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

Optimal. Leaf size=1125 \[ -\frac {\sqrt {2} \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \Pi \left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right )|\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right ) e^2}{\sqrt {c} (e f-d g)^3 \sqrt {c x^2+b x+a}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} g \sqrt {f+g x} \sqrt {-\frac {c \left (c x^2+b x+a\right )}{b^2-4 a c}} 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right ) e}{(e f-d g)^2 \left (c f^2-b g f+a g^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {c x^2+b x+a}}+\frac {2 g^2 \sqrt {c x^2+b x+a} e}{(e f-d g)^2 \left (c f^2-b g f+a g^2\right ) \sqrt {f+g x}}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} g (2 c f-b g) \sqrt {f+g x} \sqrt {-\frac {c \left (c x^2+b x+a\right )}{b^2-4 a c}} 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{3 (e f-d g) \left (c f^2-b g f+a g^2\right )^2 \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {c x^2+b x+a}}+\frac {2 \sqrt {2} \sqrt {b^2-4 a c} g \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {-\frac {c \left (c x^2+b x+a\right )}{b^2-4 a c}} 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{3 (e f-d g) \left (c f^2-b g f+a g^2\right ) \sqrt {f+g x} \sqrt {c x^2+b x+a}}+\frac {4 g^2 (2 c f-b g) \sqrt {c x^2+b x+a}}{3 (e f-d g) \left (c f^2-b g f+a g^2\right )^2 \sqrt {f+g x}}+\frac {2 g^2 \sqrt {c x^2+b x+a}}{3 (e f-d g) \left (c f^2-b g f+a g^2\right ) (f+g x)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.31, antiderivative size = 1125, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 12, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {957, 744, 834, 843, 718, 424, 419, 21, 934, 169, 538, 537} \[ -\frac {\sqrt {2} \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \Pi \left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right )|\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right ) e^2}{\sqrt {c} (e f-d g)^3 \sqrt {c x^2+b x+a}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} g \sqrt {f+g x} \sqrt {-\frac {c \left (c x^2+b x+a\right )}{b^2-4 a c}} 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right ) e}{(e f-d g)^2 \left (c f^2-b g f+a g^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {c x^2+b x+a}}+\frac {2 g^2 \sqrt {c x^2+b x+a} e}{(e f-d g)^2 \left (c f^2-b g f+a g^2\right ) \sqrt {f+g x}}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} g (2 c f-b g) \sqrt {f+g x} \sqrt {-\frac {c \left (c x^2+b x+a\right )}{b^2-4 a c}} 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{3 (e f-d g) \left (c f^2-b g f+a g^2\right )^2 \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {c x^2+b x+a}}+\frac {2 \sqrt {2} \sqrt {b^2-4 a c} g \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {-\frac {c \left (c x^2+b x+a\right )}{b^2-4 a c}} 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{3 (e f-d g) \left (c f^2-b g f+a g^2\right ) \sqrt {f+g x} \sqrt {c x^2+b x+a}}+\frac {4 g^2 (2 c f-b g) \sqrt {c x^2+b x+a}}{3 (e f-d g) \left (c f^2-b g f+a g^2\right )^2 \sqrt {f+g x}}+\frac {2 g^2 \sqrt {c x^2+b x+a}}{3 (e f-d g) \left (c f^2-b g f+a g^2\right ) (f+g x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*g^2*Sqrt[a + b*x + c*x^2])/(3*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*(f + g*x)^(3/2)) + (4*g^2*(2*c*f - b*g)*S
qrt[a + b*x + c*x^2])/(3*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)^2*Sqrt[f + g*x]) + (2*e*g^2*Sqrt[a + b*x + c*x^2]
)/((e*f - d*g)^2*(c*f^2 - b*f*g + a*g^2)*Sqrt[f + g*x]) - (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*g*(2*c*f - b*g)*Sqrt[f
+ g*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]*g)/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)])/(3*(e*f - d*g)*(c*f^2
- b*f*g + a*g^2)^2*Sqrt[(c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[a + b*x + c*x^2]) - (Sqrt[2]*S
qrt[b^2 - 4*a*c]*e*g*Sqrt[f + g*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[(b + Sqr
t[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*g)/(2*c*f - (b + Sqrt[b^2 - 4*a*c])
*g)])/((e*f - d*g)^2*(c*f^2 - b*f*g + a*g^2)*Sqrt[(c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[a +
b*x + c*x^2]) + (2*Sqrt[2]*Sqrt[b^2 - 4*a*c]*g*Sqrt[(c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*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]*g)/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)])/(3*(e*f - d*g)*(c*f^2 - b*f*g + a*g^
2)*Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2]) - (Sqrt[2]*e^2*Sqrt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g]*Sqrt[1 - (2*c*(
f + g*x))/(2*c*f - (b - Sqrt[b^2 - 4*a*c])*g)]*Sqrt[1 - (2*c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*E
llipticPi[(e*(2*c*f - b*g + Sqrt[b^2 - 4*a*c]*g))/(2*c*(e*f - d*g)), ArcSin[(Sqrt[2]*Sqrt[c]*Sqrt[f + g*x])/Sq
rt[2*c*f - (b - Sqrt[b^2 - 4*a*c])*g]], (b - Sqrt[b^2 - 4*a*c] - (2*c*f)/g)/(b + Sqrt[b^2 - 4*a*c] - (2*c*f)/g
)])/(Sqrt[c]*(e*f - d*g)^3*Sqrt[a + b*x + c*x^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 169

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] &&  !SimplerQ[e
+ f*x, c + d*x] &&  !SimplerQ[g + h*x, c + d*x]

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 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 538

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d*x^2)/c]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[c, 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]

Rule 934

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x])/Sqrt[a + b*x + c*x^2], Int[1/((d +
e*x)*Sqrt[f + g*x]*Sqrt[b - q + 2*c*x]*Sqrt[b + q + 2*c*x]), x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 957

Int[((f_.) + (g_.)*(x_))^(n_)/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Int
[ExpandIntegrand[1/(Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2]), (f + g*x)^(n + 1/2)/(d + e*x), x], x] /; FreeQ[{a, b
, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[
n + 1/2]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x) (f+g x)^{5/2} \sqrt {a+b x+c x^2}} \, dx &=\int \left (-\frac {g}{(e f-d g) (f+g x)^{5/2} \sqrt {a+b x+c x^2}}-\frac {e g}{(e f-d g)^2 (f+g x)^{3/2} \sqrt {a+b x+c x^2}}+\frac {e^2}{(e f-d g)^2 (d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}}\right ) \, dx\\ &=\frac {e^2 \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^2}-\frac {(e g) \int \frac {1}{(f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^2}-\frac {g \int \frac {1}{(f+g x)^{5/2} \sqrt {a+b x+c x^2}} \, dx}{e f-d g}\\ &=\frac {2 g^2 \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^{3/2}}+\frac {2 e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}+\frac {(2 e g) \int \frac {-\frac {c f}{2}-\frac {c g x}{2}}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right )}+\frac {(2 g) \int \frac {\frac {1}{2} (-3 c f+2 b g)+\frac {c g x}{2}}{(f+g x)^{3/2} \sqrt {a+b x+c x^2}} \, dx}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )}+\frac {\left (e^2 \sqrt {b-\sqrt {b^2-4 a c}+2 c x} \sqrt {b+\sqrt {b^2-4 a c}+2 c x}\right ) \int \frac {1}{\sqrt {b-\sqrt {b^2-4 a c}+2 c x} \sqrt {b+\sqrt {b^2-4 a c}+2 c x} (d+e x) \sqrt {f+g x}} \, dx}{(e f-d g)^2 \sqrt {a+b x+c x^2}}\\ &=\frac {2 g^2 \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^{3/2}}+\frac {4 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 \sqrt {f+g x}}+\frac {2 e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}-\frac {(4 g) \int \frac {\frac {1}{4} c \left (3 c f^2-g (b f+a g)\right )+\frac {1}{2} c g (2 c f-b g) x}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )^2}-\frac {(c e g) \int \frac {\sqrt {f+g x}}{\sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right )}-\frac {\left (2 e^2 \sqrt {b-\sqrt {b^2-4 a c}+2 c x} \sqrt {b+\sqrt {b^2-4 a c}+2 c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e f-d g-e x^2\right ) \sqrt {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}+\frac {2 c x^2}{g}} \sqrt {b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}+\frac {2 c x^2}{g}}} \, dx,x,\sqrt {f+g x}\right )}{(e f-d g)^2 \sqrt {a+b x+c x^2}}\\ &=\frac {2 g^2 \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^{3/2}}+\frac {4 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 \sqrt {f+g x}}+\frac {2 e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}-\frac {(2 c g (2 c f-b g)) \int \frac {\sqrt {f+g x}}{\sqrt {a+b x+c x^2}} \, dx}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )^2}+\frac {(c g) \int \frac {1}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )}-\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} e g \sqrt {f+g 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} g x^2}{2 c f-b g-\sqrt {b^2-4 a c} g}}}{\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 )}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-b g-\sqrt {b^2-4 a c} g}} \sqrt {a+b x+c x^2}}-\frac {\left (2 e^2 \sqrt {b+\sqrt {b^2-4 a c}+2 c x} \sqrt {1+\frac {2 c (f+g x)}{\left (b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}\right ) g}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e f-d g-e x^2\right ) \sqrt {b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}+\frac {2 c x^2}{g}} \sqrt {1+\frac {2 c x^2}{\left (b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}\right ) g}}} \, dx,x,\sqrt {f+g x}\right )}{(e f-d g)^2 \sqrt {a+b x+c x^2}}\\ &=\frac {2 g^2 \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^{3/2}}+\frac {4 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 \sqrt {f+g x}}+\frac {2 e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} e g \sqrt {f+g 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {a+b x+c x^2}}-\frac {\left (2 \sqrt {2} \sqrt {b^2-4 a c} g (2 c f-b g) \sqrt {f+g 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} g x^2}{2 c f-b g-\sqrt {b^2-4 a c} g}}}{\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 f-d g) \left (c f^2-b f g+a g^2\right )^2 \sqrt {\frac {c (f+g x)}{2 c f-b g-\sqrt {b^2-4 a c} g}} \sqrt {a+b x+c x^2}}+\frac {\left (2 \sqrt {2} \sqrt {b^2-4 a c} g \sqrt {\frac {c (f+g x)}{2 c f-b g-\sqrt {b^2-4 a c} g}} \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} g x^2}{2 c f-b g-\sqrt {b^2-4 a c} g}}} \, 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 f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\left (2 e^2 \sqrt {1+\frac {2 c (f+g x)}{\left (b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}\right ) g}} \sqrt {1+\frac {2 c (f+g x)}{\left (b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}\right ) g}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e f-d g-e x^2\right ) \sqrt {1+\frac {2 c x^2}{\left (b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}\right ) g}} \sqrt {1+\frac {2 c x^2}{\left (b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}\right ) g}}} \, dx,x,\sqrt {f+g x}\right )}{(e f-d g)^2 \sqrt {a+b x+c x^2}}\\ &=\frac {2 g^2 \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^{3/2}}+\frac {4 g^2 (2 c f-b g) \sqrt {a+b x+c x^2}}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 \sqrt {f+g x}}+\frac {2 e g^2 \sqrt {a+b x+c x^2}}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}-\frac {2 \sqrt {2} \sqrt {b^2-4 a c} g (2 c f-b g) \sqrt {f+g 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{3 (e f-d g) \left (c f^2-b f g+a g^2\right )^2 \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} \sqrt {b^2-4 a c} e g \sqrt {f+g 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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {a+b x+c x^2}}+\frac {2 \sqrt {2} \sqrt {b^2-4 a c} g \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \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} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{3 (e f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} e^2 \sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \Pi \left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right )|\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right )}{\sqrt {c} (e f-d g)^3 \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 15.52, size = 14759, normalized size = 13.12 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(g*x+f)^(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)*(g*x + f)^(5/2)), x)

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maple [B]  time = 0.21, size = 27597, normalized size = 24.53 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(g*x+f)^(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)*(g*x + f)^(5/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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