3.1631 \(\int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=548 \[ \frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e^3 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 \sqrt {a+b x+c x^2} \left (e x \left (-2 c e (5 b d-3 a e)+b^2 e^2+10 c^2 d^2\right )-c d e (7 b d-4 a e)+a b e^3+8 c^2 d^3\right )}{3 e^2 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac {16 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e^3 \sqrt {d+e x} \sqrt {a+b x+c x^2}} \]

[Out]

-2/3*(8*c^2*d^3+a*b*e^3-c*d*e*(-4*a*e+7*b*d)+e*(10*c^2*d^2+b^2*e^2-2*c*e*(-3*a*e+5*b*d))*x)*(c*x^2+b*x+a)^(1/2
)/e^2/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)+1/3*(16*c^2*d^2+b^2*e^2-4*c*e*(-3*a*e+4*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)*(-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^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)-16/3*(-b*e+2*c*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)/e^3/(e*x+d)^(1/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 548, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {810, 843, 718, 424, 419} \[ -\frac {2 \sqrt {a+b x+c x^2} \left (e x \left (-2 c e (5 b d-3 a e)+b^2 e^2+10 c^2 d^2\right )-c d e (7 b d-4 a e)+a b e^3+8 c^2 d^3\right )}{3 e^2 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e^3 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {16 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e^3 \sqrt {d+e x} \sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(8*c^2*d^3 + a*b*e^3 - c*d*e*(7*b*d - 4*a*e) + e*(10*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 3*a*e))*x)*Sqrt[a
+ b*x + c*x^2])/(3*e^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*(16*c^2*d^2 + b^2
*e^2 - 4*c*e*(4*b*d - 3*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^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]) - (16*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(2*c*d - b*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*e^3*Sqrt[d + e*x]*S
qrt[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 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 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 {a+b x+c x^2}}{(d+e x)^{5/2}} \, dx &=-\frac {2 \left (8 c^2 d^3+a b e^3-c d e (7 b d-4 a e)+e \left (10 c^2 d^2+b^2 e^2-2 c e (5 b d-3 a 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 ) (d+e x)^{3/2}}-\frac {2 \int \frac {\frac {1}{2} c \left (7 b^2 d e+4 a c d e-8 b \left (c d^2+a e^2\right )\right )-\frac {1}{2} c \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x}{\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 {2 \left (8 c^2 d^3+a b e^3-c d e (7 b d-4 a e)+e \left (10 c^2 d^2+b^2 e^2-2 c e (5 b d-3 a 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 ) (d+e x)^{3/2}}-\frac {(8 c (2 c d-b e)) \int \frac {1}{\sqrt {d+e x} \sqrt {a+b x+c x^2}} \, dx}{3 e^3}+\frac {\left (c \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x+c x^2}} \, dx}{3 e^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {2 \left (8 c^2 d^3+a b e^3-c d e (7 b d-4 a e)+e \left (10 c^2 d^2+b^2 e^2-2 c e (5 b d-3 a 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 ) (d+e x)^{3/2}}+\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 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 e^3 \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 (16 \sqrt {2} \sqrt {b^2-4 a c} (2 c d-b 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 e^3 \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 \left (8 c^2 d^3+a b e^3-c d e (7 b d-4 a e)+e \left (10 c^2 d^2+b^2 e^2-2 c e (5 b d-3 a 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 ) (d+e x)^{3/2}}+\frac {\sqrt {2} \sqrt {b^2-4 a c} \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 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 e^3 \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 {16 \sqrt {2} \sqrt {b^2-4 a c} (2 c d-b 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 e^3 \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 6.33, size = 1343, normalized size = 2.45 \[ \sqrt {d+e x} \sqrt {a+x (b+c x)} \left (-\frac {2 (b e-2 c d)}{3 e^2 (d+e x)^2}-\frac {2 \left (10 c^2 d^2-10 b c e d+b^2 e^2+6 a c e^2\right )}{3 e^2 \left (c d^2-b e d+a e^2\right ) (d+e x)}\right )-\frac {(d+e x)^{3/2} \sqrt {a+x (b+c x)} \left (-4 \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}}} \left (16 d^2 \left (\frac {d}{d+e x}-1\right )^2 c^3+4 e \left (a e \left (\frac {7 d^2}{(d+e x)^2}-\frac {6 d}{d+e x}+3\right )-4 b d \left (\frac {2 d^2}{(d+e x)^2}-\frac {3 d}{d+e x}+1\right )\right ) c^2+e^2 \left (\left (\frac {17 d^2}{(d+e x)^2}-\frac {18 d}{d+e x}+1\right ) b^2-\frac {4 a e \left (\frac {7 d}{d+e x}-3\right ) b}{d+e x}+\frac {12 a^2 e^2}{(d+e x)^2}\right ) c+\frac {b^2 e^3 \left (-\frac {d b}{d+e x}+b+\frac {a e}{d+e x}\right )}{d+e x}\right )+\frac {i \sqrt {2} \left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) \left (16 c^2 d^2+b^2 e^2+4 c e (3 a e-4 b d)\right ) \sqrt {\frac {-\frac {2 a e^2}{d+e x}+b \left (\frac {2 d}{d+e x}-1\right ) e-2 c d \left (\frac {d}{d+e x}-1\right )+\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}}} \sqrt {\frac {\frac {2 a e^2}{d+e x}+2 c d \left (\frac {d}{d+e x}-1\right )+b \left (e-\frac {2 d e}{d+e x}\right )+\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}}} 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^3 e^3+b^2 \left (2 c d+\sqrt {\left (b^2-4 a c\right ) e^2}\right ) e^2+4 b \left (a c e^3-4 c d e \sqrt {\left (b^2-4 a c\right ) e^2}\right )+4 c \left (4 c \sqrt {\left (b^2-4 a c\right ) e^2} d^2+a e^2 \left (3 \sqrt {\left (b^2-4 a c\right ) e^2}-2 c d\right )\right )\right ) \sqrt {\frac {-\frac {2 a e^2}{d+e x}+b \left (\frac {2 d}{d+e x}-1\right ) e-2 c d \left (\frac {d}{d+e x}-1\right )+\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}}} \sqrt {\frac {\frac {2 a e^2}{d+e x}+2 c d \left (\frac {d}{d+e x}-1\right )+b \left (e-\frac {2 d e}{d+e x}\right )+\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}}} 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 )}{6 e^4 \left (c d^2-b e d+a e^2\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}}} \sqrt {c x^2+b x+a} \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}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[d + e*x]*Sqrt[a + x*(b + c*x)]*((-2*(-2*c*d + b*e))/(3*e^2*(d + e*x)^2) - (2*(10*c^2*d^2 - 10*b*c*d*e + b
^2*e^2 + 6*a*c*e^2))/(3*e^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x))) - ((d + e*x)^(3/2)*Sqrt[a + x*(b + c*x)]*(-4*S
qrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*(16*c^3*d^2*(-1 + d/(d + e*x))^2 + (b
^2*e^3*(b - (b*d)/(d + e*x) + (a*e)/(d + e*x)))/(d + e*x) + 4*c^2*e*(a*e*(3 + (7*d^2)/(d + e*x)^2 - (6*d)/(d +
e*x)) - 4*b*d*(1 + (2*d^2)/(d + e*x)^2 - (3*d)/(d + e*x))) + c*e^2*((12*a^2*e^2)/(d + e*x)^2 + b^2*(1 + (17*d
^2)/(d + e*x)^2 - (18*d)/(d + e*x)) - (4*a*b*e*(-3 + (7*d)/(d + e*x)))/(d + e*x))) + (I*Sqrt[2]*(2*c*d - b*e +
Sqrt[(b^2 - 4*a*c)*e^2])*(16*c^2*d^2 + b^2*e^2 + 4*c*e*(-4*b*d + 3*a*e))*Sqrt[(Sqrt[(b^2 - 4*a*c)*e^2] - (2*a
*e^2)/(d + e*x) - 2*c*d*(-1 + d/(d + e*x)) + b*e*(-1 + (2*d)/(d + e*x)))/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2
])]*Sqrt[(Sqrt[(b^2 - 4*a*c)*e^2] + (2*a*e^2)/(d + e*x) + 2*c*d*(-1 + d/(d + e*x)) + b*(e - (2*d*e)/(d + e*x))
)/(-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 + S
qrt[(b^2 - 4*a*c)*e^2]))])/Sqrt[d + e*x] - (I*Sqrt[2]*(-(b^3*e^3) + b^2*e^2*(2*c*d + Sqrt[(b^2 - 4*a*c)*e^2])
+ 4*b*(a*c*e^3 - 4*c*d*e*Sqrt[(b^2 - 4*a*c)*e^2]) + 4*c*(4*c*d^2*Sqrt[(b^2 - 4*a*c)*e^2] + a*e^2*(-2*c*d + 3*S
qrt[(b^2 - 4*a*c)*e^2])))*Sqrt[(Sqrt[(b^2 - 4*a*c)*e^2] - (2*a*e^2)/(d + e*x) - 2*c*d*(-1 + d/(d + e*x)) + b*e
*(-1 + (2*d)/(d + e*x)))/(2*c*d - b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*Sqrt[(Sqrt[(b^2 - 4*a*c)*e^2] + (2*a*e^2)/(d
+ e*x) + 2*c*d*(-1 + d/(d + e*x)) + b*(e - (2*d*e)/(d + e*x)))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*Elli
pticF[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]))/(6*e^
4*(c*d^2 - b*d*e + a*e^2)*Sqrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*Sqrt[a + b
*x + c*x^2]*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 = 1.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {e x + d}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(d*exp(1)+x*
exp(1)^2)]integrate()  Bad Argument Typeintegrate()  Bad Argument TypeEvaluation time: 3.64Unable to transpose
Error: Bad Argument Value

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maple [B]  time = 0.23, size = 8612, normalized size = 15.72 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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