\(\int \frac {1}{2 x+13 x^2+15 x^3} \, dx\) [2250]

Optimal result

Integrand size = 16, antiderivative size = 27 \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=\frac {\log (x)}{2}+\frac {3}{14} \log (2+3 x)-\frac {5}{7} \log (1+5 x) \]

[Out]

1/2*ln(x)+3/14*ln(2+3*x)-5/7*ln(1+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1608, 719, 29, 646, 31} \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=\frac {\log (x)}{2}+\frac {3}{14} \log (3 x+2)-\frac {5}{7} \log (5 x+1) \]

[In]

Int[(2*x + 13*x^2 + 15*x^3)^(-1),x]

[Out]

Log[x]/2 + (3*Log[2 + 3*x])/14 - (5*Log[1 + 5*x])/7

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 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 719

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 1608

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx \\ & = \frac {1}{2} \int \frac {1}{x} \, dx+\frac {1}{2} \int \frac {-13-15 x}{2+13 x+15 x^2} \, dx \\ & = \frac {\log (x)}{2}+\frac {45}{14} \int \frac {1}{10+15 x} \, dx-\frac {75}{7} \int \frac {1}{3+15 x} \, dx \\ & = \frac {\log (x)}{2}+\frac {3}{14} \log (2+3 x)-\frac {5}{7} \log (1+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=\frac {\log (x)}{2}+\frac {3}{14} \log (2+3 x)-\frac {5}{7} \log (1+5 x) \]

[In]

Integrate[(2*x + 13*x^2 + 15*x^3)^(-1),x]

[Out]

Log[x]/2 + (3*Log[2 + 3*x])/14 - (5*Log[1 + 5*x])/7

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {\ln \left (x \right )}{2}-\frac {5 \ln \left (x +\frac {1}{5}\right )}{7}+\frac {3 \ln \left (x +\frac {2}{3}\right )}{14}\)	\(18\)
default	\(\frac {\ln \left (x \right )}{2}+\frac {3 \ln \left (2+3 x \right )}{14}-\frac {5 \ln \left (1+5 x \right )}{7}\)	\(22\)
norman	\(\frac {\ln \left (x \right )}{2}+\frac {3 \ln \left (2+3 x \right )}{14}-\frac {5 \ln \left (1+5 x \right )}{7}\)	\(22\)
risch	\(\frac {\ln \left (x \right )}{2}+\frac {3 \ln \left (2+3 x \right )}{14}-\frac {5 \ln \left (1+5 x \right )}{7}\)	\(22\)

[In]

int(1/(15*x^3+13*x^2+2*x),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x)-5/7*ln(x+1/5)+3/14*ln(x+2/3)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=-\frac {5}{7} \, \log \left (5 \, x + 1\right ) + \frac {3}{14} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate(1/(15*x^3+13*x^2+2*x),x, algorithm="fricas")

[Out]

-5/7*log(5*x + 1) + 3/14*log(3*x + 2) + 1/2*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=\frac {\log {\left (x \right )}}{2} - \frac {5 \log {\left (x + \frac {1}{5} \right )}}{7} + \frac {3 \log {\left (x + \frac {2}{3} \right )}}{14} \]

[In]

integrate(1/(15*x**3+13*x**2+2*x),x)

[Out]

log(x)/2 - 5*log(x + 1/5)/7 + 3*log(x + 2/3)/14

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=-\frac {5}{7} \, \log \left (5 \, x + 1\right ) + \frac {3}{14} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate(1/(15*x^3+13*x^2+2*x),x, algorithm="maxima")

[Out]

-5/7*log(5*x + 1) + 3/14*log(3*x + 2) + 1/2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=-\frac {5}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {3}{14} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/(15*x^3+13*x^2+2*x),x, algorithm="giac")

[Out]

-5/7*log(abs(5*x + 1)) + 3/14*log(abs(3*x + 2)) + 1/2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {1}{2 x+13 x^2+15 x^3} \, dx=\frac {3\,\ln \left (x+\frac {2}{3}\right )}{14}-\frac {5\,\ln \left (x+\frac {1}{5}\right )}{7}+\frac {\ln \left (x\right )}{2} \]

[In]

int(1/(2*x + 13*x^2 + 15*x^3),x)

[Out]

(3*log(x + 2/3))/14 - (5*log(x + 1/5))/7 + log(x)/2