3.21.92 \(\int \frac {36-120 e^{2 x}+100 e^{4 x}+(9+e^{2 x} (-15+30 x)) (i \pi +\log (-5+5 e^3))}{36-120 e^{2 x}+100 e^{4 x}} \, dx\)

Optimal. Leaf size=34 \[ x+\frac {3 x \left (i \pi +\log \left (-5+5 e^3\right )\right )}{20 \left (\frac {3}{5}-e^{2 x}\right )} \]

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Rubi [B]  time = 0.79, antiderivative size = 205, normalized size of antiderivative = 6.03, number of steps used = 18, number of rules used = 13, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {6741, 12, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 31, 29} \begin {gather*} \frac {1}{4} x^2 \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )+x-\frac {1}{16} (1-2 x)^2 \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )-\frac {1}{8} (1-2 x) \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{8} \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (3-5 e^{2 x}\right )-\frac {1}{4} x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {3 x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right )}{4 \left (3-5 e^{2 x}\right )}-\frac {1}{4} x \left (\log \left (-5 \left (1-e^3\right )\right )+i \pi \right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(36 - 120*E^(2*x) + 100*E^(4*x) + (9 + E^(2*x)*(-15 + 30*x))*(I*Pi + Log[-5 + 5*E^3]))/(36 - 120*E^(2*x) +
100*E^(4*x)),x]

[Out]

x - ((1 - 2*x)^2*(I*Pi + Log[-5*(1 - E^3)]))/16 - (x*(I*Pi + Log[-5*(1 - E^3)]))/4 + (3*x*(I*Pi + Log[-5*(1 -
E^3)]))/(4*(3 - 5*E^(2*x))) + (x^2*(I*Pi + Log[-5*(1 - E^3)]))/4 + ((I*Pi + Log[-5*(1 - E^3)])*Log[3 - 5*E^(2*
x)])/8 - ((1 - 2*x)*(I*Pi + Log[-5*(1 - E^3)])*Log[1 - (5*E^(2*x))/3])/8 - (x*(I*Pi + Log[-5*(1 - E^3)])*Log[1
- (5*E^(2*x))/3])/4

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2191

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6741

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

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 {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{4 \left (3-5 e^{2 x}\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {36-120 e^{2 x}+100 e^{4 x}+\left (9+e^{2 x} (-15+30 x)\right ) \left (i \pi +\log \left (-5+5 e^3\right )\right )}{\left (3-5 e^{2 x}\right )^2} \, dx\\ &=\frac {1}{4} \int \left (4+\frac {3 (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{3-5 e^{2 x}}+\frac {18 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{\left (3-5 e^{2 x}\right )^2}\right ) \, dx\\ &=x+\frac {1}{4} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {1-2 x}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (9 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {x}{\left (3-5 e^{2 x}\right )^2} \, dx\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {1}{4} \left (5 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {e^{2 x} (1-2 x)}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {x}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (15 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {e^{2 x} x}{\left (3-5 e^{2 x}\right )^2} \, dx\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{4} \left (-i \pi -\log \left (-5 \left (1-e^3\right )\right )\right ) \int \log \left (1-\frac {5 e^{2 x}}{3}\right ) \, dx-\frac {1}{4} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {1}{3-5 e^{2 x}} \, dx+\frac {1}{2} \left (5 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \int \frac {e^{2 x} x}{3-5 e^{2 x}} \, dx\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{8} \left (-i \pi -\log \left (-5 \left (1-e^3\right )\right )\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {5 x}{3}\right )}{x} \, dx,x,e^{2 x}\right )+\frac {1}{4} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \int \log \left (1-\frac {5 e^{2 x}}{3}\right ) \, dx-\frac {1}{8} \left (3 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(3-5 x) x} \, dx,x,e^{2 x}\right )\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )+\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \text {Li}_2\left (\frac {5 e^{2 x}}{3}\right )-\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )+\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {5 x}{3}\right )}{x} \, dx,x,e^{2 x}\right )-\frac {1}{8} \left (5 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{3-5 x} \, dx,x,e^{2 x}\right )\\ &=x-\frac {1}{16} (1-2 x)^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {3 x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )}{4 \left (3-5 e^{2 x}\right )}+\frac {1}{4} x^2 \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right )+\frac {1}{8} \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (3-5 e^{2 x}\right )-\frac {1}{8} (1-2 x) \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )-\frac {1}{4} x \left (i \pi +\log \left (-5 \left (1-e^3\right )\right )\right ) \log \left (1-\frac {5 e^{2 x}}{3}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 35, normalized size = 1.03 \begin {gather*} \frac {1}{4} x \left (4+\frac {3 i \pi +\log (125)+3 \log \left (-1+e^3\right )}{3-5 e^{2 x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(36 - 120*E^(2*x) + 100*E^(4*x) + (9 + E^(2*x)*(-15 + 30*x))*(I*Pi + Log[-5 + 5*E^3]))/(36 - 120*E^(
2*x) + 100*E^(4*x)),x]

[Out]

(x*(4 + ((3*I)*Pi + Log[125] + 3*Log[-1 + E^3])/(3 - 5*E^(2*x))))/4

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fricas [A]  time = 0.98, size = 33, normalized size = 0.97 \begin {gather*} \frac {20 \, x e^{\left (2 \, x\right )} - 3 \, x \log \left (-5 \, e^{3} + 5\right ) - 12 \, x}{4 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((30*x-15)*exp(x)^2+9)*log(-5*exp(3)+5)+100*exp(x)^4-120*exp(x)^2+36)/(100*exp(x)^4-120*exp(x)^2+36
),x, algorithm="fricas")

[Out]

1/4*(20*x*e^(2*x) - 3*x*log(-5*e^3 + 5) - 12*x)/(5*e^(2*x) - 3)

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giac [A]  time = 0.30, size = 33, normalized size = 0.97 \begin {gather*} \frac {20 \, x e^{\left (2 \, x\right )} - 3 \, x \log \left (-5 \, e^{3} + 5\right ) - 12 \, x}{4 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((30*x-15)*exp(x)^2+9)*log(-5*exp(3)+5)+100*exp(x)^4-120*exp(x)^2+36)/(100*exp(x)^4-120*exp(x)^2+36
),x, algorithm="giac")

[Out]

1/4*(20*x*e^(2*x) - 3*x*log(-5*e^3 + 5) - 12*x)/(5*e^(2*x) - 3)

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maple [A]  time = 0.10, size = 26, normalized size = 0.76

method	result	size
risch	\(x -\frac {3 x \left (\ln \relax (5)+\ln \left (-{\mathrm e}^{3}+1\right )\right )}{4 \left (5 \,{\mathrm e}^{2 x}-3\right )}\)	\(26\)
norman	\(\frac {\left (-3-\frac {3 \ln \relax (5)}{4}-\frac {3 \ln \left (-{\mathrm e}^{3}+1\right )}{4}\right ) x +5 x \,{\mathrm e}^{2 x}}{5 \,{\mathrm e}^{2 x}-3}\)	\(37\)
default	\(\ln \left ({\mathrm e}^{x}\right )+\frac {\ln \relax (5) \ln \left ({\mathrm e}^{x}\right )}{4}+\frac {\ln \left (-{\mathrm e}^{3}+1\right ) \ln \left ({\mathrm e}^{x}\right )}{4}-\frac {5 \ln \left (-{\mathrm e}^{3}+1\right ) x \,{\mathrm e}^{2 x}}{4 \left (5 \,{\mathrm e}^{2 x}-3\right )}-\frac {5 \ln \relax (5) x \,{\mathrm e}^{2 x}}{4 \left (5 \,{\mathrm e}^{2 x}-3\right )}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((30*x-15)*exp(x)^2+9)*ln(-5*exp(3)+5)+100*exp(x)^4-120*exp(x)^2+36)/(100*exp(x)^4-120*exp(x)^2+36),x,met
hod=_RETURNVERBOSE)

[Out]

x-3/4*x*(ln(5)+ln(-exp(3)+1))/(5*exp(2*x)-3)

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maxima [B]  time = 0.40, size = 93, normalized size = 2.74 \begin {gather*} \frac {1}{8} \, {\left (2 \, x - \frac {3}{5 \, e^{\left (2 \, x\right )} - 3} - \log \left (5 \, e^{\left (2 \, x\right )} - 3\right )\right )} \log \left (-5 \, e^{3} + 5\right ) - \frac {1}{8} \, {\left (\frac {10 \, x e^{\left (2 \, x\right )}}{5 \, e^{\left (2 \, x\right )} - 3} - \log \left (e^{\left (2 \, x\right )} - \frac {3}{5}\right )\right )} \log \left (-5 \, e^{3} + 5\right ) + x + \frac {3 \, \log \left (-5 \, e^{3} + 5\right )}{8 \, {\left (5 \, e^{\left (2 \, x\right )} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((30*x-15)*exp(x)^2+9)*log(-5*exp(3)+5)+100*exp(x)^4-120*exp(x)^2+36)/(100*exp(x)^4-120*exp(x)^2+36
),x, algorithm="maxima")

[Out]

1/8*(2*x - 3/(5*e^(2*x) - 3) - log(5*e^(2*x) - 3))*log(-5*e^3 + 5) - 1/8*(10*x*e^(2*x)/(5*e^(2*x) - 3) - log(e
^(2*x) - 3/5))*log(-5*e^3 + 5) + x + 3/8*log(-5*e^3 + 5)/(5*e^(2*x) - 3)

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mupad [B]  time = 1.21, size = 22, normalized size = 0.65 \begin {gather*} x-\frac {3\,x\,\ln \left (5-5\,{\mathrm {e}}^3\right )}{20\,{\mathrm {e}}^{2\,x}-12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*exp(4*x) - 120*exp(2*x) + log(5 - 5*exp(3))*(exp(2*x)*(30*x - 15) + 9) + 36)/(100*exp(4*x) - 120*exp(
2*x) + 36),x)

[Out]

x - (3*x*log(5 - 5*exp(3)))/(20*exp(2*x) - 12)

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sympy [A]  time = 0.22, size = 32, normalized size = 0.94 \begin {gather*} x + \frac {3 x \log {\relax (5 )} + 3 x \log {\left (-1 + e^{3} \right )} + 3 i \pi x}{12 - 20 e^{2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((30*x-15)*exp(x)**2+9)*ln(-5*exp(3)+5)+100*exp(x)**4-120*exp(x)**2+36)/(100*exp(x)**4-120*exp(x)**
2+36),x)

[Out]

x + (3*x*log(5) + 3*x*log(-1 + exp(3)) + 3*I*pi*x)/(12 - 20*exp(2*x))

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