\(\int \frac {-60-8 x^2+(-75+40 x^2+4 x^3) \log (5)}{10 x^2 \log (5)} \, dx\) [7976]

Optimal result

Integrand size = 33, antiderivative size = 27 \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=x-\left (-4+\frac {3}{2 x}-\frac {x}{5}\right ) \left (-5+x-\frac {4}{\log (5)}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {12, 14} \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=\frac {x^2}{5}-\frac {4 x (1-5 \log (5))}{5 \log (5)}+\frac {3 (4+5 \log (5))}{2 x \log (5)} \]

[In]

Int[(-60 - 8*x^2 + (-75 + 40*x^2 + 4*x^3)*Log[5])/(10*x^2*Log[5]),x]

[Out]

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

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 {\int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{x^2} \, dx}{10 \log (5)} \\ & = \frac {\int \left (4 x \log (5)+8 (-1+5 \log (5))-\frac {15 (4+5 \log (5))}{x^2}\right ) \, dx}{10 \log (5)} \\ & = \frac {x^2}{5}-\frac {4 x (1-5 \log (5))}{5 \log (5)}+\frac {3 (4+5 \log (5))}{2 x \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=\frac {15}{2 x}+4 x+\frac {x^2}{5}+\frac {6}{x \log (5)}-\frac {4 x}{5 \log (5)} \]

[In]

Integrate[(-60 - 8*x^2 + (-75 + 40*x^2 + 4*x^3)*Log[5])/(10*x^2*Log[5]),x]

[Out]

15/(2*x) + 4*x + x^2/5 + 6/(x*Log[5]) - (4*x)/(5*Log[5])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {x^{2}}{5}+4 x -\frac {4 x}{5 \ln \left (5\right )}+\frac {15}{2 x}+\frac {6}{x \ln \left (5\right )}\)	\(31\)
default	\(\frac {2 x^{2} \ln \left (5\right )+40 x \ln \left (5\right )-8 x -\frac {-75 \ln \left (5\right )-60}{x}}{10 \ln \left (5\right )}\)	\(34\)
gosper	\(\frac {2 x^{3} \ln \left (5\right )+40 x^{2} \ln \left (5\right )-8 x^{2}+75 \ln \left (5\right )+60}{10 x \ln \left (5\right )}\)	\(35\)
parallelrisch	\(\frac {2 x^{3} \ln \left (5\right )+40 x^{2} \ln \left (5\right )-8 x^{2}+75 \ln \left (5\right )+60}{10 x \ln \left (5\right )}\)	\(35\)
norman	\(\frac {\frac {x^{3}}{5}+\frac {\frac {15 \ln \left (5\right )}{2}+6}{\ln \left (5\right )}+\frac {4 \left (5 \ln \left (5\right )-1\right ) x^{2}}{5 \ln \left (5\right )}}{x}\)	\(38\)

[In]

int(1/10*((4*x^3+40*x^2-75)*ln(5)-8*x^2-60)/x^2/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/5*x^2+4*x-4/5*x/ln(5)+15/2/x+6/x/ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=-\frac {8 \, x^{2} - {\left (2 \, x^{3} + 40 \, x^{2} + 75\right )} \log \left (5\right ) - 60}{10 \, x \log \left (5\right )} \]

[In]

integrate(1/10*((4*x^3+40*x^2-75)*log(5)-8*x^2-60)/x^2/log(5),x, algorithm="fricas")

[Out]

-1/10*(8*x^2 - (2*x^3 + 40*x^2 + 75)*log(5) - 60)/(x*log(5))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=\frac {2 x^{2} \log {\left (5 \right )} + x \left (-8 + 40 \log {\left (5 \right )}\right ) + \frac {60 + 75 \log {\left (5 \right )}}{x}}{10 \log {\left (5 \right )}} \]

[In]

integrate(1/10*((4*x**3+40*x**2-75)*ln(5)-8*x**2-60)/x**2/ln(5),x)

[Out]

(2*x**2*log(5) + x*(-8 + 40*log(5)) + (60 + 75*log(5))/x)/(10*log(5))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=\frac {2 \, x^{2} \log \left (5\right ) + 8 \, x {\left (5 \, \log \left (5\right ) - 1\right )} + \frac {15 \, {\left (5 \, \log \left (5\right ) + 4\right )}}{x}}{10 \, \log \left (5\right )} \]

[In]

integrate(1/10*((4*x^3+40*x^2-75)*log(5)-8*x^2-60)/x^2/log(5),x, algorithm="maxima")

[Out]

1/10*(2*x^2*log(5) + 8*x*(5*log(5) - 1) + 15*(5*log(5) + 4)/x)/log(5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=\frac {2 \, x^{2} \log \left (5\right ) + 40 \, x \log \left (5\right ) - 8 \, x + \frac {15 \, {\left (5 \, \log \left (5\right ) + 4\right )}}{x}}{10 \, \log \left (5\right )} \]

[In]

integrate(1/10*((4*x^3+40*x^2-75)*log(5)-8*x^2-60)/x^2/log(5),x, algorithm="giac")

[Out]

1/10*(2*x^2*log(5) + 40*x*log(5) - 8*x + 15*(5*log(5) + 4)/x)/log(5)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-60-8 x^2+\left (-75+40 x^2+4 x^3\right ) \log (5)}{10 x^2 \log (5)} \, dx=\frac {x^2}{5}+\frac {15\,\ln \left (5\right )+12}{2\,x\,\ln \left (5\right )}+\frac {x\,\left (40\,\ln \left (5\right )-8\right )}{10\,\ln \left (5\right )} \]

[In]

int(-((4*x^2)/5 - (log(5)*(40*x^2 + 4*x^3 - 75))/10 + 6)/(x^2*log(5)),x)

[Out]

x^2/5 + (15*log(5) + 12)/(2*x*log(5)) + (x*(40*log(5) - 8))/(10*log(5))