### 3.262 $$\int \frac{\sqrt{a+b x+c x^2} (A+B x+C x^2)}{(d+e x)^{5/2}} \, dx$$

Optimal. Leaf size=712 $-\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \left (e (-2 a C e-3 b B e+8 b C d)-2 c \left (8 C d^2-e (4 B d-A e)\right )\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{3 c e^4 \sqrt{d+e x} \sqrt{a+b x+c x^2}}+\frac{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (2 \left (4 c d-\frac{b e}{2}\right ) \left (-a C e-A c e+b C d+B c d-\frac{2 c C d^2}{e}\right )+6 c (e (a B e-a C d+A c d)+b d (C d-B e))\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c e^3 \sqrt{a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{2 \left (a+b x+c x^2\right )^{3/2} \left (C d^2-e (B d-A e)\right )}{3 e (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac{2 \sqrt{a+b x+c x^2} \left (e^2 x \left (-a C e-A c e+b C d+B c d-\frac{2 c C d^2}{e}\right )+e (b d-a e) (7 C d-3 B e)-c d \left (8 C d^2-e (4 B d-A e)\right )\right )}{3 e^3 \sqrt{d+e x} \left (a e^2-b d e+c d^2\right )}$

[Out]

(-2*(e*(b*d - a*e)*(7*C*d - 3*B*e) - c*d*(8*C*d^2 - e*(4*B*d - A*e)) + e^2*(B*c*d + b*C*d - (2*c*C*d^2)/e - A*
c*e - a*C*e)*x)*Sqrt[a + b*x + c*x^2])/(3*e^3*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x]) - (2*(C*d^2 - e*(B*d - A*
e))*(a + b*x + c*x^2)^(3/2))/(3*e*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*(4*
c*d - (b*e)/2)*(B*c*d + b*C*d - (2*c*C*d^2)/e - A*c*e - a*C*e) + 6*c*(b*d*(C*d - B*e) + e*(A*c*d - a*C*d + a*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*e^3*
(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[a + b*x + c*x^2]) - (2*Sq
rt[2]*Sqrt[b^2 - 4*a*c]*(e*(8*b*C*d - 3*b*B*e - 2*a*C*e) - 2*c*(8*C*d^2 - 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[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*e^4*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 1.27107, antiderivative size = 711, 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 = {1650, 812, 843, 718, 424, 419} $-\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \left (e (-2 a C e-3 b B e+8 b C d)-2 c \left (8 C d^2-e (4 B d-A e)\right )\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c e^4 \sqrt{d+e x} \sqrt{a+b x+c x^2}}+\frac{\sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (2 \left (4 c d-\frac{b e}{2}\right ) \left (-a C e-A c e+b C d+B c d-\frac{2 c C d^2}{e}\right )+6 c (e (a B e-a C d+A c d)+b d (C d-B e))\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c e^3 \sqrt{a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{2 \left (a+b x+c x^2\right )^{3/2} \left (C d^2-e (B d-A e)\right )}{3 e (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac{2 \sqrt{a+b x+c x^2} \left (-e x \left (-a C e-A c e+b C d+B c d-\frac{2 c C d^2}{e}\right )-(b d-a e) (7 C d-3 B e)-c d (4 B d-A e)+\frac{8 c C d^3}{e}\right )}{3 e^2 \sqrt{d+e x} \left (a e^2-b d e+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*((8*c*C*d^3)/e - c*d*(4*B*d - A*e) - (b*d - a*e)*(7*C*d - 3*B*e) - e*(B*c*d + b*C*d - (2*c*C*d^2)/e - A*c*e
- a*C*e)*x)*Sqrt[a + b*x + c*x^2])/(3*e^2*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x]) - (2*(C*d^2 - e*(B*d - A*e))
*(a + b*x + c*x^2)^(3/2))/(3*e*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*(4*c*d
- (b*e)/2)*(B*c*d + b*C*d - (2*c*C*d^2)/e - A*c*e - a*C*e) + 6*c*(b*d*(C*d - B*e) + e*(A*c*d - a*C*d + a*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*e^3*(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[a + b*x + c*x^2]) - (2*Sqrt[
2]*Sqrt[b^2 - 4*a*c]*(e*(8*b*C*d - 3*b*B*e - 2*a*C*e) - 2*c*(8*C*d^2 - 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[Sqrt[(b + Sqr
t[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 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

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{\sqrt{a+b x+c x^2} \left (A+B x+C x^2\right )}{(d+e x)^{5/2}} \, dx &=-\frac{2 \left (C d^2-e (B d-A e)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{2 \int \frac{\left (-\frac{3 (b d (C d-B e)+e (A c d-a C d+a B e))}{2 e}+\frac{3}{2} \left (B c d+b C d-\frac{2 c C d^2}{e}-A c e-a C e\right ) x\right ) \sqrt{a+b x+c x^2}}{(d+e x)^{3/2}} \, dx}{3 \left (c d^2-b d e+a e^2\right )}\\ &=\frac{2 \left (\frac{8 c C d^3}{e}-c d (4 B d-A e)-(b d-a e) (7 C d-3 B e)-e \left (B c d+b C d-\frac{2 c C d^2}{e}-A c e-a C e\right ) x\right ) \sqrt{a+b x+c x^2}}{3 e^2 \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x}}-\frac{2 \left (C d^2-e (B d-A e)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac{4 \int \frac{\frac{3}{4} \left (2 (2 b d-a e) \left (B c d+b C d-\frac{2 c C d^2}{e}-A c e-a C e\right )+3 b (b d (C d-B e)+e (A c d-a C d+a B e))\right )+\frac{3}{4} \left (2 \left (4 c d-\frac{b e}{2}\right ) \left (B c d+b C d-\frac{2 c C d^2}{e}-A c e-a C e\right )+6 c (b d (C d-B e)+e (A c d-a C d+a B e))\right ) x}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{9 e^2 \left (c d^2-b d e+a e^2\right )}\\ &=\frac{2 \left (\frac{8 c C d^3}{e}-c d (4 B d-A e)-(b d-a e) (7 C d-3 B e)-e \left (B c d+b C d-\frac{2 c C d^2}{e}-A c e-a C e\right ) x\right ) \sqrt{a+b x+c x^2}}{3 e^2 \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x}}-\frac{2 \left (C d^2-e (B d-A e)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{\left (e (8 b C d-3 b B e-2 a C e)-2 c \left (8 C d^2-e (4 B d-A e)\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{3 e^4}-\frac{\left (b C e^2 (b d-a e)+2 c^2 \left (8 C d^3-d e (4 B d-A e)\right )-c e \left (16 b C d^2-b e (7 B d-A e)-2 a e (7 C d-3 B e)\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x+c x^2}} \, dx}{3 e^4 \left (c d^2-b d e+a e^2\right )}\\ &=\frac{2 \left (\frac{8 c C d^3}{e}-c d (4 B d-A e)-(b d-a e) (7 C d-3 B e)-e \left (B c d+b C d-\frac{2 c C d^2}{e}-A c e-a C e\right ) x\right ) \sqrt{a+b x+c x^2}}{3 e^2 \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x}}-\frac{2 \left (C d^2-e (B d-A e)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{\left (\sqrt{2} \sqrt{b^2-4 a c} \left (b C e^2 (b d-a e)+2 c^2 \left (8 C d^3-d e (4 B d-A e)\right )-c e \left (16 b C d^2-b e (7 B d-A e)-2 a e (7 C d-3 B 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 )}{3 c e^4 \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{a+b x+c x^2}}-\frac{\left (2 \sqrt{2} \sqrt{b^2-4 a c} \left (e (8 b C d-3 b B e-2 a C e)-2 c \left (8 C d^2-e (4 B d-A e)\right )\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 \left (\frac{8 c C d^3}{e}-c d (4 B d-A e)-(b d-a e) (7 C d-3 B e)-e \left (B c d+b C d-\frac{2 c C d^2}{e}-A c e-a C e\right ) x\right ) \sqrt{a+b x+c x^2}}{3 e^2 \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x}}-\frac{2 \left (C d^2-e (B d-A e)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac{\sqrt{2} \sqrt{b^2-4 a c} \left (b C e^2 (b d-a e)+2 c^2 \left (8 C d^3-d e (4 B d-A e)\right )-c e \left (16 b C d^2-b e (7 B d-A e)-2 a e (7 C d-3 B 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 )}{3 c e^4 \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{a+b x+c x^2}}-\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \left (e (8 b C d-3 b B e-2 a C e)-2 c \left (8 C d^2-e (4 B d-A e)\right )\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 = 14.4505, size = 8456, normalized size = 11.88 $\text{Result too large to show}$

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sqrt[a + b*x + c*x^2]*(A + B*x + C*x^2))/(d + e*x)^(5/2),x]

[Out]

Result too large to show

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Maple [B]  time = 0.452, size = 21038, normalized size = 29.6 \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)*(c*x^2+b*x+a)^(1/2)/(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 )} \sqrt{c x^{2} + b x + a}}{{\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)*(c*x^2+b*x+a)^(1/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*x^2 + B*x + A)*sqrt(c*x^2 + b*x + a)/(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 )} \sqrt{c x^{2} + b x + a} \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)*(c*x^2+b*x+a)^(1/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

integral((C*x^2 + B*x + A)*sqrt(c*x^2 + b*x + a)*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 ) \sqrt{a + b x + c x^{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)*(c*x**2+b*x+a)**(1/2)/(e*x+d)**(5/2),x)

[Out]

Integral((A + B*x + C*x**2)*sqrt(a + b*x + c*x**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)*(c*x^2+b*x+a)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

Timed out