Integrand size = 24, antiderivative size = 47 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\frac {x \sqrt {1-x^2}}{1+x^2}+2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right ) \] Output:
2*arctan(x*2^(1/2)/(-x^2+1)^(1/2))*2^(1/2)+x*(-x^2+1)^(1/2)/(x^2+1)
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\frac {x \sqrt {1-x^2}}{1+x^2}+2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right ) \] Input:
Integrate[(5 + x^2)/(Sqrt[1 - x^2]*(1 + x^2)^2),x]
Output:
(x*Sqrt[1 - x^2])/(1 + x^2) + 2*Sqrt[2]*ArcTan[(Sqrt[2]*x)/Sqrt[1 - x^2]]
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {402, 27, 291, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^2+5}{\sqrt {1-x^2} \left (x^2+1\right )^2} \, dx\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {x \sqrt {1-x^2}}{x^2+1}-\frac {1}{4} \int -\frac {16}{\sqrt {1-x^2} \left (x^2+1\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {1}{\sqrt {1-x^2} \left (x^2+1\right )}dx+\frac {\sqrt {1-x^2} x}{x^2+1}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle 4 \int \frac {1}{\frac {2 x^2}{1-x^2}+1}d\frac {x}{\sqrt {1-x^2}}+\frac {\sqrt {1-x^2} x}{x^2+1}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle 2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )+\frac {\sqrt {1-x^2} x}{x^2+1}\) |
Input:
Int[(5 + x^2)/(Sqrt[1 - x^2]*(1 + x^2)^2),x]
Output:
(x*Sqrt[1 - x^2])/(1 + x^2) + 2*Sqrt[2]*ArcTan[(Sqrt[2]*x)/Sqrt[1 - x^2]]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Time = 0.47 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(\frac {\left (-2 x^{2}-2\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}}{2 x}\right )+x \sqrt {-x^{2}+1}}{x^{2}+1}\) | \(50\) |
risch | \(-\frac {x \left (x^{2}-1\right )}{\left (x^{2}+1\right ) \sqrt {-x^{2}+1}}-2 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}\, x}{x^{2}-1}\right )\) | \(53\) |
trager | \(\frac {x \sqrt {-x^{2}+1}}{x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}+4 x \sqrt {-x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{x^{2}+1}\right )\) | \(66\) |
default | \(-2 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}\, x}{x^{2}-1}\right )-\frac {\sqrt {-x^{2}+1}\, x}{2 \left (x^{2}-1\right ) \left (\frac {\left (-x^{2}+1\right ) x^{2}}{\left (x^{2}-1\right )^{2}}+\frac {1}{2}\right )}\) | \(70\) |
Input:
int((x^2+5)/(x^2+1)^2/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
((-2*x^2-2)*2^(1/2)*arctan(1/2/x*2^(1/2)*(-x^2+1)^(1/2))+x*(-x^2+1)^(1/2)) /(x^2+1)
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=-\frac {2 \, \sqrt {2} {\left (x^{2} + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-x^{2} + 1} x}{x^{2} - 1}\right ) - \sqrt {-x^{2} + 1} x}{x^{2} + 1} \] Input:
integrate((x^2+5)/(x^2+1)^2/(-x^2+1)^(1/2),x, algorithm="fricas")
Output:
-(2*sqrt(2)*(x^2 + 1)*arctan(sqrt(2)*sqrt(-x^2 + 1)*x/(x^2 - 1)) - sqrt(-x ^2 + 1)*x)/(x^2 + 1)
\[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\int \frac {x^{2} + 5}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 1\right )^{2}}\, dx \] Input:
integrate((x**2+5)/(x**2+1)**2/(-x**2+1)**(1/2),x)
Output:
Integral((x**2 + 5)/(sqrt(-(x - 1)*(x + 1))*(x**2 + 1)**2), x)
\[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\int { \frac {x^{2} + 5}{{\left (x^{2} + 1\right )}^{2} \sqrt {-x^{2} + 1}} \,d x } \] Input:
integrate((x^2+5)/(x^2+1)^2/(-x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate((x^2 + 5)/((x^2 + 1)^2*sqrt(-x^2 + 1)), x)
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (39) = 78\).
Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.62 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\sqrt {2} {\left (\pi \mathrm {sgn}\left (x\right ) + 2 \, \arctan \left (-\frac {\sqrt {2} x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{4 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} - \frac {2 \, {\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}}{{\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}^{2} + 8} \] Input:
integrate((x^2+5)/(x^2+1)^2/(-x^2+1)^(1/2),x, algorithm="giac")
Output:
sqrt(2)*(pi*sgn(x) + 2*arctan(-1/4*sqrt(2)*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1))) - 2*(x/(sqrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1) - 1)/x)/((x/(sqrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1) - 1)/x)^2 + 8)
Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.45 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (-1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\mathrm {i}}\right )\,1{}\mathrm {i}-\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+1{}\mathrm {i}}\right )\,1{}\mathrm {i}+\frac {\sqrt {1-x^2}}{2\,\left (x-\mathrm {i}\right )}+\frac {\sqrt {1-x^2}}{2\,\left (x+1{}\mathrm {i}\right )} \] Input:
int((x^2 + 5)/((1 - x^2)^(1/2)*(x^2 + 1)^2),x)
Output:
2^(1/2)*log(((2^(1/2)*(x*1i - 1)*1i)/2 - (1 - x^2)^(1/2)*1i)/(x - 1i))*1i - 2^(1/2)*log(((2^(1/2)*(x*1i + 1)*1i)/2 + (1 - x^2)^(1/2)*1i)/(x + 1i))*1 i + (1 - x^2)^(1/2)/(2*(x - 1i)) + (1 - x^2)^(1/2)/(2*(x + 1i))
Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.85 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\frac {2 \sqrt {2}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )}{\sqrt {2}+1}\right ) x^{2}+2 \sqrt {2}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )}{\sqrt {2}+1}\right )+\sqrt {-x^{2}+1}\, x -\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )+i \right ) i \,x^{2}-\sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )+i \right ) i +\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )-i \right ) i \,x^{2}+\sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (x \right )}{2}\right )-i \right ) i}{x^{2}+1} \] Input:
int((x^2+5)/(x^2+1)^2/(-x^2+1)^(1/2),x)
Output:
(2*sqrt(2)*atan(tan(asin(x)/2)/(sqrt(2) + 1))*x**2 + 2*sqrt(2)*atan(tan(as in(x)/2)/(sqrt(2) + 1)) + sqrt( - x**2 + 1)*x - sqrt(2)*log( - sqrt(2)*i + tan(asin(x)/2) + i)*i*x**2 - sqrt(2)*log( - sqrt(2)*i + tan(asin(x)/2) + i)*i + sqrt(2)*log(sqrt(2)*i + tan(asin(x)/2) - i)*i*x**2 + sqrt(2)*log(sq rt(2)*i + tan(asin(x)/2) - i)*i)/(x**2 + 1)