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

Optimal. Leaf size=110 \[ -\frac {2 \sqrt {\frac {\sqrt {-c} (f+g x)}{\sqrt {-c} f+g}} \Pi \left (\frac {2 e}{\sqrt {-c} d+e};\sin ^{-1}\left (\frac {\sqrt {1-\sqrt {-c} x}}{\sqrt {2}}\right )|\frac {2 g}{\sqrt {-c} f+g}\right )}{\left (\sqrt {-c} d+e\right ) \sqrt {f+g x}} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {932, 168, 538, 537} \[ -\frac {2 \sqrt {\frac {\sqrt {-c} (f+g x)}{\sqrt {-c} f+g}} \Pi \left (\frac {2 e}{\sqrt {-c} d+e};\sin ^{-1}\left (\frac {\sqrt {1-\sqrt {-c} x}}{\sqrt {2}}\right )|\frac {2 g}{\sqrt {-c} f+g}\right )}{\left (\sqrt {-c} d+e\right ) \sqrt {f+g x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[(Sqrt[-c]*(f + g*x))/(Sqrt[-c]*f + g)]*EllipticPi[(2*e)/(Sqrt[-c]*d + e), ArcSin[Sqrt[1 - Sqrt[-c]*x]
/Sqrt[2]], (2*g)/(Sqrt[-c]*f + g)])/((Sqrt[-c]*d + e)*Sqrt[f + g*x])

Rule 168

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] && GtQ[(d*e - c
*f)/d, 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 932

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

Rubi steps

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

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Mathematica [C]  time = 0.95, size = 261, normalized size = 2.37 \[ -\frac {2 i (f+g x) \sqrt {\frac {g \left (x+\frac {i}{\sqrt {c}}\right )}{f+g x}} \sqrt {-\frac {-g x+\frac {i g}{\sqrt {c}}}{f+g x}} \left (F\left (i \sinh ^{-1}\left (\frac {\sqrt {-f-\frac {i g}{\sqrt {c}}}}{\sqrt {f+g x}}\right )|\frac {\sqrt {c} f-i g}{\sqrt {c} f+i g}\right )-\Pi \left (\frac {\sqrt {c} (e f-d g)}{e \left (\sqrt {c} f+i g\right )};i \sinh ^{-1}\left (\frac {\sqrt {-f-\frac {i g}{\sqrt {c}}}}{\sqrt {f+g x}}\right )|\frac {\sqrt {c} f-i g}{\sqrt {c} f+i g}\right )\right )}{\sqrt {c x^2+1} \sqrt {-f-\frac {i g}{\sqrt {c}}} (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[(g*(I/Sqrt[c] + x))/(f + g*x)]*Sqrt[-(((I*g)/Sqrt[c] - g*x)/(f + g*x))]*(f + g*x)*(EllipticF[I*Ar
cSinh[Sqrt[-f - (I*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*g)/(Sqrt[c]*f + I*g)] - EllipticPi[(Sqrt[c]*(e*f
- d*g))/(e*(Sqrt[c]*f + I*g)), I*ArcSinh[Sqrt[-f - (I*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*g)/(Sqrt[c]*
f + I*g)]))/(Sqrt[-f - (I*g)/Sqrt[c]]*(e*f - d*g)*Sqrt[1 + c*x^2])

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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+1)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 215, normalized size = 1.95 \[ \frac {2 \left (\sqrt {-c}\, f +g \right ) \sqrt {-\frac {\left (\sqrt {-c}\, x -1\right ) g}{\sqrt {-c}\, f +g}}\, \sqrt {-\frac {\left (\sqrt {-c}\, x +1\right ) g}{\sqrt {-c}\, f -g}}\, \sqrt {\frac {\left (g x +f \right ) \sqrt {-c}}{\sqrt {-c}\, f +g}}\, \sqrt {c \,x^{2}+1}\, \sqrt {g x +f}\, \EllipticPi \left (\sqrt {\frac {\left (g x +f \right ) \sqrt {-c}}{\sqrt {-c}\, f +g}}, -\frac {\left (\sqrt {-c}\, f +g \right ) e}{\sqrt {-c}\, \left (d g -e f \right )}, \sqrt {\frac {\sqrt {-c}\, f +g}{\sqrt {-c}\, f -g}}\right )}{\sqrt {-c}\, \left (d g -e f \right ) \left (c g \,x^{3}+c f \,x^{2}+g x +f \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {f+g\,x}\,\sqrt {c\,x^2+1}\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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