3.85.20 \(\int (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))) \, dx\)

Optimal. Leaf size=23 \[ -5^{e^{4+e^{(-1+4 x+\log (100))^2}}}+x^2 \]

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Rubi [A]  time = 4.88, antiderivative size = 36, normalized size of antiderivative = 1.57, number of steps used = 2, number of rules used = 1, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6706} \begin {gather*} x^2-5^{e^{100^{8 x-2} e^{16 x^2-8 x+1+\log ^2(100)}+4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*x + 5^E^(4 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2))*E^(5 + E^(1 - 8*x + 16*x^2 + (-2 +
8*x)*Log[100] + Log[100]^2) - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2)*((8 - 32*x)*Log[5] - 8*Log[5]*
Log[100]),x]

[Out]

-5^E^(4 + 100^(-2 + 8*x)*E^(1 - 8*x + 16*x^2 + Log[100]^2)) + x^2

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x^2+\int 5^{\exp \left (4+\exp \left (1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)\right )\right )} \exp \left (5+\exp \left (1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)\right )-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)\right ) ((8-32 x) \log (5)-8 \log (5) \log (100)) \, dx\\ &=-5^{e^{4+100^{-2+8 x} e^{1-8 x+16 x^2+\log ^2(100)}}}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 5.06, size = 0, normalized size = 0.00 \begin {gather*} \int \left (2 x+5^{e^{4+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}}} e^{5+e^{1-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)}-8 x+16 x^2+(-2+8 x) \log (100)+\log ^2(100)} ((8-32 x) \log (5)-8 \log (5) \log (100))\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[2*x + 5^E^(4 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2))*E^(5 + E^(1 - 8*x + 16*x^2 +
(-2 + 8*x)*Log[100] + Log[100]^2) - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2)*((8 - 32*x)*Log[5] - 8*L
og[5]*Log[100]),x]

[Out]

Integrate[2*x + 5^E^(4 + E^(1 - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2))*E^(5 + E^(1 - 8*x + 16*x^2 +
(-2 + 8*x)*Log[100] + Log[100]^2) - 8*x + 16*x^2 + (-2 + 8*x)*Log[100] + Log[100]^2)*((8 - 32*x)*Log[5] - 8*L
og[5]*Log[100]), x]

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fricas [A]  time = 0.60, size = 37, normalized size = 1.61 \begin {gather*} x^{2} - 5^{e^{\left (e^{\left (16 \, x^{2} + 4 \, {\left (4 \, x - 1\right )} \log \left (10\right ) + 4 \, \log \left (10\right )^{2} - 8 \, x + 1\right )} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)*log(10)+(-32*x+8)*log(5))*exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)*exp(exp(4*log(
10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4)*exp(log(5)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4))+2*
x,x, algorithm="fricas")

[Out]

x^2 - 5^e^(e^(16*x^2 + 4*(4*x - 1)*log(10) + 4*log(10)^2 - 8*x + 1) + 4)

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giac [A]  time = 0.20, size = 37, normalized size = 1.61 \begin {gather*} x^{2} - 5^{e^{\left (e^{\left (16 \, x^{2} + 16 \, x \log \left (10\right ) + 4 \, \log \left (10\right )^{2} - 8 \, x - 4 \, \log \left (10\right ) + 1\right )} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)*log(10)+(-32*x+8)*log(5))*exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)*exp(exp(4*log(
10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4)*exp(log(5)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4))+2*
x,x, algorithm="giac")

[Out]

x^2 - 5^e^(e^(16*x^2 + 16*x*log(10) + 4*log(10)^2 - 8*x - 4*log(10) + 1) + 4)

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maple [A]  time = 0.25, size = 28, normalized size = 1.22

method	result	size
risch	\(-5^{{\mathrm e}^{{\mathrm e}^{\left (2 \ln \relax (2)+2 \ln \relax (5)+4 x -1\right )^{2}}+4}}+x^{2}\)	\(28\)
default	\(-{\mathrm e}^{\ln \relax (5) {\mathrm e}^{{\mathrm e}^{4 \ln \left (10\right )^{2}+2 \left (8 x -2\right ) \ln \left (10\right )+16 x^{2}-8 x +1}+4}}+x^{2}\)	\(40\)
norman	\(-{\mathrm e}^{\ln \relax (5) {\mathrm e}^{{\mathrm e}^{4 \ln \left (10\right )^{2}+2 \left (8 x -2\right ) \ln \left (10\right )+16 x^{2}-8 x +1}+4}}+x^{2}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*ln(5)*ln(10)+(-32*x+8)*ln(5))*exp(4*ln(10)^2+2*(8*x-2)*ln(10)+16*x^2-8*x+1)*exp(exp(4*ln(10)^2+2*(8*x
-2)*ln(10)+16*x^2-8*x+1)+4)*exp(ln(5)*exp(exp(4*ln(10)^2+2*(8*x-2)*ln(10)+16*x^2-8*x+1)+4))+2*x,x,method=_RETU
RNVERBOSE)

[Out]

-5^exp(exp((2*ln(2)+2*ln(5)+4*x-1)^2)+4)+x^2

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maxima [B]  time = 1.07, size = 54, normalized size = 2.35 \begin {gather*} x^{2} - 5^{e^{\left (\frac {1}{625} \cdot 2^{8 \, \log \relax (5) - 4} e^{\left (16 \, x^{2} + 16 \, x \log \relax (5) + 4 \, \log \relax (5)^{2} + 16 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} - 8 \, x + 1\right )} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(5)*log(10)+(-32*x+8)*log(5))*exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)*exp(exp(4*log(
10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4)*exp(log(5)*exp(exp(4*log(10)^2+2*(8*x-2)*log(10)+16*x^2-8*x+1)+4))+2*
x,x, algorithm="maxima")

[Out]

x^2 - 5^e^(1/625*2^(8*log(5) - 4)*e^(16*x^2 + 16*x*log(5) + 4*log(5)^2 + 16*x*log(2) + 4*log(2)^2 - 8*x + 1) +
4)

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mupad [B]  time = 0.44, size = 37, normalized size = 1.61 \begin {gather*} x^2-5^{{\mathrm {e}}^{\frac {{10}^{16\,x}\,{\mathrm {e}}^{-8\,x}\,\mathrm {e}\,{\mathrm {e}}^{4\,{\ln \left (10\right )}^2}\,{\mathrm {e}}^{16\,x^2}}{10000}+4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - exp(2*log(10)*(8*x - 2) - 8*x + 4*log(10)^2 + 16*x^2 + 1)*exp(exp(2*log(10)*(8*x - 2) - 8*x + 4*log(
10)^2 + 16*x^2 + 1) + 4)*exp(exp(exp(2*log(10)*(8*x - 2) - 8*x + 4*log(10)^2 + 16*x^2 + 1) + 4)*log(5))*(log(5
)*(32*x - 8) + 16*log(5)*log(10)),x)

[Out]

x^2 - 5^exp((10^(16*x)*exp(-8*x)*exp(1)*exp(4*log(10)^2)*exp(16*x^2))/10000 + 4)

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sympy [A]  time = 0.60, size = 37, normalized size = 1.61 \begin {gather*} x^{2} - e^{e^{e^{16 x^{2} - 8 x + \left (16 x - 4\right ) \log {\left (10 \right )} + 1 + 4 \log {\left (10 \right )}^{2}} + 4} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*ln(5)*ln(10)+(-32*x+8)*ln(5))*exp(4*ln(10)**2+2*(8*x-2)*ln(10)+16*x**2-8*x+1)*exp(exp(4*ln(10)*
*2+2*(8*x-2)*ln(10)+16*x**2-8*x+1)+4)*exp(ln(5)*exp(exp(4*ln(10)**2+2*(8*x-2)*ln(10)+16*x**2-8*x+1)+4))+2*x,x)

[Out]

x**2 - exp(exp(exp(16*x**2 - 8*x + (16*x - 4)*log(10) + 1 + 4*log(10)**2) + 4)*log(5))

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