3.15.42 \(\int \frac {e^{10+2 e^{4-x}+10 x} (10-2 e^{4-x})+2 x \log ^2(4)+e^{5+e^{4-x}+5 x} (-2 e^{4-x} x \log (4)+(2+10 x) \log (4))}{\log ^2(4)} \, dx\)

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

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Rubi [F]  time = 0.60, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{10+2 e^{4-x}+10 x} \left (10-2 e^{4-x}\right )+2 x \log ^2(4)+e^{5+e^{4-x}+5 x} \left (-2 e^{4-x} x \log (4)+(2+10 x) \log (4)\right )}{\log ^2(4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(10 + 2*E^(4 - x) + 10*x)*(10 - 2*E^(4 - x)) + 2*x*Log[4]^2 + E^(5 + E^(4 - x) + 5*x)*(-2*E^(4 - x)*x*L
og[4] + (2 + 10*x)*Log[4]))/Log[4]^2,x]

[Out]

x^2 + (10240*E^50*Gamma[-10, -2*E^(4 - x)])/Log[4]^2 + (1024*E^50*Gamma[-9, -2*E^(4 - x)])/Log[4]^2 - (2*E^25*
Gamma[-5, -E^(4 - x)])/Log[4] - (2*Defer[Int][E^(9 + E^(4 - x) + 4*x)*x, x])/Log[4] + (10*Defer[Int][E^(5 + E^
(4 - x) + 5*x)*x, x])/Log[4]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (e^{10+2 e^{4-x}+10 x} \left (10-2 e^{4-x}\right )+2 x \log ^2(4)+e^{5+e^{4-x}+5 x} \left (-2 e^{4-x} x \log (4)+(2+10 x) \log (4)\right )\right ) \, dx}{\log ^2(4)}\\ &=x^2+\frac {\int e^{10+2 e^{4-x}+10 x} \left (10-2 e^{4-x}\right ) \, dx}{\log ^2(4)}+\frac {\int e^{5+e^{4-x}+5 x} \left (-2 e^{4-x} x \log (4)+(2+10 x) \log (4)\right ) \, dx}{\log ^2(4)}\\ &=x^2+\frac {\int \left (-2 e^{9+e^{4-x}+4 x} x \log (4)+2 e^{5+e^{4-x}+5 x} (1+5 x) \log (4)\right ) \, dx}{\log ^2(4)}+\frac {\operatorname {Subst}\left (\int 2 e^{10+\frac {2 e^4}{x}} x^8 \left (-e^4+5 x\right ) \, dx,x,e^x\right )}{\log ^2(4)}\\ &=x^2+\frac {2 \operatorname {Subst}\left (\int e^{10+\frac {2 e^4}{x}} x^8 \left (-e^4+5 x\right ) \, dx,x,e^x\right )}{\log ^2(4)}-\frac {2 \int e^{9+e^{4-x}+4 x} x \, dx}{\log (4)}+\frac {2 \int e^{5+e^{4-x}+5 x} (1+5 x) \, dx}{\log (4)}\\ &=x^2+\frac {2 \operatorname {Subst}\left (\int \left (-e^{14+\frac {2 e^4}{x}} x^8+5 e^{10+\frac {2 e^4}{x}} x^9\right ) \, dx,x,e^x\right )}{\log ^2(4)}-\frac {2 \int e^{9+e^{4-x}+4 x} x \, dx}{\log (4)}+\frac {2 \int \left (e^{5+e^{4-x}+5 x}+5 e^{5+e^{4-x}+5 x} x\right ) \, dx}{\log (4)}\\ &=x^2-\frac {2 \operatorname {Subst}\left (\int e^{14+\frac {2 e^4}{x}} x^8 \, dx,x,e^x\right )}{\log ^2(4)}+\frac {10 \operatorname {Subst}\left (\int e^{10+\frac {2 e^4}{x}} x^9 \, dx,x,e^x\right )}{\log ^2(4)}+\frac {2 \int e^{5+e^{4-x}+5 x} \, dx}{\log (4)}-\frac {2 \int e^{9+e^{4-x}+4 x} x \, dx}{\log (4)}+\frac {10 \int e^{5+e^{4-x}+5 x} x \, dx}{\log (4)}\\ &=x^2+\frac {10240 e^{50} \Gamma \left (-10,-2 e^{4-x}\right )}{\log ^2(4)}+\frac {1024 e^{50} \Gamma \left (-9,-2 e^{4-x}\right )}{\log ^2(4)}-\frac {2 \int e^{9+e^{4-x}+4 x} x \, dx}{\log (4)}+\frac {2 \operatorname {Subst}\left (\int e^{5+\frac {e^4}{x}} x^4 \, dx,x,e^x\right )}{\log (4)}+\frac {10 \int e^{5+e^{4-x}+5 x} x \, dx}{\log (4)}\\ &=x^2+\frac {10240 e^{50} \Gamma \left (-10,-2 e^{4-x}\right )}{\log ^2(4)}+\frac {1024 e^{50} \Gamma \left (-9,-2 e^{4-x}\right )}{\log ^2(4)}-\frac {2 e^{25} \Gamma \left (-5,-e^{4-x}\right )}{\log (4)}-\frac {2 \int e^{9+e^{4-x}+4 x} x \, dx}{\log (4)}+\frac {10 \int e^{5+e^{4-x}+5 x} x \, dx}{\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 26, normalized size = 1.13 \begin {gather*} \frac {\left (e^{5+e^{4-x}+5 x}+x \log (4)\right )^2}{\log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(10 + 2*E^(4 - x) + 10*x)*(10 - 2*E^(4 - x)) + 2*x*Log[4]^2 + E^(5 + E^(4 - x) + 5*x)*(-2*E^(4 -
x)*x*Log[4] + (2 + 10*x)*Log[4]))/Log[4]^2,x]

[Out]

(E^(5 + E^(4 - x) + 5*x) + x*Log[4])^2/Log[4]^2

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fricas [A]  time = 0.68, size = 47, normalized size = 2.04 \begin {gather*} \frac {4 \, x^{2} \log \relax (2)^{2} + 4 \, x e^{\left (5 \, x + e^{\left (-x + 4\right )} + 5\right )} \log \relax (2) + e^{\left (10 \, x + 2 \, e^{\left (-x + 4\right )} + 10\right )}}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, x \log \relax (2)^{2} - {\left (e^{\left (-x + 4\right )} - 5\right )} e^{\left (10 \, x + 2 \, e^{\left (-x + 4\right )} + 10\right )} - 2 \, {\left (x e^{\left (-x + 4\right )} \log \relax (2) - {\left (5 \, x + 1\right )} \log \relax (2)\right )} e^{\left (5 \, x + e^{\left (-x + 4\right )} + 5\right )}}{2 \, \log \relax (2)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/2*(4*x*log(2)^2 - (e^(-x + 4) - 5)*e^(10*x + 2*e^(-x + 4) + 10) - 2*(x*e^(-x + 4)*log(2) - (5*x +
1)*log(2))*e^(5*x + e^(-x + 4) + 5))/log(2)^2, x)

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maple [A]  time = 0.15, size = 43, normalized size = 1.87

method	result	size
risch	\(x^{2}+\frac {{\mathrm e}^{10 x +10+2 \,{\mathrm e}^{-x +4}}}{4 \ln \relax (2)^{2}}+\frac {x \,{\mathrm e}^{5+{\mathrm e}^{-x +4}+5 x}}{\ln \relax (2)}\)	\(43\)
default	\(\frac {4 \ln \relax (2) {\mathrm e}^{25} x \,{\mathrm e}^{{\mathrm e}^{-x +4}} {\mathrm e}^{5 x -20}-10 \,{\mathrm e}^{10} {\mathrm e}^{40} \left (-\frac {{\mathrm e}^{10 x} {\mathrm e}^{-40} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{10}-\frac {{\mathrm e}^{9 x} {\mathrm e}^{-36} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{45}-\frac {{\mathrm e}^{8 x} {\mathrm e}^{-32} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{180}-\frac {{\mathrm e}^{7 x} {\mathrm e}^{-28} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{630}-\frac {{\mathrm e}^{6 x} {\mathrm e}^{-24} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{1890}-\frac {{\mathrm e}^{5 x} {\mathrm e}^{-20} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{4725}-\frac {{\mathrm e}^{4 x} {\mathrm e}^{-16} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{9450}-\frac {{\mathrm e}^{3 x} {\mathrm e}^{-12} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{14175}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{-x +4}} {\mathrm e}^{2 x} {\mathrm e}^{-8}}{14175}-\frac {2 \,{\mathrm e}^{x} {\mathrm e}^{-4} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{14175}-\frac {4 \expIntegralEi \left (1, -2 \,{\mathrm e}^{-x} {\mathrm e}^{4}\right )}{14175}\right )+2 \,{\mathrm e}^{10} {\mathrm e}^{40} \left (-\frac {{\mathrm e}^{9 x} {\mathrm e}^{-36} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{9}-\frac {{\mathrm e}^{8 x} {\mathrm e}^{-32} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{36}-\frac {{\mathrm e}^{7 x} {\mathrm e}^{-28} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{126}-\frac {{\mathrm e}^{6 x} {\mathrm e}^{-24} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{378}-\frac {{\mathrm e}^{5 x} {\mathrm e}^{-20} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{945}-\frac {{\mathrm e}^{4 x} {\mathrm e}^{-16} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{1890}-\frac {{\mathrm e}^{3 x} {\mathrm e}^{-12} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{2835}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{-x +4}} {\mathrm e}^{2 x} {\mathrm e}^{-8}}{2835}-\frac {2 \,{\mathrm e}^{x} {\mathrm e}^{-4} {\mathrm e}^{2 \,{\mathrm e}^{-x +4}}}{2835}-\frac {4 \expIntegralEi \left (1, -2 \,{\mathrm e}^{-x} {\mathrm e}^{4}\right )}{2835}\right )+4 x^{2} \ln \relax (2)^{2}}{4 \ln \relax (2)^{2}}\)	\(461\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-2*exp(-x+4)+10)*exp(5*x+5)^2*exp(exp(-x+4))^2+(-4*x*ln(2)*exp(-x+4)+2*(10*x+2)*ln(2))*exp(5*x+5)*ex
p(exp(-x+4))+8*x*ln(2)^2)/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

x^2+1/4/ln(2)^2*exp(10*x+10+2*exp(-x+4))+x/ln(2)*exp(5+exp(-x+4)+5*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4 \, x^{2} \log \relax (2)^{2} + 4 \, x e^{\left (5 \, x + e^{\left (-x + 4\right )} + 5\right )} \log \relax (2) + e^{\left (10 \, x + 2 \, e^{\left (-x + 4\right )} + 10\right )}}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x^2*log(2)^2 + e^(10*x + 2*e^(-x + 4) + 10) + 2*integrate(-2*(x*e^(4*x + 9)*log(2) - (5*x*e^5*log(2) +
e^5*log(2))*e^(5*x))*e^(e^(-x + 4)), x))/log(2)^2

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mupad [B]  time = 0.15, size = 27, normalized size = 1.17 \begin {gather*} \frac {{\left ({\mathrm {e}}^{5\,x+{\mathrm {e}}^{-x}\,{\mathrm {e}}^4+5}+2\,x\,\ln \relax (2)\right )}^2}{4\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*log(2)^2 + (exp(exp(4 - x))*exp(5*x + 5)*(2*log(2)*(10*x + 2) - 4*x*exp(4 - x)*log(2)))/4 - (exp(2*ex
p(4 - x))*exp(10*x + 10)*(2*exp(4 - x) - 10))/4)/log(2)^2,x)

[Out]

(exp(5*x + exp(-x)*exp(4) + 5) + 2*x*log(2))^2/(4*log(2)^2)

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sympy [B]  time = 0.30, size = 63, normalized size = 2.74 \begin {gather*} x^{2} + \frac {\left (4 x e^{25} e^{40 - 10 x} e^{e^{4 - x}} \log {\relax (2 )}^{2} + e^{50} e^{20 - 5 x} e^{2 e^{4 - x}} \log {\relax (2 )}\right ) e^{15 x - 60}}{4 \log {\relax (2 )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2 + (4*x*exp(25)*exp(40 - 10*x)*exp(exp(4 - x))*log(2)**2 + exp(50)*exp(20 - 5*x)*exp(2*exp(4 - x))*log(2))
*exp(15*x - 60)/(4*log(2)**3)

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