3.2467 \(\int \frac {1}{(d+e x)^{5/2} \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=523 \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2 \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {4 e \sqrt {a+b x+c x^2} (2 c d-b e)}{3 \sqrt {d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac {2 e \sqrt {a+b x+c x^2}}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )} \]

[Out]

-2/3*e*(c*x^2+b*x+a)^(1/2)/(a*e^2-b*d*e+c*d^2)/(e*x+d)^(3/2)-4/3*e*(-b*e+2*c*d)*(c*x^2+b*x+a)^(1/2)/(a*e^2-b*d
*e+c*d^2)^2/(e*x+d)^(1/2)+2/3*(-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)*(-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)/(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*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)/(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 = 0.41, antiderivative size = 523, 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 = {744, 834, 843, 718, 424, 419} \[ -\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {d+e x} \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac {2 \sqrt {2} \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )^2 \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {4 e \sqrt {a+b x+c x^2} (2 c d-b e)}{3 \sqrt {d+e x} \left (a e^2-b d e+c d^2\right )^2}-\frac {2 e \sqrt {a+b x+c x^2}}{3 (d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*Sqrt[a + b*x + c*x^2])/(3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(3/2)) - (4*e*(2*c*d - b*e)*Sqrt[a + b*x + c
*x^2])/(3*(c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x]) + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*(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*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]*Sqrt[b^2 - 4*a*c]*
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*(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 744

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))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 834

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

________________________________________________________________________________________

Mathematica [C]  time = 6.05, size = 643, normalized size = 1.23 \[ \frac {2 \left (\frac {i (d+e x)^{5/2} \sqrt {1-\frac {2 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt {e^2 \left (b^2-4 a c\right )}-b e+2 c d\right )}} \sqrt {\frac {2 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt {e^2 \left (b^2-4 a c\right )}+b e-2 c d\right )}+1} \left (\left (c \left (2 d \sqrt {e^2 \left (b^2-4 a c\right )}-a e^2-3 b d e\right )+b e \left (b e-\sqrt {e^2 \left (b^2-4 a c\right )}\right )+3 c^2 d^2\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )+(b e-2 c d) \left (\sqrt {e^2 \left (b^2-4 a c\right )}-b e+2 c d\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c d^2-b e d+a e^2}{-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}}}{\sqrt {d+e x}}\right )|-\frac {-2 c d+b e+\sqrt {\left (b^2-4 a c\right ) e^2}}{2 c d-b e+\sqrt {\left (b^2-4 a c\right ) e^2}}\right )\right )}{\sqrt {2} \sqrt {\frac {e (a e-b d)+c d^2}{\sqrt {e^2 \left (b^2-4 a c\right )}+b e-2 c d}}}+2 e^2 (d+e x) (a+x (b+c x)) (2 c d-b e)+e^2 (a+x (b+c x)) (e (-a e+3 b d+2 b e x)-c d (5 d+4 e x))\right )}{3 e (d+e x)^{3/2} \sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{c e^{3} x^{5} + {\left (3 \, c d e^{2} + b e^{3}\right )} x^{4} + a d^{3} + {\left (3 \, c d^{2} e + 3 \, b d e^{2} + a e^{3}\right )} x^{3} + {\left (c d^{3} + 3 \, b d^{2} e + 3 \, a d e^{2}\right )} x^{2} + {\left (b d^{3} + 3 \, a d^{2} e\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.22, size = 5822, normalized size = 11.13 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________