3.45.78 \(\int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{5 x^3} \, dx\)

Optimal. Leaf size=24 \[ 1+e^x+x+x \left (1+\frac {e^3 x}{5}\right )+\frac {\log (x)}{x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14, 2194, 2304} \begin {gather*} \frac {e^3 x^2}{5}+\frac {\log (x)}{x^2}+2 x+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + 10*x^3 + 5*E^x*x^3 + 2*E^3*x^4 - 10*Log[x])/(5*x^3),x]

[Out]

E^x + 2*x + (E^3*x^2)/5 + Log[x]/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 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {5+10 x^3+5 e^x x^3+2 e^3 x^4-10 \log (x)}{x^3} \, dx\\ &=\frac {1}{5} \int \left (5 e^x+\frac {5+10 x^3+2 e^3 x^4-10 \log (x)}{x^3}\right ) \, dx\\ &=\frac {1}{5} \int \frac {5+10 x^3+2 e^3 x^4-10 \log (x)}{x^3} \, dx+\int e^x \, dx\\ &=e^x+\frac {1}{5} \int \left (\frac {5+10 x^3+2 e^3 x^4}{x^3}-\frac {10 \log (x)}{x^3}\right ) \, dx\\ &=e^x+\frac {1}{5} \int \frac {5+10 x^3+2 e^3 x^4}{x^3} \, dx-2 \int \frac {\log (x)}{x^3} \, dx\\ &=e^x+\frac {1}{2 x^2}+\frac {\log (x)}{x^2}+\frac {1}{5} \int \left (10+\frac {5}{x^3}+2 e^3 x\right ) \, dx\\ &=e^x+2 x+\frac {e^3 x^2}{5}+\frac {\log (x)}{x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 23, normalized size = 0.96 \begin {gather*} e^x+2 x+\frac {e^3 x^2}{5}+\frac {\log (x)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 10*x^3 + 5*E^x*x^3 + 2*E^3*x^4 - 10*Log[x])/(5*x^3),x]

[Out]

E^x + 2*x + (E^3*x^2)/5 + Log[x]/x^2

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 28, normalized size = 1.17 \begin {gather*} \frac {x^{4} e^{3} + 10 \, x^{3} + 5 \, x^{2} e^{x} + 5 \, \log \relax (x)}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*log(x)+5*exp(x)*x^3+2*x^4*exp(3)+10*x^3+5)/x^3,x, algorithm="fricas")

[Out]

1/5*(x^4*e^3 + 10*x^3 + 5*x^2*e^x + 5*log(x))/x^2

________________________________________________________________________________________

giac [A]  time = 0.17, size = 28, normalized size = 1.17 \begin {gather*} \frac {x^{4} e^{3} + 10 \, x^{3} + 5 \, x^{2} e^{x} + 5 \, \log \relax (x)}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*log(x)+5*exp(x)*x^3+2*x^4*exp(3)+10*x^3+5)/x^3,x, algorithm="giac")

[Out]

1/5*(x^4*e^3 + 10*x^3 + 5*x^2*e^x + 5*log(x))/x^2

________________________________________________________________________________________

maple [A]  time = 0.05, size = 20, normalized size = 0.83

method	result	size
default	\(2 x +\frac {x^{2} {\mathrm e}^{3}}{5}+\frac {\ln \relax (x )}{x^{2}}+{\mathrm e}^{x}\)	\(20\)
risch	\(2 x +\frac {x^{2} {\mathrm e}^{3}}{5}+\frac {\ln \relax (x )}{x^{2}}+{\mathrm e}^{x}\)	\(20\)
norman	\(\frac {{\mathrm e}^{x} x^{2}+2 x^{3}+\frac {x^{4} {\mathrm e}^{3}}{5}+\ln \relax (x )}{x^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-10*ln(x)+5*exp(x)*x^3+2*x^4*exp(3)+10*x^3+5)/x^3,x,method=_RETURNVERBOSE)

[Out]

2*x+1/5*x^2*exp(3)+ln(x)/x^2+exp(x)

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 19, normalized size = 0.79 \begin {gather*} \frac {1}{5} \, x^{2} e^{3} + 2 \, x + \frac {\log \relax (x)}{x^{2}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*log(x)+5*exp(x)*x^3+2*x^4*exp(3)+10*x^3+5)/x^3,x, algorithm="maxima")

[Out]

1/5*x^2*e^3 + 2*x + log(x)/x^2 + e^x

________________________________________________________________________________________

mupad [B]  time = 3.49, size = 19, normalized size = 0.79 \begin {gather*} 2\,x+{\mathrm {e}}^x+\frac {\ln \relax (x)}{x^2}+\frac {x^2\,{\mathrm {e}}^3}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*exp(x) - 2*log(x) + (2*x^4*exp(3))/5 + 2*x^3 + 1)/x^3,x)

[Out]

2*x + exp(x) + log(x)/x^2 + (x^2*exp(3))/5

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 20, normalized size = 0.83 \begin {gather*} \frac {x^{2} e^{3}}{5} + 2 x + e^{x} + \frac {\log {\relax (x )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*ln(x)+5*exp(x)*x**3+2*x**4*exp(3)+10*x**3+5)/x**3,x)

[Out]

x**2*exp(3)/5 + 2*x + exp(x) + log(x)/x**2

________________________________________________________________________________________