2}\right )^{\frac{3}{2}}}{\left (d + e x\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(e*x+d)**(5/2),x)

[Out]

Integral((a + b*x + c*x**2)**(3/2)/(d + e*x)**(5/2), 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((c*x^2+b*x+a)^(3/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

Timed out