3.92.31 \(\int \frac {1}{2} (30+9 x+e^x (108+66 x+9 x^2) \log (19)+e^{2 x} (54+45 x+9 x^2) \log ^2(19)) \, dx\)

Optimal. Leaf size=23 \[ \left (2+\frac {3 (2+x) \left (x+e^x x \log (19)\right )}{2 x}\right )^2 \]

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Rubi [B]  time = 0.11, antiderivative size = 77, normalized size of antiderivative = 3.35, number of steps used = 18, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 2196, 2194, 2176} \begin {gather*} \frac {9 x^2}{4}+\frac {9}{4} e^{2 x} x^2 \log ^2(19)+\frac {9}{2} e^x x^2 \log (19)+15 x+9 e^{2 x} x \log ^2(19)+9 e^{2 x} \log ^2(19)+24 e^x x \log (19)+30 e^x \log (19) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(30 + 9*x + E^x*(108 + 66*x + 9*x^2)*Log[19] + E^(2*x)*(54 + 45*x + 9*x^2)*Log[19]^2)/2,x]

[Out]

15*x + (9*x^2)/4 + 30*E^x*Log[19] + 24*E^x*x*Log[19] + (9*E^x*x^2*Log[19])/2 + 9*E^(2*x)*Log[19]^2 + 9*E^(2*x)
*x*Log[19]^2 + (9*E^(2*x)*x^2*Log[19]^2)/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} &=\frac {1}{2} \int \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx\\ &=15 x+\frac {9 x^2}{4}+\frac {1}{2} \log (19) \int e^x \left (108+66 x+9 x^2\right ) \, dx+\frac {1}{2} \log ^2(19) \int e^{2 x} \left (54+45 x+9 x^2\right ) \, dx\\ &=15 x+\frac {9 x^2}{4}+\frac {1}{2} \log (19) \int \left (108 e^x+66 e^x x+9 e^x x^2\right ) \, dx+\frac {1}{2} \log ^2(19) \int \left (54 e^{2 x}+45 e^{2 x} x+9 e^{2 x} x^2\right ) \, dx\\ &=15 x+\frac {9 x^2}{4}+\frac {1}{2} (9 \log (19)) \int e^x x^2 \, dx+(33 \log (19)) \int e^x x \, dx+(54 \log (19)) \int e^x \, dx+\frac {1}{2} \left (9 \log ^2(19)\right ) \int e^{2 x} x^2 \, dx+\frac {1}{2} \left (45 \log ^2(19)\right ) \int e^{2 x} x \, dx+\left (27 \log ^2(19)\right ) \int e^{2 x} \, dx\\ &=15 x+\frac {9 x^2}{4}+54 e^x \log (19)+33 e^x x \log (19)+\frac {9}{2} e^x x^2 \log (19)+\frac {27}{2} e^{2 x} \log ^2(19)+\frac {45}{4} e^{2 x} x \log ^2(19)+\frac {9}{4} e^{2 x} x^2 \log ^2(19)-(9 \log (19)) \int e^x x \, dx-(33 \log (19)) \int e^x \, dx-\frac {1}{2} \left (9 \log ^2(19)\right ) \int e^{2 x} x \, dx-\frac {1}{4} \left (45 \log ^2(19)\right ) \int e^{2 x} \, dx\\ &=15 x+\frac {9 x^2}{4}+21 e^x \log (19)+24 e^x x \log (19)+\frac {9}{2} e^x x^2 \log (19)+\frac {63}{8} e^{2 x} \log ^2(19)+9 e^{2 x} x \log ^2(19)+\frac {9}{4} e^{2 x} x^2 \log ^2(19)+(9 \log (19)) \int e^x \, dx+\frac {1}{4} \left (9 \log ^2(19)\right ) \int e^{2 x} \, dx\\ &=15 x+\frac {9 x^2}{4}+30 e^x \log (19)+24 e^x x \log (19)+\frac {9}{2} e^x x^2 \log (19)+9 e^{2 x} \log ^2(19)+9 e^{2 x} x \log ^2(19)+\frac {9}{4} e^{2 x} x^2 \log ^2(19)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 55, normalized size = 2.39 \begin {gather*} \frac {1}{2} \left (30 x+\frac {9 x^2}{2}+3 e^x \left (20+16 x+3 x^2\right ) \log (19)+9 e^{2 x} \left (2+2 x+\frac {x^2}{2}\right ) \log ^2(19)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30 + 9*x + E^x*(108 + 66*x + 9*x^2)*Log[19] + E^(2*x)*(54 + 45*x + 9*x^2)*Log[19]^2)/2,x]

[Out]

(30*x + (9*x^2)/2 + 3*E^x*(20 + 16*x + 3*x^2)*Log[19] + 9*E^(2*x)*(2 + 2*x + x^2/2)*Log[19]^2)/2

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fricas [A]  time = 0.92, size = 43, normalized size = 1.87 \begin {gather*} \frac {9}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{\left (2 \, x\right )} \log \left (19\right )^{2} + \frac {3}{2} \, {\left (3 \, x^{2} + 16 \, x + 20\right )} e^{x} \log \left (19\right ) + \frac {9}{4} \, x^{2} + 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(9*x^2+45*x+54)*log(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*log(19)*exp(x)+9/2*x+15,x, algorithm="fr
icas")

[Out]

9/4*(x^2 + 4*x + 4)*e^(2*x)*log(19)^2 + 3/2*(3*x^2 + 16*x + 20)*e^x*log(19) + 9/4*x^2 + 15*x

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giac [A]  time = 0.16, size = 43, normalized size = 1.87 \begin {gather*} \frac {9}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{\left (2 \, x\right )} \log \left (19\right )^{2} + \frac {3}{2} \, {\left (3 \, x^{2} + 16 \, x + 20\right )} e^{x} \log \left (19\right ) + \frac {9}{4} \, x^{2} + 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(9*x^2+45*x+54)*log(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*log(19)*exp(x)+9/2*x+15,x, algorithm="gi
ac")

[Out]

9/4*(x^2 + 4*x + 4)*e^(2*x)*log(19)^2 + 3/2*(3*x^2 + 16*x + 20)*e^x*log(19) + 9/4*x^2 + 15*x

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maple [B]  time = 0.05, size = 46, normalized size = 2.00

method	result	size
risch	\(\frac {\ln \left (19\right )^{2} \left (18+18 x +\frac {9}{2} x^{2}\right ) {\mathrm e}^{2 x}}{2}+\frac {\ln \left (19\right ) \left (9 x^{2}+48 x +60\right ) {\mathrm e}^{x}}{2}+\frac {9 x^{2}}{4}+15 x\)	\(46\)
default	\(15 x +\frac {9 x^{2}}{4}+\frac {9 \,{\mathrm e}^{2 x} \ln \left (19\right )^{2} x^{2}}{4}+9 \ln \left (19\right )^{2} {\mathrm e}^{2 x} x +9 \ln \left (19\right )^{2} {\mathrm e}^{2 x}+\frac {9 \,{\mathrm e}^{x} \ln \left (19\right ) x^{2}}{2}+24 x \,{\mathrm e}^{x} \ln \left (19\right )+30 \ln \left (19\right ) {\mathrm e}^{x}\)	\(66\)
norman	\(15 x +\frac {9 x^{2}}{4}+\frac {9 \,{\mathrm e}^{2 x} \ln \left (19\right )^{2} x^{2}}{4}+9 \ln \left (19\right )^{2} {\mathrm e}^{2 x} x +9 \ln \left (19\right )^{2} {\mathrm e}^{2 x}+\frac {9 \,{\mathrm e}^{x} \ln \left (19\right ) x^{2}}{2}+24 x \,{\mathrm e}^{x} \ln \left (19\right )+30 \ln \left (19\right ) {\mathrm e}^{x}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(9*x^2+45*x+54)*ln(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*ln(19)*exp(x)+9/2*x+15,x,method=_RETURNVERBOSE)

[Out]

1/2*ln(19)^2*(18+18*x+9/2*x^2)*exp(2*x)+1/2*ln(19)*(9*x^2+48*x+60)*exp(x)+9/4*x^2+15*x

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maxima [A]  time = 0.35, size = 43, normalized size = 1.87 \begin {gather*} \frac {9}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{\left (2 \, x\right )} \log \left (19\right )^{2} + \frac {3}{2} \, {\left (3 \, x^{2} + 16 \, x + 20\right )} e^{x} \log \left (19\right ) + \frac {9}{4} \, x^{2} + 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(9*x^2+45*x+54)*log(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*log(19)*exp(x)+9/2*x+15,x, algorithm="ma
xima")

[Out]

9/4*(x^2 + 4*x + 4)*e^(2*x)*log(19)^2 + 3/2*(3*x^2 + 16*x + 20)*e^x*log(19) + 9/4*x^2 + 15*x

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mupad [B]  time = 0.11, size = 65, normalized size = 2.83 \begin {gather*} 15\,x+9\,{\mathrm {e}}^{2\,x}\,{\ln \left (19\right )}^2+30\,{\mathrm {e}}^x\,\ln \left (19\right )+\frac {9\,x^2}{4}+24\,x\,{\mathrm {e}}^x\,\ln \left (19\right )+\frac {9\,x^2\,{\mathrm {e}}^{2\,x}\,{\ln \left (19\right )}^2}{4}+\frac {9\,x^2\,{\mathrm {e}}^x\,\ln \left (19\right )}{2}+9\,x\,{\mathrm {e}}^{2\,x}\,{\ln \left (19\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x)/2 + (exp(x)*log(19)*(66*x + 9*x^2 + 108))/2 + (exp(2*x)*log(19)^2*(45*x + 9*x^2 + 54))/2 + 15,x)

[Out]

15*x + 9*exp(2*x)*log(19)^2 + 30*exp(x)*log(19) + (9*x^2)/4 + 24*x*exp(x)*log(19) + (9*x^2*exp(2*x)*log(19)^2)
/4 + (9*x^2*exp(x)*log(19))/2 + 9*x*exp(2*x)*log(19)^2

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sympy [B]  time = 0.20, size = 66, normalized size = 2.87 \begin {gather*} \frac {9 x^{2}}{4} + 15 x + \frac {\left (36 x^{2} \log {\left (19 \right )} + 192 x \log {\left (19 \right )} + 240 \log {\left (19 \right )}\right ) e^{x}}{8} + \frac {\left (18 x^{2} \log {\left (19 \right )}^{2} + 72 x \log {\left (19 \right )}^{2} + 72 \log {\left (19 \right )}^{2}\right ) e^{2 x}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(9*x**2+45*x+54)*ln(19)**2*exp(x)**2+1/2*(9*x**2+66*x+108)*ln(19)*exp(x)+9/2*x+15,x)

[Out]

9*x**2/4 + 15*x + (36*x**2*log(19) + 192*x*log(19) + 240*log(19))*exp(x)/8 + (18*x**2*log(19)**2 + 72*x*log(19
)**2 + 72*log(19)**2)*exp(2*x)/8

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