\(\int \frac {1}{2+13 x+15 x^2} \, dx\) [2240]

Optimal result

Integrand size = 12, antiderivative size = 21 \[ \int \frac {1}{2+13 x+15 x^2} \, dx=-\frac {1}{7} \log (2+3 x)+\frac {1}{7} \log (1+5 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {630, 31} \[ \int \frac {1}{2+13 x+15 x^2} \, dx=\frac {1}{7} \log (5 x+1)-\frac {1}{7} \log (3 x+2) \]

[In]

Int[(2 + 13*x + 15*x^2)^(-1),x]

[Out]

-1/7*Log[2 + 3*x] + Log[1 + 5*x]/7

Rule 31

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

Rule 630

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2+13 x+15 x^2} \, dx=-\frac {1}{7} \log (2+3 x)+\frac {1}{7} \log (1+5 x) \]

[In]

Integrate[(2 + 13*x + 15*x^2)^(-1),x]

[Out]

-1/7*Log[2 + 3*x] + Log[1 + 5*x]/7

Maple [A] (verified)

Time = 25.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{2+13 x+15 x^2} \, dx=\frac {1}{7} \, \log \left (5 \, x + 1\right ) - \frac {1}{7} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/7*log(5*x + 1) - 1/7*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{2+13 x+15 x^2} \, dx=\frac {\log {\left (x + \frac {1}{5} \right )}}{7} - \frac {\log {\left (x + \frac {2}{3} \right )}}{7} \]

[In]

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

[Out]

log(x + 1/5)/7 - log(x + 2/3)/7

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{2+13 x+15 x^2} \, dx=\frac {1}{7} \, \log \left (5 \, x + 1\right ) - \frac {1}{7} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/7*log(5*x + 1) - 1/7*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{2+13 x+15 x^2} \, dx=\frac {1}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {1}{7} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38 \[ \int \frac {1}{2+13 x+15 x^2} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {30\,x}{7}+\frac {13}{7}\right )}{7} \]

[In]

int(1/(13*x + 15*x^2 + 2),x)

[Out]

-(2*atanh((30*x)/7 + 13/7))/7