3.44.19 \(\int \frac {1+32 e^{2 x} x+288 x^2+e^x (-96 x-96 x^2)}{x} \, dx\)

Optimal. Leaf size=19 \[ \log \left (\frac {1}{2} e^{16 \left (e^x-3 x\right )^2} x\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.47, number of steps used = 7, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {14, 2194, 2176} \begin {gather*} 144 x^2+96 e^x+16 e^{2 x}-96 e^x (x+1)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 32*E^(2*x)*x + 288*x^2 + E^x*(-96*x - 96*x^2))/x,x]

[Out]

96*E^x + 16*E^(2*x) + 144*x^2 - 96*E^x*(1 + 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (32 e^{2 x}-96 e^x (1+x)+\frac {1+288 x^2}{x}\right ) \, dx\\ &=32 \int e^{2 x} \, dx-96 \int e^x (1+x) \, dx+\int \frac {1+288 x^2}{x} \, dx\\ &=16 e^{2 x}-96 e^x (1+x)+96 \int e^x \, dx+\int \left (\frac {1}{x}+288 x\right ) \, dx\\ &=96 e^x+16 e^{2 x}+144 x^2-96 e^x (1+x)+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.11 \begin {gather*} 16 e^{2 x}-96 e^x x+144 x^2+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 32*E^(2*x)*x + 288*x^2 + E^x*(-96*x - 96*x^2))/x,x]

[Out]

16*E^(2*x) - 96*E^x*x + 144*x^2 + Log[x]

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fricas [A]  time = 0.79, size = 19, normalized size = 1.00 \begin {gather*} 144 \, x^{2} - 96 \, x e^{x} + 16 \, e^{\left (2 \, x\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x*exp(x)^2+(-96*x^2-96*x)*exp(x)+288*x^2+1)/x,x, algorithm="fricas")

[Out]

144*x^2 - 96*x*e^x + 16*e^(2*x) + log(x)

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giac [A]  time = 0.21, size = 19, normalized size = 1.00 \begin {gather*} 144 \, x^{2} - 96 \, x e^{x} + 16 \, e^{\left (2 \, x\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x*exp(x)^2+(-96*x^2-96*x)*exp(x)+288*x^2+1)/x,x, algorithm="giac")

[Out]

144*x^2 - 96*x*e^x + 16*e^(2*x) + log(x)

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maple [A]  time = 0.02, size = 20, normalized size = 1.05

method	result	size
default	\(\ln \relax (x )+144 x^{2}+16 \,{\mathrm e}^{2 x}-96 \,{\mathrm e}^{x} x\)	\(20\)
norman	\(\ln \relax (x )+144 x^{2}+16 \,{\mathrm e}^{2 x}-96 \,{\mathrm e}^{x} x\)	\(20\)
risch	\(\ln \relax (x )+144 x^{2}+16 \,{\mathrm e}^{2 x}-96 \,{\mathrm e}^{x} x\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x*exp(x)^2+(-96*x^2-96*x)*exp(x)+288*x^2+1)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)+144*x^2+16*exp(x)^2-96*exp(x)*x

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maxima [A]  time = 0.35, size = 25, normalized size = 1.32 \begin {gather*} 144 \, x^{2} - 96 \, {\left (x - 1\right )} e^{x} + 16 \, e^{\left (2 \, x\right )} - 96 \, e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x*exp(x)^2+(-96*x^2-96*x)*exp(x)+288*x^2+1)/x,x, algorithm="maxima")

[Out]

144*x^2 - 96*(x - 1)*e^x + 16*e^(2*x) - 96*e^x + log(x)

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mupad [B]  time = 0.06, size = 19, normalized size = 1.00 \begin {gather*} 16\,{\mathrm {e}}^{2\,x}+\ln \relax (x)-96\,x\,{\mathrm {e}}^x+144\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x*exp(2*x) - exp(x)*(96*x + 96*x^2) + 288*x^2 + 1)/x,x)

[Out]

16*exp(2*x) + log(x) - 96*x*exp(x) + 144*x^2

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sympy [A]  time = 0.12, size = 20, normalized size = 1.05 \begin {gather*} 144 x^{2} - 96 x e^{x} + 16 e^{2 x} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x*exp(x)**2+(-96*x**2-96*x)*exp(x)+288*x**2+1)/x,x)

[Out]

144*x**2 - 96*x*exp(x) + 16*exp(2*x) + log(x)

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