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

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

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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} $$\begin{array}{c}x\log \left(x^{1+\frac{1}{{\log }^2(2)}}\right)+x^5+3x+\frac{x}{{\log }^2(2)}-x(1+\frac{1}{{\log }^2(2)})\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int (1+(3+5x^4){\log }^2(2)+{\log }^2(2)\log \left(x^{1+\frac{1}{{\log }^2(2)}}\right))dx}{{\log }^2(2)}\\ & =\frac{x}{{\log }^2(2)}+\int (3+5x^4)dx+\int \log \left(x^{1+\frac{1}{{\log }^2(2)}}\right)dx\\ & =3x+x^5-x(1+\frac{1}{{\log }^2(2)})+\frac{x}{{\log }^2(2)}+x\log \left(x^{1+\frac{1}{{\log }^2(2)}}\right)\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{(x^5+2x)\log (2{)}^2+(x\log (2{)}^2+x)\log (x)}{\log (2{)}^2}\end{array}$$

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 $$\begin{array}{c}\frac{(x\log (x)-x)(\frac{1}{\log (2{)}^2}+1)\log (2{)}^2+(x^5+3x)\log (2{)}^2+x}{\log (2{)}^2}\end{array}$$

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	$\frac{x^5\ln (2{)}^2+2x\ln (2{)}^2+\ln (2{)}^2\ln \left(x{x}^{\frac{1}{\ln (2{)}^2}}\right)x}{\ln (2{)}^2}$	$37$
risch	$x\ln \left(x^{\frac{1}{\ln (2{)}^2}}\right)+x\ln (x)-\frac{i\pi x\mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i{x}^{\frac{1}{\ln (2{)}^2}}\right)\mathrm{csgn}\left(ix{x}^{\frac{1}{\ln (2{)}^2}}\right)}{2}+\frac{i\pi x\mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(ix{x}^{\frac{1}{\ln (2{)}^2}}\right)}^2}{2}+\frac{i\pi x\mathrm{csgn}\left(i{x}^{\frac{1}{\ln (2{)}^2}}\right)\mathrm{csgn}{\left(ix{x}^{\frac{1}{\ln (2{)}^2}}\right)}^2}{2}-\frac{i\pi x\mathrm{csgn}{\left(ix{x}^{\frac{1}{\ln (2{)}^2}}\right)}^3}{2}+x^5+2x$	$121$

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 $$\begin{array}{c}-\frac{x(\frac{1}{\log (2{)}^2}+1)\log (2{)}^2-x\log (2{)}^2\log \left(x^{\frac{1}{\log (2{)}^2}+1}\right)-(x^5+3x)\log (2{)}^2-x}{\log (2{)}^2}\end{array}$$

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 $$\begin{array}{c}x(\ln \left(x^{\frac{1}{{\ln (2)}^2}+1}\right)+x^4+2)\end{array}$$

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 $$\begin{array}{c}\frac{x^5\log {(2)}^2+x\log {(2)}^2\log (x)+x\log (x)+2x\log {(2)}^2}{\log {(2)}^2}\end{array}$$

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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