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

Optimal. Leaf size=25 \[ 5-25 e^x-\left (-4+\frac {5}{x}-\frac {\log (4)}{x}\right )^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {14, 2194, 37} \begin {gather*} -\frac {(-4 x+5-\log (4))^2}{x^2}-25 e^x \end {gather*}

Antiderivative was successfully verified.

[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 2194

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

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.16 \begin {gather*} -25 e^x-\frac {8 (-5+\log (4))}{x}+\frac {(5-\log (4)) (-5+\log (4))}{x^2} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.64, size = 32, normalized size = 1.28 \begin {gather*} -\frac {25 \, x^{2} e^{x} + 4 \, {\left (4 \, x - 5\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} - 40 \, x + 25}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.13, size = 32, normalized size = 1.28 \begin {gather*} -\frac {25 \, x^{2} e^{x} + 16 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} - 40 \, x - 20 \, \log \relax (2) + 25}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 30, normalized size = 1.20

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.40, size = 38, normalized size = 1.52 \begin {gather*} -\frac {16 \, \log \relax (2)}{x} - \frac {4 \, \log \relax (2)^{2}}{x^{2}} + \frac {40}{x} + \frac {20 \, \log \relax (2)}{x^{2}} - \frac {25}{x^{2}} - 25 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.10, size = 27, normalized size = 1.08 \begin {gather*} -25\,{\mathrm {e}}^x-\frac {x\,\left (16\,\ln \relax (2)-40\right )+{\left (2\,\ln \relax (2)-5\right )}^2}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.39, size = 31, normalized size = 1.24 \begin {gather*} - 25 e^{x} - \frac {x \left (-40 + 16 \log {\relax (2 )}\right ) - 20 \log {\relax (2 )} + 4 \log {\relax (2 )}^{2} + 25}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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