3.58.67 \(\int \frac {100 x+800 x^5+(-125-750 x^4) \log (2)+(-25 x-150 x^5+(25+150 x^4) \log (2)) \log (\frac {1}{5} (-x+\log (2)))}{-100 x^3+100 x^2 \log (2)+(40 x^3-40 x^2 \log (2)) \log (\frac {1}{5} (-x+\log (2)))+(-4 x^3+4 x^2 \log (2)) \log ^2(\frac {1}{5} (-x+\log (2)))} \, dx\)

Optimal. Leaf size=29 \[ \frac {25 \left (-2+4 x^4\right )}{8 x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \]

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Rubi [F]  time = 1.60, 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 {100 x+800 x^5+\left (-125-750 x^4\right ) \log (2)+\left (-25 x-150 x^5+\left (25+150 x^4\right ) \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )}{-100 x^3+100 x^2 \log (2)+\left (40 x^3-40 x^2 \log (2)\right ) \log \left (\frac {1}{5} (-x+\log (2))\right )+\left (-4 x^3+4 x^2 \log (2)\right ) \log ^2\left (\frac {1}{5} (-x+\log (2))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(100*x + 800*x^5 + (-125 - 750*x^4)*Log[2] + (-25*x - 150*x^5 + (25 + 150*x^4)*Log[2])*Log[(-x + Log[2])/5
])/(-100*x^3 + 100*x^2*Log[2] + (40*x^3 - 40*x^2*Log[2])*Log[(-x + Log[2])/5] + (-4*x^3 + 4*x^2*Log[2])*Log[(-
x + Log[2])/5]^2),x]

[Out]

625*E^10*ExpIntegralEi[-2*(5 - Log[(-x + Log[2])/5])]*Log[2] - (125*E^5*ExpIntegralEi[-5 + Log[(-x + Log[2])/5
]]*Log[2]^2)/2 - (625*E^10*ExpIntegralEi[-2*(5 - Log[(-x + Log[2])/5])]*Log[4])/2 + (125*E^5*ExpIntegralEi[-5
+ Log[(-x + Log[2])/5]]*Log[2]*Log[4])/4 - (25*x^2*(x - Log[2]))/(2*(5 - Log[(-x + Log[2])/5])) - (25*(x - Log
[2])*Log[2]^2)/(2*(5 - Log[(-x + Log[2])/5])) + (25*(1 - 2*Log[2]^4))/(4*Log[2]*(5 - Log[(-x + Log[2])/5])) -
(25*x*(x - Log[2])*Log[4])/(4*(5 - Log[(-x + Log[2])/5])) - (25*Defer[Int][1/(x*(-5 + Log[(-x + Log[2])/5])^2)
, x])/(4*Log[2]) + (25*Defer[Int][1/(x^2*(-5 + Log[(-x + Log[2])/5])), x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 \left (-4 x-32 x^5+30 x^4 \log (2)+\log (32)+\left (1+6 x^4\right ) (x-\log (2)) \log \left (\frac {1}{5} (-x+\log (2))\right )\right )}{4 x^2 (x-\log (2)) \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx\\ &=\frac {25}{4} \int \frac {-4 x-32 x^5+30 x^4 \log (2)+\log (32)+\left (1+6 x^4\right ) (x-\log (2)) \log \left (\frac {1}{5} (-x+\log (2))\right )}{x^2 (x-\log (2)) \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx\\ &=\frac {25}{4} \int \left (\frac {1-2 x^4}{x (x-\log (2)) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}+\frac {1+6 x^4}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}\right ) \, dx\\ &=\frac {25}{4} \int \frac {1-2 x^4}{x (x-\log (2)) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx+\frac {25}{4} \int \frac {1+6 x^4}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx\\ &=\frac {25}{4} \int \left (-\frac {2 x^2}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}-\frac {1}{x \log (2) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}-\frac {2 \log ^2(2)}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}+\frac {1-2 \log ^4(2)}{(x-\log (2)) \log (2) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}-\frac {x \log (4)}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2}\right ) \, dx+\frac {25}{4} \int \left (\frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {6 x^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx\\ &=\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx-\frac {25}{2} \int \frac {x^2}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx+\frac {75}{2} \int \frac {x^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}-\frac {1}{2} \left (25 \log ^2(2)\right ) \int \frac {1}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx+\frac {\left (25 \left (1-2 \log ^4(2)\right )\right ) \int \frac {1}{(x-\log (2)) \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}-\frac {1}{4} (25 \log (4)) \int \frac {x}{\left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx\\ &=-\frac {25 x^2 (x-\log (2))}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 x (x-\log (2)) \log (4)}{4 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx+\frac {75}{2} \int \left (\frac {\log ^2(2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}+\frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {(-x+\log (2)) \log (4)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx-\frac {75}{2} \int \frac {x^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}+(25 \log (2)) \int \frac {x}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{2} \left (25 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-5+\log \left (\frac {x}{5}\right )\right )^2} \, dx,x,-x+\log (2)\right )+\frac {\left (25 \left (1-2 \log ^4(2)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-5+\log \left (\frac {x}{5}\right )\right )^2} \, dx,x,-x+\log (2)\right )}{4 \log (2)}-\frac {1}{2} (25 \log (4)) \int \frac {x}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{4} (25 \log (2) \log (4)) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx\\ &=-\frac {25 x^2 (x-\log (2))}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 (x-\log (2)) \log ^2(2)}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 x (x-\log (2)) \log (4)}{4 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx-\frac {75}{2} \int \left (\frac {\log ^2(2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}+\frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {(-x+\log (2)) \log (4)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx+\frac {75}{2} \int \frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}+(25 \log (2)) \int \left (\frac {\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx+\frac {1}{2} \left (25 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )+\frac {1}{2} \left (75 \log ^2(2)\right ) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {\left (25 \left (1-2 \log ^4(2)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}{4 \log (2)}-\frac {1}{2} (25 \log (4)) \int \left (\frac {\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}-\frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )}\right ) \, dx-\frac {1}{2} (75 \log (4)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{4} (25 \log (2) \log (4)) \operatorname {Subst}\left (\int \frac {1}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )\\ &=-\frac {25 x^2 (x-\log (2))}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 (x-\log (2)) \log ^2(2)}{2 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25 \left (1-2 \log ^4(2)\right )}{4 \log (2) \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}-\frac {25 x (x-\log (2)) \log (4)}{4 \left (5-\log \left (\frac {1}{5} (-x+\log (2))\right )\right )}+\frac {25}{4} \int \frac {1}{x^2 \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \, dx-\frac {75}{2} \int \frac {(-x+\log (2))^2}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {75}{2} \operatorname {Subst}\left (\int \frac {x^2}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )-\frac {25 \int \frac {1}{x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )^2} \, dx}{4 \log (2)}-(25 \log (2)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\left (25 \log ^2(2)\right ) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{2} \left (75 \log ^2(2)\right ) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{2} \left (75 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )+\frac {1}{2} \left (125 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {e^x}{-5+x} \, dx,x,\log \left (\frac {1}{5} (-x+\log (2))\right )\right )+\frac {1}{2} (25 \log (4)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{2} (75 \log (4)) \int \frac {-x+\log (2)}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx+\frac {1}{2} (75 \log (4)) \operatorname {Subst}\left (\int \frac {x}{-5+\log \left (\frac {x}{5}\right )} \, dx,x,-x+\log (2)\right )-\frac {1}{2} (25 \log (2) \log (4)) \int \frac {1}{-5+\log \left (\frac {1}{5} (-x+\log (2))\right )} \, dx-\frac {1}{4} (125 \log (2) \log (4)) \operatorname {Subst}\left (\int \frac {e^x}{-5+x} \, dx,x,\log \left (\frac {1}{5} (-x+\log (2))\right )\right )\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.30, size = 29, normalized size = 1.00 \begin {gather*} \frac {25 \left (-1+2 x^4\right )}{4 x \left (-5+\log \left (\frac {1}{5} (-x+\log (2))\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100*x + 800*x^5 + (-125 - 750*x^4)*Log[2] + (-25*x - 150*x^5 + (25 + 150*x^4)*Log[2])*Log[(-x + Log
[2])/5])/(-100*x^3 + 100*x^2*Log[2] + (40*x^3 - 40*x^2*Log[2])*Log[(-x + Log[2])/5] + (-4*x^3 + 4*x^2*Log[2])*
Log[(-x + Log[2])/5]^2),x]

[Out]

(25*(-1 + 2*x^4))/(4*x*(-5 + Log[(-x + Log[2])/5]))

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fricas [A]  time = 0.72, size = 26, normalized size = 0.90 \begin {gather*} \frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, \log \relax (2)\right ) - 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((150*x^4+25)*log(2)-150*x^5-25*x)*log(1/5*log(2)-1/5*x)+(-750*x^4-125)*log(2)+800*x^5+100*x)/((4*x
^2*log(2)-4*x^3)*log(1/5*log(2)-1/5*x)^2+(-40*x^2*log(2)+40*x^3)*log(1/5*log(2)-1/5*x)+100*x^2*log(2)-100*x^3)
,x, algorithm="fricas")

[Out]

25/4*(2*x^4 - 1)/(x*log(-1/5*x + 1/5*log(2)) - 5*x)

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giac [A]  time = 0.23, size = 29, normalized size = 1.00 \begin {gather*} -\frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x \log \relax (5) - x \log \left (-x + \log \relax (2)\right ) + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((150*x^4+25)*log(2)-150*x^5-25*x)*log(1/5*log(2)-1/5*x)+(-750*x^4-125)*log(2)+800*x^5+100*x)/((4*x
^2*log(2)-4*x^3)*log(1/5*log(2)-1/5*x)^2+(-40*x^2*log(2)+40*x^3)*log(1/5*log(2)-1/5*x)+100*x^2*log(2)-100*x^3)
,x, algorithm="giac")

[Out]

-25/4*(2*x^4 - 1)/(x*log(5) - x*log(-x + log(2)) + 5*x)

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maple [A]  time = 0.22, size = 25, normalized size = 0.86

method	result	size
norman	\(\frac {-\frac {25}{4}+\frac {25 x^{4}}{2}}{x \left (\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )-5\right )}\)	\(25\)
risch	\(\frac {-\frac {25}{4}+\frac {25 x^{4}}{2}}{x \left (\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )-5\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((150*x^4+25)*ln(2)-150*x^5-25*x)*ln(1/5*ln(2)-1/5*x)+(-750*x^4-125)*ln(2)+800*x^5+100*x)/((4*x^2*ln(2)-4
*x^3)*ln(1/5*ln(2)-1/5*x)^2+(-40*x^2*ln(2)+40*x^3)*ln(1/5*ln(2)-1/5*x)+100*x^2*ln(2)-100*x^3),x,method=_RETURN
VERBOSE)

[Out]

(-25/4+25/2*x^4)/x/(ln(1/5*ln(2)-1/5*x)-5)

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maxima [A]  time = 0.50, size = 28, normalized size = 0.97 \begin {gather*} -\frac {25 \, {\left (2 \, x^{4} - 1\right )}}{4 \, {\left (x {\left (\log \relax (5) + 5\right )} - x \log \left (-x + \log \relax (2)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((150*x^4+25)*log(2)-150*x^5-25*x)*log(1/5*log(2)-1/5*x)+(-750*x^4-125)*log(2)+800*x^5+100*x)/((4*x
^2*log(2)-4*x^3)*log(1/5*log(2)-1/5*x)^2+(-40*x^2*log(2)+40*x^3)*log(1/5*log(2)-1/5*x)+100*x^2*log(2)-100*x^3)
,x, algorithm="maxima")

[Out]

-25/4*(2*x^4 - 1)/(x*(log(5) + 5) - x*log(-x + log(2)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {100\,x-\ln \relax (2)\,\left (750\,x^4+125\right )-\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )\,\left (25\,x-\ln \relax (2)\,\left (150\,x^4+25\right )+150\,x^5\right )+800\,x^5}{\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )\,\left (40\,x^2\,\ln \relax (2)-40\,x^3\right )-{\ln \left (\frac {\ln \relax (2)}{5}-\frac {x}{5}\right )}^2\,\left (4\,x^2\,\ln \relax (2)-4\,x^3\right )-100\,x^2\,\ln \relax (2)+100\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(100*x - log(2)*(750*x^4 + 125) - log(log(2)/5 - x/5)*(25*x - log(2)*(150*x^4 + 25) + 150*x^5) + 800*x^5)
/(log(log(2)/5 - x/5)*(40*x^2*log(2) - 40*x^3) - log(log(2)/5 - x/5)^2*(4*x^2*log(2) - 4*x^3) - 100*x^2*log(2)
+ 100*x^3),x)

[Out]

int(-(100*x - log(2)*(750*x^4 + 125) - log(log(2)/5 - x/5)*(25*x - log(2)*(150*x^4 + 25) + 150*x^5) + 800*x^5)
/(log(log(2)/5 - x/5)*(40*x^2*log(2) - 40*x^3) - log(log(2)/5 - x/5)^2*(4*x^2*log(2) - 4*x^3) - 100*x^2*log(2)
+ 100*x^3), x)

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sympy [A]  time = 0.15, size = 22, normalized size = 0.76 \begin {gather*} \frac {50 x^{4} - 25}{4 x \log {\left (- \frac {x}{5} + \frac {\log {\relax (2 )}}{5} \right )} - 20 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((150*x**4+25)*ln(2)-150*x**5-25*x)*ln(1/5*ln(2)-1/5*x)+(-750*x**4-125)*ln(2)+800*x**5+100*x)/((4*x
**2*ln(2)-4*x**3)*ln(1/5*ln(2)-1/5*x)**2+(-40*x**2*ln(2)+40*x**3)*ln(1/5*ln(2)-1/5*x)+100*x**2*ln(2)-100*x**3)
,x)

[Out]

(50*x**4 - 25)/(4*x*log(-x/5 + log(2)/5) - 20*x)

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