3.91.95 \(\int \frac {30+30 x+e^4 (10+25 x+15 x^2)+(3 x+e^4 (-15 x-15 x^2)) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 (15 x+15 x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \log (x)+\frac {-\frac {2}{3}-x+\frac {-2+\frac {1}{5} \log (1+x)}{e^4}}{\log (x)} \]

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Rubi [F]  time = 1.31, 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 {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(30 + 30*x + E^4*(10 + 25*x + 15*x^2) + (3*x + E^4*(-15*x - 15*x^2))*Log[x] + E^4*(15 + 15*x)*Log[x]^2 + (
-3 - 3*x)*Log[1 + x])/(E^4*(15*x + 15*x^2)*Log[x]^2),x]

[Out]

(-2*(3 + E^4))/(3*E^4*Log[x]) - x/Log[x] + Log[x] + LogIntegral[x] - Defer[Int][(-1 + 5*E^4 + 5*E^4*x)/((1 + x
)*Log[x]), x]/(5*E^4) - Defer[Int][Log[1 + x]/(x*Log[x]^2), x]/(5*E^4)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{\left (15 x+15 x^2\right ) \log ^2(x)} \, dx}{e^4}\\ &=\frac {\int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{x (15+15 x) \log ^2(x)} \, dx}{e^4}\\ &=\frac {\int \left (\frac {30 \left (1+\frac {e^4}{3}\right )+30 \left (1+\frac {5 e^4}{6}\right ) x+15 e^4 x^2+3 \left (1-5 e^4\right ) x \log (x)-15 e^4 x^2 \log (x)+15 e^4 \log ^2(x)+15 e^4 x \log ^2(x)}{15 x (1+x) \log ^2(x)}-\frac {\log (1+x)}{5 x \log ^2(x)}\right ) \, dx}{e^4}\\ &=\frac {\int \frac {30 \left (1+\frac {e^4}{3}\right )+30 \left (1+\frac {5 e^4}{6}\right ) x+15 e^4 x^2+3 \left (1-5 e^4\right ) x \log (x)-15 e^4 x^2 \log (x)+15 e^4 \log ^2(x)+15 e^4 x \log ^2(x)}{x (1+x) \log ^2(x)} \, dx}{15 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ &=\frac {\int \frac {5 \left (6 (1+x)+e^4 \left (2+5 x+3 x^2\right )\right )-3 x \left (-1+5 e^4 (1+x)\right ) \log (x)+15 e^4 (1+x) \log ^2(x)}{x (1+x) \log ^2(x)} \, dx}{15 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ &=\frac {\int \left (\frac {15 e^4}{x}+\frac {5 \left (6+2 e^4+3 e^4 x\right )}{x \log ^2(x)}-\frac {3 \left (-1+5 e^4+5 e^4 x\right )}{(1+x) \log (x)}\right ) \, dx}{15 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ &=\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\int \frac {6+2 e^4+3 e^4 x}{x \log ^2(x)} \, dx}{3 e^4}\\ &=\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\int \left (\frac {3 e^4}{\log ^2(x)}+\frac {2 \left (3+e^4\right )}{x \log ^2(x)}\right ) \, dx}{3 e^4}\\ &=\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\left (2 \left (3+e^4\right )\right ) \int \frac {1}{x \log ^2(x)} \, dx}{3 e^4}+\int \frac {1}{\log ^2(x)} \, dx\\ &=-\frac {x}{\log (x)}+\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\left (2 \left (3+e^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )}{3 e^4}+\int \frac {1}{\log (x)} \, dx\\ &=-\frac {2 \left (3+e^4\right )}{3 e^4 \log (x)}-\frac {x}{\log (x)}+\log (x)+\text {li}(x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 42, normalized size = 1.45 \begin {gather*} -\frac {6+2 e^4+3 e^4 x}{3 e^4 \log (x)}+\log (x)+\frac {\log (1+x)}{5 e^4 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30 + 30*x + E^4*(10 + 25*x + 15*x^2) + (3*x + E^4*(-15*x - 15*x^2))*Log[x] + E^4*(15 + 15*x)*Log[x]
^2 + (-3 - 3*x)*Log[1 + x])/(E^4*(15*x + 15*x^2)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.63, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left (15 \, e^{4} \log \relax (x)^{2} - 5 \, {\left (3 \, x + 2\right )} e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-3)*log(x+1)+(15*x+15)*exp(4)*log(x)^2+((-15*x^2-15*x)*exp(4)+3*x)*log(x)+(15*x^2+25*x+10)*exp
(4)+30*x+30)/(15*x^2+15*x)/exp(4)/log(x)^2,x, algorithm="fricas")

[Out]

1/15*(15*e^4*log(x)^2 - 5*(3*x + 2)*e^4 + 3*log(x + 1) - 30)*e^(-4)/log(x)

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giac [A]  time = 0.16, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left (15 \, e^{4} \log \relax (x)^{2} - 15 \, x e^{4} - 10 \, e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-3)*log(x+1)+(15*x+15)*exp(4)*log(x)^2+((-15*x^2-15*x)*exp(4)+3*x)*log(x)+(15*x^2+25*x+10)*exp
(4)+30*x+30)/(15*x^2+15*x)/exp(4)/log(x)^2,x, algorithm="giac")

[Out]

1/15*(15*e^4*log(x)^2 - 15*x*e^4 - 10*e^4 + 3*log(x + 1) - 30)*e^(-4)/log(x)

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maple [A]  time = 0.48, size = 41, normalized size = 1.41

method	result	size
risch	\(\frac {{\mathrm e}^{-4} \ln \left (x +1\right )}{5 \ln \relax (x )}+\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \relax (x )^{2}-3 x \,{\mathrm e}^{4}-2 \,{\mathrm e}^{4}-6\right )}{3 \ln \relax (x )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x-3)*ln(x+1)+(15*x+15)*exp(4)*ln(x)^2+((-15*x^2-15*x)*exp(4)+3*x)*ln(x)+(15*x^2+25*x+10)*exp(4)+30*x+
30)/(15*x^2+15*x)/exp(4)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/5*exp(-4)/ln(x)*ln(x+1)+1/3*exp(-4)*(3*exp(4)*ln(x)^2-3*x*exp(4)-2*exp(4)-6)/ln(x)

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maxima [A]  time = 0.39, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left (15 \, e^{4} \log \relax (x)^{2} - 15 \, x e^{4} - 10 \, e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-3)*log(x+1)+(15*x+15)*exp(4)*log(x)^2+((-15*x^2-15*x)*exp(4)+3*x)*log(x)+(15*x^2+25*x+10)*exp
(4)+30*x+30)/(15*x^2+15*x)/exp(4)/log(x)^2,x, algorithm="maxima")

[Out]

1/15*(15*e^4*log(x)^2 - 15*x*e^4 - 10*e^4 + 3*log(x + 1) - 30)*e^(-4)/log(x)

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mupad [B]  time = 5.62, size = 28, normalized size = 0.97 \begin {gather*} \ln \relax (x)-\frac {{\mathrm {e}}^{-4}\,\left (10\,{\mathrm {e}}^4-3\,\ln \left (x+1\right )+15\,x\,{\mathrm {e}}^4+30\right )}{15\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4)*(30*x + exp(4)*(25*x + 15*x^2 + 10) + log(x)*(3*x - exp(4)*(15*x + 15*x^2)) - log(x + 1)*(3*x + 3
) + exp(4)*log(x)^2*(15*x + 15) + 30))/(log(x)^2*(15*x + 15*x^2)),x)

[Out]

log(x) - (exp(-4)*(10*exp(4) - 3*log(x + 1) + 15*x*exp(4) + 30))/(15*log(x))

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sympy [A]  time = 0.40, size = 39, normalized size = 1.34 \begin {gather*} \frac {- 3 x e^{4} - 2 e^{4} - 6}{3 e^{4} \log {\relax (x )}} + \log {\relax (x )} + \frac {\log {\left (x + 1 \right )}}{5 e^{4} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-3)*ln(x+1)+(15*x+15)*exp(4)*ln(x)**2+((-15*x**2-15*x)*exp(4)+3*x)*ln(x)+(15*x**2+25*x+10)*exp
(4)+30*x+30)/(15*x**2+15*x)/exp(4)/ln(x)**2,x)

[Out]

(-3*x*exp(4) - 2*exp(4) - 6)*exp(-4)/(3*log(x)) + log(x) + exp(-4)*log(x + 1)/(5*log(x))

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