3.45.69 \(\int \frac {12 x+16 x^2+(-14 x-8 x^2) \log (x)+(3 x+10 x^2+8 x^3) \log ^3(x)+(12+16 x) \log (3+4 x)}{(3 x+4 x^2) \log ^3(x)} \, dx\)

Optimal. Leaf size=20 \[ -2+x+x^2-\frac {2 (x+\log (3+4 x))}{\log ^2(x)} \]

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Rubi [F]  time = 0.39, 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 {12 x+16 x^2+\left (-14 x-8 x^2\right ) \log (x)+\left (3 x+10 x^2+8 x^3\right ) \log ^3(x)+(12+16 x) \log (3+4 x)}{\left (3 x+4 x^2\right ) \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(12*x + 16*x^2 + (-14*x - 8*x^2)*Log[x] + (3*x + 10*x^2 + 8*x^3)*Log[x]^3 + (12 + 16*x)*Log[3 + 4*x])/((3*
x + 4*x^2)*Log[x]^3),x]

[Out]

x + x^2 - (2*x)/Log[x]^2 - (2*x)/Log[x] + 2*LogIntegral[x] - 2*Defer[Int][(7 + 4*x)/((3 + 4*x)*Log[x]^2), x] +
4*Defer[Int][Log[3 + 4*x]/(x*Log[x]^3), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12 x+16 x^2+\left (-14 x-8 x^2\right ) \log (x)+\left (3 x+10 x^2+8 x^3\right ) \log ^3(x)+(12+16 x) \log (3+4 x)}{x (3+4 x) \log ^3(x)} \, dx\\ &=\int \left (1+2 x-\frac {2 (7+4 x)}{(3+4 x) \log ^2(x)}+\frac {4 (x+\log (3+4 x))}{x \log ^3(x)}\right ) \, dx\\ &=x+x^2-2 \int \frac {7+4 x}{(3+4 x) \log ^2(x)} \, dx+4 \int \frac {x+\log (3+4 x)}{x \log ^3(x)} \, dx\\ &=x+x^2-2 \int \frac {7+4 x}{(3+4 x) \log ^2(x)} \, dx+4 \int \left (\frac {1}{\log ^3(x)}+\frac {\log (3+4 x)}{x \log ^3(x)}\right ) \, dx\\ &=x+x^2-2 \int \frac {7+4 x}{(3+4 x) \log ^2(x)} \, dx+4 \int \frac {1}{\log ^3(x)} \, dx+4 \int \frac {\log (3+4 x)}{x \log ^3(x)} \, dx\\ &=x+x^2-\frac {2 x}{\log ^2(x)}+2 \int \frac {1}{\log ^2(x)} \, dx-2 \int \frac {7+4 x}{(3+4 x) \log ^2(x)} \, dx+4 \int \frac {\log (3+4 x)}{x \log ^3(x)} \, dx\\ &=x+x^2-\frac {2 x}{\log ^2(x)}-\frac {2 x}{\log (x)}-2 \int \frac {7+4 x}{(3+4 x) \log ^2(x)} \, dx+2 \int \frac {1}{\log (x)} \, dx+4 \int \frac {\log (3+4 x)}{x \log ^3(x)} \, dx\\ &=x+x^2-\frac {2 x}{\log ^2(x)}-\frac {2 x}{\log (x)}+2 \text {li}(x)-2 \int \frac {7+4 x}{(3+4 x) \log ^2(x)} \, dx+4 \int \frac {\log (3+4 x)}{x \log ^3(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.65, size = 24, normalized size = 1.20 \begin {gather*} x+x^2-\frac {2 x}{\log ^2(x)}-\frac {2 \log (3+4 x)}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*x + 16*x^2 + (-14*x - 8*x^2)*Log[x] + (3*x + 10*x^2 + 8*x^3)*Log[x]^3 + (12 + 16*x)*Log[3 + 4*x]
)/((3*x + 4*x^2)*Log[x]^3),x]

[Out]

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

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fricas [A]  time = 0.83, size = 27, normalized size = 1.35 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \relax (x)^{2} - 2 \, x - 2 \, \log \left (4 \, x + 3\right )}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+12)*log(3+4*x)+(8*x^3+10*x^2+3*x)*log(x)^3+(-8*x^2-14*x)*log(x)+16*x^2+12*x)/(4*x^2+3*x)/log(
x)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 24, normalized size = 1.20 \begin {gather*} x^{2} + x - \frac {2 \, x}{\log \relax (x)^{2}} - \frac {2 \, \log \left (4 \, x + 3\right )}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+12)*log(3+4*x)+(8*x^3+10*x^2+3*x)*log(x)^3+(-8*x^2-14*x)*log(x)+16*x^2+12*x)/(4*x^2+3*x)/log(
x)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 32, normalized size = 1.60

method	result	size
risch	\(-\frac {2 \ln \left (3+4 x \right )}{\ln \relax (x )^{2}}+\frac {x \left (x \ln \relax (x )^{2}+\ln \relax (x )^{2}-2\right )}{\ln \relax (x )^{2}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x+12)*ln(3+4*x)+(8*x^3+10*x^2+3*x)*ln(x)^3+(-8*x^2-14*x)*ln(x)+16*x^2+12*x)/(4*x^2+3*x)/ln(x)^3,x,met
hod=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 27, normalized size = 1.35 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \relax (x)^{2} - 2 \, x - 2 \, \log \left (4 \, x + 3\right )}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+12)*log(3+4*x)+(8*x^3+10*x^2+3*x)*log(x)^3+(-8*x^2-14*x)*log(x)+16*x^2+12*x)/(4*x^2+3*x)/log(
x)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.38, size = 24, normalized size = 1.20 \begin {gather*} x-\frac {2\,x}{{\ln \relax (x)}^2}-\frac {2\,\ln \left (4\,x+3\right )}{{\ln \relax (x)}^2}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x + log(4*x + 3)*(16*x + 12) + log(x)^3*(3*x + 10*x^2 + 8*x^3) - log(x)*(14*x + 8*x^2) + 16*x^2)/(log(
x)^3*(3*x + 4*x^2)),x)

[Out]

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

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sympy [A]  time = 0.36, size = 26, normalized size = 1.30 \begin {gather*} x^{2} + x - \frac {2 x}{\log {\relax (x )}^{2}} - \frac {2 \log {\left (4 x + 3 \right )}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x+12)*ln(3+4*x)+(8*x**3+10*x**2+3*x)*ln(x)**3+(-8*x**2-14*x)*ln(x)+16*x**2+12*x)/(4*x**2+3*x)/l
n(x)**3,x)

[Out]

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

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