\(\int \frac {1+x^4}{x (1+x^2)^2} \, dx\) [133]

Optimal result

Integrand size = 16, antiderivative size = 10 \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\frac {1}{1+x^2}+\log (x) \]

[Out]

1/(x^2+1)+ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1266, 908} \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\frac {1}{x^2+1}+\log (x) \]

[In]

Int[(1 + x^4)/(x*(1 + x^2)^2),x]

[Out]

(1 + x^2)^(-1) + Log[x]

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1266

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\frac {1}{1+x^2}+\log (x) \]

[In]

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

[Out]

(1 + x^2)^(-1) + Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {1}{x^{2}+1}+\ln \left (x \right )\)	\(11\)
norman	\(\frac {1}{x^{2}+1}+\ln \left (x \right )\)	\(11\)
risch	\(\frac {1}{x^{2}+1}+\ln \left (x \right )\)	\(11\)
parallelrisch	\(\frac {x^{2} \ln \left (x \right )+1+\ln \left (x \right )}{x^{2}+1}\)	\(19\)
meijerg	\(-\frac {x^{2}}{2 \left (x^{2}+1\right )}+\frac {1}{2}+\ln \left (x \right )-\frac {x^{2}}{2 x^{2}+2}\)	\(31\)

[In]

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

[Out]

1/(x^2+1)+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.80 \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\frac {{\left (x^{2} + 1\right )} \log \left (x\right ) + 1}{x^{2} + 1} \]

[In]

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

[Out]

((x^2 + 1)*log(x) + 1)/(x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\log {\left (x \right )} + \frac {1}{x^{2} + 1} \]

[In]

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

[Out]

log(x) + 1/(x**2 + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.40 \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\frac {1}{x^{2} + 1} + \frac {1}{2} \, \log \left (x^{2}\right ) \]

[In]

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

[Out]

1/(x^2 + 1) + 1/2*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.40 \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\frac {1}{x^{2} + 1} + \frac {1}{2} \, \log \left (x^{2}\right ) \]

[In]

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

[Out]

1/(x^2 + 1) + 1/2*log(x^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^4}{x \left (1+x^2\right )^2} \, dx=\ln \left (x\right )+\frac {1}{x^2+1} \]

[In]

int((x^4 + 1)/(x*(x^2 + 1)^2),x)

[Out]

log(x) + 1/(x^2 + 1)