3.40.50 \(\int \frac {-80-15 x+10 x^2+5 x^5+(3 x^3-5 x^5) \log (2)}{5 x^5} \, dx\)

Optimal. Leaf size=33 \[ x-\frac {\frac {-1-\frac {4}{x}+x}{x^2}+\left (\frac {3}{5}-x+x^2\right ) \log (2)}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 14} \begin {gather*} \frac {4}{x^4}+\frac {1}{x^3}-\frac {1}{x^2}+x (1-\log (2))-\frac {\log (8)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80 - 15*x + 10*x^2 + 5*x^5 + (3*x^3 - 5*x^5)*Log[2])/(5*x^5),x]

[Out]

4/x^4 + x^(-3) - x^(-2) + x*(1 - Log[2]) - Log[8]/(5*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-80-15 x+10 x^2+5 x^5+\left (3 x^3-5 x^5\right ) \log (2)}{x^5} \, dx\\ &=\frac {1}{5} \int \left (-\frac {80}{x^5}-\frac {15}{x^4}+\frac {10}{x^3}-5 (-1+\log (2))+\frac {\log (8)}{x^2}\right ) \, dx\\ &=\frac {4}{x^4}+\frac {1}{x^3}-\frac {1}{x^2}+x (1-\log (2))-\frac {\log (8)}{5 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.88 \begin {gather*} \frac {4}{x^4}+\frac {1}{x^3}-\frac {1}{x^2}+x-x \log (2)-\frac {\log (8)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80 - 15*x + 10*x^2 + 5*x^5 + (3*x^3 - 5*x^5)*Log[2])/(5*x^5),x]

[Out]

4/x^4 + x^(-3) - x^(-2) + x - x*Log[2] - Log[8]/(5*x)

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fricas [A]  time = 1.25, size = 35, normalized size = 1.06 \begin {gather*} \frac {5 \, x^{5} - 5 \, x^{2} - {\left (5 \, x^{5} + 3 \, x^{3}\right )} \log \relax (2) + 5 \, x + 20}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^5+3*x^3)*log(2)+5*x^5+10*x^2-15*x-80)/x^5,x, algorithm="fricas")

[Out]

1/5*(5*x^5 - 5*x^2 - (5*x^5 + 3*x^3)*log(2) + 5*x + 20)/x^4

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giac [A]  time = 0.20, size = 29, normalized size = 0.88 \begin {gather*} -x \log \relax (2) + x - \frac {3 \, x^{3} \log \relax (2) + 5 \, x^{2} - 5 \, x - 20}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^5+3*x^3)*log(2)+5*x^5+10*x^2-15*x-80)/x^5,x, algorithm="giac")

[Out]

-x*log(2) + x - 1/5*(3*x^3*log(2) + 5*x^2 - 5*x - 20)/x^4

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maple [A]  time = 0.04, size = 28, normalized size = 0.85

method	result	size
default	\(-x \ln \relax (2)+x +\frac {1}{x^{3}}-\frac {1}{x^{2}}-\frac {3 \ln \relax (2)}{5 x}+\frac {4}{x^{4}}\)	\(28\)
norman	\(\frac {4+x +\left (1-\ln \relax (2)\right ) x^{5}-x^{2}-\frac {3 x^{3} \ln \relax (2)}{5}}{x^{4}}\)	\(30\)
risch	\(-x \ln \relax (2)+x +\frac {-3 x^{3} \ln \relax (2)-5 x^{2}+5 x +20}{5 x^{4}}\)	\(30\)
gosper	\(-\frac {5 x^{5} \ln \relax (2)-5 x^{5}+3 x^{3} \ln \relax (2)+5 x^{2}-5 x -20}{5 x^{4}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-5*x^5+3*x^3)*ln(2)+5*x^5+10*x^2-15*x-80)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.38, size = 30, normalized size = 0.91 \begin {gather*} -x {\left (\log \relax (2) - 1\right )} - \frac {3 \, x^{3} \log \relax (2) + 5 \, x^{2} - 5 \, x - 20}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^5+3*x^3)*log(2)+5*x^5+10*x^2-15*x-80)/x^5,x, algorithm="maxima")

[Out]

-x*(log(2) - 1) - 1/5*(3*x^3*log(2) + 5*x^2 - 5*x - 20)/x^4

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mupad [B]  time = 0.07, size = 29, normalized size = 0.88 \begin {gather*} \frac {-\frac {3\,\ln \relax (2)\,x^3}{5}-x^2+x+4}{x^4}-x\,\left (\frac {\ln \left (32\right )}{5}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(2)*(3*x^3 - 5*x^5))/5 - 3*x + 2*x^2 + x^5 - 16)/x^5,x)

[Out]

(x - (3*x^3*log(2))/5 - x^2 + 4)/x^4 - x*(log(32)/5 - 1)

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sympy [A]  time = 0.27, size = 32, normalized size = 0.97 \begin {gather*} \frac {x \left (5 - 5 \log {\relax (2 )}\right )}{5} + \frac {- 3 x^{3} \log {\relax (2 )} - 5 x^{2} + 5 x + 20}{5 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x**5+3*x**3)*ln(2)+5*x**5+10*x**2-15*x-80)/x**5,x)

[Out]

x*(5 - 5*log(2))/5 + (-3*x**3*log(2) - 5*x**2 + 5*x + 20)/(5*x**4)

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