3.639 \(\int \frac {\sqrt {f+g x}}{\sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {2 \sqrt {-a} \sqrt {\frac {c x^2}{a}+1} \sqrt {f+g x} E\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{\sqrt {c} \sqrt {a+c x^2} \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}}} \]

[Out]

-2*EllipticE(1/2*(1-x*c^(1/2)/(-a)^(1/2))^(1/2)*2^(1/2),(-2*a*g/(-a*g+f*(-a)^(1/2)*c^(1/2)))^(1/2))*(-a)^(1/2)
*(g*x+f)^(1/2)*(c*x^2/a+1)^(1/2)/c^(1/2)/(c*x^2+a)^(1/2)/((g*x+f)*c^(1/2)/(g*(-a)^(1/2)+f*c^(1/2)))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {719, 424} \[ -\frac {2 \sqrt {-a} \sqrt {\frac {c x^2}{a}+1} \sqrt {f+g x} E\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{\sqrt {c} \sqrt {a+c x^2} \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[f + g*x]/Sqrt[a + c*x^2],x]

[Out]

(-2*Sqrt[-a]*Sqrt[f + g*x]*Sqrt[1 + (c*x^2)/a]*EllipticE[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a
*g)/(Sqrt[-a]*Sqrt[c]*f - a*g)])/(Sqrt[c]*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[a + c*x^2])

Rule 424

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

Rule 719

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*a*Rt[-(c/a), 2]*(d + e*x)^m*Sqrt[
1 + (c*x^2)/a])/(c*Sqrt[a + c*x^2]*((c*(d + e*x))/(c*d - a*e*Rt[-(c/a), 2]))^m), Subst[Int[(1 + (2*a*e*Rt[-(c/
a), 2]*x^2)/(c*d - a*e*Rt[-(c/a), 2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(1 - Rt[-(c/a), 2]*x)/2]], x] /; FreeQ[{a,
c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m^2, 1/4]

Rubi steps

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

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Mathematica [C]  time = 0.47, size = 294, normalized size = 2.16 \[ \frac {2 i \sqrt {f+g x} \left (\sqrt {c} f+i \sqrt {a} g\right ) \sqrt {\frac {g \left (\sqrt {a}+i \sqrt {c} x\right )}{\sqrt {a} g-i \sqrt {c} f}} \left (E\left (i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c} (f+g x)}{\sqrt {c} f-i \sqrt {a} g}}\right )|\frac {\sqrt {c} f-i \sqrt {a} g}{\sqrt {c} f+i \sqrt {a} g}\right )-F\left (i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c} (f+g x)}{\sqrt {c} f-i \sqrt {a} g}}\right )|\frac {\sqrt {c} f-i \sqrt {a} g}{\sqrt {c} f+i \sqrt {a} g}\right )\right )}{\sqrt {c} g \sqrt {a+c x^2} \sqrt {\frac {\sqrt {c} (f+g x)}{g \left (\sqrt {c} x+i \sqrt {a}\right )}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[f + g*x]/Sqrt[a + c*x^2],x]

[Out]

((2*I)*(Sqrt[c]*f + I*Sqrt[a]*g)*Sqrt[(g*(Sqrt[a] + I*Sqrt[c]*x))/((-I)*Sqrt[c]*f + Sqrt[a]*g)]*Sqrt[f + g*x]*
(EllipticE[I*ArcSinh[Sqrt[-((Sqrt[c]*(f + g*x))/(Sqrt[c]*f - I*Sqrt[a]*g))]], (Sqrt[c]*f - I*Sqrt[a]*g)/(Sqrt[
c]*f + I*Sqrt[a]*g)] - EllipticF[I*ArcSinh[Sqrt[-((Sqrt[c]*(f + g*x))/(Sqrt[c]*f - I*Sqrt[a]*g))]], (Sqrt[c]*f
- I*Sqrt[a]*g)/(Sqrt[c]*f + I*Sqrt[a]*g)]))/(Sqrt[c]*g*Sqrt[(Sqrt[c]*(f + g*x))/(g*(I*Sqrt[a] + Sqrt[c]*x))]*
Sqrt[a + c*x^2])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {g x + f}}{\sqrt {c x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^(1/2)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(g*x + f)/sqrt(c*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {g x + f}}{\sqrt {c x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 396, normalized size = 2.91 \[ \frac {2 \sqrt {g x +f}\, \sqrt {c \,x^{2}+a}\, \left (c f -\sqrt {-a c}\, g \right ) \sqrt {-\frac {\left (g x +f \right ) c}{-c f +\sqrt {-a c}\, g}}\, \sqrt {\frac {\left (-c x +\sqrt {-a c}\right ) g}{c f +\sqrt {-a c}\, g}}\, \sqrt {\frac {\left (c x +\sqrt {-a c}\right ) g}{-c f +\sqrt {-a c}\, g}}\, \left (-c f \EllipticE \left (\sqrt {-\frac {\left (g x +f \right ) c}{-c f +\sqrt {-a c}\, g}}, \sqrt {-\frac {-c f +\sqrt {-a c}\, g}{c f +\sqrt {-a c}\, g}}\right )+c f \EllipticF \left (\sqrt {-\frac {\left (g x +f \right ) c}{-c f +\sqrt {-a c}\, g}}, \sqrt {-\frac {-c f +\sqrt {-a c}\, g}{c f +\sqrt {-a c}\, g}}\right )-\sqrt {-a c}\, g \EllipticE \left (\sqrt {-\frac {\left (g x +f \right ) c}{-c f +\sqrt {-a c}\, g}}, \sqrt {-\frac {-c f +\sqrt {-a c}\, g}{c f +\sqrt {-a c}\, g}}\right )+\sqrt {-a c}\, g \EllipticF \left (\sqrt {-\frac {\left (g x +f \right ) c}{-c f +\sqrt {-a c}\, g}}, \sqrt {-\frac {-c f +\sqrt {-a c}\, g}{c f +\sqrt {-a c}\, g}}\right )\right )}{\left (c g \,x^{3}+c f \,x^{2}+a g x +a f \right ) c^{2} g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(g*x+f)^(1/2)*(c*x^2+a)^(1/2)*(c*f-(-a*c)^(1/2)*g)*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c)^(1
/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((-a*c)^(1/2)*EllipticF(
(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*g+f*EllipticF((-
(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*c-EllipticE((-(g*x
+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*(-a*c)^(1/2)*g-Ellipti
cE((-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*c*f)/g/(c*g*x
^3+c*f*x^2+a*g*x+a*f)/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {f + g x}}{\sqrt {a + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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