3.72.13 \(\int \frac {-36+e^2 (5 x+40 x^2+10 x^3)}{5 e^2 x^2} \, dx\)

Optimal. Leaf size=21 \[ -1+\frac {36}{5 e^2 x}+(4+x)^2+\log (3 x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 14} \begin {gather*} x^2+8 x+\frac {36}{5 e^2 x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-36 + E^2*(5*x + 40*x^2 + 10*x^3))/(5*E^2*x^2),x]

[Out]

36/(5*E^2*x) + 8*x + x^2 + Log[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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-36+e^2 \left (5 x+40 x^2+10 x^3\right )}{x^2} \, dx}{5 e^2}\\ &=\frac {\int \left (40 e^2-\frac {36}{x^2}+\frac {5 e^2}{x}+10 e^2 x\right ) \, dx}{5 e^2}\\ &=\frac {36}{5 e^2 x}+8 x+x^2+\log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 19, normalized size = 0.90 \begin {gather*} \frac {36}{5 e^2 x}+8 x+x^2+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36 + E^2*(5*x + 40*x^2 + 10*x^3))/(5*E^2*x^2),x]

[Out]

36/(5*E^2*x) + 8*x + x^2 + Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 29, normalized size = 1.38 \begin {gather*} \frac {{\left (5 \, x e^{2} \log \relax (x) + 5 \, {\left (x^{3} + 8 \, x^{2}\right )} e^{2} + 36\right )} e^{\left (-2\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^3+40*x^2+5*x)*exp(2)-36)/x^2/exp(2),x, algorithm="fricas")

[Out]

1/5*(5*x*e^2*log(x) + 5*(x^3 + 8*x^2)*e^2 + 36)*e^(-2)/x

________________________________________________________________________________________

giac [A]  time = 0.24, size = 29, normalized size = 1.38 \begin {gather*} \frac {1}{5} \, {\left (5 \, x^{2} e^{2} + 40 \, x e^{2} + 5 \, e^{2} \log \left ({\left | x \right |}\right ) + \frac {36}{x}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^3+40*x^2+5*x)*exp(2)-36)/x^2/exp(2),x, algorithm="giac")

[Out]

1/5*(5*x^2*e^2 + 40*x*e^2 + 5*e^2*log(abs(x)) + 36/x)*e^(-2)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 17, normalized size = 0.81

method	result	size
risch	\(x^{2}+8 x +\ln \relax (x )+\frac {36 \,{\mathrm e}^{-2}}{5 x}\)	\(17\)
norman	\(\frac {x^{3}+8 x^{2}+\frac {36 \,{\mathrm e}^{-2}}{5}}{x}+\ln \relax (x )\)	\(23\)
default	\(\frac {{\mathrm e}^{-2} \left (5 x^{2} {\mathrm e}^{2}+40 \,{\mathrm e}^{2} x +5 \,{\mathrm e}^{2} \ln \relax (x )+\frac {36}{x}\right )}{5}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((10*x^3+40*x^2+5*x)*exp(2)-36)/x^2/exp(2),x,method=_RETURNVERBOSE)

[Out]

x^2+8*x+ln(x)+36/5/x*exp(-2)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 28, normalized size = 1.33 \begin {gather*} \frac {1}{5} \, {\left (5 \, x^{2} e^{2} + 40 \, x e^{2} + 5 \, e^{2} \log \relax (x) + \frac {36}{x}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x^3+40*x^2+5*x)*exp(2)-36)/x^2/exp(2),x, algorithm="maxima")

[Out]

1/5*(5*x^2*e^2 + 40*x*e^2 + 5*e^2*log(x) + 36/x)*e^(-2)

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 16, normalized size = 0.76 \begin {gather*} 8\,x+\ln \relax (x)+\frac {36\,{\mathrm {e}}^{-2}}{5\,x}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2)*((exp(2)*(5*x + 40*x^2 + 10*x^3))/5 - 36/5))/x^2,x)

[Out]

8*x + log(x) + (36*exp(-2))/(5*x) + x^2

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 31, normalized size = 1.48 \begin {gather*} \frac {5 x^{2} e^{2} + 40 x e^{2} + 5 e^{2} \log {\relax (x )} + \frac {36}{x}}{5 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x**3+40*x**2+5*x)*exp(2)-36)/x**2/exp(2),x)

[Out]

(5*x**2*exp(2) + 40*x*exp(2) + 5*exp(2)*log(x) + 36/x)*exp(-2)/5

________________________________________________________________________________________