3.93.33 \(\int \frac {640+45 x^2-144 e^{x^2} x^4}{90 x^3} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{2} \left (-\frac {8 e^{x^2}}{5}-\frac {64}{9 x^2}+\log (x)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 2209} \begin {gather*} -\frac {4 e^{x^2}}{5}-\frac {32}{9 x^2}+\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(640 + 45*x^2 - 144*E^x^2*x^4)/(90*x^3),x]

[Out]

(-4*E^x^2)/5 - 32/(9*x^2) + Log[x]/2

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{90} \int \frac {640+45 x^2-144 e^{x^2} x^4}{x^3} \, dx\\ &=\frac {1}{90} \int \left (-144 e^{x^2} x+\frac {5 \left (128+9 x^2\right )}{x^3}\right ) \, dx\\ &=\frac {1}{18} \int \frac {128+9 x^2}{x^3} \, dx-\frac {8}{5} \int e^{x^2} x \, dx\\ &=-\frac {4 e^{x^2}}{5}+\frac {1}{18} \int \left (\frac {128}{x^3}+\frac {9}{x}\right ) \, dx\\ &=-\frac {4 e^{x^2}}{5}-\frac {32}{9 x^2}+\frac {\log (x)}{2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \begin {gather*} -\frac {4 e^{x^2}}{5}-\frac {32}{9 x^2}+\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(640 + 45*x^2 - 144*E^x^2*x^4)/(90*x^3),x]

[Out]

(-4*E^x^2)/5 - 32/(9*x^2) + Log[x]/2

________________________________________________________________________________________

fricas [A]  time = 0.99, size = 23, normalized size = 1.00 \begin {gather*} -\frac {72 \, x^{2} e^{\left (x^{2}\right )} - 45 \, x^{2} \log \relax (x) + 320}{90 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/90*(-144*x^4*exp(x^2)+45*x^2+640)/x^3,x, algorithm="fricas")

[Out]

-1/90*(72*x^2*e^(x^2) - 45*x^2*log(x) + 320)/x^2

________________________________________________________________________________________

giac [A]  time = 0.15, size = 25, normalized size = 1.09 \begin {gather*} -\frac {144 \, x^{2} e^{\left (x^{2}\right )} - 45 \, x^{2} \log \left (x^{2}\right ) + 640}{180 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/90*(-144*x^4*exp(x^2)+45*x^2+640)/x^3,x, algorithm="giac")

[Out]

-1/180*(144*x^2*e^(x^2) - 45*x^2*log(x^2) + 640)/x^2

________________________________________________________________________________________

maple [A]  time = 0.02, size = 17, normalized size = 0.74

method	result	size
default	\(\frac {\ln \relax (x )}{2}-\frac {32}{9 x^{2}}-\frac {4 \,{\mathrm e}^{x^{2}}}{5}\)	\(17\)
risch	\(\frac {\ln \relax (x )}{2}-\frac {32}{9 x^{2}}-\frac {4 \,{\mathrm e}^{x^{2}}}{5}\)	\(17\)
norman	\(\frac {-\frac {32}{9}-\frac {4 x^{2} {\mathrm e}^{x^{2}}}{5}}{x^{2}}+\frac {\ln \relax (x )}{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/90*(-144*x^4*exp(x^2)+45*x^2+640)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x)-32/9/x^2-4/5*exp(x^2)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 16, normalized size = 0.70 \begin {gather*} -\frac {32}{9 \, x^{2}} - \frac {4}{5} \, e^{\left (x^{2}\right )} + \frac {1}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/90*(-144*x^4*exp(x^2)+45*x^2+640)/x^3,x, algorithm="maxima")

[Out]

-32/9/x^2 - 4/5*e^(x^2) + 1/2*log(x)

________________________________________________________________________________________

mupad [B]  time = 0.12, size = 21, normalized size = 0.91 \begin {gather*} \frac {\ln \relax (x)}{2}-\frac {72\,x^2\,{\mathrm {e}}^{x^2}+320}{90\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2/2 - (8*x^4*exp(x^2))/5 + 64/9)/x^3,x)

[Out]

log(x)/2 - (72*x^2*exp(x^2) + 320)/(90*x^2)

________________________________________________________________________________________

sympy [A]  time = 0.25, size = 19, normalized size = 0.83 \begin {gather*} - \frac {4 e^{x^{2}}}{5} + \frac {\log {\relax (x )}}{2} - \frac {32}{9 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/90*(-144*x**4*exp(x**2)+45*x**2+640)/x**3,x)

[Out]

-4*exp(x**2)/5 + log(x)/2 - 32/(9*x**2)

________________________________________________________________________________________