3.89.70 \(\int \frac {1+(3+5 x^4) \log ^2(2)+\log ^2(2) \log (x^{1+\frac {1}{\log ^2(2)}})}{\log ^2(2)} \, dx\)

Optimal. Leaf size=16 \[ x \left (2+x^4+\log \left (x^{1+\frac {1}{\log ^2(2)}}\right )\right ) \]

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Rubi [B]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 2.06, number of steps used = 4, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {12, 2295} \begin {gather*} x \log \left (x^{1+\frac {1}{\log ^2(2)}}\right )+x^5+3 x+\frac {x}{\log ^2(2)}-x \left (1+\frac {1}{\log ^2(2)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} x \left (2+x^4+\log \left (x^{1+\frac {1}{\log ^2(2)}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 29, normalized size = 1.81 \begin {gather*} \frac {{\left (x^{5} + 2 \, x\right )} \log \relax (2)^{2} + {\left (x \log \relax (2)^{2} + x\right )} \log \relax (x)}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)^2*log(x*exp(log(x)/log(2)^2))+(5*x^4+3)*log(2)^2+1)/log(2)^2,x, algorithm="fricas")

[Out]

((x^5 + 2*x)*log(2)^2 + (x*log(2)^2 + x)*log(x))/log(2)^2

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giac [B]  time = 0.17, size = 38, normalized size = 2.38 \begin {gather*} \frac {{\left (x \log \relax (x) - x\right )} {\left (\frac {1}{\log \relax (2)^{2}} + 1\right )} \log \relax (2)^{2} + {\left (x^{5} + 3 \, x\right )} \log \relax (2)^{2} + x}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)^2*log(x*exp(log(x)/log(2)^2))+(5*x^4+3)*log(2)^2+1)/log(2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.19, size = 37, normalized size = 2.31

method	result	size
default	\(\frac {x^{5} \ln \relax (2)^{2}+2 x \ln \relax (2)^{2}+\ln \relax (2)^{2} \ln \left (x \,x^{\frac {1}{\ln \relax (2)^{2}}}\right ) x}{\ln \relax (2)^{2}}\)	\(37\)
risch	\(x \ln \left (x^{\frac {1}{\ln \relax (2)^{2}}}\right )+x \ln \relax (x )-\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{\frac {1}{\ln \relax (2)^{2}}}\right ) \mathrm {csgn}\left (i x \,x^{\frac {1}{\ln \relax (2)^{2}}}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,x^{\frac {1}{\ln \relax (2)^{2}}}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i x^{\frac {1}{\ln \relax (2)^{2}}}\right ) \mathrm {csgn}\left (i x \,x^{\frac {1}{\ln \relax (2)^{2}}}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (i x \,x^{\frac {1}{\ln \relax (2)^{2}}}\right )^{3}}{2}+x^{5}+2 x\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(2)^2*ln(x*exp(ln(x)/ln(2)^2))+(5*x^4+3)*ln(2)^2+1)/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(2)^2*(x^5*ln(2)^2+2*x*ln(2)^2+ln(2)^2*ln(x*x^(1/ln(2)^2))*x)

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maxima [B]  time = 0.37, size = 51, normalized size = 3.19 \begin {gather*} -\frac {x {\left (\frac {1}{\log \relax (2)^{2}} + 1\right )} \log \relax (2)^{2} - x \log \relax (2)^{2} \log \left (x^{\frac {1}{\log \relax (2)^{2}} + 1}\right ) - {\left (x^{5} + 3 \, x\right )} \log \relax (2)^{2} - x}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)^2*log(x*exp(log(x)/log(2)^2))+(5*x^4+3)*log(2)^2+1)/log(2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.22, size = 16, normalized size = 1.00 \begin {gather*} x\,\left (\ln \left (x^{\frac {1}{{\ln \relax (2)}^2}+1}\right )+x^4+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)^2*log(x*x^(1/log(2)^2)) + log(2)^2*(5*x^4 + 3) + 1)/log(2)^2,x)

[Out]

x*(log(x^(1/log(2)^2 + 1)) + x^4 + 2)

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sympy [B]  time = 1.12, size = 36, normalized size = 2.25 \begin {gather*} \frac {x^{5} \log {\relax (2 )}^{2} + x \log {\relax (2 )}^{2} \log {\relax (x )} + x \log {\relax (x )} + 2 x \log {\relax (2 )}^{2}}{\log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(2)**2*ln(x*exp(ln(x)/ln(2)**2))+(5*x**4+3)*ln(2)**2+1)/ln(2)**2,x)

[Out]

(x**5*log(2)**2 + x*log(2)**2*log(x) + x*log(x) + 2*x*log(2)**2)/log(2)**2

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