\(\int \frac {-80-15 x+10 x^2+5 x^5+(3 x^3-5 x^5) \log (2)}{5 x^5} \, dx\) [3950]

Optimal result

Integrand size = 36, antiderivative size = 33 \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=x-\frac {\frac {-1-\frac {4}{x}+x}{x^2}+\left (\frac {3}{5}-x+x^2\right ) \log (2)}{x} \]

[Out]

x-((x-1-4/x)/x^2+ln(2)*(3/5+x^2-x))/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 14} \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=\frac {4}{x^4}+\frac {1}{x^3}-\frac {1}{x^2}+x (1-\log (2))-\frac {\log (8)}{5 x} \]

[In]

Int[(-80 - 15*x + 10*x^2 + 5*x^5 + (3*x^3 - 5*x^5)*Log[2])/(5*x^5),x]

[Out]

4/x^4 + x^(-3) - x^(-2) + x*(1 - Log[2]) - Log[8]/(5*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{x^5} \, dx \\ & = \frac {1}{5} \int \left (-\frac {80}{x^5}-\frac {15}{x^4}+\frac {10}{x^3}-5 (-1+\log (2))+\frac {\log (8)}{x^2}\right ) \, dx \\ & = \frac {4}{x^4}+\frac {1}{x^3}-\frac {1}{x^2}+x (1-\log (2))-\frac {\log (8)}{5 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=\frac {4}{x^4}+\frac {1}{x^3}-\frac {1}{x^2}+x-x \log (2)-\frac {\log (8)}{5 x} \]

[In]

Integrate[(-80 - 15*x + 10*x^2 + 5*x^5 + (3*x^3 - 5*x^5)*Log[2])/(5*x^5),x]

[Out]

4/x^4 + x^(-3) - x^(-2) + x - x*Log[2] - Log[8]/(5*x)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85

method	result	size
default	\(-x \ln \left (2\right )+x +\frac {4}{x^{4}}-\frac {3 \ln \left (2\right )}{5 x}-\frac {1}{x^{2}}+\frac {1}{x^{3}}\)	\(28\)
norman	\(\frac {4+x +\left (1-\ln \left (2\right )\right ) x^{5}-x^{2}-\frac {3 x^{3} \ln \left (2\right )}{5}}{x^{4}}\)	\(30\)
risch	\(-x \ln \left (2\right )+x +\frac {-3 x^{3} \ln \left (2\right )-5 x^{2}+5 x +20}{5 x^{4}}\)	\(30\)
gosper	\(-\frac {5 x^{5} \ln \left (2\right )-5 x^{5}+3 x^{3} \ln \left (2\right )+5 x^{2}-5 x -20}{5 x^{4}}\)	\(35\)
parallelrisch	\(-\frac {5 x^{5} \ln \left (2\right )-5 x^{5}+3 x^{3} \ln \left (2\right )+5 x^{2}-5 x -20}{5 x^{4}}\)	\(35\)

[In]

int(1/5*((-5*x^5+3*x^3)*ln(2)+5*x^5+10*x^2-15*x-80)/x^5,x,method=_RETURNVERBOSE)

[Out]

-x*ln(2)+x+4/x^4-3/5*ln(2)/x-1/x^2+1/x^3

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=\frac {5 \, x^{5} - 5 \, x^{2} - {\left (5 \, x^{5} + 3 \, x^{3}\right )} \log \left (2\right ) + 5 \, x + 20}{5 \, x^{4}} \]

[In]

integrate(1/5*((-5*x^5+3*x^3)*log(2)+5*x^5+10*x^2-15*x-80)/x^5,x, algorithm="fricas")

[Out]

1/5*(5*x^5 - 5*x^2 - (5*x^5 + 3*x^3)*log(2) + 5*x + 20)/x^4

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=\frac {x \left (5 - 5 \log {\left (2 \right )}\right )}{5} + \frac {- 3 x^{3} \log {\left (2 \right )} - 5 x^{2} + 5 x + 20}{5 x^{4}} \]

[In]

integrate(1/5*((-5*x**5+3*x**3)*ln(2)+5*x**5+10*x**2-15*x-80)/x**5,x)

[Out]

x*(5 - 5*log(2))/5 + (-3*x**3*log(2) - 5*x**2 + 5*x + 20)/(5*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=-x {\left (\log \left (2\right ) - 1\right )} - \frac {3 \, x^{3} \log \left (2\right ) + 5 \, x^{2} - 5 \, x - 20}{5 \, x^{4}} \]

[In]

integrate(1/5*((-5*x^5+3*x^3)*log(2)+5*x^5+10*x^2-15*x-80)/x^5,x, algorithm="maxima")

[Out]

-x*(log(2) - 1) - 1/5*(3*x^3*log(2) + 5*x^2 - 5*x - 20)/x^4

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=-x \log \left (2\right ) + x - \frac {3 \, x^{3} \log \left (2\right ) + 5 \, x^{2} - 5 \, x - 20}{5 \, x^{4}} \]

[In]

integrate(1/5*((-5*x^5+3*x^3)*log(2)+5*x^5+10*x^2-15*x-80)/x^5,x, algorithm="giac")

[Out]

-x*log(2) + x - 1/5*(3*x^3*log(2) + 5*x^2 - 5*x - 20)/x^4

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{5 x^5} \, dx=\frac {-\frac {3\,\ln \left (2\right )\,x^3}{5}-x^2+x+4}{x^4}-x\,\left (\frac {\ln \left (32\right )}{5}-1\right ) \]

[In]

int(((log(2)*(3*x^3 - 5*x^5))/5 - 3*x + 2*x^2 + x^5 - 16)/x^5,x)

[Out]

(x - (3*x^3*log(2))/5 - x^2 + 4)/x^4 - x*(log(32)/5 - 1)