3.64.64 \(\int \frac {25+10 x+(2+5\ 2^{2/5}+10 x) \log (\frac {1}{5} (2+5\ 2^{2/5}+10 x))}{2+5\ 2^{2/5}+10 x} \, dx\)

Optimal. Leaf size=22 \[ \frac {1}{2} (5+2 x) \log \left (\frac {2}{5}+2^{2/5}+2 x\right ) \]

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Rubi [B]  time = 0.17, antiderivative size = 61, normalized size of antiderivative = 2.77, number of steps used = 6, number of rules used = 4, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6688, 43, 2389, 2295} \begin {gather*} \frac {1}{10} \left (10 x+5\ 2^{2/5}+2\right ) \log \left (2 x+\frac {1}{5} \left (2+5\ 2^{2/5}\right )\right )+\frac {1}{10} \left (23-5\ 2^{2/5}\right ) \log \left (10 x+5\ 2^{2/5}+2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + 10*x + (2 + 5*2^(2/5) + 10*x)*Log[(2 + 5*2^(2/5) + 10*x)/5])/(2 + 5*2^(2/5) + 10*x),x]

[Out]

((2 + 5*2^(2/5) + 10*x)*Log[(2 + 5*2^(2/5))/5 + 2*x])/10 + ((23 - 5*2^(2/5))*Log[2 + 5*2^(2/5) + 10*x])/10

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

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

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {5 (5+2 x)}{2+5\ 2^{2/5}+10 x}+\log \left (\frac {2}{5}+2^{2/5}+2 x\right )\right ) \, dx\\ &=5 \int \frac {5+2 x}{2+5\ 2^{2/5}+10 x} \, dx+\int \log \left (\frac {2}{5}+2^{2/5}+2 x\right ) \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \log (x) \, dx,x,\frac {2}{5}+2^{2/5}+2 x\right )+5 \int \left (\frac {1}{5}+\frac {23-5\ 2^{2/5}}{5 \left (2+5\ 2^{2/5}+10 x\right )}\right ) \, dx\\ &=\frac {1}{10} \left (2+5\ 2^{2/5}+10 x\right ) \log \left (\frac {1}{5} \left (2+5\ 2^{2/5}\right )+2 x\right )+\frac {1}{10} \left (23-5\ 2^{2/5}\right ) \log \left (2+5\ 2^{2/5}+10 x\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 51, normalized size = 2.32 \begin {gather*} \frac {1}{10} \left (2+5\ 2^{2/5}-\log (25)-2^{2/5} \log (3125)-x \log (9765625)+5 (5+2 x) \log \left (2+5\ 2^{2/5}+10 x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 10*x + (2 + 5*2^(2/5) + 10*x)*Log[(2 + 5*2^(2/5) + 10*x)/5])/(2 + 5*2^(2/5) + 10*x),x]

[Out]

(2 + 5*2^(2/5) - Log[25] - 2^(2/5)*Log[3125] - x*Log[9765625] + 5*(5 + 2*x)*Log[2 + 5*2^(2/5) + 10*x])/10

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fricas [A]  time = 0.67, size = 16, normalized size = 0.73 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + 5\right )} \log \left (2 \, x + 2^{\frac {2}{5}} + \frac {2}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*2^(2/5)+10*x+2)*log(2^(2/5)+2*x+2/5)+10*x+25)/(5*2^(2/5)+10*x+2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.19, size = 58, normalized size = 2.64 \begin {gather*} -\frac {1}{2} \cdot 2^{\frac {2}{5}} \log \relax (5) \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) - x \log \relax (5) + \frac {1}{2} \, {\left (2^{\frac {2}{5}} \log \relax (5) + 5\right )} \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) + x \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*2^(2/5)+10*x+2)*log(2^(2/5)+2*x+2/5)+10*x+25)/(5*2^(2/5)+10*x+2),x, algorithm="giac")

[Out]

-1/2*2^(2/5)*log(5)*log(10*x + 5*2^(2/5) + 2) - x*log(5) + 1/2*(2^(2/5)*log(5) + 5)*log(10*x + 5*2^(2/5) + 2)
+ x*log(10*x + 5*2^(2/5) + 2)

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maple [A]  time = 0.51, size = 24, normalized size = 1.09

method	result	size
norman	\(\ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right ) x +\frac {5 \ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right )}{2}\)	\(24\)
risch	\(\ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right ) x +\frac {5 \ln \left (5 \,2^{\frac {2}{5}}+10 x +2\right )}{2}\)	\(26\)
derivativedivides	\(\frac {\left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right ) \ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right )}{2}-\frac {2^{\frac {2}{5}} \ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right )}{2}+\frac {23 \ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right )}{10}\)	\(46\)
default	\(\frac {\left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right ) \ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right )}{2}-\frac {2^{\frac {2}{5}} \ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right )}{2}+\frac {23 \ln \left (2^{\frac {2}{5}}+2 x +\frac {2}{5}\right )}{10}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*2^(2/5)+10*x+2)*ln(2^(2/5)+2*x+2/5)+10*x+25)/(5*2^(2/5)+10*x+2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.46, size = 183, normalized size = 8.32 \begin {gather*} \frac {1}{20} \, {\left (5 \cdot 2^{\frac {2}{5}} + 2\right )} \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right )^{2} + \frac {1}{2} \cdot 2^{\frac {2}{5}} \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) \log \left (2 \, x + 2^{\frac {2}{5}} + \frac {2}{5}\right ) - \frac {1}{5} \, \log \relax (5) \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) + \frac {1}{10} \, \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right )^{2} - \frac {1}{10} \, {\left ({\left (5 \cdot 2^{\frac {2}{5}} + 2\right )} \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) - 10 \, x\right )} \log \left (2 \, x + 2^{\frac {2}{5}} + \frac {2}{5}\right ) - \frac {1}{4} \cdot 2^{\frac {2}{5}} {\left (2 \, \log \relax (5) \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) - \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right )^{2} + 2 \, \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) \log \left (2 \, x + 2^{\frac {2}{5}} + \frac {2}{5}\right )\right )} + \frac {5}{2} \, \log \left (10 \, x + 5 \cdot 2^{\frac {2}{5}} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*2^(2/5)+10*x+2)*log(2^(2/5)+2*x+2/5)+10*x+25)/(5*2^(2/5)+10*x+2),x, algorithm="maxima")

[Out]

1/20*(5*2^(2/5) + 2)*log(10*x + 5*2^(2/5) + 2)^2 + 1/2*2^(2/5)*log(10*x + 5*2^(2/5) + 2)*log(2*x + 2^(2/5) + 2
/5) - 1/5*log(5)*log(10*x + 5*2^(2/5) + 2) + 1/10*log(10*x + 5*2^(2/5) + 2)^2 - 1/10*((5*2^(2/5) + 2)*log(10*x
+ 5*2^(2/5) + 2) - 10*x)*log(2*x + 2^(2/5) + 2/5) - 1/4*2^(2/5)*(2*log(5)*log(10*x + 5*2^(2/5) + 2) - log(10*
x + 5*2^(2/5) + 2)^2 + 2*log(10*x + 5*2^(2/5) + 2)*log(2*x + 2^(2/5) + 2/5)) + 5/2*log(10*x + 5*2^(2/5) + 2)

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mupad [B]  time = 4.66, size = 16, normalized size = 0.73 \begin {gather*} \frac {\ln \left (2\,x+2^{2/5}+\frac {2}{5}\right )\,\left (2\,x+5\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + log(2*x + 2^(2/5) + 2/5)*(10*x + 5*2^(2/5) + 2) + 25)/(10*x + 5*2^(2/5) + 2),x)

[Out]

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

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sympy [A]  time = 0.26, size = 31, normalized size = 1.41 \begin {gather*} x \log {\left (2 x + \frac {2}{5} + 2^{\frac {2}{5}} \right )} + \frac {5 \log {\left (10 x + 2 + 5 \cdot 2^{\frac {2}{5}} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*2**(2/5)+10*x+2)*ln(2**(2/5)+2*x+2/5)+10*x+25)/(5*2**(2/5)+10*x+2),x)

[Out]

x*log(2*x + 2/5 + 2**(2/5)) + 5*log(10*x + 2 + 5*2**(2/5))/2

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