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

Optimal. Leaf size=684 \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} \left (3 C e (b d-a e)-c \left (e (B d-A e)+2 C d^2\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (e \left (3 a e (2 C d-B e)-b \left (-2 A e^2-B d e+4 C d^2\right )\right )+c d \left (e (B d-4 A e)+2 C d^2\right )\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e^2 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2 \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {a+b x+c x^2} \left (C d^2-e (B d-A e)\right )}{3 e (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac {2 \sqrt {a+b x+c x^2} \left (e \left (3 a e (2 C d-B e)-b \left (-2 A e^2-B d e+4 C d^2\right )\right )+c d \left (e (B d-4 A e)+2 C d^2\right )\right )}{3 e \sqrt {d+e x} \left (a e^2-b d e+c d^2\right )^2} \]

[Out]

-2/3*(C*d^2-e*(-A*e+B*d))*(c*x^2+b*x+a)^(1/2)/e/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)+2/3*(c*d*(2*C*d^2+e*(-4*A*e+
B*d))+e*(3*a*e*(-B*e+2*C*d)-b*(-2*A*e^2-B*d*e+4*C*d^2)))*(c*x^2+b*x+a)^(1/2)/e/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)^(
1/2)-1/3*(c*d*(2*C*d^2+e*(-4*A*e+B*d))+e*(3*a*e*(-B*e+2*C*d)-b*(-2*A*e^2-B*d*e+4*C*d^2)))*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)/e^2/(a*e^2-b*d*e+
c*d^2)^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*(3*C*e*(-a*e+b*d)-c*(2*C*d
^2+e*(-A*e+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^2/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(1/2)/(c*x^2+b*x+a
)^(1/2)

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Rubi [A]  time = 1.19, antiderivative size = 680, normalized size of antiderivative = 0.99, 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, 834, 843, 718, 424, 419} \[ \frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} \left (-3 C e (b d-a e)+c e (B d-A e)+2 c C d^2\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (2 A e+B d)\right )+c d e (B d-4 A e)+2 c C d^3\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e^2 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2 \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {a+b x+c x^2} \left (C d^2-e (B d-A e)\right )}{3 e (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac {2 \sqrt {a+b x+c x^2} \left (3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (2 A e+B d)\right )+c d e (B d-4 A e)+2 c C d^3\right )}{3 e \sqrt {d+e x} \left (a e^2-b d e+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(C*d^2 - e*(B*d - A*e))*Sqrt[a + b*x + c*x^2])/(3*e*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (2*(2*c*C*d
^3 + c*d*e*(B*d - 4*A*e) + 3*a*e^2*(2*C*d - B*e) - b*e*(4*C*d^2 - e*(B*d + 2*A*e)))*Sqrt[a + b*x + c*x^2])/(3*
e*(c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x]) - (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*c*C*d^3 + c*d*e*(B*d - 4*A*e) + 3*a
*e^2*(2*C*d - B*e) - b*e*(4*C*d^2 - e*(B*d + 2*A*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^2*(c*d^2 - b*d*e + a*e^2)^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]*(2*c*C*d^2 - 3*C*e*(b*d - a*e) + c*e*(
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^2*(c*d^2 - b*d*e + a*e^2)*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 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(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*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (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 {A+B x+C x^2}{(d+e x)^{5/2} \sqrt {a+b x+c x^2}} \, dx &=-\frac {2 \left (C d^2-e (B d-A e)\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}-\frac {2 \int \frac {-\frac {b C d^2-b e (B d+2 A e)+3 e (A c d-a C d+a B e)}{2 e}-\frac {1}{2} \left (B c d-3 b C d+\frac {2 c C d^2}{e}-A c e+3 a C e\right ) x}{(d+e x)^{3/2} \sqrt {a+b x+c x^2}} \, dx}{3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {2 \left (C d^2-e (B d-A e)\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac {2 \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A e)\right )\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right )^2 \sqrt {d+e x}}+\frac {4 \int \frac {\frac {3 b^2 C d^2 e-b d \left (c C d^2+6 a C e^2+c e (2 B d+A e)\right )+e \left (A c \left (3 c d^2-a e^2\right )+a \left (3 a C e^2-c d (C d-4 B e)\right )\right )}{4 e}-\frac {c \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A e)\right )\right ) x}{4 e}}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {2 \left (C d^2-e (B d-A e)\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac {2 \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A e)\right )\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right )^2 \sqrt {d+e x}}+\frac {\left (2 c C d^2-3 C e (b d-a e)+c e (B d-A e)\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 e^2 \left (c d^2-b d e+a e^2\right )}-\frac {\left (c \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A e)\right )\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{3 e^2 \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {2 \left (C d^2-e (B d-A e)\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac {2 \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A e)\right )\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right )^2 \sqrt {d+e x}}-\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A 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 e^2 \left (c d^2-b d e+a e^2\right )^2 \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 (2 c C d^2-3 C e (b d-a e)+c e (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^2 \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 \left (C d^2-e (B d-A e)\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right ) (d+e x)^{3/2}}+\frac {2 \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A e)\right )\right ) \sqrt {a+b x+c x^2}}{3 e \left (c d^2-b d e+a e^2\right )^2 \sqrt {d+e x}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} \left (2 c C d^3+c d e (B d-4 A e)+3 a e^2 (2 C d-B e)-b e \left (4 C d^2-e (B d+2 A 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 e^2 \left (c d^2-b d e+a e^2\right )^2 \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 (2 c C d^2-3 C e (b d-a e)+c e (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^2 \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 12.16, size = 1194, normalized size = 1.75 \[ \frac {2 \sqrt {c x^2+b x+a} \left (\frac {i \sqrt {1-\frac {2 \left (c d^2+e (a e-b d)\right )}{\left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}} \sqrt {\frac {2 \left (c d^2+e (a e-b d)\right )}{\left (-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) (d+e x)}+1} \left (\left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) \left (c d \left (2 C d^2+e (B d-4 A e)\right )+e \left (-4 b C d^2+b e (B d+2 A e)-3 a e (B e-2 C d)\right )\right ) 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 )-\left (-6 a^2 C e^4-8 a B c d e^3-3 a B \sqrt {\left (b^2-4 a c\right ) e^2} e^3+2 a c C d^2 e^2+2 A c \left (-3 c d^2-2 \sqrt {\left (b^2-4 a c\right ) e^2} d+a e^2\right ) e^2-b^2 \left (2 C d^2+e (B d+2 A e)\right ) e^2+6 a C d \sqrt {\left (b^2-4 a c\right ) e^2} e^2+b \left (2 A \left (3 c d+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) e^2+B \left (3 c d^2+\sqrt {\left (b^2-4 a c\right ) e^2} d+3 a e^2\right ) e+2 C d \left (3 a e^2-2 d \sqrt {\left (b^2-4 a c\right ) e^2}\right )\right ) e+B c d^2 \sqrt {\left (b^2-4 a c\right ) e^2} e+2 c C d^3 \sqrt {\left (b^2-4 a c\right ) e^2}\right ) F\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 )\right )}{2 \sqrt {2} \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}}} \sqrt {d+e x}}-\left (2 c C d^3+c e (B d-4 A e) d-3 a e^2 (B e-2 C d)+b e \left (e (B d+2 A e)-4 C d^2\right )\right ) \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 )\right ) (d+e x)^{3/2}}{3 e^3 \left (c d^2-b e d+a e^2\right )^2 \sqrt {a+x (b+c x)} \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}}}+\frac {\left (c x^2+b x+a\right ) \left (-\frac {2 \left (C d^2-B e d+A e^2\right )}{3 e \left (c d^2-b e d+a e^2\right ) (d+e x)^2}-\frac {2 \left (-2 c C d^3-B c e d^2+4 b C e d^2-b B e^2 d+4 A c e^2 d-6 a C e^2 d-2 A b e^3+3 a B e^3\right )}{3 e \left (c d^2-b e d+a e^2\right )^2 (d+e x)}\right ) \sqrt {d+e x}}{\sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(a + b*x + c*x^2)*((-2*(C*d^2 - B*d*e + A*e^2))/(3*e*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) - (2*
(-2*c*C*d^3 - B*c*d^2*e + 4*b*C*d^2*e - b*B*d*e^2 + 4*A*c*d*e^2 - 6*a*C*d*e^2 - 2*A*b*e^3 + 3*a*B*e^3))/(3*e*(
c*d^2 - b*d*e + a*e^2)^2*(d + e*x))))/Sqrt[a + x*(b + c*x)] + (2*(d + e*x)^(3/2)*Sqrt[a + b*x + c*x^2]*(-((2*c
*C*d^3 + c*d*e*(B*d - 4*A*e) - 3*a*e^2*(-2*C*d + B*e) + b*e*(-4*C*d^2 + e*(B*d + 2*A*e)))*(c*(-1 + d/(d + e*x)
)^2 + (e*(b - (b*d)/(d + e*x) + (a*e)/(d + e*x)))/(d + e*x))) + ((I/2)*Sqrt[1 - (2*(c*d^2 + e*(-(b*d) + a*e)))
/((2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*Sqrt[1 + (2*(c*d^2 + e*(-(b*d) + a*e)))/((-2*c*d + b*e +
Sqrt[(b^2 - 4*a*c)*e^2])*(d + e*x))]*((2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])*(c*d*(2*C*d^2 + e*(B*d - 4*A*e)
) + e*(-4*b*C*d^2 + b*e*(B*d + 2*A*e) - 3*a*e*(-2*C*d + B*e)))*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]))] - (2*a*c*C*d^2*e^2 - 8*a*B*c*d*e^3 - 6*a^2*C*e^4 + 2*c*C*d^3*Sqrt
[(b^2 - 4*a*c)*e^2] + B*c*d^2*e*Sqrt[(b^2 - 4*a*c)*e^2] + 6*a*C*d*e^2*Sqrt[(b^2 - 4*a*c)*e^2] - 3*a*B*e^3*Sqrt
[(b^2 - 4*a*c)*e^2] + 2*A*c*e^2*(-3*c*d^2 + a*e^2 - 2*d*Sqrt[(b^2 - 4*a*c)*e^2]) - b^2*e^2*(2*C*d^2 + e*(B*d +
2*A*e)) + b*e*(2*A*e^2*(3*c*d + Sqrt[(b^2 - 4*a*c)*e^2]) + 2*C*d*(3*a*e^2 - 2*d*Sqrt[(b^2 - 4*a*c)*e^2]) + B*
e*(3*c*d^2 + 3*a*e^2 + d*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]))]))/(Sqrt[2]*Sqrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c
)*e^2])]*Sqrt[d + e*x])))/(3*e^3*(c*d^2 - b*d*e + a*e^2)^2*Sqrt[a + x*(b + c*x)]*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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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C x^{2} + B x + A\right )} \sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{c e^{3} x^{5} + {\left (3 \, c d e^{2} + b e^{3}\right )} x^{4} + a d^{3} + {\left (3 \, c d^{2} e + 3 \, b d e^{2} + a e^{3}\right )} x^{3} + {\left (c d^{3} + 3 \, b d^{2} e + 3 \, a d e^{2}\right )} x^{2} + {\left (b d^{3} + 3 \, a d^{2} e\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x+a)^(1/2),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 = 20481, normalized size = 29.94 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x+a)^(1/2),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 \[ \int \frac {C\,x^2+B\,x+A}{{\left (d+e\,x\right )}^{5/2}\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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