3.41.95 \(\int (1-i \pi -2 x-6 x^2-\log (1+e^5)-4 x \log (4 x+e^4 x)) \, dx\)

Optimal. Leaf size=30 \[ x \left (1-i \pi -\log \left (1+e^5\right )-2 x \left (x+\log \left (\left (4+e^4\right ) x\right )\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2421, 2304} \begin {gather*} -2 x^3-2 x^2 \log \left (\left (4+e^4\right ) x\right )+x \left (1-i \pi -\log \left (1+e^5\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - I*Pi - 2*x - 6*x^2 - Log[1 + E^5] - 4*x*Log[4*x + E^4*x],x]

[Out]

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2421

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.17 \begin {gather*} x-i \pi x-2 x^3-x \log \left (1+e^5\right )-2 x^2 \log \left (\left (4+e^4\right ) x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - I*Pi - 2*x - 6*x^2 - Log[1 + E^5] - 4*x*Log[4*x + E^4*x],x]

[Out]

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

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fricas [A]  time = 0.71, size = 31, normalized size = 1.03 \begin {gather*} -2 \, x^{3} - 2 \, x^{2} \log \left (x e^{4} + 4 \, x\right ) - x \log \left (-e^{5} - 1\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*x*log(x*exp(2)^2+4*x)-log(-exp(5)-1)-6*x^2-2*x+1,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.13, size = 74, normalized size = 2.47 \begin {gather*} -2 \, x^{3} - x^{2} - \frac {2 \, {\left (x e^{4} + 4 \, x\right )}^{2} \log \left (x e^{4} + 4 \, x\right )}{e^{8} + 8 \, e^{4} + 16} - x \log \left (-e^{5} - 1\right ) + x + \frac {{\left (x e^{4} + 4 \, x\right )}^{2}}{e^{8} + 8 \, e^{4} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*x*log(x*exp(2)^2+4*x)-log(-exp(5)-1)-6*x^2-2*x+1,x, algorithm="giac")

[Out]

-2*x^3 - x^2 - 2*(x*e^4 + 4*x)^2*log(x*e^4 + 4*x)/(e^8 + 8*e^4 + 16) - x*log(-e^5 - 1) + x + (x*e^4 + 4*x)^2/(
e^8 + 8*e^4 + 16)

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maple [A]  time = 0.04, size = 32, normalized size = 1.07

method	result	size
risch	\(-2 x^{2} \ln \left (x \,{\mathrm e}^{4}+4 x \right )-\ln \left (-{\mathrm e}^{5}-1\right ) x -2 x^{3}+x\)	\(32\)
norman	\(\left (-\ln \left (-{\mathrm e}^{5}-1\right )+1\right ) x -2 x^{3}-2 x^{2} \ln \left (x \,{\mathrm e}^{4}+4 x \right )\)	\(36\)
derivativedivides	\(\frac {x \left (4+{\mathrm e}^{4}\right )-2 x^{3} \left (4+{\mathrm e}^{4}\right )-2 \left (4+{\mathrm e}^{4}\right ) x^{2} \ln \left (x \left (4+{\mathrm e}^{4}\right )\right )-\ln \left (-{\mathrm e}^{5}-1\right ) x \left (4+{\mathrm e}^{4}\right )}{4+{\mathrm e}^{4}}\)	\(66\)
default	\(-2 x^{3}-x^{2}+x -\frac {2 \,{\mathrm e}^{8} \ln \left (x \left (4+{\mathrm e}^{4}\right )\right ) x^{2}}{\left (4+{\mathrm e}^{4}\right )^{2}}-\frac {16 \,{\mathrm e}^{4} \ln \left (x \left (4+{\mathrm e}^{4}\right )\right ) x^{2}}{\left (4+{\mathrm e}^{4}\right )^{2}}-\frac {32 \ln \left (x \left (4+{\mathrm e}^{4}\right )\right ) x^{2}}{\left (4+{\mathrm e}^{4}\right )^{2}}+\frac {{\mathrm e}^{8} x^{2}}{\left (4+{\mathrm e}^{4}\right )^{2}}+\frac {8 \,{\mathrm e}^{4} x^{2}}{\left (4+{\mathrm e}^{4}\right )^{2}}+\frac {16 x^{2}}{\left (4+{\mathrm e}^{4}\right )^{2}}-\ln \left (-{\mathrm e}^{5}-1\right ) x\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4*x*ln(x*exp(2)^2+4*x)-ln(-exp(5)-1)-6*x^2-2*x+1,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 31, normalized size = 1.03 \begin {gather*} -2 \, x^{3} - 2 \, x^{2} \log \left (x e^{4} + 4 \, x\right ) - x \log \left (-e^{5} - 1\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*x*log(x*exp(2)^2+4*x)-log(-exp(5)-1)-6*x^2-2*x+1,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.08, size = 29, normalized size = 0.97 \begin {gather*} -x\,\left (\ln \left (-{\mathrm {e}}^5-1\right )+2\,x\,\ln \left (4\,x+x\,{\mathrm {e}}^4\right )+2\,x^2-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - log(- exp(5) - 1) - 4*x*log(4*x + x*exp(4)) - 6*x^2 - 2*x,x)

[Out]

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

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sympy [A]  time = 0.13, size = 37, normalized size = 1.23 \begin {gather*} - 2 x^{3} - 2 x^{2} \log {\relax (x )} - 2 x^{2} \log {\left (4 + e^{4} \right )} + x \left (- \log {\left (1 + e^{5} \right )} + 1 - i \pi \right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*x*ln(x*exp(2)**2+4*x)-ln(-exp(5)-1)-6*x**2-2*x+1,x)

[Out]

-2*x**3 - 2*x**2*log(x) - 2*x**2*log(4 + exp(4)) + x*(-log(1 + exp(5)) + 1 - I*pi)

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