3.98.92 \(\int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx\)

Optimal. Leaf size=16 \[ 5 \left (9-\left (-3+\frac {5 x}{e^2}\right )^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {14} \begin {gather*} \frac {150 x}{e^2}-\frac {125 x^2}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((150*x)/E^2 - (250*x^2)/E^4)/x,x]

[Out]

(150*x)/E^2 - (125*x^2)/E^4

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} &=\int \left (\frac {150}{e^2}-\frac {250 x}{e^4}\right ) \, dx\\ &=\frac {150 x}{e^2}-\frac {125 x^2}{e^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.19 \begin {gather*} \frac {50 \left (3 e^2 x-\frac {5 x^2}{2}\right )}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((150*x)/E^2 - (250*x^2)/E^4)/x,x]

[Out]

(50*(3*E^2*x - (5*x^2)/2))/E^4

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fricas [A]  time = 0.83, size = 21, normalized size = 1.31 \begin {gather*} -5 \, x^{2} e^{\left (2 \, \log \relax (5) - 4\right )} + 30 \, x e^{\left (\log \relax (5) - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(log(x)+log(5)-2)^2+30*exp(log(x)+log(5)-2))/x,x, algorithm="fricas")

[Out]

-5*x^2*e^(2*log(5) - 4) + 30*x*e^(log(5) - 2)

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giac [A]  time = 0.14, size = 17, normalized size = 1.06 \begin {gather*} -25 \, {\left (5 \, x^{2} e^{2} - 6 \, x e^{4}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(log(x)+log(5)-2)^2+30*exp(log(x)+log(5)-2))/x,x, algorithm="giac")

[Out]

-25*(5*x^2*e^2 - 6*x*e^4)*e^(-6)

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maple [A]  time = 0.04, size = 16, normalized size = 1.00

method	result	size
risch	\(-125 \,{\mathrm e}^{-4} x^{2}+150 x \,{\mathrm e}^{-2}\)	\(16\)
norman	\(\left (150 x -125 x^{2} {\mathrm e}^{-2}\right ) {\mathrm e}^{-2}\)	\(19\)
derivativedivides	\(-125 \,{\mathrm e}^{-4} x^{2}+30 \,{\mathrm e}^{\ln \relax (x )+\ln \relax (5)-2}\)	\(22\)
default	\(-125 \,{\mathrm e}^{-4} x^{2}+30 \,{\mathrm e}^{\ln \relax (x )+\ln \relax (5)-2}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*exp(ln(x)+ln(5)-2)^2+30*exp(ln(x)+ln(5)-2))/x,x,method=_RETURNVERBOSE)

[Out]

-125*exp(-2)^2*x^2+150*x*exp(-2)

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maxima [A]  time = 0.34, size = 15, normalized size = 0.94 \begin {gather*} -25 \, {\left (5 \, x^{2} - 6 \, x e^{2}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(log(x)+log(5)-2)^2+30*exp(log(x)+log(5)-2))/x,x, algorithm="maxima")

[Out]

-25*(5*x^2 - 6*x*e^2)*e^(-4)

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mupad [B]  time = 5.73, size = 13, normalized size = 0.81 \begin {gather*} -25\,x\,{\mathrm {e}}^{-4}\,\left (5\,x-6\,{\mathrm {e}}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*exp(log(5) + log(x) - 2) - 10*exp(2*log(5) + 2*log(x) - 4))/x,x)

[Out]

-25*x*exp(-4)*(5*x - 6*exp(2))

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sympy [A]  time = 0.06, size = 14, normalized size = 0.88 \begin {gather*} - \frac {125 x^{2}}{e^{4}} + \frac {150 x}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(ln(x)+ln(5)-2)**2+30*exp(ln(x)+ln(5)-2))/x,x)

[Out]

-125*x**2*exp(-4) + 150*x*exp(-2)

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