3.51.16 \(\int \frac {-100 x^3-200 e^2 x^3-100 e^4 x^3+(60 x+100 x^3+100 e^4 x^3+e^4 (20 x+20 e^2 x)+e^2 (60 x+200 x^3)) \log (x^2)+(-60 x-60 e^2 x+e^4 (-20 x-20 e^2 x)) \log ^2(x^2)}{(9+6 e^4+e^8) \log ^3(x^2)} \, dx\)

Optimal. Leaf size=30 \[ \left (1+\frac {\left (5+5 e^2\right ) x^2}{\left (-3-e^4\right ) \log \left (x^2\right )}\right )^2 \]

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Rubi [A]  time = 0.21, antiderivative size = 49, normalized size of antiderivative = 1.63, number of steps used = 7, number of rules used = 4, integrand size = 121, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6, 12, 6688, 6712} \begin {gather*} \frac {25 \left (1+e^2\right )^2 x^4}{\left (3+e^4\right )^2 \log ^2\left (x^2\right )}-\frac {10 \left (1+e^2\right ) x^2}{\left (3+e^4\right ) \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-100*x^3 - 200*E^2*x^3 - 100*E^4*x^3 + (60*x + 100*x^3 + 100*E^4*x^3 + E^4*(20*x + 20*E^2*x) + E^2*(60*x
+ 200*x^3))*Log[x^2] + (-60*x - 60*E^2*x + E^4*(-20*x - 20*E^2*x))*Log[x^2]^2)/((9 + 6*E^4 + E^8)*Log[x^2]^3),
x]

[Out]

(25*(1 + E^2)^2*x^4)/((3 + E^4)^2*Log[x^2]^2) - (10*(1 + E^2)*x^2)/((3 + E^4)*Log[x^2])

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 6688

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

Rule 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-100 e^4 x^3+\left (-100-200 e^2\right ) x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx\\ &=\int \frac {\left (-100-200 e^2-100 e^4\right ) x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\left (9+6 e^4+e^8\right ) \log ^3\left (x^2\right )} \, dx\\ &=\frac {\int \frac {\left (-100-200 e^2-100 e^4\right ) x^3+\left (60 x+100 x^3+100 e^4 x^3+e^4 \left (20 x+20 e^2 x\right )+e^2 \left (60 x+200 x^3\right )\right ) \log \left (x^2\right )+\left (-60 x-60 e^2 x+e^4 \left (-20 x-20 e^2 x\right )\right ) \log ^2\left (x^2\right )}{\log ^3\left (x^2\right )} \, dx}{\left (3+e^4\right )^2}\\ &=\frac {\int \frac {20 \left (1+e^2\right ) x \left (1-\log \left (x^2\right )\right ) \left (-5 \left (1+e^2\right ) x^2+\left (3+e^4\right ) \log \left (x^2\right )\right )}{\log ^3\left (x^2\right )} \, dx}{\left (3+e^4\right )^2}\\ &=\frac {\left (20 \left (1+e^2\right )\right ) \int \frac {x \left (1-\log \left (x^2\right )\right ) \left (-5 \left (1+e^2\right ) x^2+\left (3+e^4\right ) \log \left (x^2\right )\right )}{\log ^3\left (x^2\right )} \, dx}{\left (3+e^4\right )^2}\\ &=-\frac {\left (10 \left (1+e^2\right )\right ) \operatorname {Subst}\left (\int \left (3+e^4-5 \left (1+e^2\right ) x\right ) \, dx,x,\frac {x^2}{\log \left (x^2\right )}\right )}{\left (3+e^4\right )^2}\\ &=\frac {25 \left (1+e^2\right )^2 x^4}{\left (3+e^4\right )^2 \log ^2\left (x^2\right )}-\frac {10 \left (1+e^2\right ) x^2}{\left (3+e^4\right ) \log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 51, normalized size = 1.70 \begin {gather*} -\frac {20 \left (1+e^2\right ) \left (-\frac {5 \left (1+e^2\right ) x^4}{4 \log ^2\left (x^2\right )}+\frac {\left (3+e^4\right ) x^2}{2 \log \left (x^2\right )}\right )}{\left (3+e^4\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100*x^3 - 200*E^2*x^3 - 100*E^4*x^3 + (60*x + 100*x^3 + 100*E^4*x^3 + E^4*(20*x + 20*E^2*x) + E^2*
(60*x + 200*x^3))*Log[x^2] + (-60*x - 60*E^2*x + E^4*(-20*x - 20*E^2*x))*Log[x^2]^2)/((9 + 6*E^4 + E^8)*Log[x^
2]^3),x]

[Out]

(-20*(1 + E^2)*((-5*(1 + E^2)*x^4)/(4*Log[x^2]^2) + ((3 + E^4)*x^2)/(2*Log[x^2])))/(3 + E^4)^2

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fricas [B]  time = 0.79, size = 69, normalized size = 2.30 \begin {gather*} \frac {5 \, {\left (5 \, x^{4} e^{4} + 10 \, x^{4} e^{2} + 5 \, x^{4} - 2 \, {\left (x^{2} e^{6} + x^{2} e^{4} + 3 \, x^{2} e^{2} + 3 \, x^{2}\right )} \log \left (x^{2}\right )\right )}}{{\left (e^{8} + 6 \, e^{4} + 9\right )} \log \left (x^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*log(x^2)^2+((20*exp(2)*x+20*x)*exp(4)+100*x^3*exp(2)^
2+(200*x^3+60*x)*exp(2)+100*x^3+60*x)*log(x^2)-100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/
log(x^2)^3,x, algorithm="fricas")

[Out]

5*(5*x^4*e^4 + 10*x^4*e^2 + 5*x^4 - 2*(x^2*e^6 + x^2*e^4 + 3*x^2*e^2 + 3*x^2)*log(x^2))/((e^8 + 6*e^4 + 9)*log
(x^2)^2)

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giac [B]  time = 0.22, size = 100, normalized size = 3.33 \begin {gather*} \frac {5 \, {\left (\frac {5 \, x^{4} e^{4}}{\log \left (x^{2}\right )^{2}} + \frac {10 \, x^{4} e^{2}}{\log \left (x^{2}\right )^{2}} + \frac {5 \, x^{4}}{\log \left (x^{2}\right )^{2}} - \frac {2 \, x^{2} e^{6}}{\log \left (x^{2}\right )} - \frac {2 \, x^{2} e^{4}}{\log \left (x^{2}\right )} - \frac {6 \, x^{2} e^{2}}{\log \left (x^{2}\right )} - \frac {6 \, x^{2}}{\log \left (x^{2}\right )}\right )}}{e^{8} + 6 \, e^{4} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*log(x^2)^2+((20*exp(2)*x+20*x)*exp(4)+100*x^3*exp(2)^
2+(200*x^3+60*x)*exp(2)+100*x^3+60*x)*log(x^2)-100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/
log(x^2)^3,x, algorithm="giac")

[Out]

5*(5*x^4*e^4/log(x^2)^2 + 10*x^4*e^2/log(x^2)^2 + 5*x^4/log(x^2)^2 - 2*x^2*e^6/log(x^2) - 2*x^2*e^4/log(x^2) -
6*x^2*e^2/log(x^2) - 6*x^2/log(x^2))/(e^8 + 6*e^4 + 9)

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maple [A]  time = 0.12, size = 50, normalized size = 1.67

method	result	size
norman	\(\frac {\left (-10 \,{\mathrm e}^{2}-10\right ) x^{2} \ln \left (x^{2}\right )+\frac {25 \left (2 \,{\mathrm e}^{2}+{\mathrm e}^{4}+1\right ) x^{4}}{{\mathrm e}^{4}+3}}{\left ({\mathrm e}^{4}+3\right ) \ln \left (x^{2}\right )^{2}}\)	\(50\)
risch	\(-\frac {5 x^{2} \left (2 \,{\mathrm e}^{6} \ln \left (x^{2}\right )-5 x^{2} {\mathrm e}^{4}+2 \,{\mathrm e}^{4} \ln \left (x^{2}\right )-10 x^{2} {\mathrm e}^{2}+6 \,{\mathrm e}^{2} \ln \left (x^{2}\right )-5 x^{2}+6 \ln \left (x^{2}\right )\right )}{\left ({\mathrm e}^{8}+6 \,{\mathrm e}^{4}+9\right ) \ln \left (x^{2}\right )^{2}}\)	\(72\)
default	\(\frac {-\frac {30 x^{2}}{\ln \left (x^{2}\right )}+\frac {25 x^{4}}{\ln \left (x^{2}\right )^{2}}+\frac {25 \,{\mathrm e}^{4} x^{4}}{\ln \left (x^{2}\right )^{2}}+\frac {50 \,{\mathrm e}^{2} x^{4}}{\ln \left (x^{2}\right )^{2}}-\frac {10 \,{\mathrm e}^{6} x^{2}}{\ln \left (x^{2}\right )}-\frac {30 \,{\mathrm e}^{2} x^{2}}{\ln \left (x^{2}\right )}-\frac {10 \,{\mathrm e}^{4} x^{2}}{\ln \left (x^{2}\right )}}{{\mathrm e}^{8}+6 \,{\mathrm e}^{4}+9}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*ln(x^2)^2+((20*exp(2)*x+20*x)*exp(4)+100*x^3*exp(2)^2+(200*
x^3+60*x)*exp(2)+100*x^3+60*x)*ln(x^2)-100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/ln(x^2)^
3,x,method=_RETURNVERBOSE)

[Out]

((-10*exp(2)-10)*x^2*ln(x^2)+25*(exp(2)^2+2*exp(2)+1)/(exp(4)+3)*x^4)/(exp(4)+3)/ln(x^2)^2

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maxima [A]  time = 0.40, size = 47, normalized size = 1.57 \begin {gather*} \frac {5 \, {\left (5 \, x^{4} {\left (e^{4} + 2 \, e^{2} + 1\right )} - 4 \, x^{2} {\left (e^{6} + e^{4} + 3 \, e^{2} + 3\right )} \log \relax (x)\right )}}{4 \, {\left (e^{8} + 6 \, e^{4} + 9\right )} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*log(x^2)^2+((20*exp(2)*x+20*x)*exp(4)+100*x^3*exp(2)^
2+(200*x^3+60*x)*exp(2)+100*x^3+60*x)*log(x^2)-100*x^3*exp(2)^2-200*x^3*exp(2)-100*x^3)/(exp(4)^2+6*exp(4)+9)/
log(x^2)^3,x, algorithm="maxima")

[Out]

5/4*(5*x^4*(e^4 + 2*e^2 + 1) - 4*x^2*(e^6 + e^4 + 3*e^2 + 3)*log(x))/((e^8 + 6*e^4 + 9)*log(x)^2)

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mupad [B]  time = 3.31, size = 55, normalized size = 1.83 \begin {gather*} \frac {25\,x^5\,{\left ({\mathrm {e}}^2+1\right )}^2-x^3\,\ln \left (x^2\right )\,\left (30\,{\mathrm {e}}^2+10\,{\mathrm {e}}^4+10\,{\mathrm {e}}^6+30\right )}{x\,{\ln \left (x^2\right )}^2\,\left (6\,{\mathrm {e}}^4+{\mathrm {e}}^8+9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(200*x^3*exp(2) - log(x^2)*(60*x + exp(2)*(60*x + 200*x^3) + exp(4)*(20*x + 20*x*exp(2)) + 100*x^3*exp(4)
+ 100*x^3) + 100*x^3*exp(4) + log(x^2)^2*(60*x + exp(4)*(20*x + 20*x*exp(2)) + 60*x*exp(2)) + 100*x^3)/(log(x
^2)^3*(6*exp(4) + exp(8) + 9)),x)

[Out]

(25*x^5*(exp(2) + 1)^2 - x^3*log(x^2)*(30*exp(2) + 10*exp(4) + 10*exp(6) + 30))/(x*log(x^2)^2*(6*exp(4) + exp(
8) + 9))

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sympy [B]  time = 0.17, size = 75, normalized size = 2.50 \begin {gather*} \frac {25 x^{4} + 50 x^{4} e^{2} + 25 x^{4} e^{4} + \left (- 10 x^{2} e^{6} - 10 x^{2} e^{4} - 30 x^{2} e^{2} - 30 x^{2}\right ) \log {\left (x^{2} \right )}}{\left (9 + 6 e^{4} + e^{8}\right ) \log {\left (x^{2} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*exp(2)*x-20*x)*exp(4)-60*exp(2)*x-60*x)*ln(x**2)**2+((20*exp(2)*x+20*x)*exp(4)+100*x**3*exp(2
)**2+(200*x**3+60*x)*exp(2)+100*x**3+60*x)*ln(x**2)-100*x**3*exp(2)**2-200*x**3*exp(2)-100*x**3)/(exp(4)**2+6*
exp(4)+9)/ln(x**2)**3,x)

[Out]

(25*x**4 + 50*x**4*exp(2) + 25*x**4*exp(4) + (-10*x**2*exp(6) - 10*x**2*exp(4) - 30*x**2*exp(2) - 30*x**2)*log
(x**2))/((9 + 6*exp(4) + exp(8))*log(x**2)**2)

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