3.74 \(\int \frac {\sqrt {a+b x} (A+B x+C x^2)}{\sqrt {c+d x} \sqrt {e+f x}} \, dx\)

Optimal. Leaf size=528 \[ -\frac {2 \sqrt {a d-b c} (b e-a f) \sqrt {\frac {b (c+d x)}{b c-a d}} \sqrt {\frac {b (e+f x)}{b e-a f}} \left (a C d f (d e-c f)-b \left (5 d f (-3 A d f+B c f+2 B d e)-C \left (4 c^2 f^2+3 c d e f+8 d^2 e^2\right )\right )\right ) \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {a d-b c}}\right ),\frac {f (b c-a d)}{d (b e-a f)}\right )}{15 b^2 d^{5/2} f^3 \sqrt {c+d x} \sqrt {e+f x}}-\frac {2 \sqrt {e+f x} \sqrt {a d-b c} \sqrt {\frac {b (c+d x)}{b c-a d}} (3 b d f (a c C f+a C d e-5 A b d f+3 b c C e)+(a d f-2 b (c f+d e)) (2 a C d f-b (5 B d f-4 C (c f+d e)))) E\left (\sin ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {a d-b c}}\right )|\frac {(b c-a d) f}{d (b e-a f)}\right )}{15 b^2 d^{5/2} f^3 \sqrt {c+d x} \sqrt {\frac {b (e+f x)}{b e-a f}}}-\frac {2 \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x} (2 a C d f-b (5 B d f-4 C (c f+d e)))}{15 b d^2 f^2}+\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f} \]

[Out]

2/5*C*(b*x+a)^(3/2)*(d*x+c)^(1/2)*(f*x+e)^(1/2)/b/d/f-2/15*(2*a*C*d*f-b*(5*B*d*f-4*C*(c*f+d*e)))*(b*x+a)^(1/2)
*(d*x+c)^(1/2)*(f*x+e)^(1/2)/b/d^2/f^2-2/15*(3*b*d*f*(-5*A*b*d*f+C*a*c*f+C*a*d*e+3*C*b*c*e)+(a*d*f-2*b*(c*f+d*
e))*(2*a*C*d*f-b*(5*B*d*f-4*C*(c*f+d*e))))*EllipticE(d^(1/2)*(b*x+a)^(1/2)/(a*d-b*c)^(1/2),((-a*d+b*c)*f/d/(-a
*f+b*e))^(1/2))*(a*d-b*c)^(1/2)*(b*(d*x+c)/(-a*d+b*c))^(1/2)*(f*x+e)^(1/2)/b^2/d^(5/2)/f^3/(d*x+c)^(1/2)/(b*(f
*x+e)/(-a*f+b*e))^(1/2)-2/15*(-a*f+b*e)*(a*C*d*f*(-c*f+d*e)-b*(5*d*f*(-3*A*d*f+B*c*f+2*B*d*e)-C*(4*c^2*f^2+3*c
*d*e*f+8*d^2*e^2)))*EllipticF(d^(1/2)*(b*x+a)^(1/2)/(a*d-b*c)^(1/2),((-a*d+b*c)*f/d/(-a*f+b*e))^(1/2))*(a*d-b*
c)^(1/2)*(b*(d*x+c)/(-a*d+b*c))^(1/2)*(b*(f*x+e)/(-a*f+b*e))^(1/2)/b^2/d^(5/2)/f^3/(d*x+c)^(1/2)/(f*x+e)^(1/2)

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Rubi [A]  time = 1.03, antiderivative size = 524, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {1615, 154, 158, 114, 113, 121, 120} \[ -\frac {2 \sqrt {a d-b c} (b e-a f) \sqrt {\frac {b (c+d x)}{b c-a d}} \sqrt {\frac {b (e+f x)}{b e-a f}} \left (a C d f (d e-c f)+5 b d f (3 A d f-B (c f+2 d e))+b C \left (4 c^2 f^2+3 c d e f+8 d^2 e^2\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {a d-b c}}\right )|\frac {(b c-a d) f}{d (b e-a f)}\right )}{15 b^2 d^{5/2} f^3 \sqrt {c+d x} \sqrt {e+f x}}-\frac {2 \sqrt {e+f x} \sqrt {a d-b c} \sqrt {\frac {b (c+d x)}{b c-a d}} (3 b d f (a c C f+a C d e-5 A b d f+3 b c C e)-(a d f-2 b (c f+d e)) (-2 a C d f+5 b B d f-4 b C (c f+d e))) E\left (\sin ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {a d-b c}}\right )|\frac {(b c-a d) f}{d (b e-a f)}\right )}{15 b^2 d^{5/2} f^3 \sqrt {c+d x} \sqrt {\frac {b (e+f x)}{b e-a f}}}+\frac {2 \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x} (-2 a C d f+5 b B d f-4 b C (c f+d e))}{15 b d^2 f^2}+\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(A + B*x + C*x^2))/(Sqrt[c + d*x]*Sqrt[e + f*x]),x]

[Out]

(2*(5*b*B*d*f - 2*a*C*d*f - 4*b*C*(d*e + c*f))*Sqrt[a + b*x]*Sqrt[c + d*x]*Sqrt[e + f*x])/(15*b*d^2*f^2) + (2*
C*(a + b*x)^(3/2)*Sqrt[c + d*x]*Sqrt[e + f*x])/(5*b*d*f) - (2*Sqrt[-(b*c) + a*d]*(3*b*d*f*(3*b*c*C*e + a*C*d*e
+ a*c*C*f - 5*A*b*d*f) - (a*d*f - 2*b*(d*e + c*f))*(5*b*B*d*f - 2*a*C*d*f - 4*b*C*(d*e + c*f)))*Sqrt[(b*(c +
d*x))/(b*c - a*d)]*Sqrt[e + f*x]*EllipticE[ArcSin[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[-(b*c) + a*d]], ((b*c - a*d)*f)
/(d*(b*e - a*f))])/(15*b^2*d^(5/2)*f^3*Sqrt[c + d*x]*Sqrt[(b*(e + f*x))/(b*e - a*f)]) - (2*Sqrt[-(b*c) + a*d]*
(b*e - a*f)*(a*C*d*f*(d*e - c*f) + b*C*(8*d^2*e^2 + 3*c*d*e*f + 4*c^2*f^2) + 5*b*d*f*(3*A*d*f - B*(2*d*e + c*f
)))*Sqrt[(b*(c + d*x))/(b*c - a*d)]*Sqrt[(b*(e + f*x))/(b*e - a*f)]*EllipticF[ArcSin[(Sqrt[d]*Sqrt[a + b*x])/S
qrt[-(b*c) + a*d]], ((b*c - a*d)*f)/(d*(b*e - a*f))])/(15*b^2*d^(5/2)*f^3*Sqrt[c + d*x]*Sqrt[e + f*x])

Rule 113

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

Rule 114

Int[Sqrt[(e_.) + (f_.)*(x_)]/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*
x]*Sqrt[(b*(c + d*x))/(b*c - a*d)])/(Sqrt[c + d*x]*Sqrt[(b*(e + f*x))/(b*e - a*f)]), Int[Sqrt[(b*e)/(b*e - a*f
) + (b*f*x)/(b*e - a*f)]/(Sqrt[a + b*x]*Sqrt[(b*c)/(b*c - a*d) + (b*d*x)/(b*c - a*d)]), x], x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !(GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0]) &&  !LtQ[-((b*c - a*d)/d), 0]

Rule 120

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

Rule 121

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Dist[Sqrt[(b*(c
+ d*x))/(b*c - a*d)]/Sqrt[c + d*x], Int[1/(Sqrt[a + b*x]*Sqrt[(b*c)/(b*c - a*d) + (b*d*x)/(b*c - a*d)]*Sqrt[e
+ f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&  !GtQ[(b*c - a*d)/b, 0] && SimplerQ[a + b*x, c + d*x] && Si
mplerQ[a + b*x, e + f*x]

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 158

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

Rule 1615

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[
{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[(k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*(e + f*x)^
(p + 1))/(d*f*b^(q - 1)*(m + n + p + q + 1)), x] + Dist[1/(d*f*b^q*(m + n + p + q + 1)), Int[(a + b*x)^m*(c +
d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a +
b*x)^(q - 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*
(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; F
reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && IntegersQ[2*m, 2*n, 2*p]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x} \left (A+B x+C x^2\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx &=\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f}+\frac {2 \int \frac {\sqrt {a+b x} \left (-\frac {1}{2} b (3 b c C e+a C d e+a c C f-5 A b d f)+\frac {1}{2} b (5 b B d f-2 a C d f-4 b C (d e+c f)) x\right )}{\sqrt {c+d x} \sqrt {e+f x}} \, dx}{5 b^2 d f}\\ &=\frac {2 (5 b B d f-2 a C d f-4 b C (d e+c f)) \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x}}{15 b d^2 f^2}+\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f}+\frac {4 \int \frac {-\frac {1}{4} b (3 a d f (3 b c C e+a C d e+a c C f-5 A b d f)+(b c e+a d e+a c f) (5 b B d f-2 a C d f-4 b C (d e+c f)))-\frac {1}{4} b (3 b d f (3 b c C e+a C d e+a c C f-5 A b d f)-(a d f-2 b (d e+c f)) (5 b B d f-2 a C d f-4 b C (d e+c f))) x}{\sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x}} \, dx}{15 b^2 d^2 f^2}\\ &=\frac {2 (5 b B d f-2 a C d f-4 b C (d e+c f)) \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x}}{15 b d^2 f^2}+\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f}-\frac {(3 b d f (3 b c C e+a C d e+a c C f-5 A b d f)-(a d f-2 b (d e+c f)) (5 b B d f-2 a C d f-4 b C (d e+c f))) \int \frac {\sqrt {e+f x}}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{15 b d^2 f^3}-\frac {\left ((b e-a f) \left (a C d f (d e-c f)+b C \left (8 d^2 e^2+3 c d e f+4 c^2 f^2\right )+5 b d f (3 A d f-B (2 d e+c f))\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x}} \, dx}{15 b d^2 f^3}\\ &=\frac {2 (5 b B d f-2 a C d f-4 b C (d e+c f)) \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x}}{15 b d^2 f^2}+\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f}-\frac {\left ((b e-a f) \left (a C d f (d e-c f)+b C \left (8 d^2 e^2+3 c d e f+4 c^2 f^2\right )+5 b d f (3 A d f-B (2 d e+c f))\right ) \sqrt {\frac {b (c+d x)}{b c-a d}}\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}} \sqrt {e+f x}} \, dx}{15 b d^2 f^3 \sqrt {c+d x}}-\frac {\left ((3 b d f (3 b c C e+a C d e+a c C f-5 A b d f)-(a d f-2 b (d e+c f)) (5 b B d f-2 a C d f-4 b C (d e+c f))) \sqrt {\frac {b (c+d x)}{b c-a d}} \sqrt {e+f x}\right ) \int \frac {\sqrt {\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}}}{\sqrt {a+b x} \sqrt {\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}}} \, dx}{15 b d^2 f^3 \sqrt {c+d x} \sqrt {\frac {b (e+f x)}{b e-a f}}}\\ &=\frac {2 (5 b B d f-2 a C d f-4 b C (d e+c f)) \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x}}{15 b d^2 f^2}+\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f}-\frac {2 \sqrt {-b c+a d} (3 b d f (3 b c C e+a C d e+a c C f-5 A b d f)-(a d f-2 b (d e+c f)) (5 b B d f-2 a C d f-4 b C (d e+c f))) \sqrt {\frac {b (c+d x)}{b c-a d}} \sqrt {e+f x} E\left (\sin ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {-b c+a d}}\right )|\frac {(b c-a d) f}{d (b e-a f)}\right )}{15 b^2 d^{5/2} f^3 \sqrt {c+d x} \sqrt {\frac {b (e+f x)}{b e-a f}}}-\frac {\left ((b e-a f) \left (a C d f (d e-c f)+b C \left (8 d^2 e^2+3 c d e f+4 c^2 f^2\right )+5 b d f (3 A d f-B (2 d e+c f))\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \sqrt {\frac {b (e+f x)}{b e-a f}}\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}} \sqrt {\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}}} \, dx}{15 b d^2 f^3 \sqrt {c+d x} \sqrt {e+f x}}\\ &=\frac {2 (5 b B d f-2 a C d f-4 b C (d e+c f)) \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x}}{15 b d^2 f^2}+\frac {2 C (a+b x)^{3/2} \sqrt {c+d x} \sqrt {e+f x}}{5 b d f}-\frac {2 \sqrt {-b c+a d} (3 b d f (3 b c C e+a C d e+a c C f-5 A b d f)-(a d f-2 b (d e+c f)) (5 b B d f-2 a C d f-4 b C (d e+c f))) \sqrt {\frac {b (c+d x)}{b c-a d}} \sqrt {e+f x} E\left (\sin ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {-b c+a d}}\right )|\frac {(b c-a d) f}{d (b e-a f)}\right )}{15 b^2 d^{5/2} f^3 \sqrt {c+d x} \sqrt {\frac {b (e+f x)}{b e-a f}}}-\frac {2 \sqrt {-b c+a d} (b e-a f) \left (a C d f (d e-c f)+b C \left (8 d^2 e^2+3 c d e f+4 c^2 f^2\right )+5 b d f (3 A d f-B (2 d e+c f))\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \sqrt {\frac {b (e+f x)}{b e-a f}} F\left (\sin ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {-b c+a d}}\right )|\frac {(b c-a d) f}{d (b e-a f)}\right )}{15 b^2 d^{5/2} f^3 \sqrt {c+d x} \sqrt {e+f x}}\\ \end {align*}

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Mathematica [C]  time = 8.03, size = 615, normalized size = 1.16 \[ -\frac {2 \left (i b f (a+b x)^{3/2} (b c-a d) \sqrt {\frac {b (c+d x)}{d (a+b x)}} \sqrt {\frac {b (e+f x)}{f (a+b x)}} \left (a C d f (c f-d e)+5 b d f (3 A d f-B (2 c f+d e))+b C \left (8 c^2 f^2+3 c d e f+4 d^2 e^2\right )\right ) \operatorname {EllipticF}\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b c}{d}-a}}{\sqrt {a+b x}}\right ),\frac {b d e-a d f}{b c f-a d f}\right )+b^2 (c+d x) (e+f x) \sqrt {\frac {b c}{d}-a} \left (2 a^2 C d^2 f^2+a b d f (3 C (c f+d e)-5 B d f)-\left (b^2 \left (5 d f (3 A d f-2 B (c f+d e))+C \left (8 c^2 f^2+7 c d e f+8 d^2 e^2\right )\right )\right )\right )+i f (a+b x)^{3/2} (b c-a d) \sqrt {\frac {b (c+d x)}{d (a+b x)}} \sqrt {\frac {b (e+f x)}{f (a+b x)}} \left (2 a^2 C d^2 f^2+a b d f (3 C (c f+d e)-5 B d f)-\left (b^2 \left (5 d f (3 A d f-2 B (c f+d e))+C \left (8 c^2 f^2+7 c d e f+8 d^2 e^2\right )\right )\right )\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b c}{d}-a}}{\sqrt {a+b x}}\right )|\frac {b d e-a d f}{b c f-a d f}\right )-b^2 d f (a+b x) (c+d x) (e+f x) \sqrt {\frac {b c}{d}-a} (a C d f+5 b B d f+b C (-4 c f-4 d e+3 d f x))\right )}{15 b^3 d^3 f^3 \sqrt {a+b x} \sqrt {c+d x} \sqrt {e+f x} \sqrt {\frac {b c}{d}-a}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x]*(A + B*x + C*x^2))/(Sqrt[c + d*x]*Sqrt[e + f*x]),x]

[Out]

(-2*(b^2*Sqrt[-a + (b*c)/d]*(2*a^2*C*d^2*f^2 + a*b*d*f*(-5*B*d*f + 3*C*(d*e + c*f)) - b^2*(C*(8*d^2*e^2 + 7*c*
d*e*f + 8*c^2*f^2) + 5*d*f*(3*A*d*f - 2*B*(d*e + c*f))))*(c + d*x)*(e + f*x) - b^2*Sqrt[-a + (b*c)/d]*d*f*(a +
b*x)*(c + d*x)*(e + f*x)*(5*b*B*d*f + a*C*d*f + b*C*(-4*d*e - 4*c*f + 3*d*f*x)) + I*(b*c - a*d)*f*(2*a^2*C*d^
2*f^2 + a*b*d*f*(-5*B*d*f + 3*C*(d*e + c*f)) - b^2*(C*(8*d^2*e^2 + 7*c*d*e*f + 8*c^2*f^2) + 5*d*f*(3*A*d*f - 2
*B*(d*e + c*f))))*(a + b*x)^(3/2)*Sqrt[(b*(c + d*x))/(d*(a + b*x))]*Sqrt[(b*(e + f*x))/(f*(a + b*x))]*Elliptic
E[I*ArcSinh[Sqrt[-a + (b*c)/d]/Sqrt[a + b*x]], (b*d*e - a*d*f)/(b*c*f - a*d*f)] + I*b*(b*c - a*d)*f*(a*C*d*f*(
-(d*e) + c*f) + b*C*(4*d^2*e^2 + 3*c*d*e*f + 8*c^2*f^2) + 5*b*d*f*(3*A*d*f - B*(d*e + 2*c*f)))*(a + b*x)^(3/2)
*Sqrt[(b*(c + d*x))/(d*(a + b*x))]*Sqrt[(b*(e + f*x))/(f*(a + b*x))]*EllipticF[I*ArcSinh[Sqrt[-a + (b*c)/d]/Sq
rt[a + b*x]], (b*d*e - a*d*f)/(b*c*f - a*d*f)]))/(15*b^3*Sqrt[-a + (b*c)/d]*d^3*f^3*Sqrt[a + b*x]*Sqrt[c + d*x
]*Sqrt[e + f*x])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C x^{2} + B x + A\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {f x + e}}{d f x^{2} + c e + {\left (d e + c f\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*x^2 + B*x + A)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(f*x + e)/(d*f*x^2 + c*e + (d*e + c*f)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*x^2 + B*x + A)*sqrt(b*x + a)/(sqrt(d*x + c)*sqrt(f*x + e)), x)

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maple [B]  time = 0.04, size = 6174, normalized size = 11.69 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)*(C*x^2+B*x+A)/(d*x+c)^(1/2)/(f*x+e)^(1/2),x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*x^2 + B*x + A)*sqrt(b*x + a)/(sqrt(d*x + c)*sqrt(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(1/2)*(A + B*x + C*x^2))/((e + f*x)^(1/2)*(c + d*x)^(1/2)),x)

[Out]

int(((a + b*x)^(1/2)*(A + B*x + C*x^2))/((e + f*x)^(1/2)*(c + d*x)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)*(C*x**2+B*x+A)/(d*x+c)**(1/2)/(f*x+e)**(1/2),x)

[Out]

Integral(sqrt(a + b*x)*(A + B*x + C*x**2)/(sqrt(c + d*x)*sqrt(e + f*x)), x)

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