3.10.13 \(\int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx\) [913]

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

[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 = 0.82, 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 = {948, 175, 552, 551} \begin {gather*} -\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)};\text {ArcSin}\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)} \end {gather*}

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 175

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 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], 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 552

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

Rule 948

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 ) \text {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 ) \text {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 ) \text {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] Result contains complex when optimal does not.
time = 21.16, size = 499, normalized size = 1.78 \begin {gather*} \frac {i (f+g x) \sqrt {2-\frac {4 \left (c f^2+g (-b f+a g)\right )}{\left (2 c f-b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}} \sqrt {1+\frac {2 \left (c f^2+g (-b f+a g)\right )}{\left (-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}\right ) (f+g x)}} \left (F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b f g+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 (-b f+a g)\right )};i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c f^2-b f g+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 )}{(-e f+d g) \sqrt {\frac {c f^2+g (-b f+a g)}{-2 c f+b g+\sqrt {\left (b^2-4 a c\right ) g^2}}} \sqrt {a+x (b+c x)}} \end {gather*}

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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Maple [A]
time = 0.14, size = 330, normalized size = 1.18

method	result	size
default	\(\frac {\left (-g \sqrt {-4 a c +b^{2}}-b g +2 c f \right ) \EllipticPi \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{g \sqrt {-4 a c +b^{2}}+b g -2 c f}}, \frac {\left (g \sqrt {-4 a c +b^{2}}+b g -2 c f \right ) e}{2 c \left (d g -e f \right )}, \sqrt {-\frac {g \sqrt {-4 a c +b^{2}}+b g -2 c f}{2 c f -b g +g \sqrt {-4 a c +b^{2}}}}\right ) \sqrt {\frac {\left (b +2 c x +\sqrt {-4 a c +b^{2}}\right ) g}{g \sqrt {-4 a c +b^{2}}+b g -2 c f}}\, \sqrt {\frac {\left (-b -2 c x +\sqrt {-4 a c +b^{2}}\right ) g}{2 c f -b g +g \sqrt {-4 a c +b^{2}}}}\, \sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{g \sqrt {-4 a c +b^{2}}+b g -2 c f}}\, \sqrt {c \,x^{2}+b x +a}\, \sqrt {g x +f}}{c \left (d g -e f \right ) \left (c g \,x^{3}+b g \,x^{2}+c f \,x^{2}+a g x +b f x +f a \right )}\)	\(330\)
elliptic	\(\frac {2 \sqrt {\left (g x +f \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {f}{g}\right ) \sqrt {\frac {x +\frac {f}{g}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {f}{g}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {f}{g}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {f}{g}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \EllipticPi \left (\sqrt {\frac {x +\frac {f}{g}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {f}{g}}}, \frac {-\frac {f}{g}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {f}{g}+\frac {d}{e}}, \sqrt {\frac {-\frac {f}{g}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {f}{g}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\right )}{\sqrt {g x +f}\, \sqrt {c \,x^{2}+b x +a}\, e \sqrt {c g \,x^{3}+b g \,x^{2}+c f \,x^{2}+a g x +b f x +f a}\, \left (-\frac {f}{g}+\frac {d}{e}\right )}\)	\(377\)

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,method=_RETURNVERBOSE)

[Out]

(-g*(-4*a*c+b^2)^(1/2)-b*g+2*c*f)*EllipticPi(2^(1/2)*(-(g*x+f)*c/(g*(-4*a*c+b^2)^(1/2)+b*g-2*c*f))^(1/2),1/2*(
g*(-4*a*c+b^2)^(1/2)+b*g-2*c*f)*e/c/(d*g-e*f),(-(g*(-4*a*c+b^2)^(1/2)+b*g-2*c*f)/(2*c*f-b*g+g*(-4*a*c+b^2)^(1/
2)))^(1/2))*((b+2*c*x+(-4*a*c+b^2)^(1/2))*g/(g*(-4*a*c+b^2)^(1/2)+b*g-2*c*f))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1
/2))*g/(2*c*f-b*g+g*(-4*a*c+b^2)^(1/2)))^(1/2)*2^(1/2)*(-(g*x+f)*c/(g*(-4*a*c+b^2)^(1/2)+b*g-2*c*f))^(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)*sqrt(g*x + f)*(x*e + d)), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*sqrt(g*x + f)*(x*e + d)), x)

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

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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