3.23.68 \(\int \frac {e^4 (30+10 e^3)+e^4 (-30-10 e^3) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+(210 x^2+70 e^3 x^2) \log (x)+175 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {2 e^4}{7 x \left (1+\frac {3+e^3}{5 \log (x)}\right )} \]

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Rubi [A]  time = 0.58, antiderivative size = 37, normalized size of antiderivative = 1.37, number of steps used = 11, number of rules used = 7, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6, 6688, 12, 6742, 2306, 2309, 2178} \begin {gather*} \frac {2 e^4}{7 x}-\frac {2 e^4 \left (3+e^3\right )}{7 x \left (5 \log (x)+e^3+3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*(30 + 10*E^3) + E^4*(-30 - 10*E^3)*Log[x] - 50*E^4*Log[x]^2)/(63*x^2 + 42*E^3*x^2 + 7*E^6*x^2 + (210*
x^2 + 70*E^3*x^2)*Log[x] + 175*x^2*Log[x]^2),x]

[Out]

(2*E^4)/(7*x) - (2*E^4*(3 + E^3))/(7*x*(3 + E^3 + 5*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 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 2309

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{7 e^6 x^2+\left (63+42 e^3\right ) x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx\\ &=\int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{\left (63+42 e^3+7 e^6\right ) x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx\\ &=\int \frac {10 e^4 \left (3 \left (1+\frac {e^3}{3}\right )-\left (3+e^3\right ) \log (x)-5 \log ^2(x)\right )}{7 x^2 \left (3 \left (1+\frac {e^3}{3}\right )+5 \log (x)\right )^2} \, dx\\ &=\frac {1}{7} \left (10 e^4\right ) \int \frac {3 \left (1+\frac {e^3}{3}\right )-\left (3+e^3\right ) \log (x)-5 \log ^2(x)}{x^2 \left (3 \left (1+\frac {e^3}{3}\right )+5 \log (x)\right )^2} \, dx\\ &=\frac {1}{7} \left (10 e^4\right ) \int \left (-\frac {1}{5 x^2}+\frac {3+e^3}{x^2 \left (3 \left (1+\frac {e^3}{3}\right )+5 \log (x)\right )^2}+\frac {3+e^3}{5 x^2 \left (3 \left (1+\frac {e^3}{3}\right )+5 \log (x)\right )}\right ) \, dx\\ &=\frac {2 e^4}{7 x}+\frac {1}{7} \left (2 e^4 \left (3+e^3\right )\right ) \int \frac {1}{x^2 \left (3 \left (1+\frac {e^3}{3}\right )+5 \log (x)\right )} \, dx+\frac {1}{7} \left (10 e^4 \left (3+e^3\right )\right ) \int \frac {1}{x^2 \left (3 \left (1+\frac {e^3}{3}\right )+5 \log (x)\right )^2} \, dx\\ &=\frac {2 e^4}{7 x}-\frac {2 e^4 \left (3+e^3\right )}{7 x \left (3+e^3+5 \log (x)\right )}-\frac {1}{7} \left (2 e^4 \left (3+e^3\right )\right ) \int \frac {1}{x^2 \left (3 \left (1+\frac {e^3}{3}\right )+5 \log (x)\right )} \, dx+\frac {1}{7} \left (2 e^4 \left (3+e^3\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{3 \left (1+\frac {e^3}{3}\right )+5 x} \, dx,x,\log (x)\right )\\ &=\frac {2 e^4}{7 x}+\frac {2}{35} e^{\frac {1}{5} \left (23+e^3\right )} \left (3+e^3\right ) \text {Ei}\left (\frac {1}{5} \left (-3-e^3-5 \log (x)\right )\right )-\frac {2 e^4 \left (3+e^3\right )}{7 x \left (3+e^3+5 \log (x)\right )}-\frac {1}{7} \left (2 e^4 \left (3+e^3\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{3 \left (1+\frac {e^3}{3}\right )+5 x} \, dx,x,\log (x)\right )\\ &=\frac {2 e^4}{7 x}-\frac {2 e^4 \left (3+e^3\right )}{7 x \left (3+e^3+5 \log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 23, normalized size = 0.85 \begin {gather*} \frac {10 e^4 \log (x)}{7 x \left (3+e^3+5 \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(30 + 10*E^3) + E^4*(-30 - 10*E^3)*Log[x] - 50*E^4*Log[x]^2)/(63*x^2 + 42*E^3*x^2 + 7*E^6*x^2 +
(210*x^2 + 70*E^3*x^2)*Log[x] + 175*x^2*Log[x]^2),x]

[Out]

(10*E^4*Log[x])/(7*x*(3 + E^3 + 5*Log[x]))

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fricas [A]  time = 0.66, size = 21, normalized size = 0.78 \begin {gather*} \frac {10 \, e^{4} \log \relax (x)}{7 \, {\left (x e^{3} + 5 \, x \log \relax (x) + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(4)*log(x)^2+(-10*exp(3)-30)*exp(4)*log(x)+(10*exp(3)+30)*exp(4))/(175*x^2*log(x)^2+(70*x^2*
exp(3)+210*x^2)*log(x)+7*x^2*exp(3)^2+42*x^2*exp(3)+63*x^2),x, algorithm="fricas")

[Out]

10/7*e^4*log(x)/(x*e^3 + 5*x*log(x) + 3*x)

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giac [A]  time = 0.33, size = 21, normalized size = 0.78 \begin {gather*} \frac {10 \, e^{4} \log \relax (x)}{7 \, {\left (x e^{3} + 5 \, x \log \relax (x) + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(4)*log(x)^2+(-10*exp(3)-30)*exp(4)*log(x)+(10*exp(3)+30)*exp(4))/(175*x^2*log(x)^2+(70*x^2*
exp(3)+210*x^2)*log(x)+7*x^2*exp(3)^2+42*x^2*exp(3)+63*x^2),x, algorithm="giac")

[Out]

10/7*e^4*log(x)/(x*e^3 + 5*x*log(x) + 3*x)

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maple [A]  time = 0.06, size = 20, normalized size = 0.74

method	result	size
norman	\(\frac {10 \,{\mathrm e}^{4} \ln \relax (x )}{7 x \left (3+5 \ln \relax (x )+{\mathrm e}^{3}\right )}\)	\(20\)
risch	\(\frac {2 \,{\mathrm e}^{4}}{7 x}-\frac {2 \,{\mathrm e}^{4} \left ({\mathrm e}^{3}+3\right )}{7 x \left (3+5 \ln \relax (x )+{\mathrm e}^{3}\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-50*exp(4)*ln(x)^2+(-10*exp(3)-30)*exp(4)*ln(x)+(10*exp(3)+30)*exp(4))/(175*x^2*ln(x)^2+(70*x^2*exp(3)+21
0*x^2)*ln(x)+7*x^2*exp(3)^2+42*x^2*exp(3)+63*x^2),x,method=_RETURNVERBOSE)

[Out]

10/7*exp(4)*ln(x)/x/(3+5*ln(x)+exp(3))

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maxima [A]  time = 0.65, size = 20, normalized size = 0.74 \begin {gather*} \frac {10 \, e^{4} \log \relax (x)}{7 \, {\left (x {\left (e^{3} + 3\right )} + 5 \, x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(4)*log(x)^2+(-10*exp(3)-30)*exp(4)*log(x)+(10*exp(3)+30)*exp(4))/(175*x^2*log(x)^2+(70*x^2*
exp(3)+210*x^2)*log(x)+7*x^2*exp(3)^2+42*x^2*exp(3)+63*x^2),x, algorithm="maxima")

[Out]

10/7*e^4*log(x)/(x*(e^3 + 3) + 5*x*log(x))

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mupad [B]  time = 1.61, size = 19, normalized size = 0.70 \begin {gather*} \frac {10\,{\mathrm {e}}^4\,\ln \relax (x)}{7\,x\,\left ({\mathrm {e}}^3+5\,\ln \relax (x)+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*exp(4)*log(x)^2 - exp(4)*(10*exp(3) + 30) + exp(4)*log(x)*(10*exp(3) + 30))/(log(x)*(70*x^2*exp(3) +
210*x^2) + 175*x^2*log(x)^2 + 42*x^2*exp(3) + 7*x^2*exp(6) + 63*x^2),x)

[Out]

(10*exp(4)*log(x))/(7*x*(exp(3) + 5*log(x) + 3))

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sympy [A]  time = 0.15, size = 36, normalized size = 1.33 \begin {gather*} \frac {- 2 e^{7} - 6 e^{4}}{35 x \log {\relax (x )} + 21 x + 7 x e^{3}} + \frac {2 e^{4}}{7 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(4)*ln(x)**2+(-10*exp(3)-30)*exp(4)*ln(x)+(10*exp(3)+30)*exp(4))/(175*x**2*ln(x)**2+(70*x**2
*exp(3)+210*x**2)*ln(x)+7*x**2*exp(3)**2+42*x**2*exp(3)+63*x**2),x)

[Out]

(-2*exp(7) - 6*exp(4))/(35*x*log(x) + 21*x + 7*x*exp(3)) + 2*exp(4)/(7*x)

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