\(\int \frac {2+x+x^2}{x^2 (1+x^2)^{3/4}} \, dx\) [500]

Optimal result

Integrand size = 19, antiderivative size = 39 \[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt [4]{1+x^2}}{x}-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[Out]

-2*(x^2+1)^(1/4)/x-arctan((x^2+1)^(1/4))-arctanh((x^2+1)^(1/4))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1821, 272, 65, 218, 212, 209} \[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=-\arctan \left (\sqrt [4]{x^2+1}\right )-\text {arctanh}\left (\sqrt [4]{x^2+1}\right )-\frac {2 \sqrt [4]{x^2+1}}{x} \]

[In]

Int[(2 + x + x^2)/(x^2*(1 + x^2)^(3/4)),x]

[Out]

(-2*(1 + x^2)^(1/4))/x - ArcTan[(1 + x^2)^(1/4)] - ArcTanh[(1 + x^2)^(1/4)]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 212

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

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 272

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

Rule 1821

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt [4]{1+x^2}}{x}+\int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx \\ & = -\frac {2 \sqrt [4]{1+x^2}}{x}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (1+x)^{3/4}} \, dx,x,x^2\right ) \\ & = -\frac {2 \sqrt [4]{1+x^2}}{x}+2 \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{1+x^2}\right ) \\ & = -\frac {2 \sqrt [4]{1+x^2}}{x}-\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^2}\right )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^2}\right ) \\ & = -\frac {2 \sqrt [4]{1+x^2}}{x}-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt [4]{1+x^2}}{x}-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[In]

Integrate[(2 + x + x^2)/(x^2*(1 + x^2)^(3/4)),x]

[Out]

(-2*(1 + x^2)^(1/4))/x - ArcTan[(1 + x^2)^(1/4)] - ArcTanh[(1 + x^2)^(1/4)]

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.58 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44

method	result	size
risch	\(-\frac {2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x}+\frac {-\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{2}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{2 \Gamma \left (\frac {3}{4}\right )}\)	\(56\)
meijerg	\(x \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], -x^{2}\right )+\frac {-\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{2}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{2 \Gamma \left (\frac {3}{4}\right )}-\frac {2 \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {1}{2}\right ], -x^{2}\right )}{x}\)	\(73\)
trager	\(-\frac {2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x}-\frac {\ln \left (-\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}+2 \sqrt {x^{2}+1}+x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}+2}{x^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x^{2}}\right )}{2}\)	\(120\)

[In]

int((x^2+x+2)/x^2/(x^2+1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-2*(x^2+1)^(1/4)/x+1/2/GAMMA(3/4)*(-3/4*GAMMA(3/4)*x^2*hypergeom([1,1,7/4],[2,2],-x^2)+(-3*ln(2)+1/2*Pi+2*ln(x
))*GAMMA(3/4))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (33) = 66\).

Time = 0.66 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.97 \[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=\frac {x \arctan \left (\frac {2 \, {\left ({\left (x^{2} + 1\right )}^{\frac {3}{4}} + {\left (x^{2} + 1\right )}^{\frac {1}{4}}\right )}}{x^{2}}\right ) + x \log \left (\frac {x^{2} - 2 \, {\left (x^{2} + 1\right )}^{\frac {3}{4}} + 2 \, \sqrt {x^{2} + 1} - 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} + 2}{x^{2}}\right ) - 4 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{2 \, x} \]

[In]

integrate((x^2+x+2)/x^2/(x^2+1)^(3/4),x, algorithm="fricas")

[Out]

1/2*(x*arctan(2*((x^2 + 1)^(3/4) + (x^2 + 1)^(1/4))/x^2) + x*log((x^2 - 2*(x^2 + 1)^(3/4) + 2*sqrt(x^2 + 1) -
2*(x^2 + 1)^(1/4) + 2)/x^2) - 4*(x^2 + 1)^(1/4))/x

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.58 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.74 \[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {x^{2} e^{i \pi }} \right )} - \frac {2 {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {x^{2} e^{i \pi }} \right )}}{x} - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate((x**2+x+2)/x**2/(x**2+1)**(3/4),x)

[Out]

x*hyper((1/2, 3/4), (3/2,), x**2*exp_polar(I*pi)) - 2*hyper((-1/2, 3/4), (1/2,), x**2*exp_polar(I*pi))/x - gam
ma(3/4)*hyper((3/4, 3/4), (7/4,), exp_polar(I*pi)/x**2)/(2*x**(3/2)*gamma(7/4))

Maxima [F]

\[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=\int { \frac {x^{2} + x + 2}{{\left (x^{2} + 1\right )}^{\frac {3}{4}} x^{2}} \,d x } \]

[In]

integrate((x^2+x+2)/x^2/(x^2+1)^(3/4),x, algorithm="maxima")

[Out]

integrate((x^2 + x + 2)/((x^2 + 1)^(3/4)*x^2), x)

Giac [F]

\[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=\int { \frac {x^{2} + x + 2}{{\left (x^{2} + 1\right )}^{\frac {3}{4}} x^{2}} \,d x } \]

[In]

integrate((x^2+x+2)/x^2/(x^2+1)^(3/4),x, algorithm="giac")

[Out]

integrate((x^2 + x + 2)/((x^2 + 1)^(3/4)*x^2), x)

Mupad [B] (verification not implemented)

Time = 5.55 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.23 \[ \int \frac {2+x+x^2}{x^2 \left (1+x^2\right )^{3/4}} \, dx=x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {3}{2};\ -x^2\right )-\mathrm {atanh}\left ({\left (x^2+1\right )}^{1/4}\right )-\mathrm {atan}\left ({\left (x^2+1\right )}^{1/4}\right )-\frac {2\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{4};\ \frac {1}{2};\ -x^2\right )}{x} \]

[In]

int((x + x^2 + 2)/(x^2*(x^2 + 1)^(3/4)),x)

[Out]

x*hypergeom([1/2, 3/4], 3/2, -x^2) - atanh((x^2 + 1)^(1/4)) - atan((x^2 + 1)^(1/4)) - (2*hypergeom([-1/2, 3/4]
, 1/2, -x^2))/x