\(\int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx\) [5239]

Optimal result

Integrand size = 37, antiderivative size = 22 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x \left (5-9 x+\frac {x}{\log (4 x)}\right )}{5 \log (4)} \]

[Out]

1/10*(x/ln(4*x)-9*x+5)/ln(2)*x

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {12, 6874, 2343, 2346, 2209} \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x^2}{5 \log (4) \log (4 x)}-\frac {9 x^2}{5 \log (4)}+\frac {x}{\log (4)} \]

[In]

Int[(-x + 2*x*Log[4*x] + (5 - 18*x)*Log[4*x]^2)/(5*Log[4]*Log[4*x]^2),x]

[Out]

x/Log[4] - (9*x^2)/(5*Log[4]) + x^2/(5*Log[4]*Log[4*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 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log
[c*x^n])^(p + 1)/(b*d*n*(p + 1))), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2346

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{\log ^2(4 x)} \, dx}{5 \log (4)} \\ & = \frac {\int \left (5-18 x-\frac {x}{\log ^2(4 x)}+\frac {2 x}{\log (4 x)}\right ) \, dx}{5 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}-\frac {\int \frac {x}{\log ^2(4 x)} \, dx}{5 \log (4)}+\frac {2 \int \frac {x}{\log (4 x)} \, dx}{5 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}+\frac {x^2}{5 \log (4) \log (4 x)}+\frac {\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (4 x)\right )}{40 \log (4)}-\frac {2 \int \frac {x}{\log (4 x)} \, dx}{5 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}+\frac {\text {Ei}(2 \log (4 x))}{40 \log (4)}+\frac {x^2}{5 \log (4) \log (4 x)}-\frac {\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (4 x)\right )}{40 \log (4)} \\ & = \frac {x}{\log (4)}-\frac {9 x^2}{5 \log (4)}+\frac {x^2}{5 \log (4) \log (4 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {5 x-9 x^2+\frac {x^2}{\log (4 x)}}{5 \log (4)} \]

[In]

Integrate[(-x + 2*x*Log[4*x] + (5 - 18*x)*Log[4*x]^2)/(5*Log[4]*Log[4*x]^2),x]

[Out]

(5*x - 9*x^2 + x^2/Log[4*x])/(5*Log[4])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {-9 x^{2}+5 x +\frac {x^{2}}{\ln \left (4 x \right )}}{10 \ln \left (2\right )}\)	\(26\)
derivativedivides	\(\frac {-144 x^{2}+80 x +\frac {16 x^{2}}{\ln \left (4 x \right )}}{160 \ln \left (2\right )}\)	\(27\)
risch	\(-\frac {9 x^{2}}{10 \ln \left (2\right )}+\frac {x}{2 \ln \left (2\right )}+\frac {x^{2}}{10 \ln \left (2\right ) \ln \left (4 x \right )}\)	\(33\)
parallelrisch	\(\frac {-9 x^{2} \ln \left (4 x \right )+x^{2}+5 x \ln \left (4 x \right )}{10 \ln \left (2\right ) \ln \left (4 x \right )}\)	\(33\)
norman	\(\frac {\frac {x^{2}}{10 \ln \left (2\right )}+\frac {x \ln \left (4 x \right )}{2 \ln \left (2\right )}-\frac {9 x^{2} \ln \left (4 x \right )}{10 \ln \left (2\right )}}{\ln \left (4 x \right )}\)	\(42\)
parts	\(-\frac {9 x^{2}-5 x}{10 \ln \left (2\right )}-\frac {-\frac {x^{2}}{\ln \left (4 x \right )}-\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (4 x \right )\right )}{8}}{10 \ln \left (2\right )}-\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (4 x \right )\right )}{80 \ln \left (2\right )}\)	\(59\)

[In]

int(1/10*((-18*x+5)*ln(4*x)^2+2*x*ln(4*x)-x)/ln(2)/ln(4*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/10/ln(2)*(-9*x^2+5*x+x^2/ln(4*x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x^{2} - {\left (9 \, x^{2} - 5 \, x\right )} \log \left (4 \, x\right )}{10 \, \log \left (2\right ) \log \left (4 \, x\right )} \]

[In]

integrate(1/10*((-18*x+5)*log(4*x)^2+2*x*log(4*x)-x)/log(2)/log(4*x)^2,x, algorithm="fricas")

[Out]

1/10*(x^2 - (9*x^2 - 5*x)*log(4*x))/(log(2)*log(4*x))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=- \frac {9 x^{2}}{10 \log {\left (2 \right )}} + \frac {x^{2}}{10 \log {\left (2 \right )} \log {\left (4 x \right )}} + \frac {x}{2 \log {\left (2 \right )}} \]

[In]

integrate(1/10*((-18*x+5)*ln(4*x)**2+2*x*ln(4*x)-x)/ln(2)/ln(4*x)**2,x)

[Out]

-9*x**2/(10*log(2)) + x**2/(10*log(2)*log(4*x)) + x/(2*log(2))

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=-\frac {72 \, x^{2} - 40 \, x - {\rm Ei}\left (2 \, \log \left (4 \, x\right )\right ) + \Gamma \left (-1, -2 \, \log \left (4 \, x\right )\right )}{80 \, \log \left (2\right )} \]

[In]

integrate(1/10*((-18*x+5)*log(4*x)^2+2*x*log(4*x)-x)/log(2)/log(4*x)^2,x, algorithm="maxima")

[Out]

-1/80*(72*x^2 - 40*x - Ei(2*log(4*x)) + gamma(-1, -2*log(4*x)))/log(2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=-\frac {9 \, x^{2} - 5 \, x - \frac {x^{2}}{\log \left (4 \, x\right )}}{10 \, \log \left (2\right )} \]

[In]

integrate(1/10*((-18*x+5)*log(4*x)^2+2*x*log(4*x)-x)/log(2)/log(4*x)^2,x, algorithm="giac")

[Out]

-1/10*(9*x^2 - 5*x - x^2/log(4*x))/log(2)

Mupad [B] (verification not implemented)

Time = 10.64 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-x+2 x \log (4 x)+(5-18 x) \log ^2(4 x)}{5 \log (4) \log ^2(4 x)} \, dx=\frac {x^2}{10\,\ln \left (4\,x\right )\,\ln \left (2\right )}-\frac {x\,\left (9\,x-5\right )}{10\,\ln \left (2\right )} \]

[In]

int(-(x/10 - (x*log(4*x))/5 + (log(4*x)^2*(18*x - 5))/10)/(log(4*x)^2*log(2)),x)

[Out]

x^2/(10*log(4*x)*log(2)) - (x*(9*x - 5))/(10*log(2))