\(\int \frac {-5+32 e^x x^2}{x^2} \, dx\) [5358]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=-\frac {17}{6}+32 e^x+\frac {5}{x} \]

[Out]

-17/6+5/x+exp(5*ln(2)+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2225} \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=32 e^x+\frac {5}{x} \]

[In]

Int[(-5 + 32*E^x*x^2)/x^2,x]

[Out]

32*E^x + 5/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 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 (32 e^x-\frac {5}{x^2}\right ) \, dx \\ & = \frac {5}{x}+32 \int e^x \, dx \\ & = 32 e^x+\frac {5}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=32 e^x+\frac {5}{x} \]

[In]

Integrate[(-5 + 32*E^x*x^2)/x^2,x]

[Out]

32*E^x + 5/x

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {5}{x}+32 \,{\mathrm e}^{x}\)	\(11\)
parts	\(\frac {5}{x}+{\mathrm e}^{5 \ln \left (2\right )+x}\)	\(14\)
norman	\(\frac {5+x \,{\mathrm e}^{5 \ln \left (2\right )+x}}{x}\)	\(16\)
parallelrisch	\(\frac {5+x \,{\mathrm e}^{5 \ln \left (2\right )+x}}{x}\)	\(16\)
derivativedivides	\({\mathrm e}^{5 \ln \left (2\right )+x}+50 \ln \left (2\right )^{2} \left (-\frac {{\mathrm e}^{5 \ln \left (2\right )+x}}{x}-32 \,\operatorname {Ei}_{1}\left (-x \right )\right )-320 \ln \left (2\right ) \operatorname {Ei}_{1}\left (-x \right )+\frac {5}{x}-10 \ln \left (2\right ) \left (5 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{5 \ln \left (2\right )+x}}{x}-32 \,\operatorname {Ei}_{1}\left (-x \right )\right )-32 \,\operatorname {Ei}_{1}\left (-x \right )\right )\)	\(85\)
default	\({\mathrm e}^{5 \ln \left (2\right )+x}+50 \ln \left (2\right )^{2} \left (-\frac {{\mathrm e}^{5 \ln \left (2\right )+x}}{x}-32 \,\operatorname {Ei}_{1}\left (-x \right )\right )-320 \ln \left (2\right ) \operatorname {Ei}_{1}\left (-x \right )+\frac {5}{x}-10 \ln \left (2\right ) \left (5 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{5 \ln \left (2\right )+x}}{x}-32 \,\operatorname {Ei}_{1}\left (-x \right )\right )-32 \,\operatorname {Ei}_{1}\left (-x \right )\right )\)	\(85\)

[In]

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

[Out]

5/x+32*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=\frac {x e^{\left (x + 5 \, \log \left (2\right )\right )} + 5}{x} \]

[In]

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

[Out]

(x*e^(x + 5*log(2)) + 5)/x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.50 \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=32 e^{x} + \frac {5}{x} \]

[In]

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

[Out]

32*exp(x) + 5/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=\frac {5}{x} + 32 \, e^{x} \]

[In]

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

[Out]

5/x + 32*e^x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=\frac {32 \, x e^{x} + 5}{x} \]

[In]

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

[Out]

(32*x*e^x + 5)/x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {-5+32 e^x x^2}{x^2} \, dx=32\,{\mathrm {e}}^x+\frac {5}{x} \]

[In]

int((x^2*exp(x + 5*log(2)) - 5)/x^2,x)

[Out]

32*exp(x) + 5/x