3.97.57 \(\int \frac {15+(6 x+2 e^2 x) \log ^2(15)}{(6 x^2+2 e^2 x^2) \log ^2(15)} \, dx\)

Optimal. Leaf size=24 \[ \frac {5 (-3+x)}{2 \left (3+e^2\right ) x \log ^2(15)}+\log (x) \]

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6, 12, 186, 43} \begin {gather*} \log (x)-\frac {15}{2 \left (3+e^2\right ) x \log ^2(15)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + (6*x + 2*E^2*x)*Log[15]^2)/((6*x^2 + 2*E^2*x^2)*Log[15]^2),x]

[Out]

-15/(2*(3 + E^2)*x*Log[15]^2) + Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15+\left (6 x+2 e^2 x\right ) \log ^2(15)}{\left (6+2 e^2\right ) x^2 \log ^2(15)} \, dx\\ &=\frac {\int \frac {15+\left (6 x+2 e^2 x\right ) \log ^2(15)}{x^2} \, dx}{2 \left (3+e^2\right ) \log ^2(15)}\\ &=\frac {\int \frac {15+2 \left (3+e^2\right ) x \log ^2(15)}{x^2} \, dx}{2 \left (3+e^2\right ) \log ^2(15)}\\ &=\frac {\int \left (\frac {15}{x^2}+\frac {2 \left (3+e^2\right ) \log ^2(15)}{x}\right ) \, dx}{2 \left (3+e^2\right ) \log ^2(15)}\\ &=-\frac {15}{2 \left (3+e^2\right ) x \log ^2(15)}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.88 \begin {gather*} -\frac {15}{2 \left (3+e^2\right ) x \log ^2(15)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + (6*x + 2*E^2*x)*Log[15]^2)/((6*x^2 + 2*E^2*x^2)*Log[15]^2),x]

[Out]

-15/(2*(3 + E^2)*x*Log[15]^2) + Log[x]

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fricas [A]  time = 0.64, size = 34, normalized size = 1.42 \begin {gather*} \frac {2 \, {\left (x e^{2} + 3 \, x\right )} \log \left (15\right )^{2} \log \relax (x) - 15}{2 \, {\left (x e^{2} + 3 \, x\right )} \log \left (15\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x+6*x)*log(15)^2+15)/(2*x^2*exp(2)+6*x^2)/log(15)^2,x, algorithm="fricas")

[Out]

1/2*(2*(x*e^2 + 3*x)*log(15)^2*log(x) - 15)/((x*e^2 + 3*x)*log(15)^2)

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giac [A]  time = 0.13, size = 27, normalized size = 1.12 \begin {gather*} \frac {2 \, \log \left (15\right )^{2} \log \left ({\left | x \right |}\right ) - \frac {15}{x {\left (e^{2} + 3\right )}}}{2 \, \log \left (15\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x+6*x)*log(15)^2+15)/(2*x^2*exp(2)+6*x^2)/log(15)^2,x, algorithm="giac")

[Out]

1/2*(2*log(15)^2*log(abs(x)) - 15/(x*(e^2 + 3)))/log(15)^2

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maple [A]  time = 0.10, size = 19, normalized size = 0.79

method	result	size
norman	\(-\frac {15}{2 \ln \left (15\right )^{2} \left ({\mathrm e}^{2}+3\right ) x}+\ln \relax (x )\)	\(19\)
default	\(\frac {2 \ln \left (15\right )^{2} \left ({\mathrm e}^{2}+3\right ) \ln \relax (x )-\frac {15}{x}}{2 \ln \left (15\right )^{2} \left ({\mathrm e}^{2}+3\right )}\)	\(31\)
risch	\(-\frac {15}{2 \left (\ln \relax (3)+\ln \relax (5)\right )^{2} x \left ({\mathrm e}^{2}+3\right )}+\frac {\ln \relax (x ) \ln \relax (5)^{2}}{\left (\ln \relax (3)+\ln \relax (5)\right )^{2}}+\frac {2 \ln \relax (x ) \ln \relax (5) \ln \relax (3)}{\left (\ln \relax (3)+\ln \relax (5)\right )^{2}}+\frac {\ln \relax (x ) \ln \relax (3)^{2}}{\left (\ln \relax (3)+\ln \relax (5)\right )^{2}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(2)*x+6*x)*ln(15)^2+15)/(2*x^2*exp(2)+6*x^2)/ln(15)^2,x,method=_RETURNVERBOSE)

[Out]

-15/2/ln(15)^2/(exp(2)+3)/x+ln(x)

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maxima [A]  time = 0.36, size = 26, normalized size = 1.08 \begin {gather*} \frac {2 \, \log \left (15\right )^{2} \log \relax (x) - \frac {15}{x {\left (e^{2} + 3\right )}}}{2 \, \log \left (15\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x+6*x)*log(15)^2+15)/(2*x^2*exp(2)+6*x^2)/log(15)^2,x, algorithm="maxima")

[Out]

1/2*(2*log(15)^2*log(x) - 15/(x*(e^2 + 3)))/log(15)^2

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mupad [B]  time = 0.10, size = 25, normalized size = 1.04 \begin {gather*} \ln \relax (x)-\frac {15}{x\,\left (2\,{\mathrm {e}}^2\,{\ln \left (15\right )}^2+6\,{\ln \left (15\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(15)^2*(6*x + 2*x*exp(2)) + 15)/(log(15)^2*(2*x^2*exp(2) + 6*x^2)),x)

[Out]

log(x) - 15/(x*(2*exp(2)*log(15)^2 + 6*log(15)^2))

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sympy [A]  time = 0.15, size = 34, normalized size = 1.42 \begin {gather*} \frac {2 \left (3 + e^{2}\right ) \log {\left (15 \right )}^{2} \log {\relax (x )} - \frac {15}{x}}{6 \log {\left (15 \right )}^{2} + 2 e^{2} \log {\left (15 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x+6*x)*ln(15)**2+15)/(2*x**2*exp(2)+6*x**2)/ln(15)**2,x)

[Out]

(2*(3 + exp(2))*log(15)**2*log(x) - 15/x)/(6*log(15)**2 + 2*exp(2)*log(15)**2)

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