3.259 \(\int \frac{4 x-\sqrt{1-x^2}}{5+\sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=88 \[ -4 \sqrt{1-x^2}+20 \log \left (\sqrt{1-x^2}+5\right )-\frac{25 \tan ^{-1}\left (\frac{5 x}{2 \sqrt{6} \sqrt{1-x^2}}\right )}{2 \sqrt{6}}-x+5 \sin ^{-1}(x)+\frac{25 \tan ^{-1}\left (\frac{x}{2 \sqrt{6}}\right )}{2 \sqrt{6}} \]

[Out]

-x - 4*Sqrt[1 - x^2] + 5*ArcSin[x] + (25*ArcTan[x/(2*Sqrt[6])])/(2*Sqrt[6]) - (25*ArcTan[(5*x)/(2*Sqrt[6]*Sqrt
[1 - x^2])])/(2*Sqrt[6]) + 20*Log[5 + Sqrt[1 - x^2]]

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Rubi [A]  time = 0.238756, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6742, 1591, 190, 43, 6740, 203, 402, 216, 377} \[ -4 \sqrt{1-x^2}+20 \log \left (\sqrt{1-x^2}+5\right )-\frac{25 \tan ^{-1}\left (\frac{5 x}{2 \sqrt{6} \sqrt{1-x^2}}\right )}{2 \sqrt{6}}-x+5 \sin ^{-1}(x)+\frac{25 \tan ^{-1}\left (\frac{x}{2 \sqrt{6}}\right )}{2 \sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Int[(4*x - Sqrt[1 - x^2])/(5 + Sqrt[1 - x^2]),x]

[Out]

-x - 4*Sqrt[1 - x^2] + 5*ArcSin[x] + (25*ArcTan[x/(2*Sqrt[6])])/(2*Sqrt[6]) - (25*ArcTan[(5*x)/(2*Sqrt[6]*Sqrt
[1 - x^2])])/(2*Sqrt[6]) + 20*Log[5 + Sqrt[1 - x^2]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 1591

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6740

Int[(v_)/((a_) + (b_.)*(u_)^(n_.)), x_Symbol] :> Int[ExpandIntegrand[PolynomialInSubst[v, u, x]/(a + b*x^n), x
] /. x -> u, x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && PolynomialInQ[v, u, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{4 x-\sqrt{1-x^2}}{5+\sqrt{1-x^2}} \, dx &=\int \left (\frac{4 x}{5+\sqrt{1-x^2}}-\frac{\sqrt{1-x^2}}{5+\sqrt{1-x^2}}\right ) \, dx\\ &=4 \int \frac{x}{5+\sqrt{1-x^2}} \, dx-\int \frac{\sqrt{1-x^2}}{5+\sqrt{1-x^2}} \, dx\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{5+\sqrt{x}} \, dx,x,1-x^2\right )\right )-\int \left (1-\frac{5}{5+\sqrt{1-x^2}}\right ) \, dx\\ &=-x-4 \operatorname{Subst}\left (\int \frac{x}{5+x} \, dx,x,\sqrt{1-x^2}\right )+5 \int \frac{1}{5+\sqrt{1-x^2}} \, dx\\ &=-x-4 \operatorname{Subst}\left (\int \left (1-\frac{5}{5+x}\right ) \, dx,x,\sqrt{1-x^2}\right )+5 \int \left (\frac{5}{24+x^2}-\frac{\sqrt{1-x^2}}{24+x^2}\right ) \, dx\\ &=-x-4 \sqrt{1-x^2}+20 \log \left (5+\sqrt{1-x^2}\right )-5 \int \frac{\sqrt{1-x^2}}{24+x^2} \, dx+25 \int \frac{1}{24+x^2} \, dx\\ &=-x-4 \sqrt{1-x^2}+\frac{25 \tan ^{-1}\left (\frac{x}{2 \sqrt{6}}\right )}{2 \sqrt{6}}+20 \log \left (5+\sqrt{1-x^2}\right )+5 \int \frac{1}{\sqrt{1-x^2}} \, dx-125 \int \frac{1}{\sqrt{1-x^2} \left (24+x^2\right )} \, dx\\ &=-x-4 \sqrt{1-x^2}+5 \sin ^{-1}(x)+\frac{25 \tan ^{-1}\left (\frac{x}{2 \sqrt{6}}\right )}{2 \sqrt{6}}+20 \log \left (5+\sqrt{1-x^2}\right )-125 \operatorname{Subst}\left (\int \frac{1}{24+25 x^2} \, dx,x,\frac{x}{\sqrt{1-x^2}}\right )\\ &=-x-4 \sqrt{1-x^2}+5 \sin ^{-1}(x)+\frac{25 \tan ^{-1}\left (\frac{x}{2 \sqrt{6}}\right )}{2 \sqrt{6}}-\frac{25 \tan ^{-1}\left (\frac{5 x}{2 \sqrt{6} \sqrt{1-x^2}}\right )}{2 \sqrt{6}}+20 \log \left (5+\sqrt{1-x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.177174, size = 137, normalized size = 1.56 \[ -4 \sqrt{1-x^2}+10 \log \left (x^2+24\right )-10 \log \left (\left (x^2+24\right )^2\right )+10 \log \left (\left (x^2+24\right ) \left (-x^2+10 \sqrt{1-x^2}+26\right )\right )+\frac{25 \tan ^{-1}\left (\frac{4 x^2+409 \sqrt{1-x^2} x+96}{10 \sqrt{6} \left (17 x^2-1\right )}\right )}{2 \sqrt{6}}-x+5 \sin ^{-1}(x)+\frac{25 \tan ^{-1}\left (\frac{x}{2 \sqrt{6}}\right )}{2 \sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(4*x - Sqrt[1 - x^2])/(5 + Sqrt[1 - x^2]),x]

[Out]

-x - 4*Sqrt[1 - x^2] + 5*ArcSin[x] + (25*ArcTan[x/(2*Sqrt[6])])/(2*Sqrt[6]) + (25*ArcTan[(96 + 4*x^2 + 409*x*S
qrt[1 - x^2])/(10*Sqrt[6]*(-1 + 17*x^2))])/(2*Sqrt[6]) + 10*Log[24 + x^2] - 10*Log[(24 + x^2)^2] + 10*Log[(24
+ x^2)*(26 - x^2 + 10*Sqrt[1 - x^2])]

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Maple [A]  time = 0.02, size = 82, normalized size = 0.9 \begin{align*}{\frac{25\,\sqrt{6}}{12}\arctan \left ({\frac{x\sqrt{6}}{12}} \right ) }+10\,\ln \left ({x}^{2}+24 \right ) -x+5\,\arcsin \left ( x \right ) +{\frac{25\,\sqrt{6}}{12}\arctan \left ({\frac{5\,x\sqrt{6}}{12\,{x}^{2}-12}\sqrt{-{x}^{2}+1}} \right ) }-4\,\sqrt{-{x}^{2}+1}+20\,{\it Artanh} \left ( 1/5\,\sqrt{-{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x-(-x^2+1)^(1/2))/(5+(-x^2+1)^(1/2)),x)

[Out]

25/12*arctan(1/12*x*6^(1/2))*6^(1/2)+10*ln(x^2+24)-x+5*arcsin(x)+25/12*6^(1/2)*arctan(5/12*6^(1/2)*(-x^2+1)^(1
/2)/(x^2-1)*x)-4*(-x^2+1)^(1/2)+20*arctanh(1/5*(-x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -x - 4 \, \sqrt{-x^{2} + 1} + 5 \, \int \frac{1}{\sqrt{x + 1} \sqrt{-x + 1} + 5}\,{d x} + 20 \, \log \left (\sqrt{-x^{2} + 1} + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x-(-x^2+1)^(1/2))/(5+(-x^2+1)^(1/2)),x, algorithm="maxima")

[Out]

-x - 4*sqrt(-x^2 + 1) + 5*integrate(1/(sqrt(x + 1)*sqrt(-x + 1) + 5), x) + 20*log(sqrt(-x^2 + 1) + 5)

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Fricas [B]  time = 1.98118, size = 450, normalized size = 5.11 \begin{align*} \frac{25}{12} \, \sqrt{6} \arctan \left (\frac{1}{12} \, \sqrt{6} x\right ) + \frac{25}{12} \, \sqrt{6} \arctan \left (\frac{\sqrt{6} \sqrt{-x^{2} + 1} - \sqrt{6}}{2 \, x}\right ) + \frac{25}{12} \, \sqrt{6} \arctan \left (\frac{\sqrt{6} \sqrt{-x^{2} + 1} - \sqrt{6}}{3 \, x}\right ) - x - 4 \, \sqrt{-x^{2} + 1} - 10 \, \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) + 10 \, \log \left (x^{2} + 24\right ) - 10 \, \log \left (-\frac{x^{2} + 6 \, \sqrt{-x^{2} + 1} - 6}{x^{2}}\right ) + 10 \, \log \left (\frac{x^{2} - 4 \, \sqrt{-x^{2} + 1} + 4}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x-(-x^2+1)^(1/2))/(5+(-x^2+1)^(1/2)),x, algorithm="fricas")

[Out]

25/12*sqrt(6)*arctan(1/12*sqrt(6)*x) + 25/12*sqrt(6)*arctan(1/2*(sqrt(6)*sqrt(-x^2 + 1) - sqrt(6))/x) + 25/12*
sqrt(6)*arctan(1/3*(sqrt(6)*sqrt(-x^2 + 1) - sqrt(6))/x) - x - 4*sqrt(-x^2 + 1) - 10*arctan((sqrt(-x^2 + 1) -
1)/x) + 10*log(x^2 + 24) - 10*log(-(x^2 + 6*sqrt(-x^2 + 1) - 6)/x^2) + 10*log((x^2 - 4*sqrt(-x^2 + 1) + 4)/x^2
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{4 x - \sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}} + 5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x-(-x**2+1)**(1/2))/(5+(-x**2+1)**(1/2)),x)

[Out]

Integral((4*x - sqrt(1 - x**2))/(sqrt(1 - x**2) + 5), x)

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Giac [B]  time = 1.13468, size = 182, normalized size = 2.07 \begin{align*} \frac{25}{12} \, \sqrt{6} \arctan \left (\frac{1}{12} \, \sqrt{6} x\right ) - \frac{25}{12} \, \sqrt{6} \arctan \left (-\frac{\sqrt{6}{\left (\sqrt{-x^{2} + 1} - 1\right )}}{3 \, x}\right ) - \frac{25}{12} \, \sqrt{6} \arctan \left (-\frac{\sqrt{6}{\left (\sqrt{-x^{2} + 1} - 1\right )}}{2 \, x}\right ) - x - 4 \, \sqrt{-x^{2} + 1} + 5 \, \arcsin \left (x\right ) + 10 \, \log \left (x^{2} + 24\right ) - 10 \, \log \left (\frac{3 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 2\right ) + 10 \, \log \left (\frac{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x-(-x^2+1)^(1/2))/(5+(-x^2+1)^(1/2)),x, algorithm="giac")

[Out]

25/12*sqrt(6)*arctan(1/12*sqrt(6)*x) - 25/12*sqrt(6)*arctan(-1/3*sqrt(6)*(sqrt(-x^2 + 1) - 1)/x) - 25/12*sqrt(
6)*arctan(-1/2*sqrt(6)*(sqrt(-x^2 + 1) - 1)/x) - x - 4*sqrt(-x^2 + 1) + 5*arcsin(x) + 10*log(x^2 + 24) - 10*lo
g(3*(sqrt(-x^2 + 1) - 1)^2/x^2 + 2) + 10*log(2*(sqrt(-x^2 + 1) - 1)^2/x^2 + 3)