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

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

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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} $$\begin{array}{c}-2x^3-2x^2\log \left((4+e^4)x\right)+x(1-i\pi -\log (1+e^5))\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-x^2-2x^3+x(1-i\pi -\log (1+e^5))-4\int x\log (4x+e^{4}x)dx\\ & =-x^2-2x^3+x(1-i\pi -\log (1+e^5))-4\int x\log \left((4+e^4)x\right)dx\\ & =-2x^3+x(1-i\pi -\log (1+e^5))-2x^2\log \left((4+e^4)x\right)\end{array}\end{array}$$

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

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 $$\begin{array}{c}-2x^3-2x^2\log (x{e}^4+4x)-x\log (-e^5-1)+x\end{array}$$

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 $$\begin{array}{c}-2x^3-x^2-\frac{2{(x{e}^4+4x)}^2\log (x{e}^4+4x)}{e^8+8e^4+16}-x\log (-e^5-1)+x+\frac{{(x{e}^4+4x)}^2}{e^8+8e^4+16}\end{array}$$

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

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 $$\begin{array}{c}-2x^3-2x^2\log (x{e}^4+4x)-x\log (-e^5-1)+x\end{array}$$

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 $$\begin{array}{c}-x(\ln (-{\mathrm{e}}^5-1)+2x\ln (4x+x{\mathrm{e}}^4)+2x^2-1)\end{array}$$

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 $$\begin{array}{c}-2x^3-2x^2\log (x)-2x^2\log (4+e^4)+x(-\log (1+e^5)+1-i\pi )\end{array}$$

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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