3.74.8 \(\int \frac {e^x (130+80 x)+e^x (50 x^2+80 x^3+e^5 (169 x^2+208 x^3+64 x^4)) \log (x)+e^x (-130-30 x+80 x^2) \log (x) \log (\log (x))}{(169 x^2+208 x^3+64 x^4) \log (x)} \, dx\)

Optimal. Leaf size=30 \[ e^x \left (e^5+\frac {x+\log (\log (x))}{\frac {x}{2}+\frac {4}{5} x (1+x)}\right ) \]

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Rubi [F]  time = 4.02, 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^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(130 + 80*x) + E^x*(50*x^2 + 80*x^3 + E^5*(169*x^2 + 208*x^3 + 64*x^4))*Log[x] + E^x*(-130 - 30*x + 8
0*x^2)*Log[x]*Log[Log[x]])/((169*x^2 + 208*x^3 + 64*x^4)*Log[x]),x]

[Out]

E^(5 + x) + (10*E^x)/(13 + 8*x) + (10*Defer[Int][E^x/(x^2*Log[x]), x])/13 - (80*Defer[Int][E^x/(x*Log[x]), x])
/169 + (640*Defer[Int][E^x/((13 + 8*x)*Log[x]), x])/169 - (10*Defer[Int][(E^x*Log[Log[x]])/x^2, x])/13 + (10*D
efer[Int][(E^x*Log[Log[x]])/x, x])/13 + (640*Defer[Int][(E^x*Log[Log[x]])/(13 + 8*x)^2, x])/13 - (80*Defer[Int
][(E^x*Log[Log[x]])/(13 + 8*x), x])/13

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{x^2 \left (169+208 x+64 x^2\right ) \log (x)} \, dx\\ &=\int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{x^2 (13+8 x)^2 \log (x)} \, dx\\ &=\int \frac {e^x \left (130+80 x+\log (x) \left (x^2 \left (50+80 x+e^5 (13+8 x)^2\right )+10 \left (-13-3 x+8 x^2\right ) \log (\log (x))\right )\right )}{x^2 (13+8 x)^2 \log (x)} \, dx\\ &=\int \left (\frac {e^x \left (130+80 x+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+64 e^5 x^4 \log (x)\right )}{x^2 (13+8 x)^2 \log (x)}+\frac {10 e^x \left (-13-3 x+8 x^2\right ) \log (\log (x))}{x^2 (13+8 x)^2}\right ) \, dx\\ &=10 \int \frac {e^x \left (-13-3 x+8 x^2\right ) \log (\log (x))}{x^2 (13+8 x)^2} \, dx+\int \frac {e^x \left (130+80 x+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+64 e^5 x^4 \log (x)\right )}{x^2 (13+8 x)^2 \log (x)} \, dx\\ &=10 \int \left (-\frac {e^x \log (\log (x))}{13 x^2}+\frac {e^x \log (\log (x))}{13 x}+\frac {64 e^x \log (\log (x))}{13 (13+8 x)^2}-\frac {8 e^x \log (\log (x))}{13 (13+8 x)}\right ) \, dx+\int \frac {e^x \left (130+80 x+x^2 \left (50+80 x+e^5 (13+8 x)^2\right ) \log (x)\right )}{x^2 (13+8 x)^2 \log (x)} \, dx\\ &=-\left (\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2} \, dx\right )+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x} \, dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{13+8 x} \, dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(13+8 x)^2} \, dx+\int \left (\frac {e^x \left (50+169 e^5+16 \left (5+13 e^5\right ) x+64 e^5 x^2\right )}{(13+8 x)^2}+\frac {10 e^x}{x^2 (13+8 x) \log (x)}\right ) \, dx\\ &=-\left (\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2} \, dx\right )+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x} \, dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{13+8 x} \, dx+10 \int \frac {e^x}{x^2 (13+8 x) \log (x)} \, dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(13+8 x)^2} \, dx+\int \frac {e^x \left (50+169 e^5+16 \left (5+13 e^5\right ) x+64 e^5 x^2\right )}{(13+8 x)^2} \, dx\\ &=-\left (\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2} \, dx\right )+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x} \, dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{13+8 x} \, dx+10 \int \left (\frac {e^x}{13 x^2 \log (x)}-\frac {8 e^x}{169 x \log (x)}+\frac {64 e^x}{169 (13+8 x) \log (x)}\right ) \, dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(13+8 x)^2} \, dx+\int \left (e^{5+x}-\frac {80 e^x}{(13+8 x)^2}+\frac {10 e^x}{13+8 x}\right ) \, dx\\ &=-\left (\frac {80}{169} \int \frac {e^x}{x \log (x)} \, dx\right )+\frac {10}{13} \int \frac {e^x}{x^2 \log (x)} \, dx-\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2} \, dx+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x} \, dx+\frac {640}{169} \int \frac {e^x}{(13+8 x) \log (x)} \, dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{13+8 x} \, dx+10 \int \frac {e^x}{13+8 x} \, dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(13+8 x)^2} \, dx-80 \int \frac {e^x}{(13+8 x)^2} \, dx+\int e^{5+x} \, dx\\ &=e^{5+x}+\frac {10 e^x}{13+8 x}+\frac {5 \text {Ei}\left (\frac {1}{8} (13+8 x)\right )}{4 e^{13/8}}-\frac {80}{169} \int \frac {e^x}{x \log (x)} \, dx+\frac {10}{13} \int \frac {e^x}{x^2 \log (x)} \, dx-\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2} \, dx+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x} \, dx+\frac {640}{169} \int \frac {e^x}{(13+8 x) \log (x)} \, dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{13+8 x} \, dx-10 \int \frac {e^x}{13+8 x} \, dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(13+8 x)^2} \, dx\\ &=e^{5+x}+\frac {10 e^x}{13+8 x}-\frac {80}{169} \int \frac {e^x}{x \log (x)} \, dx+\frac {10}{13} \int \frac {e^x}{x^2 \log (x)} \, dx-\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2} \, dx+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x} \, dx+\frac {640}{169} \int \frac {e^x}{(13+8 x) \log (x)} \, dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{13+8 x} \, dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(13+8 x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 33, normalized size = 1.10 \begin {gather*} \frac {e^x \left (x \left (10+e^5 (13+8 x)\right )+10 \log (\log (x))\right )}{x (13+8 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(130 + 80*x) + E^x*(50*x^2 + 80*x^3 + E^5*(169*x^2 + 208*x^3 + 64*x^4))*Log[x] + E^x*(-130 - 30
*x + 80*x^2)*Log[x]*Log[Log[x]])/((169*x^2 + 208*x^3 + 64*x^4)*Log[x]),x]

[Out]

(E^x*(x*(10 + E^5*(13 + 8*x)) + 10*Log[Log[x]]))/(x*(13 + 8*x))

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fricas [A]  time = 0.62, size = 39, normalized size = 1.30 \begin {gather*} \frac {{\left ({\left (8 \, x^{2} + 13 \, x\right )} e^{5} + 10 \, x\right )} e^{x} + 10 \, e^{x} \log \left (\log \relax (x)\right )}{8 \, x^{2} + 13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+169*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*
log(x)+(80*x+130)*exp(x))/(64*x^4+208*x^3+169*x^2)/log(x),x, algorithm="fricas")

[Out]

(((8*x^2 + 13*x)*e^5 + 10*x)*e^x + 10*e^x*log(log(x)))/(8*x^2 + 13*x)

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giac [A]  time = 0.24, size = 41, normalized size = 1.37 \begin {gather*} \frac {8 \, x^{2} e^{\left (x + 5\right )} + 13 \, x e^{\left (x + 5\right )} + 10 \, x e^{x} + 10 \, e^{x} \log \left (\log \relax (x)\right )}{8 \, x^{2} + 13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+169*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*
log(x)+(80*x+130)*exp(x))/(64*x^4+208*x^3+169*x^2)/log(x),x, algorithm="giac")

[Out]

(8*x^2*e^(x + 5) + 13*x*e^(x + 5) + 10*x*e^x + 10*e^x*log(log(x)))/(8*x^2 + 13*x)

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maple [A]  time = 0.04, size = 40, normalized size = 1.33

method	result	size
risch	\(\frac {10 \,{\mathrm e}^{x} \ln \left (\ln \relax (x )\right )}{\left (8 x +13\right ) x}+\frac {\left (8 x \,{\mathrm e}^{5}+13 \,{\mathrm e}^{5}+10\right ) {\mathrm e}^{x}}{8 x +13}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((80*x^2-30*x-130)*exp(x)*ln(x)*ln(ln(x))+((64*x^4+208*x^3+169*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*ln(x)+(80
*x+130)*exp(x))/(64*x^4+208*x^3+169*x^2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

10/(8*x+13)/x*exp(x)*ln(ln(x))+(8*x*exp(5)+13*exp(5)+10)/(8*x+13)*exp(x)

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maxima [A]  time = 0.41, size = 39, normalized size = 1.30 \begin {gather*} \frac {{\left (8 \, x^{2} e^{5} + x {\left (13 \, e^{5} + 10\right )}\right )} e^{x} + 10 \, e^{x} \log \left (\log \relax (x)\right )}{8 \, x^{2} + 13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+169*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*
log(x)+(80*x+130)*exp(x))/(64*x^4+208*x^3+169*x^2)/log(x),x, algorithm="maxima")

[Out]

((8*x^2*e^5 + x*(13*e^5 + 10))*e^x + 10*e^x*log(log(x)))/(8*x^2 + 13*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (80\,x+130\right )+{\mathrm {e}}^x\,\ln \relax (x)\,\left ({\mathrm {e}}^5\,\left (64\,x^4+208\,x^3+169\,x^2\right )+50\,x^2+80\,x^3\right )-\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^x\,\ln \relax (x)\,\left (-80\,x^2+30\,x+130\right )}{\ln \relax (x)\,\left (64\,x^4+208\,x^3+169\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(80*x + 130) + exp(x)*log(x)*(exp(5)*(169*x^2 + 208*x^3 + 64*x^4) + 50*x^2 + 80*x^3) - log(log(x))
*exp(x)*log(x)*(30*x - 80*x^2 + 130))/(log(x)*(169*x^2 + 208*x^3 + 64*x^4)),x)

[Out]

int((exp(x)*(80*x + 130) + exp(x)*log(x)*(exp(5)*(169*x^2 + 208*x^3 + 64*x^4) + 50*x^2 + 80*x^3) - log(log(x))
*exp(x)*log(x)*(30*x - 80*x^2 + 130))/(log(x)*(169*x^2 + 208*x^3 + 64*x^4)), x)

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sympy [A]  time = 0.53, size = 36, normalized size = 1.20 \begin {gather*} \frac {\left (8 x^{2} e^{5} + 10 x + 13 x e^{5} + 10 \log {\left (\log {\relax (x )} \right )}\right ) e^{x}}{8 x^{2} + 13 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((80*x**2-30*x-130)*exp(x)*ln(x)*ln(ln(x))+((64*x**4+208*x**3+169*x**2)*exp(5)+80*x**3+50*x**2)*exp(
x)*ln(x)+(80*x+130)*exp(x))/(64*x**4+208*x**3+169*x**2)/ln(x),x)

[Out]

(8*x**2*exp(5) + 10*x + 13*x*exp(5) + 10*log(log(x)))*exp(x)/(8*x**2 + 13*x)

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