3.1.87 \(\int (308+44 e^{4/3}+968 x+e^x (14+102 x+44 x^2+e^{4/3} (2+2 x)) \log (5)+e^{2 x} (2 x+2 x^2) \log ^2(5)) \, dx\)

Optimal. Leaf size=19 \[ \left (7+e^{4/3}+22 x+e^x x \log (5)\right )^2 \]

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Rubi [B]  time = 0.15, antiderivative size = 72, normalized size of antiderivative = 3.79, number of steps used = 19, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2196, 2194, 2176, 1593} \begin {gather*} 484 x^2+e^{2 x} x^2 \log ^2(5)+44 e^x x^2 \log (5)+44 \left (7+e^{4/3}\right ) x+14 e^x x \log (5)-2 e^{x+\frac {4}{3}} \log (5)+2 e^{x+\frac {4}{3}} (x+1) \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[308 + 44*E^(4/3) + 968*x + E^x*(14 + 102*x + 44*x^2 + E^(4/3)*(2 + 2*x))*Log[5] + E^(2*x)*(2*x + 2*x^2)*Lo
g[5]^2,x]

[Out]

44*(7 + E^(4/3))*x + 484*x^2 - 2*E^(4/3 + x)*Log[5] + 14*E^x*x*Log[5] + 44*E^x*x^2*Log[5] + 2*E^(4/3 + x)*(1 +
x)*Log[5] + E^(2*x)*x^2*Log[5]^2

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} &=44 \left (7+e^{4/3}\right ) x+484 x^2+\log (5) \int e^x \left (14+102 x+44 x^2+e^{4/3} (2+2 x)\right ) \, dx+\log ^2(5) \int e^{2 x} \left (2 x+2 x^2\right ) \, dx\\ &=44 \left (7+e^{4/3}\right ) x+484 x^2+\log (5) \int \left (14 e^x+102 e^x x+44 e^x x^2+2 e^{\frac {4}{3}+x} (1+x)\right ) \, dx+\log ^2(5) \int e^{2 x} x (2+2 x) \, dx\\ &=44 \left (7+e^{4/3}\right ) x+484 x^2+(2 \log (5)) \int e^{\frac {4}{3}+x} (1+x) \, dx+(14 \log (5)) \int e^x \, dx+(44 \log (5)) \int e^x x^2 \, dx+(102 \log (5)) \int e^x x \, dx+\log ^2(5) \int \left (2 e^{2 x} x+2 e^{2 x} x^2\right ) \, dx\\ &=44 \left (7+e^{4/3}\right ) x+484 x^2+14 e^x \log (5)+102 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)-(2 \log (5)) \int e^{\frac {4}{3}+x} \, dx-(88 \log (5)) \int e^x x \, dx-(102 \log (5)) \int e^x \, dx+\left (2 \log ^2(5)\right ) \int e^{2 x} x \, dx+\left (2 \log ^2(5)\right ) \int e^{2 x} x^2 \, dx\\ &=44 \left (7+e^{4/3}\right ) x+484 x^2-88 e^x \log (5)-2 e^{\frac {4}{3}+x} \log (5)+14 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)+e^{2 x} x \log ^2(5)+e^{2 x} x^2 \log ^2(5)+(88 \log (5)) \int e^x \, dx-\log ^2(5) \int e^{2 x} \, dx-\left (2 \log ^2(5)\right ) \int e^{2 x} x \, dx\\ &=44 \left (7+e^{4/3}\right ) x+484 x^2-2 e^{\frac {4}{3}+x} \log (5)+14 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)-\frac {1}{2} e^{2 x} \log ^2(5)+e^{2 x} x^2 \log ^2(5)+\log ^2(5) \int e^{2 x} \, dx\\ &=44 \left (7+e^{4/3}\right ) x+484 x^2-2 e^{\frac {4}{3}+x} \log (5)+14 e^x x \log (5)+44 e^x x^2 \log (5)+2 e^{\frac {4}{3}+x} (1+x) \log (5)+e^{2 x} x^2 \log ^2(5)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 29, normalized size = 1.53 \begin {gather*} x \left (22+e^x \log (5)\right ) \left (14+2 e^{4/3}+22 x+e^x x \log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[308 + 44*E^(4/3) + 968*x + E^x*(14 + 102*x + 44*x^2 + E^(4/3)*(2 + 2*x))*Log[5] + E^(2*x)*(2*x + 2*x
^2)*Log[5]^2,x]

[Out]

x*(22 + E^x*Log[5])*(14 + 2*E^(4/3) + 22*x + E^x*x*Log[5])

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fricas [B]  time = 0.63, size = 45, normalized size = 2.37 \begin {gather*} x^{2} e^{\left (2 \, x\right )} \log \relax (5)^{2} + 2 \, {\left (22 \, x^{2} + x e^{\frac {4}{3}} + 7 \, x\right )} e^{x} \log \relax (5) + 484 \, x^{2} + 44 \, x e^{\frac {4}{3}} + 308 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+2*x)*log(5)^2*exp(x)^2+((2*x+2)*exp(4/3)+44*x^2+102*x+14)*log(5)*exp(x)+44*exp(4/3)+968*x+308
,x, algorithm="fricas")

[Out]

x^2*e^(2*x)*log(5)^2 + 2*(22*x^2 + x*e^(4/3) + 7*x)*e^x*log(5) + 484*x^2 + 44*x*e^(4/3) + 308*x

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giac [B]  time = 0.35, size = 49, normalized size = 2.58 \begin {gather*} x^{2} e^{\left (2 \, x\right )} \log \relax (5)^{2} + 484 \, x^{2} + 44 \, x e^{\frac {4}{3}} + 2 \, {\left (x e^{\left (x + \frac {4}{3}\right )} + {\left (22 \, x^{2} + 7 \, x\right )} e^{x}\right )} \log \relax (5) + 308 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+2*x)*log(5)^2*exp(x)^2+((2*x+2)*exp(4/3)+44*x^2+102*x+14)*log(5)*exp(x)+44*exp(4/3)+968*x+308
,x, algorithm="giac")

[Out]

x^2*e^(2*x)*log(5)^2 + 484*x^2 + 44*x*e^(4/3) + 2*(x*e^(x + 4/3) + (22*x^2 + 7*x)*e^x)*log(5) + 308*x

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maple [B]  time = 0.03, size = 46, normalized size = 2.42

method	result	size
risch	\({\mathrm e}^{2 x} \ln \relax (5)^{2} x^{2}+\ln \relax (5) \left (2 x \,{\mathrm e}^{\frac {4}{3}}+44 x^{2}+14 x \right ) {\mathrm e}^{x}+44 x \,{\mathrm e}^{\frac {4}{3}}+484 x^{2}+308 x\)	\(46\)
norman	\(\left (44 \,{\mathrm e}^{\frac {4}{3}}+308\right ) x +\left (2 \ln \relax (5) {\mathrm e}^{\frac {4}{3}}+14 \ln \relax (5)\right ) x \,{\mathrm e}^{x}+{\mathrm e}^{2 x} \ln \relax (5)^{2} x^{2}+484 x^{2}+44 x^{2} \ln \relax (5) {\mathrm e}^{x}\)	\(51\)
default	\(308 x +2 \,{\mathrm e}^{x} \ln \relax (5) {\mathrm e}^{\frac {4}{3}} x +44 x^{2} \ln \relax (5) {\mathrm e}^{x}+14 x \,{\mathrm e}^{x} \ln \relax (5)+{\mathrm e}^{2 x} \ln \relax (5)^{2} x^{2}+484 x^{2}+44 x \,{\mathrm e}^{\frac {4}{3}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+2*x)*ln(5)^2*exp(x)^2+((2*x+2)*exp(4/3)+44*x^2+102*x+14)*ln(5)*exp(x)+44*exp(4/3)+968*x+308,x,metho
d=_RETURNVERBOSE)

[Out]

exp(2*x)*ln(5)^2*x^2+ln(5)*(2*x*exp(4/3)+44*x^2+14*x)*exp(x)+44*x*exp(4/3)+484*x^2+308*x

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maxima [B]  time = 0.51, size = 44, normalized size = 2.32 \begin {gather*} x^{2} e^{\left (2 \, x\right )} \log \relax (5)^{2} + 2 \, {\left (22 \, x^{2} + x {\left (e^{\frac {4}{3}} + 7\right )}\right )} e^{x} \log \relax (5) + 484 \, x^{2} + 44 \, x e^{\frac {4}{3}} + 308 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+2*x)*log(5)^2*exp(x)^2+((2*x+2)*exp(4/3)+44*x^2+102*x+14)*log(5)*exp(x)+44*exp(4/3)+968*x+308
,x, algorithm="maxima")

[Out]

x^2*e^(2*x)*log(5)^2 + 2*(22*x^2 + x*(e^(4/3) + 7))*e^x*log(5) + 484*x^2 + 44*x*e^(4/3) + 308*x

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mupad [B]  time = 0.29, size = 24, normalized size = 1.26 \begin {gather*} x\,\left ({\mathrm {e}}^x\,\ln \relax (5)+22\right )\,\left (22\,x+2\,{\mathrm {e}}^{4/3}+x\,{\mathrm {e}}^x\,\ln \relax (5)+14\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(968*x + 44*exp(4/3) + exp(x)*log(5)*(102*x + 44*x^2 + exp(4/3)*(2*x + 2) + 14) + exp(2*x)*log(5)^2*(2*x +
2*x^2) + 308,x)

[Out]

x*(exp(x)*log(5) + 22)*(22*x + 2*exp(4/3) + x*exp(x)*log(5) + 14)

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sympy [B]  time = 0.17, size = 58, normalized size = 3.05 \begin {gather*} x^{2} e^{2 x} \log {\relax (5 )}^{2} + 484 x^{2} + x \left (44 e^{\frac {4}{3}} + 308\right ) + \left (44 x^{2} \log {\relax (5 )} + 2 x e^{\frac {4}{3}} \log {\relax (5 )} + 14 x \log {\relax (5 )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+2*x)*ln(5)**2*exp(x)**2+((2*x+2)*exp(4/3)+44*x**2+102*x+14)*ln(5)*exp(x)+44*exp(4/3)+968*x+3
08,x)

[Out]

x**2*exp(2*x)*log(5)**2 + 484*x**2 + x*(44*exp(4/3) + 308) + (44*x**2*log(5) + 2*x*exp(4/3)*log(5) + 14*x*log(
5))*exp(x)

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