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

Optimal. Leaf size=280 \[ -\frac {\sqrt {2} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} \Pi \left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right )|\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right )}{\sqrt {c} \sqrt {a+b x+c x^2} (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 1.25, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {934, 169, 538, 537} \[ -\frac {\sqrt {2} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}} \Pi \left (\frac {e \left (2 c f-b g+\sqrt {b^2-4 a c} g\right )}{2 c (e f-d g)};\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {f+g x}}{\sqrt {2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g}}\right )|\frac {b-\sqrt {b^2-4 a c}-\frac {2 c f}{g}}{b+\sqrt {b^2-4 a c}-\frac {2 c f}{g}}\right )}{\sqrt {c} \sqrt {a+b x+c x^2} (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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*f - d*g)*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 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 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]

Rubi steps

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

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Mathematica [C]  time = 1.75, size = 499, normalized size = 1.78 \[ \frac {i (f+g x) \sqrt {2-\frac {4 \left (g (a g-b f)+c f^2\right )}{(f+g x) \left (\sqrt {g^2 \left (b^2-4 a c\right )}-b g+2 c f\right )}} \sqrt {\frac {2 \left (g (a g-b f)+c f^2\right )}{(f+g x) \left (\sqrt {g^2 \left (b^2-4 a c\right )}+b g-2 c f\right )}+1} \left (F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b g f+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right )|-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )-\Pi \left (\frac {(e f-d g) \left (2 c f-b g-\sqrt {\left (b^2-4 a c\right ) g^2}\right )}{2 e \left (c f^2+g (a g-b f)\right )};i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b g f+a g^2}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}}}{\sqrt {f+g x}}\right )|-\frac {-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}{2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}}\right )\right )}{\sqrt {a+x (b+c x)} (d g-e f) \sqrt {\frac {g (a g-b f)+c f^2}{\sqrt {g^2 \left (b^2-4 a c\right )}+b g-2 c f}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(f + g*x)*Sqrt[2 - (4*(c*f^2 + g*(-(b*f) + a*g)))/((2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(f + g*x))]*Sqrt
[1 + (2*(c*f^2 + g*(-(b*f) + a*g)))/((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])*(f + g*x))]*(EllipticF[I*ArcSinh
[(Sqrt[2]*Sqrt[(c*f^2 - b*f*g + a*g^2)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c*f +
b*g + Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f - b*g + Sqrt[(b^2 - 4*a*c)*g^2]))] - EllipticPi[((e*f - d*g)*(2*c*f - b*
g - Sqrt[(b^2 - 4*a*c)*g^2]))/(2*e*(c*f^2 + g*(-(b*f) + a*g))), I*ArcSinh[(Sqrt[2]*Sqrt[(c*f^2 - b*f*g + a*g^2
)/(-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])])/Sqrt[f + g*x]], -((-2*c*f + b*g + Sqrt[(b^2 - 4*a*c)*g^2])/(2*c*f
- b*g + Sqrt[(b^2 - 4*a*c)*g^2]))]))/((-(e*f) + d*g)*Sqrt[(c*f^2 + g*(-(b*f) + a*g))/(-2*c*f + b*g + Sqrt[(b^
2 - 4*a*c)*g^2])]*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(1/(e*x+d)/(g*x+f)^(1/2)/(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 \[ \int \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} \sqrt {g x + f}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 330, normalized size = 1.18 \[ \frac {\left (-b g +2 c f -\sqrt {-4 a c +b^{2}}\, g \right ) \sqrt {\frac {\left (2 c x +b +\sqrt {-4 a c +b^{2}}\right ) g}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \sqrt {\frac {\left (-2 c x -b +\sqrt {-4 a c +b^{2}}\right ) g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \sqrt {c \,x^{2}+b x +a}\, \sqrt {g x +f}\, \EllipticPi \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \frac {\left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) e}{2 \left (d g -e f \right ) c}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )}{\left (d g -e f \right ) \left (c g \,x^{3}+b g \,x^{2}+c f \,x^{2}+a g x +b f x +a f \right ) c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b*g+2*c*f-(-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)/(d*g-e*f)/c*e,(-(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))*((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)*2^(1/2)*(-(g*x+f)/(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)*c)^(1/2)/c
*(c*x^2+b*x+a)^(1/2)*(g*x+f)^(1/2)/(d*g-e*f)/(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 \[ \int \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} \sqrt {g x + f}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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