3.63.33 \(\int \frac {1+18 x+e^x (-36+e^x (-36-72 x)-36 x) x}{9 x} \, dx\)

Optimal. Leaf size=24 \[ 2 x-4 \left (3+e^x \left (1+e^x\right ) x\right )+\frac {\log (x)}{9} \]

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Rubi [A]  time = 0.05, antiderivative size = 42, normalized size of antiderivative = 1.75, number of steps used = 9, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {12, 14, 2176, 2194, 43} \begin {gather*} 2 x+4 e^x+2 e^{2 x}-4 e^x (x+1)-2 e^{2 x} (2 x+1)+\frac {\log (x)}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 18*x + E^x*(-36 + E^x*(-36 - 72*x) - 36*x)*x)/(9*x),x]

[Out]

4*E^x + 2*E^(2*x) + 2*x - 4*E^x*(1 + x) - 2*E^(2*x)*(1 + 2*x) + Log[x]/9

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}{9} \int \frac {1+18 x+e^x \left (-36+e^x (-36-72 x)-36 x\right ) x}{x} \, dx\\ &=\frac {1}{9} \int \left (-36 e^x (1+x)-36 e^{2 x} (1+2 x)+\frac {1+18 x}{x}\right ) \, dx\\ &=\frac {1}{9} \int \frac {1+18 x}{x} \, dx-4 \int e^x (1+x) \, dx-4 \int e^{2 x} (1+2 x) \, dx\\ &=-4 e^x (1+x)-2 e^{2 x} (1+2 x)+\frac {1}{9} \int \left (18+\frac {1}{x}\right ) \, dx+4 \int e^x \, dx+4 \int e^{2 x} \, dx\\ &=4 e^x+2 e^{2 x}+2 x-4 e^x (1+x)-2 e^{2 x} (1+2 x)+\frac {\log (x)}{9}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 1.00 \begin {gather*} 2 x-4 e^x x-4 e^{2 x} x+\frac {\log (x)}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 18*x + E^x*(-36 + E^x*(-36 - 72*x) - 36*x)*x)/(9*x),x]

[Out]

2*x - 4*E^x*x - 4*E^(2*x)*x + Log[x]/9

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fricas [A]  time = 0.60, size = 34, normalized size = 1.42 \begin {gather*} \frac {18 \, x^{2} - 36 \, x e^{\left (x + \log \relax (x)\right )} + x \log \relax (x) - 36 \, e^{\left (2 \, x + 2 \, \log \relax (x)\right )}}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-72*x-36)*exp(x)-36*x-36)*exp(x+log(x))+18*x+1)/x,x, algorithm="fricas")

[Out]

1/9*(18*x^2 - 36*x*e^(x + log(x)) + x*log(x) - 36*e^(2*x + 2*log(x)))/x

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giac [A]  time = 0.14, size = 20, normalized size = 0.83 \begin {gather*} -4 \, x e^{\left (2 \, x\right )} - 4 \, x e^{x} + 2 \, x + \frac {1}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-72*x-36)*exp(x)-36*x-36)*exp(x+log(x))+18*x+1)/x,x, algorithm="giac")

[Out]

-4*x*e^(2*x) - 4*x*e^x + 2*x + 1/9*log(x)

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maple [A]  time = 0.04, size = 21, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(((-72*x-36)*exp(x)-36*x-36)*exp(x+ln(x))+18*x+1)/x,x,method=_RETURNVERBOSE)

[Out]

2*x+1/9*ln(x)-4*exp(x)*x-4*x*exp(2*x)

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maxima [A]  time = 0.35, size = 36, normalized size = 1.50 \begin {gather*} -2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x - 1\right )} e^{x} + 2 \, x - 2 \, e^{\left (2 \, x\right )} - 4 \, e^{x} + \frac {1}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-72*x-36)*exp(x)-36*x-36)*exp(x+log(x))+18*x+1)/x,x, algorithm="maxima")

[Out]

-2*(2*x - 1)*e^(2*x) - 4*(x - 1)*e^x + 2*x - 2*e^(2*x) - 4*e^x + 1/9*log(x)

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mupad [B]  time = 4.32, size = 20, normalized size = 0.83 \begin {gather*} 2\,x+\frac {\ln \relax (x)}{9}-4\,x\,{\mathrm {e}}^{2\,x}-4\,x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - (exp(x + log(x))*(36*x + exp(x)*(72*x + 36) + 36))/9 + 1/9)/x,x)

[Out]

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

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sympy [A]  time = 0.13, size = 22, normalized size = 0.92 \begin {gather*} - 4 x e^{2 x} - 4 x e^{x} + 2 x + \frac {\log {\relax (x )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(((-72*x-36)*exp(x)-36*x-36)*exp(x+ln(x))+18*x+1)/x,x)

[Out]

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

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