\(\int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx\) [2247]

Optimal result

Integrand size = 20, antiderivative size = 26 \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x) \]

[Out]

1/15*x-4/63*ln(2+3*x)+1/175*ln(1+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1599, 717, 646, 31} \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {x}{15}-\frac {4}{63} \log (3 x+2)+\frac {1}{175} \log (5 x+1) \]

[In]

Int[x^3/(2*x + 13*x^2 + 15*x^3),x]

[Out]

x/15 - (4*Log[2 + 3*x])/63 + Log[1 + 5*x]/175

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 717

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

Rule 1599

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{2+13 x+15 x^2} \, dx \\ & = \frac {x}{15}+\frac {1}{15} \int \frac {-2-13 x}{2+13 x+15 x^2} \, dx \\ & = \frac {x}{15}+\frac {3}{35} \int \frac {1}{3+15 x} \, dx-\frac {20}{21} \int \frac {1}{10+15 x} \, dx \\ & = \frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x) \]

[In]

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

[Out]

x/15 - (4*Log[2 + 3*x])/63 + Log[1 + 5*x]/175

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(\frac {x}{15}-\frac {4 \ln \left (\frac {2}{3}+x \right )}{63}+\frac {\ln \left (x +\frac {1}{5}\right )}{175}\)	\(17\)
default	\(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\)	\(21\)
norman	\(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\)	\(21\)
risch	\(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\)	\(21\)

[In]

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

[Out]

1/15*x-4/63*ln(2/3+x)+1/175*ln(x+1/5)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {1}{15} \, x + \frac {1}{175} \, \log \left (5 \, x + 1\right ) - \frac {4}{63} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/15*x + 1/175*log(5*x + 1) - 4/63*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {x}{15} + \frac {\log {\left (x + \frac {1}{5} \right )}}{175} - \frac {4 \log {\left (x + \frac {2}{3} \right )}}{63} \]

[In]

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

[Out]

x/15 + log(x + 1/5)/175 - 4*log(x + 2/3)/63

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {1}{15} \, x + \frac {1}{175} \, \log \left (5 \, x + 1\right ) - \frac {4}{63} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/15*x + 1/175*log(5*x + 1) - 4/63*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {1}{15} \, x + \frac {1}{175} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {4}{63} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

1/15*x + 1/175*log(abs(5*x + 1)) - 4/63*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {x^3}{2 x+13 x^2+15 x^3} \, dx=\frac {x}{15}-\frac {4\,\ln \left (x+\frac {2}{3}\right )}{63}+\frac {\ln \left (x+\frac {1}{5}\right )}{175} \]

[In]

int(x^3/(2*x + 13*x^2 + 15*x^3),x)

[Out]

x/15 - (4*log(x + 2/3))/63 + log(x + 1/5)/175