3.26.37 \(\int \frac {-7+(28+60 x) \log (x^2) \log (\frac {7+15 x}{15 x})}{(7 x+15 x^2) \log (\frac {7+15 x}{15 x})} \, dx\)

Optimal. Leaf size=25 \[ 2+\log ^2\left (x^2\right )-\log \left (\frac {3}{\log \left (1+\frac {7}{15 x}\right )}\right ) \]

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Rubi [F]  time = 0.33, 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 {-7+(28+60 x) \log \left (x^2\right ) \log \left (\frac {7+15 x}{15 x}\right )}{\left (7 x+15 x^2\right ) \log \left (\frac {7+15 x}{15 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-7 + (28 + 60*x)*Log[x^2]*Log[(7 + 15*x)/(15*x)])/((7*x + 15*x^2)*Log[(7 + 15*x)/(15*x)]),x]

[Out]

Log[x^2]^2 - Defer[Int][1/(x*Log[1 + 7/(15*x)]), x] + 15*Defer[Int][1/((7 + 15*x)*Log[1 + 7/(15*x)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-7+(28+60 x) \log \left (x^2\right ) \log \left (\frac {7+15 x}{15 x}\right )}{x (7+15 x) \log \left (\frac {7+15 x}{15 x}\right )} \, dx\\ &=\int \frac {-\frac {7}{(7+15 x) \log \left (1+\frac {7}{15 x}\right )}+4 \log \left (x^2\right )}{x} \, dx\\ &=\int \left (-\frac {7}{x (7+15 x) \log \left (1+\frac {7}{15 x}\right )}+\frac {4 \log \left (x^2\right )}{x}\right ) \, dx\\ &=4 \int \frac {\log \left (x^2\right )}{x} \, dx-7 \int \frac {1}{x (7+15 x) \log \left (1+\frac {7}{15 x}\right )} \, dx\\ &=\log ^2\left (x^2\right )-7 \int \left (\frac {1}{7 x \log \left (1+\frac {7}{15 x}\right )}-\frac {15}{7 (7+15 x) \log \left (1+\frac {7}{15 x}\right )}\right ) \, dx\\ &=\log ^2\left (x^2\right )+15 \int \frac {1}{(7+15 x) \log \left (1+\frac {7}{15 x}\right )} \, dx-\int \frac {1}{x \log \left (1+\frac {7}{15 x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.02, size = 18, normalized size = 0.72 \begin {gather*} \log ^2\left (x^2\right )+\log \left (\log \left (1+\frac {7}{15 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-7 + (28 + 60*x)*Log[x^2]*Log[(7 + 15*x)/(15*x)])/((7*x + 15*x^2)*Log[(7 + 15*x)/(15*x)]),x]

[Out]

Log[x^2]^2 + Log[Log[1 + 7/(15*x)]]

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fricas [A]  time = 0.56, size = 19, normalized size = 0.76 \begin {gather*} \log \left (x^{2}\right )^{2} + \log \left (\log \left (\frac {15 \, x + 7}{15 \, x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x+28)*log(1/15*(15*x+7)/x)*log(x^2)-7)/(15*x^2+7*x)/log(1/15*(15*x+7)/x),x, algorithm="fricas")

[Out]

log(x^2)^2 + log(log(1/15*(15*x + 7)/x))

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giac [A]  time = 0.28, size = 23, normalized size = 0.92 \begin {gather*} 4 \, \log \relax (x)^{2} + \log \left (-\log \left (15\right ) + \log \left (15 \, x + 7\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x+28)*log(1/15*(15*x+7)/x)*log(x^2)-7)/(15*x^2+7*x)/log(1/15*(15*x+7)/x),x, algorithm="giac")

[Out]

4*log(x)^2 + log(-log(15) + log(15*x + 7) - log(x))

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maple [A]  time = 0.44, size = 20, normalized size = 0.80

method	result	size
norman	\(\ln \left (x^{2}\right )^{2}+\ln \left (\ln \left (\frac {15 x +7}{15 x}\right )\right )\)	\(20\)
default	\(\ln \left (x^{2}\right )^{2}+\ln \left (-\ln \left (15\right )+\ln \left (15+\frac {7}{x}\right )\right )\)	\(22\)
risch	\(-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+4 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2}\right )^{3}+4 \ln \relax (x )^{2}+\ln \left (\ln \left (x +\frac {7}{15}\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x +\frac {7}{15}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {7}{15}\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {7}{15}\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x +\frac {7}{15}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {7}{15}\right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x +\frac {7}{15}\right )}{x}\right )^{3}-2 i \ln \relax (x )\right )}{2}\right )\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((60*x+28)*ln(1/15*(15*x+7)/x)*ln(x^2)-7)/(15*x^2+7*x)/ln(1/15*(15*x+7)/x),x,method=_RETURNVERBOSE)

[Out]

ln(x^2)^2+ln(ln(1/15*(15*x+7)/x))

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maxima [A]  time = 0.54, size = 27, normalized size = 1.08 \begin {gather*} 4 \, \log \relax (x)^{2} + \log \left (-\log \relax (5) - \log \relax (3) + \log \left (15 \, x + 7\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x+28)*log(1/15*(15*x+7)/x)*log(x^2)-7)/(15*x^2+7*x)/log(1/15*(15*x+7)/x),x, algorithm="maxima")

[Out]

4*log(x)^2 + log(-log(5) - log(3) + log(15*x + 7) - log(x))

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mupad [B]  time = 1.65, size = 16, normalized size = 0.64 \begin {gather*} {\ln \left (x^2\right )}^2+\ln \left (\ln \left (\frac {x+\frac {7}{15}}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)*log((x + 7/15)/x)*(60*x + 28) - 7)/(log((x + 7/15)/x)*(7*x + 15*x^2)),x)

[Out]

log(log((x + 7/15)/x)) + log(x^2)^2

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sympy [A]  time = 0.31, size = 15, normalized size = 0.60 \begin {gather*} \log {\left (x^{2} \right )}^{2} + \log {\left (\log {\left (\frac {x + \frac {7}{15}}{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x+28)*ln(1/15*(15*x+7)/x)*ln(x**2)-7)/(15*x**2+7*x)/ln(1/15*(15*x+7)/x),x)

[Out]

log(x**2)**2 + log(log((x + 7/15)/x))

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