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

Optimal. Leaf size=21 \[ -x+\frac {5 e^x x}{3}-e^{10} \log (2 x) \]

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Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {12, 14, 2176, 2194, 43} \begin {gather*} -x-\frac {5 e^x}{3}+\frac {5}{3} e^x (x+1)-e^{10} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*E^x)/3 - x + (5*E^x*(1 + x))/3 - E^10*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]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.90 \begin {gather*} -x+\frac {5 e^x x}{3}-e^{10} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x + (5*E^x*x)/3 - E^10*Log[x]

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fricas [A]  time = 0.67, size = 15, normalized size = 0.71 \begin {gather*} \frac {5}{3} \, x e^{x} - e^{10} \log \relax (x) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*x*e^x - e^10*log(x) - x

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giac [A]  time = 0.16, size = 15, normalized size = 0.71 \begin {gather*} \frac {5}{3} \, x e^{x} - e^{10} \log \relax (x) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*x*e^x - e^10*log(x) - x

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maple [A]  time = 0.03, size = 16, normalized size = 0.76

method	result	size
risch	\(-x -{\mathrm e}^{10} \ln \relax (x )+\frac {5 \,{\mathrm e}^{x} x}{3}\)	\(16\)
default	\(-x -{\mathrm e}^{10} \ln \relax (x )+\frac {5 \,{\mathrm e}^{x} x}{3}\)	\(18\)
norman	\(-x -{\mathrm e}^{10} \ln \relax (x )+\frac {5 \,{\mathrm e}^{x} x}{3}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x-exp(10)*ln(x)+5/3*exp(x)*x

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maxima [A]  time = 0.35, size = 21, normalized size = 1.00 \begin {gather*} \frac {5}{3} \, {\left (x - 1\right )} e^{x} - e^{10} \log \relax (x) - x + \frac {5}{3} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 15, normalized size = 0.71 \begin {gather*} \frac {5\,x\,{\mathrm {e}}^x}{3}-{\mathrm {e}}^{10}\,\ln \relax (x)-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*x*exp(x))/3 - exp(10)*log(x) - x

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sympy [A]  time = 0.14, size = 15, normalized size = 0.71 \begin {gather*} \frac {5 x e^{x}}{3} - x - e^{10} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x*exp(x)/3 - x - exp(10)*log(x)

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