3.1652 \(\int \frac {(b+2 c x) \sqrt {d+e x}}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=517 \[ -\frac {2 e \sqrt {d+e x} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt {2} e \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {b^2-4 a c} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {4 \sqrt {2} e \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {d+e x}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 517, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {768, 740, 843, 718, 424, 419} \[ -\frac {2 e \sqrt {d+e x} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt {2} e \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {b^2-4 a c} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {4 \sqrt {2} e \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {d+e x}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x])/(3*(a + b*x + c*x^2)^(3/2)) - (2*e*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)*Sqrt[a + b*x + c*x^2]) + (Sqrt[2]*e*(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*Sqrt[b^2 - 4*a*c]*(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]) - (4*Sqrt[2]*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*Sqrt[b^2 - 4*a*c]*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 768

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

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

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Mathematica [C]  time = 7.38, size = 1016, normalized size = 1.97 \[ \frac {\sqrt {d+e x} \left (-\frac {4 e \left (e b^2+c (e x-d) b-2 c (a e+c d x)\right )}{\left (b^2-4 a c\right ) \left (e (b d-a e)-c d^2\right )}+\frac {(d+e x) \left (-\frac {4 (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 {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-b \sqrt {\left (b^2-4 a c\right ) e^2} e+2 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)}} 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 )}{\sqrt {d+e x}}\right )}{\left (b^2-4 a c\right ) \left (c d^2+e (a e-b d)\right ) \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}}}}-\frac {4}{a+x (b+c x)}\right )}{6 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(-4/(a + x*(b + c*x)) - (4*e*(b^2*e - 2*c*(a*e + c*d*x) + b*c*(-d + e*x)))/((b^2 - 4*a*c)*(-(c*
d^2) + e*(b*d - a*e))) + ((d + e*x)*((-4*e^2*(-2*c*d + b*e)*Sqrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*e + Sq
rt[(b^2 - 4*a*c)*e^2])]*(a + x*(b + c*x)))/(d + e*x)^2 - (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*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))]*Ellipti
cE[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] + (I*Sqrt[
2]*(b^2*e^2 - 4*a*c*e^2 + 2*c*d*Sqrt[(b^2 - 4*a*c)*e^2] - 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]))/((b^2 - 4*a*c)*(c*d^2 + e*(-(b*d) + a*e))*S
qrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])])))/(6*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {e x + d}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, {\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x + {\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \, {\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.20, size = 5571, normalized size = 10.78 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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