3.41.92 \(\int (141+10 x+e^x (1+x) \log ^2(5)) \, dx\)

Optimal. Leaf size=21 \[ -5+5 (14+x)^2+x \left (1+e^x \log ^2(5)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2176, 2194} \begin {gather*} 5 x^2+141 x-e^x \log ^2(5)+e^x (x+1) \log ^2(5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[141 + 10*x + E^x*(1 + x)*Log[5]^2,x]

[Out]

141*x + 5*x^2 - E^x*Log[5]^2 + E^x*(1 + x)*Log[5]^2

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} &=141 x+5 x^2+\log ^2(5) \int e^x (1+x) \, dx\\ &=141 x+5 x^2+e^x (1+x) \log ^2(5)-\log ^2(5) \int e^x \, dx\\ &=141 x+5 x^2-e^x \log ^2(5)+e^x (1+x) \log ^2(5)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.86 \begin {gather*} 141 x+5 x^2+e^x x \log ^2(5) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[141 + 10*x + E^x*(1 + x)*Log[5]^2,x]

[Out]

141*x + 5*x^2 + E^x*x*Log[5]^2

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fricas [A]  time = 1.02, size = 17, normalized size = 0.81 \begin {gather*} x e^{x} \log \relax (5)^{2} + 5 \, x^{2} + 141 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)*log(5)^2*exp(x)+10*x+141,x, algorithm="fricas")

[Out]

x*e^x*log(5)^2 + 5*x^2 + 141*x

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giac [A]  time = 0.20, size = 17, normalized size = 0.81 \begin {gather*} x e^{x} \log \relax (5)^{2} + 5 \, x^{2} + 141 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)*log(5)^2*exp(x)+10*x+141,x, algorithm="giac")

[Out]

x*e^x*log(5)^2 + 5*x^2 + 141*x

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maple [A]  time = 0.02, size = 18, normalized size = 0.86

method	result	size
default	\(141 x +x \ln \relax (5)^{2} {\mathrm e}^{x}+5 x^{2}\)	\(18\)
norman	\(141 x +x \ln \relax (5)^{2} {\mathrm e}^{x}+5 x^{2}\)	\(18\)
risch	\(141 x +x \ln \relax (5)^{2} {\mathrm e}^{x}+5 x^{2}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+1)*ln(5)^2*exp(x)+10*x+141,x,method=_RETURNVERBOSE)

[Out]

141*x+x*ln(5)^2*exp(x)+5*x^2

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maxima [A]  time = 0.37, size = 17, normalized size = 0.81 \begin {gather*} x e^{x} \log \relax (5)^{2} + 5 \, x^{2} + 141 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)*log(5)^2*exp(x)+10*x+141,x, algorithm="maxima")

[Out]

x*e^x*log(5)^2 + 5*x^2 + 141*x

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mupad [B]  time = 0.05, size = 14, normalized size = 0.67 \begin {gather*} x\,\left (5\,x+{\mathrm {e}}^x\,{\ln \relax (5)}^2+141\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*x + exp(x)*log(5)^2*(x + 1) + 141,x)

[Out]

x*(5*x + exp(x)*log(5)^2 + 141)

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sympy [A]  time = 0.09, size = 17, normalized size = 0.81 \begin {gather*} 5 x^{2} + x e^{x} \log {\relax (5 )}^{2} + 141 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)*ln(5)**2*exp(x)+10*x+141,x)

[Out]

5*x**2 + x*exp(x)*log(5)**2 + 141*x

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