\(\int \frac {1+x^4}{x^3+x^5} \, dx\) [373]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {1+x^4}{x^3+x^5} \, dx=-\frac {1}{2 x^2}-\log (x)+\log \left (1+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1607, 1266, 908} \[ \int \frac {1+x^4}{x^3+x^5} \, dx=-\frac {1}{2 x^2}+\log \left (x^2+1\right )-\log (x) \]

[In]

Int[(1 + x^4)/(x^3 + x^5),x]

[Out]

-1/2*1/x^2 - Log[x] + Log[1 + x^2]

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]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^4}{x^3+x^5} \, dx=-\frac {1}{2 x^2}-\log (x)+\log \left (1+x^2\right ) \]

[In]

Integrate[(1 + x^4)/(x^3 + x^5),x]

[Out]

-1/2*1/x^2 - Log[x] + Log[1 + x^2]

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {1}{2 x^{2}}-\ln \left (x \right )+\ln \left (x^{2}+1\right )\)	\(17\)
norman	\(-\frac {1}{2 x^{2}}-\ln \left (x \right )+\ln \left (x^{2}+1\right )\)	\(17\)
meijerg	\(-\frac {1}{2 x^{2}}-\ln \left (x \right )+\ln \left (x^{2}+1\right )\)	\(17\)
risch	\(-\frac {1}{2 x^{2}}-\ln \left (x \right )+\ln \left (x^{2}+1\right )\)	\(17\)
parallelrisch	\(-\frac {2 \ln \left (x \right ) x^{2}-2 \ln \left (x^{2}+1\right ) x^{2}+1}{2 x^{2}}\)	\(26\)

[In]

int((x^4+1)/(x^5+x^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {1+x^4}{x^3+x^5} \, dx=\frac {2 \, x^{2} \log \left (x^{2} + 1\right ) - 2 \, x^{2} \log \left (x\right ) - 1}{2 \, x^{2}} \]

[In]

integrate((x^4+1)/(x^5+x^3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1+x^4}{x^3+x^5} \, dx=- \log {\left (x \right )} + \log {\left (x^{2} + 1 \right )} - \frac {1}{2 x^{2}} \]

[In]

integrate((x**4+1)/(x**5+x**3),x)

[Out]

-log(x) + log(x**2 + 1) - 1/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1+x^4}{x^3+x^5} \, dx=-\frac {1}{2 \, x^{2}} + \log \left (x^{2} + 1\right ) - \log \left (x\right ) \]

[In]

integrate((x^4+1)/(x^5+x^3),x, algorithm="maxima")

[Out]

-1/2/x^2 + log(x^2 + 1) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {1+x^4}{x^3+x^5} \, dx=\frac {x^{2} - 1}{2 \, x^{2}} + \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x^{2}\right ) \]

[In]

integrate((x^4+1)/(x^5+x^3),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1+x^4}{x^3+x^5} \, dx=\ln \left (x^2+1\right )-\ln \left (x\right )-\frac {1}{2\,x^2} \]

[In]

int((x^4 + 1)/(x^3 + x^5),x)

[Out]

log(x^2 + 1) - log(x) - 1/(2*x^2)