\(\int \frac {1}{x^3 (2+13 x+15 x^2)} \, dx\) [2243]

Optimal result

Integrand size = 16, antiderivative size = 41 \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{4 x^2}+\frac {13}{4 x}+\frac {139 \log (x)}{8}+\frac {27}{56} \log (2+3 x)-\frac {125}{7} \log (1+5 x) \]

[Out]

-1/4/x^2+13/4/x+139/8*ln(x)+27/56*ln(2+3*x)-125/7*ln(1+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, 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} \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{4 x^2}+\frac {13}{4 x}+\frac {139 \log (x)}{8}+\frac {27}{56} \log (3 x+2)-\frac {125}{7} \log (5 x+1) \]

[In]

Int[1/(x^3*(2 + 13*x + 15*x^2)),x]

[Out]

-1/4*1/x^2 + 13/(4*x) + (139*Log[x])/8 + (27*Log[2 + 3*x])/56 - (125*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}{4 x^2}+\frac {1}{2} \int \frac {-13-15 x}{x^2 \left (2+13 x+15 x^2\right )} \, dx \\ & = -\frac {1}{4 x^2}+\frac {1}{2} \int \left (-\frac {13}{2 x^2}+\frac {139}{4 x}+\frac {81}{28 (2+3 x)}-\frac {1250}{7 (1+5 x)}\right ) \, dx \\ & = -\frac {1}{4 x^2}+\frac {13}{4 x}+\frac {139 \log (x)}{8}+\frac {27}{56} \log (2+3 x)-\frac {125}{7} \log (1+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1}{4 x^2}+\frac {13}{4 x}+\frac {139 \log (x)}{8}+\frac {27}{56} \log (2+3 x)-\frac {125}{7} \log (1+5 x) \]

[In]

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

[Out]

-1/4*1/x^2 + 13/(4*x) + (139*Log[x])/8 + (27*Log[2 + 3*x])/56 - (125*Log[1 + 5*x])/7

Maple [A] (verified)

Time = 25.47 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76

method	result	size
norman	\(\frac {-\frac {1}{4}+\frac {13 x}{4}}{x^{2}}+\frac {139 \ln \left (x \right )}{8}+\frac {27 \ln \left (2+3 x \right )}{56}-\frac {125 \ln \left (1+5 x \right )}{7}\)	\(31\)
risch	\(\frac {-\frac {1}{4}+\frac {13 x}{4}}{x^{2}}+\frac {139 \ln \left (x \right )}{8}+\frac {27 \ln \left (2+3 x \right )}{56}-\frac {125 \ln \left (1+5 x \right )}{7}\)	\(31\)
default	\(-\frac {1}{4 x^{2}}+\frac {13}{4 x}+\frac {139 \ln \left (x \right )}{8}+\frac {27 \ln \left (2+3 x \right )}{56}-\frac {125 \ln \left (1+5 x \right )}{7}\)	\(32\)
parallelrisch	\(\frac {973 \ln \left (x \right ) x^{2}+27 \ln \left (\frac {2}{3}+x \right ) x^{2}-1000 \ln \left (x +\frac {1}{5}\right ) x^{2}-14+182 x}{56 x^{2}}\)	\(36\)

[In]

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

[Out]

(-1/4+13/4*x)/x^2+139/8*ln(x)+27/56*ln(2+3*x)-125/7*ln(1+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=-\frac {1000 \, x^{2} \log \left (5 \, x + 1\right ) - 27 \, x^{2} \log \left (3 \, x + 2\right ) - 973 \, x^{2} \log \left (x\right ) - 182 \, x + 14}{56 \, x^{2}} \]

[In]

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

[Out]

-1/56*(1000*x^2*log(5*x + 1) - 27*x^2*log(3*x + 2) - 973*x^2*log(x) - 182*x + 14)/x^2

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=\frac {139 \log {\left (x \right )}}{8} - \frac {125 \log {\left (x + \frac {1}{5} \right )}}{7} + \frac {27 \log {\left (x + \frac {2}{3} \right )}}{56} + \frac {13 x - 1}{4 x^{2}} \]

[In]

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

[Out]

139*log(x)/8 - 125*log(x + 1/5)/7 + 27*log(x + 2/3)/56 + (13*x - 1)/(4*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=\frac {13 \, x - 1}{4 \, x^{2}} - \frac {125}{7} \, \log \left (5 \, x + 1\right ) + \frac {27}{56} \, \log \left (3 \, x + 2\right ) + \frac {139}{8} \, \log \left (x\right ) \]

[In]

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

[Out]

1/4*(13*x - 1)/x^2 - 125/7*log(5*x + 1) + 27/56*log(3*x + 2) + 139/8*log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=\frac {13 \, x - 1}{4 \, x^{2}} - \frac {125}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {27}{56} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {139}{8} \, \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/4*(13*x - 1)/x^2 - 125/7*log(abs(5*x + 1)) + 27/56*log(abs(3*x + 2)) + 139/8*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x^3 \left (2+13 x+15 x^2\right )} \, dx=\frac {27\,\ln \left (x+\frac {2}{3}\right )}{56}-\frac {125\,\ln \left (x+\frac {1}{5}\right )}{7}+\frac {139\,\ln \left (x\right )}{8}+\frac {\frac {13\,x}{4}-\frac {1}{4}}{x^2} \]

[In]

int(1/(x^3*(13*x + 15*x^2 + 2)),x)

[Out]

(27*log(x + 2/3))/56 - (125*log(x + 1/5))/7 + (139*log(x))/8 + ((13*x)/4 - 1/4)/x^2