3.2478 \(\int \frac {(d+e x)^{3/2}}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=542 \[ \frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {d+e x} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (4 a c e-5 b^2 e+8 c x (2 c d-b e)+8 b c d\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {8 \sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 542, 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 = {738, 822, 843, 718, 424, 419} \[ \frac {2 \sqrt {2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 c \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {d+e x} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (4 a c e-5 b^2 e+8 c x (2 c d-b e)+8 b c d\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {8 \sqrt {2} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) + (2*Sqrt[d + e*x
]*(8*b*c*d - 5*b^2*e + 4*a*c*e + 8*c*(2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2]) - (8*Sqrt[2]*
(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*(b^2 - 4*a*c)^(3/2)*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 + 3*b^2*e^2 - 4*c*e*(4*b*d - a*e))*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqr
t[-((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*(b^2 - 4*a*c)^(3/2)*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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], 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] && LtQ[p, -1] && GtQ[m, 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 {(d+e x)^{3/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 \sqrt {d+e x} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (8 c d^2-5 b d e+2 a e^2\right )+\frac {3}{2} 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 )}\\ &=-\frac {2 \sqrt {d+e x} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d-5 b^2 e+4 a c e+8 c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {4 \int \frac {-\frac {1}{4} e \left (8 b c d-3 b^2 e-4 a c e\right ) \left (c d^2-b d e+a e^2\right )-2 c e (2 c d-b e) \left (c d^2-b d e+a e^2\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 )}\\ &=-\frac {2 \sqrt {d+e x} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d-5 b^2 e+4 a c e+8 c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {(8 c (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 )^2}+\frac {\left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\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}\\ &=-\frac {2 \sqrt {d+e x} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d-5 b^2 e+4 a c e+8 c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {\left (8 \sqrt {2} (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 \left (b^2-4 a c\right )^{3/2} \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+3 b^2 e^2-4 c e (4 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 \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 \sqrt {d+e x} (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 \sqrt {d+e x} \left (8 b c d-5 b^2 e+4 a c e+8 c (2 c d-b e) x\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {8 \sqrt {2} (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 \left (b^2-4 a c\right )^{3/2} \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+3 b^2 e^2-4 c e (4 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 \left (b^2-4 a c\right )^{3/2} \sqrt {d+e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 7.30, size = 1141, normalized size = 2.11 \[ \frac {\sqrt {d+e x} \left (c x^2+b x+a\right )^3 \left (\frac {2 (-b d-2 c x d+2 a e+b e x)}{3 \left (b^2-4 a c\right ) \left (c x^2+b x+a\right )^2}-\frac {2 \left (5 e b^2-8 c d b+8 c e x b-4 a c e-16 c^2 d x\right )}{3 \left (b^2-4 a c\right )^2 \left (c x^2+b x+a\right )}\right )}{(a+x (b+c x))^{5/2}}-\frac {(d+e x)^{3/2} \left (c x^2+b x+a\right )^{5/2} \left (-16 (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}}} \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 )+\frac {4 i \sqrt {2} (b e-2 c d) \left (2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}\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^2 e^2+4 a c e^2+4 b \sqrt {\left (b^2-4 a c\right ) e^2} e-8 c d \sqrt {\left (b^2-4 a c\right ) e^2}\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 )}{3 \left (4 a c-b^2\right )^2 e \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))^{5/2} \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[(d + e*x)^(3/2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[d + e*x]*(a + b*x + c*x^2)^3*((2*(-(b*d) + 2*a*e - 2*c*d*x + b*e*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^
2) - (2*(-8*b*c*d + 5*b^2*e - 4*a*c*e - 16*c^2*d*x + 8*b*c*e*x))/(3*(b^2 - 4*a*c)^2*(a + b*x + c*x^2))))/(a +
x*(b + c*x))^(5/2) - ((d + e*x)^(3/2)*(a + b*x + c*x^2)^(5/2)*(-16*(-2*c*d + b*e)*Sqrt[(c*d^2 + e*(-(b*d) + a*
e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2])]*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d)/(d + e*x) + (a*e)/(d + e
*x)))/(d + e*x)) + ((4*I)*Sqrt[2]*(-2*c*d + b*e)*(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*(-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 + Sqrt[(b^2 - 4*a*c)*e^2]))])/Sqrt[d + e*x] - (I*Sqrt[2]*(-(b^2*e^2) + 4*a*c*e^2 - 8*c*d*Sqrt[(b^2
- 4*a*c)*e^2] + 4*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*(-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])]*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]))/(3*(-b^2 + 4*a*c)^2*e*Sqrt[(c*d^2 + e*(-(b*d) + a*e))/(-2*c*d + b*e + Sqrt[(b^2 - 4*a*c)*e^2]
)]*(a + x*(b + c*x))^(5/2)*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 = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (e x + d\right )}^{\frac {3}{2}}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, {\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x + {\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \, {\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*(e*x + d)^(3/2)/(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 \[ \int \frac {{\left (e x + d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.22, size = 8889, normalized size = 16.40 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)/(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 \[ \int \frac {{\left (e x + d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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