3.53.7 \(\int \frac {-30 x \log (x^2)+(90+60 x) \log (3+2 x)}{(3 x+2 x^2) \log ^2(3+2 x)} \, dx\)

Optimal. Leaf size=14 \[ \frac {15 \log \left (x^2\right )}{\log (3+2 x)} \]

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Rubi [F]  time = 0.32, 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 {-30 x \log \left (x^2\right )+(90+60 x) \log (3+2 x)}{\left (3 x+2 x^2\right ) \log ^2(3+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-30*x*Log[x^2] + (90 + 60*x)*Log[3 + 2*x])/((3*x + 2*x^2)*Log[3 + 2*x]^2),x]

[Out]

30*Defer[Int][1/(x*Log[3 + 2*x]), x] - 15*Defer[Subst][Defer[Int][Log[(-3/2 + x/2)^2]/(x*Log[x]^2), x], x, 3 +
2*x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-30 x \log \left (x^2\right )+(90+60 x) \log (3+2 x)}{x (3+2 x) \log ^2(3+2 x)} \, dx\\ &=\int \left (-\frac {30 \log \left (x^2\right )}{(3+2 x) \log ^2(3+2 x)}+\frac {30}{x \log (3+2 x)}\right ) \, dx\\ &=-\left (30 \int \frac {\log \left (x^2\right )}{(3+2 x) \log ^2(3+2 x)} \, dx\right )+30 \int \frac {1}{x \log (3+2 x)} \, dx\\ &=-\left (15 \operatorname {Subst}\left (\int \frac {\log \left (\left (-\frac {3}{2}+\frac {x}{2}\right )^2\right )}{x \log ^2(x)} \, dx,x,3+2 x\right )\right )+30 \int \frac {1}{x \log (3+2 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 14, normalized size = 1.00 \begin {gather*} \frac {15 \log \left (x^2\right )}{\log (3+2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30*x*Log[x^2] + (90 + 60*x)*Log[3 + 2*x])/((3*x + 2*x^2)*Log[3 + 2*x]^2),x]

[Out]

(15*Log[x^2])/Log[3 + 2*x]

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fricas [A]  time = 0.44, size = 14, normalized size = 1.00 \begin {gather*} \frac {15 \, \log \left (x^{2}\right )}{\log \left (2 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*log(x^2)+(60*x+90)*log(2*x+3))/(2*x^2+3*x)/log(2*x+3)^2,x, algorithm="fricas")

[Out]

15*log(x^2)/log(2*x + 3)

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giac [A]  time = 0.14, size = 14, normalized size = 1.00 \begin {gather*} \frac {15 \, \log \left (x^{2}\right )}{\log \left (2 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*log(x^2)+(60*x+90)*log(2*x+3))/(2*x^2+3*x)/log(2*x+3)^2,x, algorithm="giac")

[Out]

15*log(x^2)/log(2*x + 3)

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maple [C]  time = 0.19, size = 65, normalized size = 4.64

method	result	size
risch	\(\frac {30 \ln \relax (x )-\frac {15 i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}+15 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\frac {15 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}}{\ln \left (2 x +3\right )}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-30*x*ln(x^2)+(60*x+90)*ln(2*x+3))/(2*x^2+3*x)/ln(2*x+3)^2,x,method=_RETURNVERBOSE)

[Out]

15/2*(4*ln(x)-I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x^2)^3)/ln(2*x+3)

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maxima [A]  time = 0.39, size = 12, normalized size = 0.86 \begin {gather*} \frac {30 \, \log \relax (x)}{\log \left (2 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*log(x^2)+(60*x+90)*log(2*x+3))/(2*x^2+3*x)/log(2*x+3)^2,x, algorithm="maxima")

[Out]

30*log(x)/log(2*x + 3)

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mupad [B]  time = 3.39, size = 14, normalized size = 1.00 \begin {gather*} \frac {15\,\ln \left (x^2\right )}{\ln \left (2\,x+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*x*log(x^2) - log(2*x + 3)*(60*x + 90))/(log(2*x + 3)^2*(3*x + 2*x^2)),x)

[Out]

(15*log(x^2))/log(2*x + 3)

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sympy [A]  time = 0.23, size = 12, normalized size = 0.86 \begin {gather*} \frac {15 \log {\left (x^{2} \right )}}{\log {\left (2 x + 3 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*ln(x**2)+(60*x+90)*ln(2*x+3))/(2*x**2+3*x)/ln(2*x+3)**2,x)

[Out]

15*log(x**2)/log(2*x + 3)

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