\(\int \frac {e^4-18 x+(5 e^4+10 x) \log (\frac {1}{e^8+4 e^4 x+4 x^2})}{e^4+2 x} \, dx\) [959]

Optimal result

Integrand size = 45, antiderivative size = 21 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=1+2 x+x \left (-1+5 \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6820, 6874, 45, 2436, 2332} \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=x+\frac {5}{2} \left (2 x+e^4\right ) \log \left (\frac {1}{\left (2 x+e^4\right )^2}\right )+5 e^4 \log \left (2 x+e^4\right ) \]

[In]

Int[(E^4 - 18*x + (5*E^4 + 10*x)*Log[(E^8 + 4*E^4*x + 4*x^2)^(-1)])/(E^4 + 2*x),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

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

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^4-18 x+5 \left (e^4+2 x\right ) \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )}{e^4+2 x} \, dx \\ & = \int \left (\frac {e^4-18 x}{e^4+2 x}+5 \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )\right ) \, dx \\ & = 5 \int \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right ) \, dx+\int \frac {e^4-18 x}{e^4+2 x} \, dx \\ & = \frac {5}{2} \text {Subst}\left (\int \log \left (\frac {1}{x^2}\right ) \, dx,x,e^4+2 x\right )+\int \left (-9+\frac {10 e^4}{e^4+2 x}\right ) \, dx \\ & = x+\frac {5}{2} \left (e^4+2 x\right ) \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )+5 e^4 \log \left (e^4+2 x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=x+\frac {5}{2} \left (e^4+2 x\right ) \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )+5 e^4 \log \left (e^4+2 x\right ) \]

[In]

Integrate[(E^4 - 18*x + (5*E^4 + 10*x)*Log[(E^8 + 4*E^4*x + 4*x^2)^(-1)])/(E^4 + 2*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
risch	\(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\)	\(22\)
default	\(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\)	\(24\)
norman	\(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\)	\(24\)
parts	\(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\)	\(24\)
parallelrisch	\(5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x -{\mathrm e}^{4}+x\)	\(28\)

[In]

int(((5*exp(4)+10*x)*ln(1/(exp(4)^2+4*x*exp(4)+4*x^2))+exp(4)-18*x)/(2*x+exp(4)),x,method=_RETURNVERBOSE)

[Out]

x+5*ln(1/(exp(8)+4*x*exp(4)+4*x^2))*x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=5 \, x \log \left (\frac {1}{4 \, x^{2} + 4 \, x e^{4} + e^{8}}\right ) + x \]

[In]

integrate(((5*exp(4)+10*x)*log(1/(exp(4)^2+4*x*exp(4)+4*x^2))+exp(4)-18*x)/(2*x+exp(4)),x, algorithm="fricas")

[Out]

5*x*log(1/(4*x^2 + 4*x*e^4 + e^8)) + x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=5 x \log {\left (\frac {1}{4 x^{2} + 4 x e^{4} + e^{8}} \right )} + x \]

[In]

integrate(((5*exp(4)+10*x)*ln(1/(exp(4)**2+4*x*exp(4)+4*x**2))+exp(4)-18*x)/(2*x+exp(4)),x)

[Out]

5*x*log(1/(4*x**2 + 4*x*exp(4) + exp(8))) + x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (20) = 40\).

Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 5.14 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=-\frac {5}{2} \, e^{4} \log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) \log \left (2 \, x + e^{4}\right ) - \frac {5}{2} \, e^{4} \log \left (2 \, x + e^{4}\right )^{2} + \frac {5}{2} \, {\left (\log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) \log \left (2 \, x + e^{4}\right ) - \log \left (2 \, x + e^{4}\right )^{2}\right )} e^{4} + \frac {5}{2} \, {\left (e^{4} \log \left (2 \, x + e^{4}\right ) - 2 \, x\right )} \log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) + x \]

[In]

integrate(((5*exp(4)+10*x)*log(1/(exp(4)^2+4*x*exp(4)+4*x^2))+exp(4)-18*x)/(2*x+exp(4)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=-5 \, x \log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) + x \]

[In]

integrate(((5*exp(4)+10*x)*log(1/(exp(4)^2+4*x*exp(4)+4*x^2))+exp(4)-18*x)/(2*x+exp(4)),x, algorithm="giac")

[Out]

-5*x*log(4*x^2 + 4*x*e^4 + e^8) + x

Mupad [B] (verification not implemented)

Time = 9.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^4-18 x+\left (5 e^4+10 x\right ) \log \left (\frac {1}{e^8+4 e^4 x+4 x^2}\right )}{e^4+2 x} \, dx=x+5\,x\,\ln \left (\frac {1}{4\,x^2+4\,{\mathrm {e}}^4\,x+{\mathrm {e}}^8}\right ) \]

[In]

int((exp(4) - 18*x + log(1/(exp(8) + 4*x*exp(4) + 4*x^2))*(10*x + 5*exp(4)))/(2*x + exp(4)),x)

[Out]

x + 5*x*log(1/(exp(8) + 4*x*exp(4) + 4*x^2))