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

Optimal. Leaf size=18 \[ -\frac {1}{2 x^2}+\log \left (x^2+1\right )-\log (x) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1593, 1252, 894} \[ -\frac {1}{2 x^2}+\log \left (x^2+1\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)/(x^3 + x^5),x]

[Out]

-1/(2*x^2) - Log[x] + Log[1 + x^2]

Rule 894

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 1252

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 1593

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*} \int \frac {1+x^4}{x^3+x^5} \, dx &=\int \frac {1+x^4}{x^3 \left (1+x^2\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+x^2}{x^2 (1+x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {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*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.00 \[ -\frac {1}{2 x^2}+\log \left (x^2+1\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)/(x^3 + x^5),x]

[Out]

-1/2*1/x^2 - Log[x] + Log[1 + x^2]

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fricas [A]  time = 0.67, size = 25, normalized size = 1.39 \[ \frac {2 \, x^{2} \log \left (x^{2} + 1\right ) - 2 \, x^{2} \log \relax (x) - 1}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.27, size = 23, normalized size = 1.28 \[ \frac {x^{2} - 1}{2 \, x^{2}} + \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 17, normalized size = 0.94 \[ -\ln \relax (x )+\ln \left (x^{2}+1\right )-\frac {1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.92, size = 16, normalized size = 0.89 \[ -\frac {1}{2 \, x^{2}} + \log \left (x^{2} + 1\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2 + log(x^2 + 1) - log(x)

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mupad [B]  time = 0.05, size = 16, normalized size = 0.89 \[ \ln \left (x^2+1\right )-\ln \relax (x)-\frac {1}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)/(x^3 + x^5),x)

[Out]

log(x^2 + 1) - log(x) - 1/(2*x^2)

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sympy [A]  time = 0.11, size = 15, normalized size = 0.83 \[ - \log {\relax (x )} + \log {\left (x^{2} + 1 \right )} - \frac {1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x**2 + 1) - 1/(2*x**2)

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