3.3.59 \(\int \frac {4 x-\sqrt {1-x^2}}{5+\sqrt {1-x^2}} \, dx\) [259]

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

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 = {6874, 1605, 196, 45, 6872, 209, 399, 222, 385} \begin {gather*} 5 \text {ArcSin}(x)-\frac {25 \text {ArcTan}\left (\frac {5 x}{2 \sqrt {6} \sqrt {1-x^2}}\right )}{2 \sqrt {6}}+\frac {25 \text {ArcTan}\left (\frac {x}{2 \sqrt {6}}\right )}{2 \sqrt {6}}-4 \sqrt {1-x^2}+20 \log \left (\sqrt {1-x^2}+5\right )-x \end {gather*}

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 45

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 196

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 209

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

Rule 222

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 385

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]

Rule 399

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

Rule 1605

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 6872

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 6874

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

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 \text {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 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.23, size = 108, normalized size = 1.23 \begin {gather*} -x-4 \sqrt {1-x^2}+10 \tan ^{-1}\left (\frac {x}{-1+\sqrt {1-x^2}}\right )-\frac {25 \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{-1+\sqrt {1-x^2}}\right )}{\sqrt {6}}-20 \log \left (-1+\sqrt {1-x^2}\right )+20 \log \left (-4-x^2+4 \sqrt {1-x^2}\right ) \end {gather*}

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] + 10*ArcTan[x/(-1 + Sqrt[1 - x^2])] - (25*ArcTan[(Sqrt[3/2]*x)/(-1 + Sqrt[1 - x^2])])/Sqr
t[6] - 20*Log[-1 + Sqrt[1 - x^2]] + 20*Log[-4 - x^2 + 4*Sqrt[1 - x^2]]

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Maple [A]
time = 0.44, size = 82, normalized size = 0.93

method	result	size
default	\(\frac {25 \arctan \left (\frac {x \sqrt {6}}{12}\right ) \sqrt {6}}{12}+10 \ln \left (x^{2}+24\right )-x +5 \arcsin \left (x \right )+\frac {25 \sqrt {6}\, \arctan \left (\frac {5 \sqrt {6}\, \sqrt {-x^{2}+1}\, x}{12 \left (x^{2}-1\right )}\right )}{12}-4 \sqrt {-x^{2}+1}+20 \arctanh \left (\frac {\sqrt {-x^{2}+1}}{5}\right )\)	\(82\)
trager	\(\text {Expression too large to display}\)	\(883\)

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,method=_RETURNVERBOSE)

[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 = 0.00, 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((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] Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (66) = 132\).
time = 0.48, size = 160, normalized size = 1.82 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 x - \sqrt {1 - x^{2}}}{\sqrt {1 - x^{2}} + 5}\, dx \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (66) = 132\).
time = 0.57, size = 135, normalized size = 1.53 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.38, size = 159, normalized size = 1.81 \begin {gather*} 5\,\mathrm {asin}\left (x\right )-x-4\,\sqrt {1-x^2}-\frac {\sqrt {24}\,\ln \left (\frac {\frac {2\,\sqrt {6}\,x}{5}+\sqrt {1-x^2}\,1{}\mathrm {i}+\frac {1}{5}{}\mathrm {i}}{x-\sqrt {6}\,2{}\mathrm {i}}\right )\,\left (125+\sqrt {24}\,100{}\mathrm {i}\right )\,1{}\mathrm {i}}{240}-\frac {\sqrt {24}\,\ln \left (\frac {-\frac {\sqrt {24}\,x}{5}+\sqrt {1-x^2}\,1{}\mathrm {i}+\frac {1}{5}{}\mathrm {i}}{x+\sqrt {24}\,1{}\mathrm {i}}\right )\,\left (-125+\sqrt {24}\,100{}\mathrm {i}\right )\,1{}\mathrm {i}}{240}-\frac {\sqrt {24}\,\ln \left (x-\sqrt {6}\,2{}\mathrm {i}\right )\,\left (25+\sqrt {24}\,20{}\mathrm {i}\right )\,1{}\mathrm {i}}{48}-\frac {\sqrt {24}\,\ln \left (x+\sqrt {24}\,1{}\mathrm {i}\right )\,\left (-25+\sqrt {24}\,20{}\mathrm {i}\right )\,1{}\mathrm {i}}{48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*asin(x) - x - 4*(1 - x^2)^(1/2) - (24^(1/2)*log(((2*6^(1/2)*x)/5 + (1 - x^2)^(1/2)*1i + 1i/5)/(x - 6^(1/2)*2
i))*(24^(1/2)*100i + 125)*1i)/240 - (24^(1/2)*log(((1 - x^2)^(1/2)*1i - (24^(1/2)*x)/5 + 1i/5)/(x + 24^(1/2)*1
i))*(24^(1/2)*100i - 125)*1i)/240 - (24^(1/2)*log(x - 6^(1/2)*2i)*(24^(1/2)*20i + 25)*1i)/48 - (24^(1/2)*log(x
+ 24^(1/2)*1i)*(24^(1/2)*20i - 25)*1i)/48

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