\(\int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx\) [7759]

Optimal result

Integrand size = 31, antiderivative size = 25 \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=5-25 e^x-\left (-4+\frac {5}{x}-\frac {\log (4)}{x}\right )^2 \]

[Out]

5-25*exp(x)-(5/x-4-2*ln(2)/x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {14, 2225, 37} \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=-\frac {(-4 x+5-\log (4))^2}{x^2}-25 e^x \]

[In]

Int[(50 - 40*x - 25*E^x*x^3 + (-20 + 8*x)*Log[4] + 2*Log[4]^2)/x^3,x]

[Out]

-25*E^x - (5 - 4*x - Log[4])^2/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]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-25 e^x+\frac {2 (-5+\log (4)) (-5+4 x+\log (4))}{x^3}\right ) \, dx \\ & = -\left (25 \int e^x \, dx\right )+(2 (-5+\log (4))) \int \frac {-5+4 x+\log (4)}{x^3} \, dx \\ & = -25 e^x-\frac {(5-4 x-\log (4))^2}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=-25 e^x-\frac {8 (-5+\log (4))}{x}+\frac {(5-\log (4)) (-5+\log (4))}{x^2} \]

[In]

Integrate[(50 - 40*x - 25*E^x*x^3 + (-20 + 8*x)*Log[4] + 2*Log[4]^2)/x^3,x]

[Out]

-25*E^x - (8*(-5 + Log[4]))/x + ((5 - Log[4])*(-5 + Log[4]))/x^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {\left (-16 \ln \left (2\right )+40\right ) x -4 \ln \left (2\right )^{2}+20 \ln \left (2\right )-25}{x^{2}}-25 \,{\mathrm e}^{x}\)	\(30\)
parts	\(2 \left (2 \ln \left (2\right )-5\right ) \left (-\frac {2 \ln \left (2\right )-5}{2 x^{2}}-\frac {4}{x}\right )-25 \,{\mathrm e}^{x}\)	\(31\)
norman	\(\frac {\left (-16 \ln \left (2\right )+40\right ) x -25 \,{\mathrm e}^{x} x^{2}-4 \ln \left (2\right )^{2}+20 \ln \left (2\right )-25}{x^{2}}\)	\(32\)
parallelrisch	\(-\frac {25 \,{\mathrm e}^{x} x^{2}+4 \ln \left (2\right )^{2}+16 x \ln \left (2\right )-20 \ln \left (2\right )-40 x +25}{x^{2}}\)	\(33\)
default	\(-\frac {25}{x^{2}}+\frac {40}{x}+\frac {20 \ln \left (2\right )}{x^{2}}-\frac {4 \ln \left (2\right )^{2}}{x^{2}}-\frac {16 \ln \left (2\right )}{x}-25 \,{\mathrm e}^{x}\)	\(39\)

[In]

int((-25*exp(x)*x^3+8*ln(2)^2+2*(8*x-20)*ln(2)-40*x+50)/x^3,x,method=_RETURNVERBOSE)

[Out]

((-16*ln(2)+40)*x-4*ln(2)^2+20*ln(2)-25)/x^2-25*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=-\frac {25 \, x^{2} e^{x} + 4 \, {\left (4 \, x - 5\right )} \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 40 \, x + 25}{x^{2}} \]

[In]

integrate((-25*exp(x)*x^3+8*log(2)^2+2*(8*x-20)*log(2)-40*x+50)/x^3,x, algorithm="fricas")

[Out]

-(25*x^2*e^x + 4*(4*x - 5)*log(2) + 4*log(2)^2 - 40*x + 25)/x^2

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=- 25 e^{x} - \frac {x \left (-40 + 16 \log {\left (2 \right )}\right ) - 20 \log {\left (2 \right )} + 4 \log {\left (2 \right )}^{2} + 25}{x^{2}} \]

[In]

integrate((-25*exp(x)*x**3+8*ln(2)**2+2*(8*x-20)*ln(2)-40*x+50)/x**3,x)

[Out]

-25*exp(x) - (x*(-40 + 16*log(2)) - 20*log(2) + 4*log(2)**2 + 25)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=-\frac {16 \, \log \left (2\right )}{x} - \frac {4 \, \log \left (2\right )^{2}}{x^{2}} + \frac {40}{x} + \frac {20 \, \log \left (2\right )}{x^{2}} - \frac {25}{x^{2}} - 25 \, e^{x} \]

[In]

integrate((-25*exp(x)*x^3+8*log(2)^2+2*(8*x-20)*log(2)-40*x+50)/x^3,x, algorithm="maxima")

[Out]

-16*log(2)/x - 4*log(2)^2/x^2 + 40/x + 20*log(2)/x^2 - 25/x^2 - 25*e^x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=-\frac {25 \, x^{2} e^{x} + 16 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 40 \, x - 20 \, \log \left (2\right ) + 25}{x^{2}} \]

[In]

integrate((-25*exp(x)*x^3+8*log(2)^2+2*(8*x-20)*log(2)-40*x+50)/x^3,x, algorithm="giac")

[Out]

-(25*x^2*e^x + 16*x*log(2) + 4*log(2)^2 - 40*x - 20*log(2) + 25)/x^2

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {50-40 x-25 e^x x^3+(-20+8 x) \log (4)+2 \log ^2(4)}{x^3} \, dx=-25\,{\mathrm {e}}^x-\frac {x\,\left (16\,\ln \left (2\right )-40\right )+{\left (2\,\ln \left (2\right )-5\right )}^2}{x^2} \]

[In]

int((2*log(2)*(8*x - 20) - 40*x - 25*x^3*exp(x) + 8*log(2)^2 + 50)/x^3,x)

[Out]

- 25*exp(x) - (x*(16*log(2) - 40) + (2*log(2) - 5)^2)/x^2