\(\int \frac {1}{2 x+3 x^{1+n}} \, dx\) [358]

Optimal result

Integrand size = 13, antiderivative size = 22 \[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=\frac {\log (x)}{2}-\frac {\log \left (2+3 x^n\right )}{2 n} \]

[Out]

1/2*ln(x)-1/2*ln(2+3*x^n)/n

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1607, 272, 36, 29, 31} \[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=\frac {\log (x)}{2}-\frac {\log \left (3 x^n+2\right )}{2 n} \]

[In]

Int[(2*x + 3*x^(1 + n))^(-1),x]

[Out]

Log[x]/2 - Log[2 + 3*x^n]/(2*n)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 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 \left (2+3 x^n\right )} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {1}{x (2+3 x)} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,x^n\right )}{2 n}-\frac {3 \text {Subst}\left (\int \frac {1}{2+3 x} \, dx,x,x^n\right )}{2 n} \\ & = \frac {\log (x)}{2}-\frac {\log \left (2+3 x^n\right )}{2 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=\frac {\log \left (x^n\right )-\log \left (n \left (2+3 x^n\right )\right )}{2 n} \]

[In]

Integrate[(2*x + 3*x^(1 + n))^(-1),x]

[Out]

(Log[x^n] - Log[n*(2 + 3*x^n)])/(2*n)

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {n \ln \left (x \right )+\ln \left (x \right )-\ln \left (x +\frac {3 x^{1+n}}{2}\right )}{2 n}\)	\(25\)
meijerg	\(\frac {n \ln \left (x \right )+\ln \left (3\right )-\ln \left (2\right )-\ln \left (1+\frac {3 x^{n}}{2}\right )}{2 n}\)	\(27\)
risch	\(\frac {\ln \left (x \right )}{2 n}+\frac {\ln \left (x \right )}{2}-\frac {\ln \left (\frac {2 x}{3}+x^{1+n}\right )}{2 n}\)	\(28\)
norman	\(\frac {\left (1+n \right ) \ln \left (x \right )}{2 n}-\frac {\ln \left (2 x +3 \,{\mathrm e}^{\left (1+n \right ) \ln \left (x \right )}\right )}{2 n}\)	\(31\)

[In]

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

[Out]

1/2*(n*ln(x)+ln(x)-ln(x+3/2*x^(1+n)))/n

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=\frac {{\left (n + 1\right )} \log \left (x\right ) - \log \left (3 \, x^{n + 1} + 2 \, x\right )}{2 \, n} \]

[In]

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

[Out]

1/2*((n + 1)*log(x) - log(3*x^(n + 1) + 2*x))/n

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (15) = 30\).

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{2} + \frac {\log {\left (x \right )}}{2 n} - \frac {\log {\left (2 x + 3 x^{n + 1} \right )}}{2 n} & \text {for}\: n \neq 0 \\\frac {\log {\left (x \right )}}{5} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(x)/2 + log(x)/(2*n) - log(2*x + 3*x**(n + 1))/(2*n), Ne(n, 0)), (log(x)/5, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=-\frac {\log \left (x^{n} + \frac {2}{3}\right )}{2 \, n} + \frac {1}{2} \, \log \left (x\right ) \]

[In]

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

[Out]

-1/2*log(x^n + 2/3)/n + 1/2*log(x)

Giac [F]

\[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=\int { \frac {1}{3 \, x^{n + 1} + 2 \, x} \,d x } \]

[In]

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

[Out]

integrate(1/(3*x^(n + 1) + 2*x), x)

Mupad [B] (verification not implemented)

Time = 9.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{2 x+3 x^{1+n}} \, dx=\frac {\ln \left (x\right )\,\left (n+1\right )}{2\,n}-\frac {\ln \left (\frac {2\,x}{3}+x^{n+1}\right )}{2\,n} \]

[In]

int(1/(2*x + 3*x^(n + 1)),x)

[Out]

(log(x)*(n + 1))/(2*n) - log((2*x)/3 + x^(n + 1))/(2*n)