3.83.85 \(\int (8 x+e^{3+x} (-8 x-4 x^2+(-4-4 x) \log (3)) \log ^8(5)+e^{6+2 x} (2 x+2 x^2+(2+4 x) \log (3)+2 \log ^2(3)) \log ^{16}(5)) \, dx\)

Optimal. Leaf size=20 \[ \left (-2 x+e^{3+x} (x+\log (3)) \log ^8(5)\right )^2 \]

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Rubi [B]  time = 0.18, antiderivative size = 153, normalized size of antiderivative = 7.65, number of steps used = 18, number of rules used = 3, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2196, 2176, 2194} \begin {gather*} 4 x^2+e^{2 x+6} x^2 \log ^{16}(5)-4 e^{x+3} x^2 \log ^8(5)+e^{2 x+6} x \log ^{16}(5) (1+\log (9))-e^{2 x+6} x \log ^{16}(5)-\frac {1}{2} e^{2 x+6} \log ^{16}(5) (1+\log (9))+\frac {1}{2} e^{2 x+6} \log ^{16}(5)+4 e^{x+3} \log (3) \log ^8(5)-4 e^{x+3} (x+1) \log (3) \log ^8(5)+\frac {1}{2} e^{2 x+6} \log ^{16}(5) \left (2 \log ^2(3)+\log (9)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8*x + E^(3 + x)*(-8*x - 4*x^2 + (-4 - 4*x)*Log[3])*Log[5]^8 + E^(6 + 2*x)*(2*x + 2*x^2 + (2 + 4*x)*Log[3]
+ 2*Log[3]^2)*Log[5]^16,x]

[Out]

4*x^2 - 4*E^(3 + x)*x^2*Log[5]^8 + 4*E^(3 + x)*Log[3]*Log[5]^8 - 4*E^(3 + x)*(1 + x)*Log[3]*Log[5]^8 + (E^(6 +
2*x)*Log[5]^16)/2 - E^(6 + 2*x)*x*Log[5]^16 + E^(6 + 2*x)*x^2*Log[5]^16 - (E^(6 + 2*x)*Log[5]^16*(1 + Log[9])
)/2 + E^(6 + 2*x)*x*Log[5]^16*(1 + Log[9]) + (E^(6 + 2*x)*Log[5]^16*(2*Log[3]^2 + Log[9]))/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]

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} &=4 x^2+\log ^8(5) \int e^{3+x} \left (-8 x-4 x^2+(-4-4 x) \log (3)\right ) \, dx+\log ^{16}(5) \int e^{6+2 x} \left (2 x+2 x^2+(2+4 x) \log (3)+2 \log ^2(3)\right ) \, dx\\ &=4 x^2+\log ^8(5) \int \left (-8 e^{3+x} x-4 e^{3+x} x^2-4 e^{3+x} (1+x) \log (3)\right ) \, dx+\log ^{16}(5) \int \left (2 e^{6+2 x} x^2+2 e^{6+2 x} x (1+\log (9))+2 e^{6+2 x} \log ^2(3) \left (1+\frac {\log (9)}{2 \log ^2(3)}\right )\right ) \, dx\\ &=4 x^2-\left (4 \log ^8(5)\right ) \int e^{3+x} x^2 \, dx-\left (8 \log ^8(5)\right ) \int e^{3+x} x \, dx-\left (4 \log (3) \log ^8(5)\right ) \int e^{3+x} (1+x) \, dx+\left (2 \log ^{16}(5)\right ) \int e^{6+2 x} x^2 \, dx+\left (2 \log ^{16}(5) (1+\log (9))\right ) \int e^{6+2 x} x \, dx+\left (\log ^{16}(5) \left (2 \log ^2(3)+\log (9)\right )\right ) \int e^{6+2 x} \, dx\\ &=4 x^2-8 e^{3+x} x \log ^8(5)-4 e^{3+x} x^2 \log ^8(5)-4 e^{3+x} (1+x) \log (3) \log ^8(5)+e^{6+2 x} x^2 \log ^{16}(5)+e^{6+2 x} x \log ^{16}(5) (1+\log (9))+\frac {1}{2} e^{6+2 x} \log ^{16}(5) \left (2 \log ^2(3)+\log (9)\right )+\left (8 \log ^8(5)\right ) \int e^{3+x} \, dx+\left (8 \log ^8(5)\right ) \int e^{3+x} x \, dx+\left (4 \log (3) \log ^8(5)\right ) \int e^{3+x} \, dx-\left (2 \log ^{16}(5)\right ) \int e^{6+2 x} x \, dx-\left (\log ^{16}(5) (1+\log (9))\right ) \int e^{6+2 x} \, dx\\ &=4 x^2+8 e^{3+x} \log ^8(5)-4 e^{3+x} x^2 \log ^8(5)+4 e^{3+x} \log (3) \log ^8(5)-4 e^{3+x} (1+x) \log (3) \log ^8(5)-e^{6+2 x} x \log ^{16}(5)+e^{6+2 x} x^2 \log ^{16}(5)-\frac {1}{2} e^{6+2 x} \log ^{16}(5) (1+\log (9))+e^{6+2 x} x \log ^{16}(5) (1+\log (9))+\frac {1}{2} e^{6+2 x} \log ^{16}(5) \left (2 \log ^2(3)+\log (9)\right )-\left (8 \log ^8(5)\right ) \int e^{3+x} \, dx+\log ^{16}(5) \int e^{6+2 x} \, dx\\ &=4 x^2-4 e^{3+x} x^2 \log ^8(5)+4 e^{3+x} \log (3) \log ^8(5)-4 e^{3+x} (1+x) \log (3) \log ^8(5)+\frac {1}{2} e^{6+2 x} \log ^{16}(5)-e^{6+2 x} x \log ^{16}(5)+e^{6+2 x} x^2 \log ^{16}(5)-\frac {1}{2} e^{6+2 x} \log ^{16}(5) (1+\log (9))+e^{6+2 x} x \log ^{16}(5) (1+\log (9))+\frac {1}{2} e^{6+2 x} \log ^{16}(5) \left (2 \log ^2(3)+\log (9)\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 53, normalized size = 2.65 \begin {gather*} 4 x^2-4 e^{3+x} x (x+\log (3)) \log ^8(5)+\frac {1}{2} e^{6+2 x} \log ^{16}(5) \left (2 x^2+2 \log ^2(3)+x \log (81)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8*x + E^(3 + x)*(-8*x - 4*x^2 + (-4 - 4*x)*Log[3])*Log[5]^8 + E^(6 + 2*x)*(2*x + 2*x^2 + (2 + 4*x)*L
og[3] + 2*Log[3]^2)*Log[5]^16,x]

[Out]

4*x^2 - 4*E^(3 + x)*x*(x + Log[3])*Log[5]^8 + (E^(6 + 2*x)*Log[5]^16*(2*x^2 + 2*Log[3]^2 + x*Log[81]))/2

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fricas [B]  time = 0.84, size = 48, normalized size = 2.40 \begin {gather*} {\left (x^{2} + 2 \, x \log \relax (3) + \log \relax (3)^{2}\right )} e^{\left (2 \, x + 6\right )} \log \relax (5)^{16} - 4 \, {\left (x^{2} + x \log \relax (3)\right )} e^{\left (x + 3\right )} \log \relax (5)^{8} + 4 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(3)^2+(4*x+2)*log(3)+2*x^2+2*x)*log(5)^16*exp(3+x)^2+((-4*x-4)*log(3)-4*x^2-8*x)*log(5)^8*exp(
3+x)+8*x,x, algorithm="fricas")

[Out]

(x^2 + 2*x*log(3) + log(3)^2)*e^(2*x + 6)*log(5)^16 - 4*(x^2 + x*log(3))*e^(x + 3)*log(5)^8 + 4*x^2

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giac [B]  time = 0.31, size = 48, normalized size = 2.40 \begin {gather*} {\left (x^{2} + 2 \, x \log \relax (3) + \log \relax (3)^{2}\right )} e^{\left (2 \, x + 6\right )} \log \relax (5)^{16} - 4 \, {\left (x^{2} + x \log \relax (3)\right )} e^{\left (x + 3\right )} \log \relax (5)^{8} + 4 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(3)^2+(4*x+2)*log(3)+2*x^2+2*x)*log(5)^16*exp(3+x)^2+((-4*x-4)*log(3)-4*x^2-8*x)*log(5)^8*exp(
3+x)+8*x,x, algorithm="giac")

[Out]

(x^2 + 2*x*log(3) + log(3)^2)*e^(2*x + 6)*log(5)^16 - 4*(x^2 + x*log(3))*e^(x + 3)*log(5)^8 + 4*x^2

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maple [B]  time = 0.08, size = 51, normalized size = 2.55

method	result	size
risch	\(\ln \relax (5)^{16} \left (\ln \relax (3)^{2}+2 x \ln \relax (3)+x^{2}\right ) {\mathrm e}^{2 x +6}+\ln \relax (5)^{8} \left (-4 x \ln \relax (3)-4 x^{2}\right ) {\mathrm e}^{3+x}+4 x^{2}\)	\(51\)
default	\(-4 \ln \relax (5)^{8} \ln \relax (3) {\mathrm e}^{3+x} x -4 \ln \relax (5)^{8} {\mathrm e}^{3+x} x^{2}+\ln \relax (5)^{16} \ln \relax (3)^{2} {\mathrm e}^{2 x +6}+2 \ln \relax (5)^{16} \ln \relax (3) {\mathrm e}^{2 x +6} x +\ln \relax (5)^{16} {\mathrm e}^{2 x +6} x^{2}+4 x^{2}\)	\(77\)
norman	\(-4 \ln \relax (5)^{8} \ln \relax (3) {\mathrm e}^{3+x} x -4 \ln \relax (5)^{8} {\mathrm e}^{3+x} x^{2}+\ln \relax (5)^{16} \ln \relax (3)^{2} {\mathrm e}^{2 x +6}+2 \ln \relax (5)^{16} \ln \relax (3) {\mathrm e}^{2 x +6} x +\ln \relax (5)^{16} {\mathrm e}^{2 x +6} x^{2}+4 x^{2}\)	\(77\)
derivativedivides	\(-72-24 x -4 \ln \relax (5)^{8} \ln \relax (3) {\mathrm e}^{3+x} \left (3+x \right )-4 \ln \relax (5)^{8} {\mathrm e}^{3+x} \left (3+x \right )^{2}+12 \ln \relax (5)^{8} {\mathrm e}^{3+x} \ln \relax (3)+24 \ln \relax (5)^{8} {\mathrm e}^{3+x} \left (3+x \right )-36 \ln \relax (5)^{8} {\mathrm e}^{3+x}+\ln \relax (5)^{16} \ln \relax (3)^{2} {\mathrm e}^{2 x +6}+2 \ln \relax (5)^{16} \ln \relax (3) {\mathrm e}^{2 x +6} \left (3+x \right )+\ln \relax (5)^{16} {\mathrm e}^{2 x +6} \left (3+x \right )^{2}-6 \ln \relax (5)^{16} {\mathrm e}^{2 x +6} \ln \relax (3)-6 \ln \relax (5)^{16} {\mathrm e}^{2 x +6} \left (3+x \right )+9 \ln \relax (5)^{16} {\mathrm e}^{2 x +6}+4 \left (3+x \right )^{2}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(3)^2+(4*x+2)*ln(3)+2*x^2+2*x)*ln(5)^16*exp(3+x)^2+((-4*x-4)*ln(3)-4*x^2-8*x)*ln(5)^8*exp(3+x)+8*x,x,
method=_RETURNVERBOSE)

[Out]

ln(5)^16*(ln(3)^2+2*x*ln(3)+x^2)*exp(2*x+6)+ln(5)^8*(-4*x*ln(3)-4*x^2)*exp(3+x)+4*x^2

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maxima [B]  time = 0.45, size = 96, normalized size = 4.80 \begin {gather*} {\left (x^{2} e^{6} + 2 \, x e^{6} \log \relax (3) + e^{6} \log \relax (3)^{2}\right )} e^{\left (2 \, x\right )} \log \relax (5)^{16} - 4 \, {\left ({\left (x e^{3} - e^{3}\right )} e^{x} \log \relax (3) + {\left (x^{2} e^{3} - 2 \, x e^{3} + 2 \, e^{3}\right )} e^{x} + 2 \, {\left (x e^{3} - e^{3}\right )} e^{x} + e^{\left (x + 3\right )} \log \relax (3)\right )} \log \relax (5)^{8} + 4 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(3)^2+(4*x+2)*log(3)+2*x^2+2*x)*log(5)^16*exp(3+x)^2+((-4*x-4)*log(3)-4*x^2-8*x)*log(5)^8*exp(
3+x)+8*x,x, algorithm="maxima")

[Out]

(x^2*e^6 + 2*x*e^6*log(3) + e^6*log(3)^2)*e^(2*x)*log(5)^16 - 4*((x*e^3 - e^3)*e^x*log(3) + (x^2*e^3 - 2*x*e^3
+ 2*e^3)*e^x + 2*(x*e^3 - e^3)*e^x + e^(x + 3)*log(3))*log(5)^8 + 4*x^2

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mupad [B]  time = 0.22, size = 27, normalized size = 1.35 \begin {gather*} {\left ({\mathrm {e}}^{x+3}\,\ln \relax (3)\,{\ln \relax (5)}^8-2\,x+x\,{\mathrm {e}}^{x+3}\,{\ln \relax (5)}^8\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x + exp(2*x + 6)*log(5)^16*(2*x + log(3)*(4*x + 2) + 2*log(3)^2 + 2*x^2) - exp(x + 3)*log(5)^8*(8*x + lo
g(3)*(4*x + 4) + 4*x^2),x)

[Out]

(exp(x + 3)*log(3)*log(5)^8 - 2*x + x*exp(x + 3)*log(5)^8)^2

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sympy [B]  time = 0.19, size = 70, normalized size = 3.50 \begin {gather*} 4 x^{2} + \left (- 4 x^{2} \log {\relax (5 )}^{8} - 4 x \log {\relax (3 )} \log {\relax (5 )}^{8}\right ) e^{x + 3} + \left (x^{2} \log {\relax (5 )}^{16} + 2 x \log {\relax (3 )} \log {\relax (5 )}^{16} + \log {\relax (3 )}^{2} \log {\relax (5 )}^{16}\right ) e^{2 x + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(3)**2+(4*x+2)*ln(3)+2*x**2+2*x)*ln(5)**16*exp(3+x)**2+((-4*x-4)*ln(3)-4*x**2-8*x)*ln(5)**8*exp
(3+x)+8*x,x)

[Out]

4*x**2 + (-4*x**2*log(5)**8 - 4*x*log(3)*log(5)**8)*exp(x + 3) + (x**2*log(5)**16 + 2*x*log(3)*log(5)**16 + lo
g(3)**2*log(5)**16)*exp(2*x + 6)

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