3.19.43 \(\int \frac {-x^2-4 x^3+16 x^4+e^x (4 x^3+4 x^4)-2 \log ^2(4)-4 x^3 \log (x)}{x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {\log ^2(4)}{x^2}+4 x \left (e^x+2 x-\log (x)\right )-\log (x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 9, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {14, 2176, 2194, 2295} \begin {gather*} 8 x^2+\frac {\log ^2(4)}{x^2}-4 e^x+4 e^x (x+1)-4 x \log (x)-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^2 - 4*x^3 + 16*x^4 + E^x*(4*x^3 + 4*x^4) - 2*Log[4]^2 - 4*x^3*Log[x])/x^3,x]

[Out]

-4*E^x + 8*x^2 + 4*E^x*(1 + x) + Log[4]^2/x^2 - Log[x] - 4*x*Log[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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (4 e^x (1+x)+\frac {-x^2-4 x^3+16 x^4-2 \log ^2(4)-4 x^3 \log (x)}{x^3}\right ) \, dx\\ &=4 \int e^x (1+x) \, dx+\int \frac {-x^2-4 x^3+16 x^4-2 \log ^2(4)-4 x^3 \log (x)}{x^3} \, dx\\ &=4 e^x (1+x)-4 \int e^x \, dx+\int \left (\frac {-x^2-4 x^3+16 x^4-2 \log ^2(4)}{x^3}-4 \log (x)\right ) \, dx\\ &=-4 e^x+4 e^x (1+x)-4 \int \log (x) \, dx+\int \frac {-x^2-4 x^3+16 x^4-2 \log ^2(4)}{x^3} \, dx\\ &=-4 e^x+4 x+4 e^x (1+x)-4 x \log (x)+\int \left (-4-\frac {1}{x}+16 x-\frac {2 \log ^2(4)}{x^3}\right ) \, dx\\ &=-4 e^x+8 x^2+4 e^x (1+x)+\frac {\log ^2(4)}{x^2}-\log (x)-4 x \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 29, normalized size = 1.07 \begin {gather*} 4 e^x x+8 x^2+\frac {\log ^2(4)}{x^2}-\log (x)-4 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 - 4*x^3 + 16*x^4 + E^x*(4*x^3 + 4*x^4) - 2*Log[4]^2 - 4*x^3*Log[x])/x^3,x]

[Out]

4*E^x*x + 8*x^2 + Log[4]^2/x^2 - Log[x] - 4*x*Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 36, normalized size = 1.33 \begin {gather*} \frac {8 \, x^{4} + 4 \, x^{3} e^{x} + 4 \, \log \relax (2)^{2} - {\left (4 \, x^{3} + x^{2}\right )} \log \relax (x)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3*log(x)+(4*x^4+4*x^3)*exp(x)-8*log(2)^2+16*x^4-4*x^3-x^2)/x^3,x, algorithm="fricas")

[Out]

(8*x^4 + 4*x^3*e^x + 4*log(2)^2 - (4*x^3 + x^2)*log(x))/x^2

________________________________________________________________________________________

giac [A]  time = 0.25, size = 37, normalized size = 1.37 \begin {gather*} \frac {8 \, x^{4} + 4 \, x^{3} e^{x} - 4 \, x^{3} \log \relax (x) - x^{2} \log \relax (x) + 4 \, \log \relax (2)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3*log(x)+(4*x^4+4*x^3)*exp(x)-8*log(2)^2+16*x^4-4*x^3-x^2)/x^3,x, algorithm="giac")

[Out]

(8*x^4 + 4*x^3*e^x - 4*x^3*log(x) - x^2*log(x) + 4*log(2)^2)/x^2

________________________________________________________________________________________

maple [A]  time = 0.06, size = 30, normalized size = 1.11

method	result	size
default	\(4 \,{\mathrm e}^{x} x +8 x^{2}+\frac {4 \ln \relax (2)^{2}}{x^{2}}-\ln \relax (x )-4 x \ln \relax (x )\)	\(30\)
risch	\(-4 x \ln \relax (x )-\frac {-8 x^{4}-4 \,{\mathrm e}^{x} x^{3}+x^{2} \ln \relax (x )-4 \ln \relax (2)^{2}}{x^{2}}\)	\(37\)
norman	\(\frac {-x^{2} \ln \relax (x )+8 x^{4}+4 \ln \relax (2)^{2}-4 x^{3} \ln \relax (x )+4 \,{\mathrm e}^{x} x^{3}}{x^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^3*ln(x)+(4*x^4+4*x^3)*exp(x)-8*ln(2)^2+16*x^4-4*x^3-x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

4*exp(x)*x+8*x^2+4*ln(2)^2/x^2-ln(x)-4*x*ln(x)

________________________________________________________________________________________

maxima [A]  time = 0.57, size = 35, normalized size = 1.30 \begin {gather*} 8 \, x^{2} + 4 \, {\left (x - 1\right )} e^{x} - 4 \, x \log \relax (x) + \frac {4 \, \log \relax (2)^{2}}{x^{2}} + 4 \, e^{x} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3*log(x)+(4*x^4+4*x^3)*exp(x)-8*log(2)^2+16*x^4-4*x^3-x^2)/x^3,x, algorithm="maxima")

[Out]

8*x^2 + 4*(x - 1)*e^x - 4*x*log(x) + 4*log(2)^2/x^2 + 4*e^x - log(x)

________________________________________________________________________________________

mupad [B]  time = 1.21, size = 30, normalized size = 1.11 \begin {gather*} \frac {4\,{\ln \relax (2)}^2}{x^2}-\ln \relax (x)+x\,\left (4\,{\mathrm {e}}^x-4\,\ln \relax (x)\right )+8\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x^3*log(x) - exp(x)*(4*x^3 + 4*x^4) + 8*log(2)^2 + x^2 + 4*x^3 - 16*x^4)/x^3,x)

[Out]

(4*log(2)^2)/x^2 - log(x) + x*(4*exp(x) - 4*log(x)) + 8*x^2

________________________________________________________________________________________

sympy [A]  time = 0.33, size = 31, normalized size = 1.15 \begin {gather*} 8 x^{2} + 4 x e^{x} - 4 x \log {\relax (x )} - \log {\relax (x )} + \frac {4 \log {\relax (2 )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**3*ln(x)+(4*x**4+4*x**3)*exp(x)-8*ln(2)**2+16*x**4-4*x**3-x**2)/x**3,x)

[Out]

8*x**2 + 4*x*exp(x) - 4*x*log(x) - log(x) + 4*log(2)**2/x**2

________________________________________________________________________________________