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

Optimal. Leaf size=21 \[ 1+2 x+x \left (-1+5 \log \left (\frac {1}{\left (e^4+2 x\right )^2}\right )\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 36, normalized size of antiderivative = 1.71, number of steps used = 7, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6688, 6742, 43, 2389, 2295} \begin {gather*} 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 ) \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2295

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

Rule 2389

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 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^4-18 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} \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 1.71 \begin {gather*} 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 {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.64, size = 21, normalized size = 1.00 \begin {gather*} 5 \, x \log \left (\frac {1}{4 \, x^{2} + 4 \, x e^{4} + e^{8}}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.49, size = 19, normalized size = 0.90 \begin {gather*} -5 \, x \log \left (4 \, x^{2} + 4 \, x e^{4} + e^{8}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.17, size = 22, normalized size = 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	\(x +5 \ln \left (\frac {1}{{\mathrm e}^{8}+4 x \,{\mathrm e}^{4}+4 x^{2}}\right ) x\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [B]  time = 0.77, size = 108, normalized size = 5.14 \begin {gather*} -\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.28, size = 21, normalized size = 1.00 \begin {gather*} x+5\,x\,\ln \left (\frac {1}{4\,x^2+4\,{\mathrm {e}}^4\,x+{\mathrm {e}}^8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.14, size = 22, normalized size = 1.05 \begin {gather*} 5 x \log {\left (\frac {1}{4 x^{2} + 4 x e^{4} + e^{8}} \right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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