\(\int \frac {5+x^2}{\sqrt {1-x^2} (1+x^2)^2} \, dx\) [258]

Optimal result

Integrand size = 24, antiderivative size = 47 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\frac {x \sqrt {1-x^2}}{1+x^2}+2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right ) \]

[Out]

2*arctan(x*2^(1/2)/(-x^2+1)^(1/2))*2^(1/2)+x*(-x^2+1)^(1/2)/(x^2+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {541, 12, 385, 209} \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )+\frac {\sqrt {1-x^2} x}{x^2+1} \]

[In]

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

[Out]

(x*Sqrt[1 - x^2])/(1 + x^2) + 2*Sqrt[2]*ArcTan[(Sqrt[2]*x)/Sqrt[1 - x^2]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x \sqrt {1-x^2}}{1+x^2}-\frac {1}{4} \int -\frac {16}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx \\ & = \frac {x \sqrt {1-x^2}}{1+x^2}+4 \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx \\ & = \frac {x \sqrt {1-x^2}}{1+x^2}+4 \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1-x^2}}\right ) \\ & = \frac {x \sqrt {1-x^2}}{1+x^2}+2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\frac {x \sqrt {1-x^2}}{1+x^2}+2 \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right ) \]

[In]

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

[Out]

(x*Sqrt[1 - x^2])/(1 + x^2) + 2*Sqrt[2]*ArcTan[(Sqrt[2]*x)/Sqrt[1 - x^2]]

Maple [A] (verified)

Time = 0.62 (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	\(\frac {x \sqrt {-x^{2}+1}}{x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}+4 x \sqrt {-x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{x^{2}+1}\right )\)	\(69\)
default	\(-2 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}\, x}{x^{2}-1}\right )-\frac {\sqrt {-x^{2}+1}\, x}{2 \left (x^{2}-1\right ) \left (\frac {\left (-x^{2}+1\right ) x^{2}}{\left (x^{2}-1\right )^{2}}+\frac {1}{2}\right )}\)	\(70\)

[In]

int((x^2+5)/(x^2+1)^2/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

((-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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=-\frac {2 \, \sqrt {2} {\left (x^{2} + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-x^{2} + 1}}{2 \, x}\right ) - \sqrt {-x^{2} + 1} x}{x^{2} + 1} \]

[In]

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

[Out]

-(2*sqrt(2)*(x^2 + 1)*arctan(1/2*sqrt(2)*sqrt(-x^2 + 1)/x) - sqrt(-x^2 + 1)*x)/(x^2 + 1)

Sympy [F]

\[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\int \frac {x^{2} + 5}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 1\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\int { \frac {x^{2} + 5}{{\left (x^{2} + 1\right )}^{2} \sqrt {-x^{2} + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (39) = 78\).

Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.62 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\sqrt {2} {\left (\pi \mathrm {sgn}\left (x\right ) + 2 \, \arctan \left (-\frac {\sqrt {2} x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{4 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} - \frac {2 \, {\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}}{{\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )}^{2} + 8} \]

[In]

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

[Out]

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/(s
qrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1) - 1)/x)/((x/(sqrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1) - 1)/x)^2 + 8)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.45 \[ \int \frac {5+x^2}{\sqrt {1-x^2} \left (1+x^2\right )^2} \, dx=\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (-1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\mathrm {i}}\right )\,1{}\mathrm {i}-\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+1{}\mathrm {i}}\right )\,1{}\mathrm {i}+\frac {\sqrt {1-x^2}}{2\,\left (x-\mathrm {i}\right )}+\frac {\sqrt {1-x^2}}{2\,\left (x+1{}\mathrm {i}\right )} \]

[In]

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

[Out]

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)*1
i)/2 + (1 - x^2)^(1/2)*1i)/(x + 1i))*1i + (1 - x^2)^(1/2)/(2*(x - 1i)) + (1 - x^2)^(1/2)/(2*(x + 1i))