\(\int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx\) [375]

Optimal result

Integrand size = 20, antiderivative size = 42 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=19 x+\frac {3 x^2}{2}+\frac {x^3}{3}+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (2+x) \]

[Out]

19*x+3/2*x^2+1/3*x^3+3126/35*ln(5-x)-1/10*ln(x)-31/14*ln(2+x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1608, 1642} \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {x^3}{3}+\frac {3 x^2}{2}+19 x+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (x+2) \]

[In]

Int[(1 + x^5)/(-10*x - 3*x^2 + x^3),x]

[Out]

19*x + (3*x^2)/2 + x^3/3 + (3126*Log[5 - x])/35 - Log[x]/10 - (31*Log[2 + x])/14

Rule 1608

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

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1+x^5}{x \left (-10-3 x+x^2\right )} \, dx \\ & = \int \left (19+\frac {3126}{35 (-5+x)}-\frac {1}{10 x}+3 x+x^2-\frac {31}{14 (2+x)}\right ) \, dx \\ & = 19 x+\frac {3 x^2}{2}+\frac {x^3}{3}+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=19 x+\frac {3 x^2}{2}+\frac {x^3}{3}+\frac {3126}{35} \log (5-x)-\frac {\log (x)}{10}-\frac {31}{14} \log (2+x) \]

[In]

Integrate[(1 + x^5)/(-10*x - 3*x^2 + x^3),x]

[Out]

19*x + (3*x^2)/2 + x^3/3 + (3126*Log[5 - x])/35 - Log[x]/10 - (31*Log[2 + x])/14

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\)	\(31\)
norman	\(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\)	\(31\)
risch	\(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\)	\(31\)
parallelrisch	\(\frac {x^{3}}{3}+\frac {3 x^{2}}{2}+19 x -\frac {\ln \left (x \right )}{10}+\frac {3126 \ln \left (-5+x \right )}{35}-\frac {31 \ln \left (x +2\right )}{14}\)	\(31\)

[In]

int((x^5+1)/(x^3-3*x^2-10*x),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3+3/2*x^2+19*x-1/10*ln(x)+3126/35*ln(-5+x)-31/14*ln(x+2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {1}{3} \, x^{3} + \frac {3}{2} \, x^{2} + 19 \, x - \frac {31}{14} \, \log \left (x + 2\right ) + \frac {3126}{35} \, \log \left (x - 5\right ) - \frac {1}{10} \, \log \left (x\right ) \]

[In]

integrate((x^5+1)/(x^3-3*x^2-10*x),x, algorithm="fricas")

[Out]

1/3*x^3 + 3/2*x^2 + 19*x - 31/14*log(x + 2) + 3126/35*log(x - 5) - 1/10*log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {x^{3}}{3} + \frac {3 x^{2}}{2} + 19 x - \frac {\log {\left (x \right )}}{10} + \frac {3126 \log {\left (x - 5 \right )}}{35} - \frac {31 \log {\left (x + 2 \right )}}{14} \]

[In]

integrate((x**5+1)/(x**3-3*x**2-10*x),x)

[Out]

x**3/3 + 3*x**2/2 + 19*x - log(x)/10 + 3126*log(x - 5)/35 - 31*log(x + 2)/14

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {1}{3} \, x^{3} + \frac {3}{2} \, x^{2} + 19 \, x - \frac {31}{14} \, \log \left (x + 2\right ) + \frac {3126}{35} \, \log \left (x - 5\right ) - \frac {1}{10} \, \log \left (x\right ) \]

[In]

integrate((x^5+1)/(x^3-3*x^2-10*x),x, algorithm="maxima")

[Out]

1/3*x^3 + 3/2*x^2 + 19*x - 31/14*log(x + 2) + 3126/35*log(x - 5) - 1/10*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=\frac {1}{3} \, x^{3} + \frac {3}{2} \, x^{2} + 19 \, x - \frac {31}{14} \, \log \left ({\left | x + 2 \right |}\right ) + \frac {3126}{35} \, \log \left ({\left | x - 5 \right |}\right ) - \frac {1}{10} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((x^5+1)/(x^3-3*x^2-10*x),x, algorithm="giac")

[Out]

1/3*x^3 + 3/2*x^2 + 19*x - 31/14*log(abs(x + 2)) + 3126/35*log(abs(x - 5)) - 1/10*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {1+x^5}{-10 x-3 x^2+x^3} \, dx=19\,x-\frac {31\,\ln \left (x+2\right )}{14}+\frac {3126\,\ln \left (x-5\right )}{35}-\frac {\ln \left (x\right )}{10}+\frac {3\,x^2}{2}+\frac {x^3}{3} \]

[In]

int(-(x^5 + 1)/(10*x + 3*x^2 - x^3),x)

[Out]

19*x - (31*log(x + 2))/14 + (3126*log(x - 5))/35 - log(x)/10 + (3*x^2)/2 + x^3/3