3.103.19 \(\int \frac {x (i \pi +\log (3))^4-2 (i \pi +\log (3))^4 \log (x)+(x (i \pi +\log (3))^4-(i \pi +\log (3))^4 \log ^2(x)) \log (x-\log ^2(x))+(-4 e^{4+x} x^3 (i \pi +\log (3))^2+8 e^{4+x} x^2 (i \pi +\log (3))^2 \log (x)) \log ^2(x-\log ^2(x))+(e^{4+x} (-4 x^3+2 x^4) (i \pi +\log (3))^2+e^{4+x} (4 x^2-2 x^3) (i \pi +\log (3))^2 \log ^2(x)) \log ^3(x-\log ^2(x))+(-8 e^{8+2 x} x^6+8 e^{8+2 x} x^5 \log ^2(x)) \log ^5(x-\log ^2(x))}{(-4 x^6+4 x^5 \log ^2(x)) \log ^5(x-\log ^2(x))} \, dx\)

Optimal. Leaf size=38 \[ \left (-e^{4+x}+\frac {(i \pi +\log (3))^2}{4 x^2 \log ^2\left (x-\log ^2(x)\right )}\right )^2 \]

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Rubi [B]  time = 5.52, antiderivative size = 109, normalized size of antiderivative = 2.87, number of steps used = 8, number of rules used = 7, integrand size = 262, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {2561, 6688, 12, 6742, 2194, 6687, 2288} \begin {gather*} \frac {(\pi -i \log (3))^4}{16 x^4 \log ^4\left (x-\log ^2(x)\right )}+\frac {e^{x+4} (\pi -i \log (3))^2 \left (x^2 \log \left (x-\log ^2(x)\right )-x \log ^2(x) \log \left (x-\log ^2(x)\right )\right )}{2 x^3 \left (x-\log ^2(x)\right ) \log ^3\left (x-\log ^2(x)\right )}+e^{2 x+8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(I*Pi + Log[3])^4 - 2*(I*Pi + Log[3])^4*Log[x] + (x*(I*Pi + Log[3])^4 - (I*Pi + Log[3])^4*Log[x]^2)*Log
[x - Log[x]^2] + (-4*E^(4 + x)*x^3*(I*Pi + Log[3])^2 + 8*E^(4 + x)*x^2*(I*Pi + Log[3])^2*Log[x])*Log[x - Log[x
]^2]^2 + (E^(4 + x)*(-4*x^3 + 2*x^4)*(I*Pi + Log[3])^2 + E^(4 + x)*(4*x^2 - 2*x^3)*(I*Pi + Log[3])^2*Log[x]^2)
*Log[x - Log[x]^2]^3 + (-8*E^(8 + 2*x)*x^6 + 8*E^(8 + 2*x)*x^5*Log[x]^2)*Log[x - Log[x]^2]^5)/((-4*x^6 + 4*x^5
*Log[x]^2)*Log[x - Log[x]^2]^5),x]

[Out]

E^(8 + 2*x) + (Pi - I*Log[3])^4/(16*x^4*Log[x - Log[x]^2]^4) + (E^(4 + x)*(Pi - I*Log[3])^2*(x^2*Log[x - Log[x
]^2] - x*Log[x]^2*Log[x - Log[x]^2]))/(2*x^3*(x - Log[x]^2)*Log[x - Log[x]^2]^3)

Rule 12

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

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 {x (i \pi +\log (3))^4-2 (i \pi +\log (3))^4 \log (x)+\left (x (i \pi +\log (3))^4-(i \pi +\log (3))^4 \log ^2(x)\right ) \log \left (x-\log ^2(x)\right )+\left (-4 e^{4+x} x^3 (i \pi +\log (3))^2+8 e^{4+x} x^2 (i \pi +\log (3))^2 \log (x)\right ) \log ^2\left (x-\log ^2(x)\right )+\left (e^{4+x} \left (-4 x^3+2 x^4\right ) (i \pi +\log (3))^2+e^{4+x} \left (4 x^2-2 x^3\right ) (i \pi +\log (3))^2 \log ^2(x)\right ) \log ^3\left (x-\log ^2(x)\right )+\left (-8 e^{8+2 x} x^6+8 e^{8+2 x} x^5 \log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )}{x^5 \left (-4 x+4 \log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )} \, dx\\ &=\int \frac {-x (i \pi +\log (3))^4+2 (i \pi +\log (3))^4 \log (x)-(i \pi +\log (3))^4 \left (x-\log ^2(x)\right ) \log \left (x-\log ^2(x)\right )-4 e^{4+x} x^2 (\pi -i \log (3))^2 (x-2 \log (x)) \log ^2\left (x-\log ^2(x)\right )+2 e^{4+x} (-2+x) x^2 (\pi -i \log (3))^2 \left (x-\log ^2(x)\right ) \log ^3\left (x-\log ^2(x)\right )+8 e^{8+2 x} x^5 \left (x-\log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )}{4 x^5 \left (x-\log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )} \, dx\\ &=\frac {1}{4} \int \frac {-x (i \pi +\log (3))^4+2 (i \pi +\log (3))^4 \log (x)-(i \pi +\log (3))^4 \left (x-\log ^2(x)\right ) \log \left (x-\log ^2(x)\right )-4 e^{4+x} x^2 (\pi -i \log (3))^2 (x-2 \log (x)) \log ^2\left (x-\log ^2(x)\right )+2 e^{4+x} (-2+x) x^2 (\pi -i \log (3))^2 \left (x-\log ^2(x)\right ) \log ^3\left (x-\log ^2(x)\right )+8 e^{8+2 x} x^5 \left (x-\log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )}{x^5 \left (x-\log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )} \, dx\\ &=\frac {1}{4} \int \left (8 e^{8+2 x}-\frac {(\pi -i \log (3))^4 \left (x-2 \log (x)+x \log \left (x-\log ^2(x)\right )-\log ^2(x) \log \left (x-\log ^2(x)\right )\right )}{x^5 \left (x-\log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )}+\frac {2 e^{4+x} (\pi -i \log (3))^2 \left (-2 x+4 \log (x)-2 x \log \left (x-\log ^2(x)\right )+x^2 \log \left (x-\log ^2(x)\right )+2 \log ^2(x) \log \left (x-\log ^2(x)\right )-x \log ^2(x) \log \left (x-\log ^2(x)\right )\right )}{x^3 \left (x-\log ^2(x)\right ) \log ^3\left (x-\log ^2(x)\right )}\right ) \, dx\\ &=2 \int e^{8+2 x} \, dx+\frac {1}{2} (\pi -i \log (3))^2 \int \frac {e^{4+x} \left (-2 x+4 \log (x)-2 x \log \left (x-\log ^2(x)\right )+x^2 \log \left (x-\log ^2(x)\right )+2 \log ^2(x) \log \left (x-\log ^2(x)\right )-x \log ^2(x) \log \left (x-\log ^2(x)\right )\right )}{x^3 \left (x-\log ^2(x)\right ) \log ^3\left (x-\log ^2(x)\right )} \, dx-\frac {1}{4} (\pi -i \log (3))^4 \int \frac {x-2 \log (x)+x \log \left (x-\log ^2(x)\right )-\log ^2(x) \log \left (x-\log ^2(x)\right )}{x^5 \left (x-\log ^2(x)\right ) \log ^5\left (x-\log ^2(x)\right )} \, dx\\ &=e^{8+2 x}+\frac {(\pi -i \log (3))^4}{16 x^4 \log ^4\left (x-\log ^2(x)\right )}+\frac {e^{4+x} (\pi -i \log (3))^2 \left (x^2 \log \left (x-\log ^2(x)\right )-x \log ^2(x) \log \left (x-\log ^2(x)\right )\right )}{2 x^3 \left (x-\log ^2(x)\right ) \log ^3\left (x-\log ^2(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 52, normalized size = 1.37 \begin {gather*} \frac {\left ((\pi -i \log (3))^2+4 e^{4+x} x^2 \log ^2\left (x-\log ^2(x)\right )\right )^2}{16 x^4 \log ^4\left (x-\log ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(I*Pi + Log[3])^4 - 2*(I*Pi + Log[3])^4*Log[x] + (x*(I*Pi + Log[3])^4 - (I*Pi + Log[3])^4*Log[x]^
2)*Log[x - Log[x]^2] + (-4*E^(4 + x)*x^3*(I*Pi + Log[3])^2 + 8*E^(4 + x)*x^2*(I*Pi + Log[3])^2*Log[x])*Log[x -
Log[x]^2]^2 + (E^(4 + x)*(-4*x^3 + 2*x^4)*(I*Pi + Log[3])^2 + E^(4 + x)*(4*x^2 - 2*x^3)*(I*Pi + Log[3])^2*Log
[x]^2)*Log[x - Log[x]^2]^3 + (-8*E^(8 + 2*x)*x^6 + 8*E^(8 + 2*x)*x^5*Log[x]^2)*Log[x - Log[x]^2]^5)/((-4*x^6 +
4*x^5*Log[x]^2)*Log[x - Log[x]^2]^5),x]

[Out]

((Pi - I*Log[3])^2 + 4*E^(4 + x)*x^2*Log[x - Log[x]^2]^2)^2/(16*x^4*Log[x - Log[x]^2]^4)

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fricas [B]  time = 0.93, size = 111, normalized size = 2.92 \begin {gather*} \frac {16 \, x^{4} e^{\left (2 \, x + 8\right )} \log \left (-\log \relax (x)^{2} + x\right )^{4} + \pi ^{4} - 4 i \, \pi ^{3} \log \relax (3) - 6 \, \pi ^{2} \log \relax (3)^{2} + 4 i \, \pi \log \relax (3)^{3} + \log \relax (3)^{4} + 8 \, {\left (\pi ^{2} x^{2} - 2 i \, \pi x^{2} \log \relax (3) - x^{2} \log \relax (3)^{2}\right )} e^{\left (x + 4\right )} \log \left (-\log \relax (x)^{2} + x\right )^{2}}{16 \, x^{4} \log \left (-\log \relax (x)^{2} + x\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^5*exp(4+x)^2*log(x)^2-8*x^6*exp(4+x)^2)*log(-log(x)^2+x)^5+((-2*x^3+4*x^2)*(log(3)+I*pi)^2*exp
(4+x)*log(x)^2+(2*x^4-4*x^3)*(log(3)+I*pi)^2*exp(4+x))*log(-log(x)^2+x)^3+(8*x^2*(log(3)+I*pi)^2*exp(4+x)*log(
x)-4*x^3*(log(3)+I*pi)^2*exp(4+x))*log(-log(x)^2+x)^2+(-(log(3)+I*pi)^4*log(x)^2+x*(log(3)+I*pi)^4)*log(-log(x
)^2+x)-2*(log(3)+I*pi)^4*log(x)+x*(log(3)+I*pi)^4)/(4*x^5*log(x)^2-4*x^6)/log(-log(x)^2+x)^5,x, algorithm="fri
cas")

[Out]

1/16*(16*x^4*e^(2*x + 8)*log(-log(x)^2 + x)^4 + pi^4 - 4*I*pi^3*log(3) - 6*pi^2*log(3)^2 + 4*I*pi*log(3)^3 + l
og(3)^4 + 8*(pi^2*x^2 - 2*I*pi*x^2*log(3) - x^2*log(3)^2)*e^(x + 4)*log(-log(x)^2 + x)^2)/(x^4*log(-log(x)^2 +
x)^4)

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giac [B]  time = 0.30, size = 139, normalized size = 3.66 \begin {gather*} \frac {16 \, x^{4} e^{\left (2 \, x + 8\right )} \log \left (-\log \relax (x)^{2} + x\right )^{4} + 8 \, \pi ^{2} x^{2} e^{\left (x + 4\right )} \log \left (-\log \relax (x)^{2} + x\right )^{2} - 16 i \, \pi x^{2} e^{\left (x + 4\right )} \log \relax (3) \log \left (-\log \relax (x)^{2} + x\right )^{2} - 8 \, x^{2} e^{\left (x + 4\right )} \log \relax (3)^{2} \log \left (-\log \relax (x)^{2} + x\right )^{2} + \pi ^{4} - 4 i \, \pi ^{3} \log \relax (3) - 6 \, \pi ^{2} \log \relax (3)^{2} + 4 i \, \pi \log \relax (3)^{3} + \log \relax (3)^{4}}{16 \, x^{4} \log \left (-\log \relax (x)^{2} + x\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^5*exp(4+x)^2*log(x)^2-8*x^6*exp(4+x)^2)*log(-log(x)^2+x)^5+((-2*x^3+4*x^2)*(log(3)+I*pi)^2*exp
(4+x)*log(x)^2+(2*x^4-4*x^3)*(log(3)+I*pi)^2*exp(4+x))*log(-log(x)^2+x)^3+(8*x^2*(log(3)+I*pi)^2*exp(4+x)*log(
x)-4*x^3*(log(3)+I*pi)^2*exp(4+x))*log(-log(x)^2+x)^2+(-(log(3)+I*pi)^4*log(x)^2+x*(log(3)+I*pi)^4)*log(-log(x
)^2+x)-2*(log(3)+I*pi)^4*log(x)+x*(log(3)+I*pi)^4)/(4*x^5*log(x)^2-4*x^6)/log(-log(x)^2+x)^5,x, algorithm="gia
c")

[Out]

1/16*(16*x^4*e^(2*x + 8)*log(-log(x)^2 + x)^4 + 8*pi^2*x^2*e^(x + 4)*log(-log(x)^2 + x)^2 - 16*I*pi*x^2*e^(x +
4)*log(3)*log(-log(x)^2 + x)^2 - 8*x^2*e^(x + 4)*log(3)^2*log(-log(x)^2 + x)^2 + pi^4 - 4*I*pi^3*log(3) - 6*p
i^2*log(3)^2 + 4*I*pi*log(3)^3 + log(3)^4)/(x^4*log(-log(x)^2 + x)^4)

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maple [B]  time = 0.36, size = 128, normalized size = 3.37

method	result	size
risch	\({\mathrm e}^{2 x +8}+\frac {-16 i \pi \ln \relax (3) x^{2} {\mathrm e}^{4+x} \ln \left (-\ln \relax (x )^{2}+x \right )^{2}+8 \pi ^{2} x^{2} {\mathrm e}^{4+x} \ln \left (-\ln \relax (x )^{2}+x \right )^{2}-8 \ln \relax (3)^{2} x^{2} {\mathrm e}^{4+x} \ln \left (-\ln \relax (x )^{2}+x \right )^{2}-4 i \ln \relax (3) \pi ^{3}+4 i \ln \relax (3)^{3} \pi +\pi ^{4}-6 \pi ^{2} \ln \relax (3)^{2}+\ln \relax (3)^{4}}{16 x^{4} \ln \left (-\ln \relax (x )^{2}+x \right )^{4}}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^5*exp(4+x)^2*ln(x)^2-8*x^6*exp(4+x)^2)*ln(-ln(x)^2+x)^5+((-2*x^3+4*x^2)*(ln(3)+I*Pi)^2*exp(4+x)*ln(x
)^2+(2*x^4-4*x^3)*(ln(3)+I*Pi)^2*exp(4+x))*ln(-ln(x)^2+x)^3+(8*x^2*(ln(3)+I*Pi)^2*exp(4+x)*ln(x)-4*x^3*(ln(3)+
I*Pi)^2*exp(4+x))*ln(-ln(x)^2+x)^2+(-(ln(3)+I*Pi)^4*ln(x)^2+x*(ln(3)+I*Pi)^4)*ln(-ln(x)^2+x)-2*(ln(3)+I*Pi)^4*
ln(x)+x*(ln(3)+I*Pi)^4)/(4*x^5*ln(x)^2-4*x^6)/ln(-ln(x)^2+x)^5,x,method=_RETURNVERBOSE)

[Out]

exp(2*x+8)+1/16*(-16*I*Pi*ln(3)*x^2*exp(4+x)*ln(-ln(x)^2+x)^2+8*Pi^2*x^2*exp(4+x)*ln(-ln(x)^2+x)^2-8*ln(3)^2*x
^2*exp(4+x)*ln(-ln(x)^2+x)^2-4*I*ln(3)*Pi^3+4*I*ln(3)^3*Pi+Pi^4-6*Pi^2*ln(3)^2+ln(3)^4)/x^4/ln(-ln(x)^2+x)^4

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maxima [B]  time = 0.82, size = 104, normalized size = 2.74 \begin {gather*} \frac {16 \, x^{4} e^{\left (2 \, x + 8\right )} \log \left (-\log \relax (x)^{2} + x\right )^{4} + 8 \, {\left (\pi ^{2} - 2 i \, \pi \log \relax (3) - \log \relax (3)^{2}\right )} x^{2} e^{\left (x + 4\right )} \log \left (-\log \relax (x)^{2} + x\right )^{2} + \pi ^{4} - 4 i \, \pi ^{3} \log \relax (3) - 6 \, \pi ^{2} \log \relax (3)^{2} + 4 i \, \pi \log \relax (3)^{3} + \log \relax (3)^{4}}{16 \, x^{4} \log \left (-\log \relax (x)^{2} + x\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^5*exp(4+x)^2*log(x)^2-8*x^6*exp(4+x)^2)*log(-log(x)^2+x)^5+((-2*x^3+4*x^2)*(log(3)+I*pi)^2*exp
(4+x)*log(x)^2+(2*x^4-4*x^3)*(log(3)+I*pi)^2*exp(4+x))*log(-log(x)^2+x)^3+(8*x^2*(log(3)+I*pi)^2*exp(4+x)*log(
x)-4*x^3*(log(3)+I*pi)^2*exp(4+x))*log(-log(x)^2+x)^2+(-(log(3)+I*pi)^4*log(x)^2+x*(log(3)+I*pi)^4)*log(-log(x
)^2+x)-2*(log(3)+I*pi)^4*log(x)+x*(log(3)+I*pi)^4)/(4*x^5*log(x)^2-4*x^6)/log(-log(x)^2+x)^5,x, algorithm="max
ima")

[Out]

1/16*(16*x^4*e^(2*x + 8)*log(-log(x)^2 + x)^4 + 8*(pi^2 - 2*I*pi*log(3) - log(3)^2)*x^2*e^(x + 4)*log(-log(x)^
2 + x)^2 + pi^4 - 4*I*pi^3*log(3) - 6*pi^2*log(3)^2 + 4*I*pi*log(3)^3 + log(3)^4)/(x^4*log(-log(x)^2 + x)^4)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(x)*(Pi*1i + log(3))^4 + log(x - log(x)^2)^2*(4*x^3*exp(x + 4)*(Pi*1i + log(3))^2 - 8*x^2*exp(x + 4
)*log(x)*(Pi*1i + log(3))^2) - x*(Pi*1i + log(3))^4 + log(x - log(x)^2)^3*(exp(x + 4)*(Pi*1i + log(3))^2*(4*x^
3 - 2*x^4) - exp(x + 4)*log(x)^2*(Pi*1i + log(3))^2*(4*x^2 - 2*x^3)) + log(x - log(x)^2)*(log(x)^2*(Pi*1i + lo
g(3))^4 - x*(Pi*1i + log(3))^4) + log(x - log(x)^2)^5*(8*x^6*exp(2*x + 8) - 8*x^5*exp(2*x + 8)*log(x)^2))/(log
(x - log(x)^2)^5*(4*x^5*log(x)^2 - 4*x^6)),x)

[Out]

\text{Hanged}

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**5*exp(4+x)**2*ln(x)**2-8*x**6*exp(4+x)**2)*ln(-ln(x)**2+x)**5+((-2*x**3+4*x**2)*(ln(3)+I*pi)*
*2*exp(4+x)*ln(x)**2+(2*x**4-4*x**3)*(ln(3)+I*pi)**2*exp(4+x))*ln(-ln(x)**2+x)**3+(8*x**2*(ln(3)+I*pi)**2*exp(
4+x)*ln(x)-4*x**3*(ln(3)+I*pi)**2*exp(4+x))*ln(-ln(x)**2+x)**2+(-(ln(3)+I*pi)**4*ln(x)**2+x*(ln(3)+I*pi)**4)*l
n(-ln(x)**2+x)-2*(ln(3)+I*pi)**4*ln(x)+x*(ln(3)+I*pi)**4)/(4*x**5*ln(x)**2-4*x**6)/ln(-ln(x)**2+x)**5,x)

[Out]

Timed out

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