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

Optimal. Leaf size=467 \[ \frac {2 \sqrt {2} g \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{c e \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} \Pi \left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right )|\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right )}{\sqrt {c} e \sqrt {a+b x+c x^2}} \]

[Out]

2*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)/c/e/(g*x+f)^(1/2)/(c*x^2+b*x+a)^(1/2)-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
)/e/c^(1/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 1.60, antiderivative size = 467, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {943, 718, 419, 934, 169, 538, 537} \[ \frac {2 \sqrt {2} g \sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{c e \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\sqrt {2} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} \Pi \left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right )|\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right )}{\sqrt {c} e \sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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*S
qrt[b^2 - 4*a*c]*g)/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)])/(c*e*Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2]) - (Sqrt[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)]*EllipticPi[(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])/Sqrt[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*Sqrt[a + b*x + c*x^2])

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 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 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 943

Int[Sqrt[(f_.) + (g_.)*(x_)]/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist
[g/e, Int[1/(Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2]), x], x] + Dist[(e*f - d*g)/e, Int[1/((d + e*x)*Sqrt[f + g*x]
*Sqrt[a + b*x + c*x^2]), 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {f+g x}}{(d+e x) \sqrt {a+b x+c x^2}} \, dx &=\frac {g \int \frac {1}{\sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{e}+\frac {(e f-d g) \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx}{e}\\ &=\frac {\left ((e f-d g) \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 \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 )}{c e \sqrt {f+g x} \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 )}{c e \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\left (2 (e f-d g) \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 \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 )}{c e \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\left (2 (e f-d g) \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 \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 )}{c e \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\frac {\left (2 (e f-d g) \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 \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 )}{c e \sqrt {f+g x} \sqrt {a+b x+c x^2}}-\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 )}{\sqrt {c} e \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 1.62, size = 379, normalized size = 0.81 \[ -\frac {i \sqrt {2} \sqrt {\frac {g \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}} \sqrt {1-\frac {2 c (f+g x)}{g \left (\sqrt {b^2-4 a c}-b\right )+2 c f}} \left (F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) g-2 c f}} \sqrt {f+g x}\right )|\frac {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}{2 c f+\left (\sqrt {b^2-4 a c}-b\right ) g}\right )-\Pi \left (\frac {e \left (2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g\right )}{2 c (e f-d g)};i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) g-2 c f}} \sqrt {f+g x}\right )|\frac {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}{2 c f+\left (\sqrt {b^2-4 a c}-b\right ) g}\right )\right )}{e \sqrt {a+x (b+c x)} \sqrt {\frac {c}{g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[2]*Sqrt[(g*(b + Sqrt[b^2 - 4*a*c] + 2*c*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)]*(EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(-2*c*f + (b + Sqrt[b^2 - 4*
a*c])*g)]*Sqrt[f + g*x]], (2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)/(2*c*f + (-b + Sqrt[b^2 - 4*a*c])*g)] - Elliptic
Pi[(e*(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g))/(2*c*(e*f - d*g)), I*ArcSinh[Sqrt[2]*Sqrt[c/(-2*c*f + (b + Sqrt[b^2
- 4*a*c])*g)]*Sqrt[f + g*x]], (2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)/(2*c*f + (-b + Sqrt[b^2 - 4*a*c])*g)]))/(e*
Sqrt[c/(-2*c*f + (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[a + x*(b + c*x)])

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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((g*x+f)^(1/2)/(e*x+d)/(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 \[ \int \frac {\sqrt {g x + f}}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 834, normalized size = 1.79 \[ \frac {\left (-b g \EllipticF \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )+b g \EllipticPi \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \frac {\left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) e}{2 \left (d g -e f \right ) c}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )+2 c f \EllipticF \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )-2 c f \EllipticPi \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \frac {\left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) e}{2 \left (d g -e f \right ) c}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )-\sqrt {-4 a c +b^{2}}\, g \EllipticF \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )+\sqrt {-4 a c +b^{2}}\, g \EllipticPi \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \frac {\left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) e}{2 \left (d g -e f \right ) c}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )\right ) \sqrt {g x +f}\, \sqrt {c \,x^{2}+b x +a}\, \sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \sqrt {\frac {\left (-2 c x -b +\sqrt {-4 a c +b^{2}}\right ) g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \sqrt {\frac {\left (2 c x +b +\sqrt {-4 a c +b^{2}}\right ) g}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}}{\left (c g \,x^{3}+b g \,x^{2}+c f \,x^{2}+a g x +b f x +a f \right ) c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f + g*x)^(1/2)/((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 \[ \int \frac {\sqrt {f + g x}}{\left (d + e x\right ) \sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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