\(\int \frac {1}{x^2 (2+13 x+15 x^2)} \, dx\) [2242]

Optimal result

Integrand size = 16, antiderivative size = 34 \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \]

[Out]

-1/2/x-13/4*ln(x)-9/28*ln(2+3*x)+25/7*ln(1+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {723, 814} \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (3 x+2)+\frac {25}{7} \log (5 x+1) \]

[In]

Int[1/(x^2*(2 + 13*x + 15*x^2)),x]

[Out]

-1/2*1/x - (13*Log[x])/4 - (9*Log[2 + 3*x])/28 + (25*Log[1 + 5*x])/7

Rule 723

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

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2 x}+\frac {1}{2} \int \frac {-13-15 x}{x \left (2+13 x+15 x^2\right )} \, dx \\ & = -\frac {1}{2 x}+\frac {1}{2} \int \left (-\frac {13}{2 x}-\frac {27}{14 (2+3 x)}+\frac {250}{7 (1+5 x)}\right ) \, dx \\ & = -\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \]

[In]

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

[Out]

-1/2*1/x - (13*Log[x])/4 - (9*Log[2 + 3*x])/28 + (25*Log[1 + 5*x])/7

Maple [A] (verified)

Time = 25.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (2+3 x \right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\)	\(27\)
norman	\(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (2+3 x \right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\)	\(27\)
risch	\(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (2+3 x \right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\)	\(27\)
parallelrisch	\(-\frac {91 \ln \left (x \right ) x +9 \ln \left (\frac {2}{3}+x \right ) x -100 \ln \left (x +\frac {1}{5}\right ) x +14}{28 x}\)	\(27\)

[In]

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

[Out]

-1/2/x-13/4*ln(x)-9/28*ln(2+3*x)+25/7*ln(1+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=\frac {100 \, x \log \left (5 \, x + 1\right ) - 9 \, x \log \left (3 \, x + 2\right ) - 91 \, x \log \left (x\right ) - 14}{28 \, x} \]

[In]

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

[Out]

1/28*(100*x*log(5*x + 1) - 9*x*log(3*x + 2) - 91*x*log(x) - 14)/x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=- \frac {13 \log {\left (x \right )}}{4} + \frac {25 \log {\left (x + \frac {1}{5} \right )}}{7} - \frac {9 \log {\left (x + \frac {2}{3} \right )}}{28} - \frac {1}{2 x} \]

[In]

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

[Out]

-13*log(x)/4 + 25*log(x + 1/5)/7 - 9*log(x + 2/3)/28 - 1/(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{2 \, x} + \frac {25}{7} \, \log \left (5 \, x + 1\right ) - \frac {9}{28} \, \log \left (3 \, x + 2\right ) - \frac {13}{4} \, \log \left (x\right ) \]

[In]

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

[Out]

-1/2/x + 25/7*log(5*x + 1) - 9/28*log(3*x + 2) - 13/4*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{2 \, x} + \frac {25}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {9}{28} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {13}{4} \, \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

-1/2/x + 25/7*log(abs(5*x + 1)) - 9/28*log(abs(3*x + 2)) - 13/4*log(abs(x))

Mupad [B] (verification not implemented)

Time = 9.89 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx=\frac {25\,\ln \left (x+\frac {1}{5}\right )}{7}-\frac {9\,\ln \left (x+\frac {2}{3}\right )}{28}-\frac {13\,\ln \left (x\right )}{4}-\frac {1}{2\,x} \]

[In]

int(1/(x^2*(13*x + 15*x^2 + 2)),x)

[Out]

(25*log(x + 1/5))/7 - (9*log(x + 2/3))/28 - (13*log(x))/4 - 1/(2*x)