3.2.62 \(\int (1+e^x (4800 x+2400 x^2) \log (5)+(3+7200 x^2) \log (5)) \, dx\)

Optimal. Leaf size=19 \[ -3+x+3 \left (x+800 x^2 \left (e^x+x\right )\right ) \log (5) \]

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Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} 2400 x^3 \log (5)+2400 e^x x^2 \log (5)+x+3 x \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + E^x*(4800*x + 2400*x^2)*Log[5] + (3 + 7200*x^2)*Log[5],x]

[Out]

x + 3*x*Log[5] + 2400*E^x*x^2*Log[5] + 2400*x^3*Log[5]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x+\log (5) \int e^x \left (4800 x+2400 x^2\right ) \, dx+\log (5) \int \left (3+7200 x^2\right ) \, dx\\ &=x+3 x \log (5)+2400 x^3 \log (5)+\log (5) \int e^x x (4800+2400 x) \, dx\\ &=x+3 x \log (5)+2400 x^3 \log (5)+\log (5) \int \left (4800 e^x x+2400 e^x x^2\right ) \, dx\\ &=x+3 x \log (5)+2400 x^3 \log (5)+(2400 \log (5)) \int e^x x^2 \, dx+(4800 \log (5)) \int e^x x \, dx\\ &=x+3 x \log (5)+4800 e^x x \log (5)+2400 e^x x^2 \log (5)+2400 x^3 \log (5)-(4800 \log (5)) \int e^x \, dx-(4800 \log (5)) \int e^x x \, dx\\ &=x-4800 e^x \log (5)+3 x \log (5)+2400 e^x x^2 \log (5)+2400 x^3 \log (5)+(4800 \log (5)) \int e^x \, dx\\ &=x+3 x \log (5)+2400 e^x x^2 \log (5)+2400 x^3 \log (5)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.26 \begin {gather*} x+3 x \log (5)+2400 e^x x^2 \log (5)+2400 x^3 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + E^x*(4800*x + 2400*x^2)*Log[5] + (3 + 7200*x^2)*Log[5],x]

[Out]

x + 3*x*Log[5] + 2400*E^x*x^2*Log[5] + 2400*x^3*Log[5]

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fricas [A]  time = 0.61, size = 22, normalized size = 1.16 \begin {gather*} 2400 \, x^{2} e^{x} \log \relax (5) + 3 \, {\left (800 \, x^{3} + x\right )} \log \relax (5) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2400*x^2+4800*x)*log(5)*exp(x)+(7200*x^2+3)*log(5)+1,x, algorithm="fricas")

[Out]

2400*x^2*e^x*log(5) + 3*(800*x^3 + x)*log(5) + x

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giac [A]  time = 0.27, size = 22, normalized size = 1.16 \begin {gather*} 2400 \, x^{2} e^{x} \log \relax (5) + 3 \, {\left (800 \, x^{3} + x\right )} \log \relax (5) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2400*x^2+4800*x)*log(5)*exp(x)+(7200*x^2+3)*log(5)+1,x, algorithm="giac")

[Out]

2400*x^2*e^x*log(5) + 3*(800*x^3 + x)*log(5) + x

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maple [A]  time = 0.02, size = 24, normalized size = 1.26




method result size



default \(x +2400 x^{3} \ln \relax (5)+3 x \ln \relax (5)+2400 x^{2} \ln \relax (5) {\mathrm e}^{x}\) \(24\)
risch \(x +2400 x^{3} \ln \relax (5)+3 x \ln \relax (5)+2400 x^{2} \ln \relax (5) {\mathrm e}^{x}\) \(24\)
norman \(\left (3 \ln \relax (5)+1\right ) x +2400 x^{3} \ln \relax (5)+2400 x^{2} \ln \relax (5) {\mathrm e}^{x}\) \(26\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2400*x^2+4800*x)*ln(5)*exp(x)+(7200*x^2+3)*ln(5)+1,x,method=_RETURNVERBOSE)

[Out]

x+2400*x^3*ln(5)+3*x*ln(5)+2400*x^2*ln(5)*exp(x)

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maxima [A]  time = 0.62, size = 22, normalized size = 1.16 \begin {gather*} 2400 \, x^{2} e^{x} \log \relax (5) + 3 \, {\left (800 \, x^{3} + x\right )} \log \relax (5) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2400*x^2+4800*x)*log(5)*exp(x)+(7200*x^2+3)*log(5)+1,x, algorithm="maxima")

[Out]

2400*x^2*e^x*log(5) + 3*(800*x^3 + x)*log(5) + x

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mupad [B]  time = 0.14, size = 23, normalized size = 1.21 \begin {gather*} x\,\left (\ln \left (125\right )+1\right )+2400\,x^3\,\ln \relax (5)+2400\,x^2\,{\mathrm {e}}^x\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(5)*(7200*x^2 + 3) + exp(x)*log(5)*(4800*x + 2400*x^2) + 1,x)

[Out]

x*(log(125) + 1) + 2400*x^3*log(5) + 2400*x^2*exp(x)*log(5)

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sympy [A]  time = 0.11, size = 27, normalized size = 1.42 \begin {gather*} 2400 x^{3} \log {\relax (5 )} + 2400 x^{2} e^{x} \log {\relax (5 )} + x \left (1 + 3 \log {\relax (5 )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2400*x**2+4800*x)*ln(5)*exp(x)+(7200*x**2+3)*ln(5)+1,x)

[Out]

2400*x**3*log(5) + 2400*x**2*exp(x)*log(5) + x*(1 + 3*log(5))

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