3.2480 \(\int \frac {1}{\sqrt {d+e x} (a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=725 \[ -\frac {2 \sqrt {d+e x} \left (-2 c x (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right )-\left (2 a c e+b^2 (-e)+b c d\right ) \left (-5 c e (b d-2 a e)-2 b^2 e^2+8 c^2 d^2\right )+3 a c e (2 c d-b e)^2\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2}+\frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-5 a e)-b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {2 \sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2 \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {d+e x} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} \left (a e^2-b d e+c d^2\right )} \]

[Out]

-2/3*(b*c*d-b^2*e+2*a*c*e+c*(-b*e+2*c*d)*x)*(e*x+d)^(1/2)/(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)/(c*x^2+b*x+a)^(3/2)
-2/3*(3*a*c*e*(-b*e+2*c*d)^2-(2*a*c*e-b^2*e+b*c*d)*(8*c^2*d^2-2*b^2*e^2-5*c*e*(-2*a*e+b*d))-2*c*(-b*e+2*c*d)*(
4*c^2*d^2-b^2*e^2-4*c*e*(-2*a*e+b*d))*x)*(e*x+d)^(1/2)/(-4*a*c+b^2)^2/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)^(1/2
)-2/3*(-b*e+2*c*d)*(4*c^2*d^2-b^2*e^2-4*c*e*(-2*a*e+b*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)*(e*x+d)^(1
/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/(-4*a*c+b^2)^(3/2)/(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*(16*c^2*d^2-b^2*e^2-4*c*e*(-5*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)*(-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)
/(-4*a*c+b^2)^(3/2)/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.74, antiderivative size = 725, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {740, 822, 843, 718, 424, 419} \[ -\frac {2 \sqrt {d+e x} \left (-2 c x (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right )-\left (2 a c e+b^2 (-e)+b c d\right ) \left (-5 c e (b d-2 a e)-2 b^2 e^2+8 c^2 d^2\right )+3 a c e (2 c d-b e)^2\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2}+\frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-5 a e)-b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {2 \sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2 \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {d+e x} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d + e*x]*(a + b*x + c*x^2)^(5/2)),x]

[Out]

(-2*Sqrt[d + e*x]*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(a +
b*x + c*x^2)^(3/2)) - (2*Sqrt[d + e*x]*(3*a*c*e*(2*c*d - b*e)^2 - (b*c*d - b^2*e + 2*a*c*e)*(8*c^2*d^2 - 2*b^
2*e^2 - 5*c*e*(b*d - 2*a*e)) - 2*c*(2*c*d - b*e)*(4*c^2*d^2 - b^2*e^2 - 4*c*e*(b*d - 2*a*e))*x))/(3*(b^2 - 4*a
*c)^2*(c*d^2 - b*d*e + a*e^2)^2*Sqrt[a + b*x + c*x^2]) - (2*Sqrt[2]*(2*c*d - b*e)*(4*c^2*d^2 - b^2*e^2 - 4*c*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*(b^2 - 4*a*c)^(3/2)*(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]*(16*c^2*d^2 - b^2*e^2 - 4*c*e*(4*b*d - 5*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*(
b^2 - 4*a*c)^(3/2)*(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 740

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

Rule 822

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d+e x} \left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 \sqrt {d+e x} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (8 c^2 d^2-2 b^2 e^2-5 c e (b d-2 a e)\right )+\frac {3}{2} c e (2 c d-b e) x}{\sqrt {d+e x} \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {2 \sqrt {d+e x} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \sqrt {d+e x} \left (3 a c e (2 c d-b e)^2-\left (b c d-b^2 e+2 a c e\right ) \left (8 c^2 d^2-2 b^2 e^2-5 c e (b d-2 a e)\right )-2 c (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x\right )}{3 \left (b^2-4 a c\right )^2 \left (c d^2-b d e+a e^2\right )^2 \sqrt {a+b x+c x^2}}+\frac {4 \int \frac {-\frac {1}{4} c e \left (b^3 d e^2-4 a c e \left (c d^2+5 a e^2\right )+4 b c d \left (2 c d^2+5 a e^2\right )-b^2 \left (11 c d^2 e-a e^3\right )\right )-\frac {1}{2} c e (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )^2 \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {2 \sqrt {d+e x} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \sqrt {d+e x} \left (3 a c e (2 c d-b e)^2-\left (b c d-b^2 e+2 a c e\right ) \left (8 c^2 d^2-2 b^2 e^2-5 c e (b d-2 a e)\right )-2 c (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x\right )}{3 \left (b^2-4 a c\right )^2 \left (c d^2-b d e+a e^2\right )^2 \sqrt {a+b x+c x^2}}+\frac {\left (c \left (16 c^2 d^2-b^2 e^2-4 c e (4 b d-5 a e)\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 )^2 \left (c d^2-b d e+a e^2\right )}-\frac {\left (2 c (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )^2 \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {2 \sqrt {d+e x} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \sqrt {d+e x} \left (3 a c e (2 c d-b e)^2-\left (b c d-b^2 e+2 a c e\right ) \left (8 c^2 d^2-2 b^2 e^2-5 c e (b d-2 a e)\right )-2 c (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x\right )}{3 \left (b^2-4 a c\right )^2 \left (c d^2-b d e+a e^2\right )^2 \sqrt {a+b x+c x^2}}-\frac {\left (2 \sqrt {2} (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\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 \left (b^2-4 a c\right )^{3/2} \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} \left (16 c^2 d^2-b^2 e^2-4 c e (4 b d-5 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 \left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right ) \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 \sqrt {d+e x} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \sqrt {d+e x} \left (3 a c e (2 c d-b e)^2-\left (b c d-b^2 e+2 a c e\right ) \left (8 c^2 d^2-2 b^2 e^2-5 c e (b d-2 a e)\right )-2 c (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x\right )}{3 \left (b^2-4 a c\right )^2 \left (c d^2-b d e+a e^2\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {2} (2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\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 \left (b^2-4 a c\right )^{3/2} \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} \left (16 c^2 d^2-b^2 e^2-4 c e (4 b d-5 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 \left (b^2-4 a c\right )^{3/2} \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 = 12.87, size = 5566, normalized size = 7.68 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d + e*x]*(a + b*x + c*x^2)^(5/2)),x]

[Out]

Result too large to show

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fricas [F]  time = 1.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{c^{3} e x^{7} + {\left (c^{3} d + 3 \, b c^{2} e\right )} x^{6} + 3 \, {\left (b c^{2} d + {\left (b^{2} c + a c^{2}\right )} e\right )} x^{5} + {\left (3 \, {\left (b^{2} c + a c^{2}\right )} d + {\left (b^{3} + 6 \, a b c\right )} e\right )} x^{4} + a^{3} d + {\left ({\left (b^{3} + 6 \, a b c\right )} d + 3 \, {\left (a b^{2} + a^{2} c\right )} e\right )} x^{3} + 3 \, {\left (a^{2} b e + {\left (a b^{2} + a^{2} c\right )} d\right )} x^{2} + {\left (3 \, a^{2} b d + a^{3} e\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*sqrt(e*x + d)/(c^3*e*x^7 + (c^3*d + 3*b*c^2*e)*x^6 + 3*(b*c^2*d + (b^2*c + a*c^
2)*e)*x^5 + (3*(b^2*c + a*c^2)*d + (b^3 + 6*a*b*c)*e)*x^4 + a^3*d + ((b^3 + 6*a*b*c)*d + 3*(a*b^2 + a^2*c)*e)*
x^3 + 3*(a^2*b*e + (a*b^2 + a^2*c)*d)*x^2 + (3*a^2*b*d + a^3*e)*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} \sqrt {e x + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^2 + b*x + a)^(5/2)*sqrt(e*x + d)), x)

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maple [B]  time = 0.38, size = 19400, normalized size = 26.76 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} \sqrt {e x + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^2 + b*x + a)^(5/2)*sqrt(e*x + d)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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