3.80.46 \(\int \frac {12-3 e^{25+x} x^2+e^{25} (-12-33 x^2)}{e^{25} x^2} \, dx\)

Optimal. Leaf size=22 \[ 3 \left (-e^x+\frac {4-\frac {4}{e^{25}}}{x}-11 x\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 14, 2194} \begin {gather*} -33 x-3 e^x+\frac {12 \left (1-\frac {1}{e^{25}}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 - 3*E^(25 + x)*x^2 + E^25*(-12 - 33*x^2))/(E^25*x^2),x]

[Out]

-3*E^x + (12*(1 - E^(-25)))/x - 33*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 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 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} &=\frac {\int \frac {12-3 e^{25+x} x^2+e^{25} \left (-12-33 x^2\right )}{x^2} \, dx}{e^{25}}\\ &=\frac {\int \left (-3 e^{25+x}-\frac {3 \left (-4+4 e^{25}+11 e^{25} x^2\right )}{x^2}\right ) \, dx}{e^{25}}\\ &=-\frac {3 \int e^{25+x} \, dx}{e^{25}}-\frac {3 \int \frac {-4+4 e^{25}+11 e^{25} x^2}{x^2} \, dx}{e^{25}}\\ &=-3 e^x-\frac {3 \int \left (11 e^{25}+\frac {4 \left (-1+e^{25}\right )}{x^2}\right ) \, dx}{e^{25}}\\ &=-3 e^x+\frac {12 \left (1-\frac {1}{e^{25}}\right )}{x}-33 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} -3 e^x+\frac {12}{x}-\frac {12}{e^{25} x}-33 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 - 3*E^(25 + x)*x^2 + E^25*(-12 - 33*x^2))/(E^25*x^2),x]

[Out]

-3*E^x + 12/x - 12/(E^25*x) - 33*x

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fricas [A]  time = 0.85, size = 25, normalized size = 1.14 \begin {gather*} -\frac {3 \, {\left ({\left (11 \, x^{2} - 4\right )} e^{25} + x e^{\left (x + 25\right )} + 4\right )} e^{\left (-25\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2*exp(25)*exp(x)+(-33*x^2-12)*exp(25)+12)/x^2/exp(25),x, algorithm="fricas")

[Out]

-3*((11*x^2 - 4)*e^25 + x*e^(x + 25) + 4)*e^(-25)/x

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giac [A]  time = 0.20, size = 26, normalized size = 1.18 \begin {gather*} -\frac {3 \, {\left (11 \, x^{2} e^{25} + x e^{\left (x + 25\right )} - 4 \, e^{25} + 4\right )} e^{\left (-25\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2*exp(25)*exp(x)+(-33*x^2-12)*exp(25)+12)/x^2/exp(25),x, algorithm="giac")

[Out]

-3*(11*x^2*e^25 + x*e^(x + 25) - 4*e^25 + 4)*e^(-25)/x

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maple [A]  time = 0.03, size = 25, normalized size = 1.14

method	result	size
risch	\(-33 x +\frac {12 \,{\mathrm e}^{-25} {\mathrm e}^{25}}{x}-\frac {12 \,{\mathrm e}^{-25}}{x}-3 \,{\mathrm e}^{x}\)	\(25\)
norman	\(\frac {-33 x^{2}+12 \left ({\mathrm e}^{25}-1\right ) {\mathrm e}^{-25}-3 \,{\mathrm e}^{x} x}{x}\)	\(26\)
default	\({\mathrm e}^{-25} \left (-\frac {12}{x}+\frac {12 \,{\mathrm e}^{25}}{x}-3 \,{\mathrm e}^{x} {\mathrm e}^{25}-33 x \,{\mathrm e}^{25}\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2*exp(25)*exp(x)+(-33*x^2-12)*exp(25)+12)/x^2/exp(25),x,method=_RETURNVERBOSE)

[Out]

-33*x+12*exp(-25)/x*exp(25)-12*exp(-25)/x-3*exp(x)

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maxima [A]  time = 0.36, size = 26, normalized size = 1.18 \begin {gather*} -3 \, {\left (11 \, x e^{25} - \frac {4 \, e^{25}}{x} + \frac {4}{x} + e^{\left (x + 25\right )}\right )} e^{\left (-25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2*exp(25)*exp(x)+(-33*x^2-12)*exp(25)+12)/x^2/exp(25),x, algorithm="maxima")

[Out]

-3*(11*x*e^25 - 4*e^25/x + 4/x + e^(x + 25))*e^(-25)

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mupad [B]  time = 5.77, size = 19, normalized size = 0.86 \begin {gather*} -33\,x-3\,{\mathrm {e}}^x-\frac {12\,{\mathrm {e}}^{-25}-12}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-25)*(exp(25)*(33*x^2 + 12) + 3*x^2*exp(25)*exp(x) - 12))/x^2,x)

[Out]

- 33*x - 3*exp(x) - (12*exp(-25) - 12)/x

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sympy [A]  time = 0.14, size = 24, normalized size = 1.09 \begin {gather*} \frac {- 33 x e^{25} - \frac {12 - 12 e^{25}}{x}}{e^{25}} - 3 e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**2*exp(25)*exp(x)+(-33*x**2-12)*exp(25)+12)/x**2/exp(25),x)

[Out]

(-33*x*exp(25) - (12 - 12*exp(25))/x)*exp(-25) - 3*exp(x)

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