\(\int \frac {-20 x+2 x^2+(-10 x+x^2) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx\) [4183]

Optimal result

Integrand size = 31, antiderivative size = 30 \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=\frac {\frac {(5-x) (2+\log (2))}{x}+\frac {9 (3-x) \log (3)}{x^2}}{x} \]

[Out]

((5-x)/x*(ln(2)+2)+9*ln(3)/x^2*(-x+3))/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {14} \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=\frac {27 \log (3)}{x^3}+\frac {10-\log \left (\frac {19683}{32}\right )}{x^2}-\frac {2+\log (2)}{x} \]

[In]

Int[(-20*x + 2*x^2 + (-10*x + x^2)*Log[2] + (-81 + 18*x)*Log[3])/x^4,x]

[Out]

-((2 + Log[2])/x) + (27*Log[3])/x^3 + (10 - Log[19683/32])/x^2

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}& = \int \left (\frac {2+\log (2)}{x^2}-\frac {81 \log (3)}{x^4}+\frac {2 \left (-10+\log \left (\frac {19683}{32}\right )\right )}{x^3}\right ) \, dx \\ & = -\frac {2+\log (2)}{x}+\frac {27 \log (3)}{x^3}+\frac {10-\log \left (\frac {19683}{32}\right )}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=\frac {-2-\log (2)}{x}+\frac {27 \log (3)}{x^3}+\frac {10-9 \log (3)+\log (32)}{x^2} \]

[In]

Integrate[(-20*x + 2*x^2 + (-10*x + x^2)*Log[2] + (-81 + 18*x)*Log[3])/x^4,x]

[Out]

(-2 - Log[2])/x + (27*Log[3])/x^3 + (10 - 9*Log[3] + Log[32])/x^2

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07

method	result	size
norman	\(\frac {\left (-2-\ln \left (2\right )\right ) x^{2}+\left (5 \ln \left (2\right )-9 \ln \left (3\right )+10\right ) x +27 \ln \left (3\right )}{x^{3}}\)	\(32\)
risch	\(\frac {\left (-2-\ln \left (2\right )\right ) x^{2}+\left (5 \ln \left (2\right )-9 \ln \left (3\right )+10\right ) x +27 \ln \left (3\right )}{x^{3}}\)	\(32\)
default	\(-\frac {\ln \left (2\right )+2}{x}-\frac {-10 \ln \left (2\right )+18 \ln \left (3\right )-20}{2 x^{2}}+\frac {27 \ln \left (3\right )}{x^{3}}\)	\(33\)
gosper	\(-\frac {x^{2} \ln \left (2\right )-5 x \ln \left (2\right )+9 x \ln \left (3\right )+2 x^{2}-27 \ln \left (3\right )-10 x}{x^{3}}\)	\(35\)
parallelrisch	\(-\frac {x^{2} \ln \left (2\right )-5 x \ln \left (2\right )+9 x \ln \left (3\right )+2 x^{2}-27 \ln \left (3\right )-10 x}{x^{3}}\)	\(35\)

[In]

int(((18*x-81)*ln(3)+(x^2-10*x)*ln(2)+2*x^2-20*x)/x^4,x,method=_RETURNVERBOSE)

[Out]

((-2-ln(2))*x^2+(5*ln(2)-9*ln(3)+10)*x+27*ln(3))/x^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=-\frac {2 \, x^{2} + 9 \, {\left (x - 3\right )} \log \left (3\right ) + {\left (x^{2} - 5 \, x\right )} \log \left (2\right ) - 10 \, x}{x^{3}} \]

[In]

integrate(((18*x-81)*log(3)+(x^2-10*x)*log(2)+2*x^2-20*x)/x^4,x, algorithm="fricas")

[Out]

-(2*x^2 + 9*(x - 3)*log(3) + (x^2 - 5*x)*log(2) - 10*x)/x^3

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=\frac {x^{2} \left (-2 - \log {\left (2 \right )}\right ) + x \left (- 9 \log {\left (3 \right )} + 5 \log {\left (2 \right )} + 10\right ) + 27 \log {\left (3 \right )}}{x^{3}} \]

[In]

integrate(((18*x-81)*ln(3)+(x**2-10*x)*ln(2)+2*x**2-20*x)/x**4,x)

[Out]

(x**2*(-2 - log(2)) + x*(-9*log(3) + 5*log(2) + 10) + 27*log(3))/x**3

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=-\frac {x^{2} {\left (\log \left (2\right ) + 2\right )} + x {\left (9 \, \log \left (3\right ) - 5 \, \log \left (2\right ) - 10\right )} - 27 \, \log \left (3\right )}{x^{3}} \]

[In]

integrate(((18*x-81)*log(3)+(x^2-10*x)*log(2)+2*x^2-20*x)/x^4,x, algorithm="maxima")

[Out]

-(x^2*(log(2) + 2) + x*(9*log(3) - 5*log(2) - 10) - 27*log(3))/x^3

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=-\frac {x^{2} \log \left (2\right ) + 2 \, x^{2} + 9 \, x \log \left (3\right ) - 5 \, x \log \left (2\right ) - 10 \, x - 27 \, \log \left (3\right )}{x^{3}} \]

[In]

integrate(((18*x-81)*log(3)+(x^2-10*x)*log(2)+2*x^2-20*x)/x^4,x, algorithm="giac")

[Out]

-(x^2*log(2) + 2*x^2 + 9*x*log(3) - 5*x*log(2) - 10*x - 27*log(3))/x^3

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-20 x+2 x^2+\left (-10 x+x^2\right ) \log (2)+(-81+18 x) \log (3)}{x^4} \, dx=-\frac {\left (\ln \left (2\right )+2\right )\,x^2+\left (\ln \left (\frac {19683}{32}\right )-10\right )\,x-\ln \left (7625597484987\right )}{x^3} \]

[In]

int(-(20*x - log(3)*(18*x - 81) + log(2)*(10*x - x^2) - 2*x^2)/x^4,x)

[Out]

-(x^2*(log(2) + 2) - log(7625597484987) + x*(log(19683/32) - 10))/x^3