\(\int \frac {x^3}{2+13 x+15 x^2} \, dx\) [2237]

Optimal result

Integrand size = 16, antiderivative size = 33 \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=-\frac {13 x}{225}+\frac {x^2}{30}+\frac {8}{189} \log (2+3 x)-\frac {1}{875} \log (1+5 x) \]

[Out]

-13/225*x+1/30*x^2+8/189*ln(2+3*x)-1/875*ln(1+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {715, 646, 31} \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=\frac {x^2}{30}-\frac {13 x}{225}+\frac {8}{189} \log (3 x+2)-\frac {1}{875} \log (5 x+1) \]

[In]

Int[x^3/(2 + 13*x + 15*x^2),x]

[Out]

(-13*x)/225 + x^2/30 + (8*Log[2 + 3*x])/189 - Log[1 + 5*x]/875

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 715

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, 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] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {13}{225}+\frac {x}{15}+\frac {26+139 x}{225 \left (2+13 x+15 x^2\right )}\right ) \, dx \\ & = -\frac {13 x}{225}+\frac {x^2}{30}+\frac {1}{225} \int \frac {26+139 x}{2+13 x+15 x^2} \, dx \\ & = -\frac {13 x}{225}+\frac {x^2}{30}-\frac {3}{175} \int \frac {1}{3+15 x} \, dx+\frac {40}{63} \int \frac {1}{10+15 x} \, dx \\ & = -\frac {13 x}{225}+\frac {x^2}{30}+\frac {8}{189} \log (2+3 x)-\frac {1}{875} \log (1+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=-\frac {13 x}{225}+\frac {x^2}{30}+\frac {8}{189} \log (2+3 x)-\frac {1}{875} \log (1+5 x) \]

[In]

Integrate[x^3/(2 + 13*x + 15*x^2),x]

[Out]

(-13*x)/225 + x^2/30 + (8*Log[2 + 3*x])/189 - Log[1 + 5*x]/875

Maple [A] (verified)

Time = 21.92 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {x^{2}}{30}-\frac {13 x}{225}+\frac {8 \ln \left (\frac {2}{3}+x \right )}{189}-\frac {\ln \left (x +\frac {1}{5}\right )}{875}\)	\(22\)
default	\(-\frac {13 x}{225}+\frac {x^{2}}{30}+\frac {8 \ln \left (2+3 x \right )}{189}-\frac {\ln \left (1+5 x \right )}{875}\)	\(26\)
norman	\(-\frac {13 x}{225}+\frac {x^{2}}{30}+\frac {8 \ln \left (2+3 x \right )}{189}-\frac {\ln \left (1+5 x \right )}{875}\)	\(26\)
risch	\(-\frac {13 x}{225}+\frac {x^{2}}{30}+\frac {8 \ln \left (2+3 x \right )}{189}-\frac {\ln \left (1+5 x \right )}{875}\)	\(26\)

[In]

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

[Out]

1/30*x^2-13/225*x+8/189*ln(2/3+x)-1/875*ln(x+1/5)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=\frac {1}{30} \, x^{2} - \frac {13}{225} \, x - \frac {1}{875} \, \log \left (5 \, x + 1\right ) + \frac {8}{189} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/30*x^2 - 13/225*x - 1/875*log(5*x + 1) + 8/189*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=\frac {x^{2}}{30} - \frac {13 x}{225} - \frac {\log {\left (x + \frac {1}{5} \right )}}{875} + \frac {8 \log {\left (x + \frac {2}{3} \right )}}{189} \]

[In]

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

[Out]

x**2/30 - 13*x/225 - log(x + 1/5)/875 + 8*log(x + 2/3)/189

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=\frac {1}{30} \, x^{2} - \frac {13}{225} \, x - \frac {1}{875} \, \log \left (5 \, x + 1\right ) + \frac {8}{189} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/30*x^2 - 13/225*x - 1/875*log(5*x + 1) + 8/189*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=\frac {1}{30} \, x^{2} - \frac {13}{225} \, x - \frac {1}{875} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {8}{189} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

1/30*x^2 - 13/225*x - 1/875*log(abs(5*x + 1)) + 8/189*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {x^3}{2+13 x+15 x^2} \, dx=\frac {8\,\ln \left (x+\frac {2}{3}\right )}{189}-\frac {13\,x}{225}-\frac {\ln \left (x+\frac {1}{5}\right )}{875}+\frac {x^2}{30} \]

[In]

int(x^3/(13*x + 15*x^2 + 2),x)

[Out]

(8*log(x + 2/3))/189 - (13*x)/225 - log(x + 1/5)/875 + x^2/30