3.61.27 \(\int \frac {(6 x-2 e^5 x) \log (x^2)}{25 \log (7)+50 \log (7) \log (x^2)+25 \log (7) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=32 \[ \frac {1}{5} \left (-2+\frac {\left (3-e^5\right ) x^2}{5 \log (7) \left (1+\log \left (x^2\right )\right )}\right ) \]

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Rubi [C]  time = 0.29, antiderivative size = 134, normalized size of antiderivative = 4.19, number of steps used = 10, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6, 12, 6688, 2306, 2310, 2178, 2366, 6482} \begin {gather*} \frac {\left (3-e^5\right ) \log \left (x^2\right ) \text {Ei}\left (\log \left (x^2\right )+1\right )}{25 e \log (7)}-\frac {\left (3-e^5\right ) \left (\log \left (x^2\right )+1\right ) \text {Ei}\left (\log \left (x^2\right )+1\right )}{25 e \log (7)}+\frac {\left (3-e^5\right ) \text {Ei}\left (\log \left (x^2\right )+1\right )}{25 e \log (7)}-\frac {\left (3-e^5\right ) x^2 \log \left (x^2\right )}{25 \log (7) \left (\log \left (x^2\right )+1\right )}+\frac {\left (3-e^5\right ) x^2}{25 \log (7)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((6*x - 2*E^5*x)*Log[x^2])/(25*Log[7] + 50*Log[7]*Log[x^2] + 25*Log[7]*Log[x^2]^2),x]

[Out]

((3 - E^5)*x^2)/(25*Log[7]) + ((3 - E^5)*ExpIntegralEi[1 + Log[x^2]])/(25*E*Log[7]) + ((3 - E^5)*ExpIntegralEi
[1 + Log[x^2]]*Log[x^2])/(25*E*Log[7]) - ((3 - E^5)*x^2*Log[x^2])/(25*Log[7]*(1 + Log[x^2])) - ((3 - E^5)*ExpI
ntegralEi[1 + Log[x^2]]*(1 + Log[x^2]))/(25*E*Log[7])

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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2310

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

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, 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 \frac {\left (6-2 e^5\right ) x \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx\\ &=\left (2 \left (3-e^5\right )\right ) \int \frac {x \log \left (x^2\right )}{25 \log (7)+50 \log (7) \log \left (x^2\right )+25 \log (7) \log ^2\left (x^2\right )} \, dx\\ &=\left (2 \left (3-e^5\right )\right ) \int \frac {x \log \left (x^2\right )}{25 \log (7) \left (1+\log \left (x^2\right )\right )^2} \, dx\\ &=\frac {\left (2 \left (3-e^5\right )\right ) \int \frac {x \log \left (x^2\right )}{\left (1+\log \left (x^2\right )\right )^2} \, dx}{25 \log (7)}\\ &=\frac {\left (3-e^5\right ) \text {Ei}\left (1+\log \left (x^2\right )\right ) \log \left (x^2\right )}{25 e \log (7)}-\frac {\left (3-e^5\right ) x^2 \log \left (x^2\right )}{25 \log (7) \left (1+\log \left (x^2\right )\right )}-\frac {\left (4 \left (3-e^5\right )\right ) \int \left (\frac {\text {Ei}\left (1+\log \left (x^2\right )\right )}{2 e x}-\frac {x}{2 \left (1+\log \left (x^2\right )\right )}\right ) \, dx}{25 \log (7)}\\ &=\frac {\left (3-e^5\right ) \text {Ei}\left (1+\log \left (x^2\right )\right ) \log \left (x^2\right )}{25 e \log (7)}-\frac {\left (3-e^5\right ) x^2 \log \left (x^2\right )}{25 \log (7) \left (1+\log \left (x^2\right )\right )}+\frac {\left (2 \left (3-e^5\right )\right ) \int \frac {x}{1+\log \left (x^2\right )} \, dx}{25 \log (7)}-\frac {\left (2 \left (3-e^5\right )\right ) \int \frac {\text {Ei}\left (1+\log \left (x^2\right )\right )}{x} \, dx}{25 e \log (7)}\\ &=\frac {\left (3-e^5\right ) \text {Ei}\left (1+\log \left (x^2\right )\right ) \log \left (x^2\right )}{25 e \log (7)}-\frac {\left (3-e^5\right ) x^2 \log \left (x^2\right )}{25 \log (7) \left (1+\log \left (x^2\right )\right )}+\frac {\left (3-e^5\right ) \operatorname {Subst}\left (\int \frac {e^x}{1+x} \, dx,x,\log \left (x^2\right )\right )}{25 \log (7)}-\frac {\left (3-e^5\right ) \operatorname {Subst}\left (\int \text {Ei}(1+x) \, dx,x,\log \left (x^2\right )\right )}{25 e \log (7)}\\ &=\frac {\left (3-e^5\right ) x^2}{25 \log (7)}+\frac {\left (3-e^5\right ) \text {Ei}\left (1+\log \left (x^2\right )\right )}{25 e \log (7)}+\frac {\left (3-e^5\right ) \text {Ei}\left (1+\log \left (x^2\right )\right ) \log \left (x^2\right )}{25 e \log (7)}-\frac {\left (3-e^5\right ) x^2 \log \left (x^2\right )}{25 \log (7) \left (1+\log \left (x^2\right )\right )}-\frac {\left (3-e^5\right ) \text {Ei}\left (1+\log \left (x^2\right )\right ) \left (1+\log \left (x^2\right )\right )}{25 e \log (7)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 24, normalized size = 0.75 \begin {gather*} -\frac {\left (-3+e^5\right ) x^2}{25 \log (7) \left (1+\log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((6*x - 2*E^5*x)*Log[x^2])/(25*Log[7] + 50*Log[7]*Log[x^2] + 25*Log[7]*Log[x^2]^2),x]

[Out]

-1/25*((-3 + E^5)*x^2)/(Log[7]*(1 + Log[x^2]))

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fricas [A]  time = 0.63, size = 26, normalized size = 0.81 \begin {gather*} -\frac {x^{2} e^{5} - 3 \, x^{2}}{25 \, {\left (\log \relax (7) \log \left (x^{2}\right ) + \log \relax (7)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(5)+6*x)*log(x^2)/(25*log(7)*log(x^2)^2+50*log(7)*log(x^2)+25*log(7)),x, algorithm="fricas"
)

[Out]

-1/25*(x^2*e^5 - 3*x^2)/(log(7)*log(x^2) + log(7))

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giac [A]  time = 0.19, size = 37, normalized size = 1.16 \begin {gather*} -\frac {x^{2} e^{5}}{25 \, {\left (\log \relax (7) \log \left (x^{2}\right ) + \log \relax (7)\right )}} + \frac {3 \, x^{2}}{25 \, {\left (\log \relax (7) \log \left (x^{2}\right ) + \log \relax (7)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(5)+6*x)*log(x^2)/(25*log(7)*log(x^2)^2+50*log(7)*log(x^2)+25*log(7)),x, algorithm="giac")

[Out]

-1/25*x^2*e^5/(log(7)*log(x^2) + log(7)) + 3/25*x^2/(log(7)*log(x^2) + log(7))

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maple [A]  time = 0.36, size = 22, normalized size = 0.69

method	result	size
default	\(-\frac {\left ({\mathrm e}^{5}-3\right ) x^{2}}{25 \ln \relax (7) \left (\ln \left (x^{2}\right )+1\right )}\)	\(22\)
norman	\(-\frac {\left ({\mathrm e}^{5}-3\right ) x^{2}}{25 \ln \relax (7) \left (\ln \left (x^{2}\right )+1\right )}\)	\(22\)
risch	\(-\frac {\left ({\mathrm e}^{5}-3\right ) x^{2}}{25 \ln \relax (7) \left (\ln \left (x^{2}\right )+1\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*exp(5)+6*x)*ln(x^2)/(25*ln(7)*ln(x^2)^2+50*ln(7)*ln(x^2)+25*ln(7)),x,method=_RETURNVERBOSE)

[Out]

-1/25*(exp(5)-3)/ln(7)*x^2/(ln(x^2)+1)

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maxima [A]  time = 0.47, size = 20, normalized size = 0.62 \begin {gather*} -\frac {x^{2} {\left (e^{5} - 3\right )}}{25 \, {\left (2 \, \log \relax (7) \log \relax (x) + \log \relax (7)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(5)+6*x)*log(x^2)/(25*log(7)*log(x^2)^2+50*log(7)*log(x^2)+25*log(7)),x, algorithm="maxima"
)

[Out]

-1/25*x^2*(e^5 - 3)/(2*log(7)*log(x) + log(7))

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mupad [B]  time = 4.45, size = 21, normalized size = 0.66 \begin {gather*} -\frac {x^2\,\left ({\mathrm {e}}^5-3\right )}{25\,\ln \relax (7)\,\left (\ln \left (x^2\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)*(6*x - 2*x*exp(5)))/(25*log(7) + 50*log(x^2)*log(7) + 25*log(x^2)^2*log(7)),x)

[Out]

-(x^2*(exp(5) - 3))/(25*log(7)*(log(x^2) + 1))

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sympy [A]  time = 0.10, size = 26, normalized size = 0.81 \begin {gather*} \frac {- x^{2} e^{5} + 3 x^{2}}{25 \log {\relax (7 )} \log {\left (x^{2} \right )} + 25 \log {\relax (7 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(5)+6*x)*ln(x**2)/(25*ln(7)*ln(x**2)**2+50*ln(7)*ln(x**2)+25*ln(7)),x)

[Out]

(-x**2*exp(5) + 3*x**2)/(25*log(7)*log(x**2) + 25*log(7))

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