3.1644 \(\int \frac {(b+2 c x) (d+e x)^{5/2}}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=468 \[ -\frac {10 \sqrt {2} e \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}+\frac {10 \sqrt {2} e \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {10 e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}{3 c}-\frac {2 (d+e x)^{5/2}}{\sqrt {a+b x+c x^2}} \]

[Out]

-2*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(1/2)+10/3*e^2*(e*x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)/c+10/3*e*(-b*e+2*c*d)*Elliptic
E(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^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)-10/3*e*(a*e^2-b*d*e+c*d^2)*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^2/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 468, 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, 742, 843, 718, 424, 419} \[ -\frac {10 \sqrt {2} e \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}+\frac {10 \sqrt {2} e \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c^2 \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {10 e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}{3 c}-\frac {2 (d+e x)^{5/2}}{\sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^(5/2))/Sqrt[a + b*x + c*x^2] + (10*e^2*Sqrt[d + e*x]*Sqrt[a + b*x + c*x^2])/(3*c) + (10*Sqrt[2]*
Sqrt[b^2 - 4*a*c]*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*c^2*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2]) - (10
*Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*(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)]*S
qrt[-((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^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 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, 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]
 && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && 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) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (d+e x)^{5/2}}{\sqrt {a+b x+c x^2}}+(5 e) \int \frac {(d+e x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 (d+e x)^{5/2}}{\sqrt {a+b x+c x^2}}+\frac {10 e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}{3 c}+\frac {(10 e) \int \frac {\frac {1}{2} \left (3 c d^2-e (b d+a e)\right )+e (2 c d-b e) x}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 c}\\ &=-\frac {2 (d+e x)^{5/2}}{\sqrt {a+b x+c x^2}}+\frac {10 e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}{3 c}+\frac {(10 e (2 c d-b e)) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{3 c}-\frac {\left (5 e \left (c d^2-b d e+a e^2\right )\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 c}\\ &=-\frac {2 (d+e x)^{5/2}}{\sqrt {a+b x+c x^2}}+\frac {10 e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}{3 c}+\frac {\left (10 \sqrt {2} \sqrt {b^2-4 a c} 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 c^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 (10 \sqrt {2} \sqrt {b^2-4 a c} e \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 {-\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^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 (d+e x)^{5/2}}{\sqrt {a+b x+c x^2}}+\frac {10 e^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}{3 c}+\frac {10 \sqrt {2} \sqrt {b^2-4 a c} 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 c^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 {10 \sqrt {2} \sqrt {b^2-4 a c} e \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 {-\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^2 \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 4.22, size = 627, normalized size = 1.34 \[ \frac {\sqrt {d+e x} \left (2 \left (\frac {e^2 (5 a+x (5 b+2 c x))}{c}-3 d^2-6 d e x\right )+\frac {10 (d+e x) \left (\frac {2 e^2 (a+x (b+c x)) (2 c d-b e)}{(d+e x)^2}+\frac {i \sqrt {1-\frac {2 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt {e^2 \left (b^2-4 a c\right )}-b e+2 c d\right )}} \sqrt {\frac {2 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt {e^2 \left (b^2-4 a c\right )}+b e-2 c d\right )}+1} \left (\left (c \left (2 d \sqrt {e^2 \left (b^2-4 a c\right )}-a e^2-3 b d e\right )+b e \left (b e-\sqrt {e^2 \left (b^2-4 a c\right )}\right )+3 c^2 d^2\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )+(b e-2 c d) \left (\sqrt {e^2 \left (b^2-4 a c\right )}-b e+2 c d\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )\right )}{\sqrt {2} \sqrt {d+e x} \sqrt {\frac {e (a e-b d)+c d^2}{\sqrt {e^2 \left (b^2-4 a c\right )}+b e-2 c d}}}\right )}{c^2}\right )}{3 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[d + e*x]*(2*(-3*d^2 - 6*d*e*x + (e^2*(5*a + x*(5*b + 2*c*x)))/c) + (10*(d + e*x)*((2*e^2*(2*c*d - b*e)*(
a + x*(b + c*x)))/(d + e*x)^2 + (I*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)*(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])*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]))] + (3*c^2*d^2 + b*e*(b*e - Sqrt[(b^2 - 4*a*c)*e^2]) + c*(-3*b*d*e - a*e
^2 + 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 + S
qrt[(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])))/c^2))/(3*Sqrt[a + x*(b + c*x)])

________________________________________________________________________________________

fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c e^{2} x^{3} + b d^{2} + {\left (4 \, c d e + b e^{2}\right )} x^{2} + 2 \, {\left (c d^{2} + b d e\right )} x\right )} \sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.15, size = 2942, normalized size = 6.29 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(3/2),x)

[Out]

-1/3*(e*x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)*(15*2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e
)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*a*b*e^3*(-(e*x+d)/(b*e
-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2
)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)-30*2^(1/2)*EllipticF(2^(1/2)*(-(e*x+
d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)
*e))^(1/2))*a*c*d*e^2*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e
+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)-
5*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a
*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticF(2^
(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a
*c+b^2)^(1/2)*e))^(1/2))*(-4*a*c+b^2)^(1/2)*a*e^3-15*2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^
2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*b^2*d*e^2*(-
(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2
)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)+45*2^(1/2)*EllipticF(2^(
1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*
c+b^2)^(1/2)*e))^(1/2))*b*c*d^2*e*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^
(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*
e)*e)^(1/2)+5*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e
+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*
EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e
+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*(-4*a*c+b^2)^(1/2)*b*d*e^2-30*2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*
c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2)
)*c^2*d^3*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a
*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)-5*2^(1/2)*(-
(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2
)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticF(2^(1/2)*(-(e*x
+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2
)*e))^(1/2))*(-4*a*c+b^2)^(1/2)*c*d^2*e-20*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*
x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-
4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*
d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*a*b*e^3+40*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*
a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b
+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+
b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*a*c*d*e^2+
20*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*
a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2
^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*
a*c+b^2)^(1/2)*e))^(1/2))*b^2*d*e^2-60*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+
(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*
c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-
4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*b*c*d^2*e+40*2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*
c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(
-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^
2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*c^2*d^3-4*c^
2*e^3*x^3-10*b*c*e^3*x^2+8*c^2*d*e^2*x^2-10*a*c*e^3*x-10*b*c*d*e^2*x+18*x*c^2*d^2*e-10*a*c*d*e^2+6*c^2*d^3)/c^
2/(c*e*x^3+b*e*x^2+c*d*x^2+a*e*x+b*d*x+a*d)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^(5/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (b+2\,c\,x\right )\,{\left (d+e\,x\right )}^{5/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^(3/2),x)

[Out]

int(((b + 2*c*x)*(d + e*x)^(5/2))/(a + b*x + c*x^2)^(3/2), x)

________________________________________________________________________________________

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)**(5/2)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________