3.90.88 \(\int \frac {-e^{1+x} x^3+e (-32+x^3)+x^3 \log (2)}{e x^3} \, dx\)

Optimal. Leaf size=20 \[ 1-e^x+\frac {16}{x^2}+x+\frac {x \log (2)}{e} \]

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2194} \begin {gather*} \frac {16}{x^2}-e^x+\frac {x (e+\log (2))}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^(1 + x)*x^3) + E*(-32 + x^3) + x^3*Log[2])/(E*x^3),x]

[Out]

-E^x + 16/x^2 + (x*(E + Log[2]))/E

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 {-e^{1+x} x^3+e \left (-32+x^3\right )+x^3 \log (2)}{x^3} \, dx}{e}\\ &=\frac {\int \left (-e^{1+x}+\frac {-32 e+x^3 (e+\log (2))}{x^3}\right ) \, dx}{e}\\ &=-\frac {\int e^{1+x} \, dx}{e}+\frac {\int \frac {-32 e+x^3 (e+\log (2))}{x^3} \, dx}{e}\\ &=-e^x+\frac {\int \left (-\frac {32 e}{x^3}+e \left (1+\frac {\log (2)}{e}\right )\right ) \, dx}{e}\\ &=-e^x+\frac {16}{x^2}+\frac {x (e+\log (2))}{e}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.95 \begin {gather*} -e^x+\frac {16}{x^2}+x+\frac {x \log (2)}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(1 + x)*x^3) + E*(-32 + x^3) + x^3*Log[2])/(E*x^3),x]

[Out]

-E^x + 16/x^2 + x + (x*Log[2])/E

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fricas [A]  time = 0.77, size = 30, normalized size = 1.50 \begin {gather*} \frac {{\left (x^{3} \log \relax (2) - x^{2} e^{\left (x + 1\right )} + {\left (x^{3} + 16\right )} e\right )} e^{\left (-1\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(1)*exp(x)+x^3*log(2)+(x^3-32)*exp(1))/x^3/exp(1),x, algorithm="fricas")

[Out]

(x^3*log(2) - x^2*e^(x + 1) + (x^3 + 16)*e)*e^(-1)/x^2

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giac [A]  time = 0.12, size = 32, normalized size = 1.60 \begin {gather*} \frac {{\left (x^{3} e + x^{3} \log \relax (2) - x^{2} e^{\left (x + 1\right )} + 16 \, e\right )} e^{\left (-1\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(1)*exp(x)+x^3*log(2)+(x^3-32)*exp(1))/x^3/exp(1),x, algorithm="giac")

[Out]

(x^3*e + x^3*log(2) - x^2*e^(x + 1) + 16*e)*e^(-1)/x^2

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maple [A]  time = 0.05, size = 23, normalized size = 1.15

method	result	size
risch	\({\mathrm e}^{-1} x \,{\mathrm e}+{\mathrm e}^{-1} x \ln \relax (2)+\frac {16}{x^{2}}-{\mathrm e}^{x}\)	\(23\)
norman	\(\frac {16+\left ({\mathrm e}+\ln \relax (2)\right ) {\mathrm e}^{-1} x^{3}-{\mathrm e}^{x} x^{2}}{x^{2}}\)	\(27\)
default	\({\mathrm e}^{-1} \left (\frac {16 \,{\mathrm e}}{x^{2}}-{\mathrm e} \,{\mathrm e}^{x}+x \,{\mathrm e}+x \ln \relax (2)\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3*exp(1)*exp(x)+x^3*ln(2)+(x^3-32)*exp(1))/x^3/exp(1),x,method=_RETURNVERBOSE)

[Out]

exp(-1)*x*exp(1)+exp(-1)*x*ln(2)+16/x^2-exp(x)

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maxima [A]  time = 0.36, size = 25, normalized size = 1.25 \begin {gather*} {\left (x e + x \log \relax (2) + \frac {16 \, e}{x^{2}} - e^{\left (x + 1\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(1)*exp(x)+x^3*log(2)+(x^3-32)*exp(1))/x^3/exp(1),x, algorithm="maxima")

[Out]

(x*e + x*log(2) + 16*e/x^2 - e^(x + 1))*e^(-1)

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mupad [B]  time = 0.12, size = 19, normalized size = 0.95 \begin {gather*} \frac {16}{x^2}-{\mathrm {e}}^x+x\,{\mathrm {e}}^{-1}\,\left (\mathrm {e}+\ln \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-1)*(x^3*log(2) + exp(1)*(x^3 - 32) - x^3*exp(1)*exp(x)))/x^3,x)

[Out]

16/x^2 - exp(x) + x*exp(-1)*(exp(1) + log(2))

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sympy [A]  time = 0.15, size = 22, normalized size = 1.10 \begin {gather*} \frac {x \left (\log {\relax (2 )} + e\right ) + \frac {16 e}{x^{2}}}{e} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3*exp(1)*exp(x)+x**3*ln(2)+(x**3-32)*exp(1))/x**3/exp(1),x)

[Out]

(x*(log(2) + E) + 16*E/x**2)*exp(-1) - exp(x)

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