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

Optimal. Leaf size=27 \[ x-\left (-4+\frac {3}{2 x}-\frac {x}{5}\right ) \left (-5+x-\frac {4}{\log (5)}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {12, 14} \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 36, normalized size = 1.33 \begin {gather*} \frac {15}{2 x}+4 x+\frac {x^2}{5}+\frac {6}{x \log (5)}-\frac {4 x}{5 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[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])

________________________________________________________________________________________

fricas [A]  time = 0.78, size = 32, normalized size = 1.19 \begin {gather*} -\frac {8 \, x^{2} - {\left (2 \, x^{3} + 40 \, x^{2} + 75\right )} \log \relax (5) - 60}{10 \, x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [A]  time = 0.12, size = 33, normalized size = 1.22 \begin {gather*} \frac {2 \, x^{2} \log \relax (5) + 40 \, x \log \relax (5) - 8 \, x + \frac {15 \, {\left (5 \, \log \relax (5) + 4\right )}}{x}}{10 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 31, normalized size = 1.15

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 34, normalized size = 1.26 \begin {gather*} \frac {2 \, x^{2} \log \relax (5) + 8 \, x {\left (5 \, \log \relax (5) - 1\right )} + \frac {15 \, {\left (5 \, \log \relax (5) + 4\right )}}{x}}{10 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 5.30, size = 34, normalized size = 1.26 \begin {gather*} \frac {x^2}{5}+\frac {15\,\ln \relax (5)+12}{2\,x\,\ln \relax (5)}+\frac {x\,\left (40\,\ln \relax (5)-8\right )}{10\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 29, normalized size = 1.07 \begin {gather*} \frac {2 x^{2} \log {\relax (5 )} + x \left (-8 + 40 \log {\relax (5 )}\right ) + \frac {60 + 75 \log {\relax (5 )}}{x}}{10 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________