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 \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} \left (e (-2 a C e-3 b B e+8 b C d)-2 c \left (8 C d^2-e (4 B d-A e)\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e^4 \sqrt {d+e x} \sqrt {a+b x+c x^2}}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (2 \left (4 c d-\frac {b e}{2}\right ) \left (-a C e-A c e+b C d+B c d-\frac {2 c C d^2}{e}\right )+6 c (e (a B e-a C d+A c d)+b d (C d-B e))\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e^3 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \left (a+b x+c x^2\right )^{3/2} \left (C d^2-e (B d-A e)\right )}{3 e (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac {2 \sqrt {a+b x+c x^2} \left (e^2 x \left (-a C e-A c e+b C d+B c d-\frac {2 c C d^2}{e}\right )+e (b d-a e) (7 C d-3 B e)-c d \left (8 C d^2-e (4 B d-A e)\right )\right )}{3 e^3 \sqrt {d+e x} \left (a e^2-b d e+c d^2\right )} \]

[Out]

-2/3*(C*d^2-e*(-A*e+B*d))*(c*x^2+b*x+a)^(3/2)/e/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)-2/3*(e*(-a*e+b*d)*(-3*B*e+7*
C*d)-c*d*(8*C*d^2-e*(-A*e+4*B*d))+e^2*(B*c*d+b*C*d-2*c*C*d^2/e-A*c*e-a*C*e)*x)*(c*x^2+b*x+a)^(1/2)/e^3/(a*e^2-
b*d*e+c*d^2)/(e*x+d)^(1/2)+1/3*(2*(4*c*d-1/2*b*e)*(B*c*d+b*C*d-2*c*C*d^2/e-A*c*e-a*C*e)+6*c*(b*d*(-B*e+C*d)+e*
(A*c*d+B*a*e-C*a*d)))*EllipticE(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),(-2*e*(-4*
a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*2^(1/2)*(-4*a*c+b^2)^(1/2)*(e*x+d)^(1/2)*(-c*(c*x^2+b*
x+a)/(-4*a*c+b^2))^(1/2)/c/e^3/(a*e^2-b*d*e+c*d^2)/(c*x^2+b*x+a)^(1/2)/(c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/
2))))^(1/2)-2/3*(e*(-3*B*b*e-2*C*a*e+8*C*b*d)-2*c*(8*C*d^2-e*(-A*e+4*B*d)))*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^
2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),(-2*e*(-4*a*c+b^2)^(1/2)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2))*
2^(1/2)*(-4*a*c+b^2)^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*(c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^
(1/2)/c/e^4/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.27, antiderivative size = 711, normalized size of antiderivative = 1.00, 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 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)])

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 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 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 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]

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*}

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Mathematica [C]  time = 14.27, size = 8456, normalized size = 11.88 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[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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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C x^{2} + B x + A\right )} \sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C x^{2} + B x + A\right )} \sqrt {c x^{2} + b x + a}}{{\left (e x + d\right )}^{\frac {5}{2}}}\,{d x} \]

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]

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

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maple [B]  time = 0.13, size = 21038, normalized size = 29.55 \[ \text {output too large to display} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C x^{2} + B x + A\right )} \sqrt {c x^{2} + b x + a}}{{\left (e x + d\right )}^{\frac {5}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,x^2+B\,x+A\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (d+e\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x + C*x^2)*(a + b*x + c*x^2)^(1/2))/(d + e*x)^(5/2),x)

[Out]

int(((A + B*x + C*x^2)*(a + b*x + c*x^2)^(1/2))/(d + e*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x + C x^{2}\right ) \sqrt {a + b x + c x^{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]

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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