3.85.14 $\int \frac{3072+2432x+268x^2+8x^3+e^{e^{x/4}}(-64+e^{x/4}(-16x-x^2))}{1024+128x+4x^2}dx$

Optimal. Leaf size=29 $$-e^4+3x+x^2-\frac{e^{e^{x/4}}x}{16+x}$$

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Rubi [F]  time = 0.50, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.000, Rules used = {} $$\begin{array}{c}\int \frac{3072+2432x+268x^2+8x^3+e^{e^{x/4}}(-64+e^{x/4}(-16x-x^2))}{1024+128x+4x^2}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(3072 + 2432*x + 268*x^2 + 8*x^3 + E^E^(x/4)*(-64 + E^(x/4)*(-16*x - x^2)))/(1024 + 128*x + 4*x^2),x]

[Out]

-E^E^(x/4) + 3*x + x^2 - 16*Defer[Int][E^E^(x/4)/(16 + x)^2, x] + 4*Defer[Int][E^((4*E^(x/4) + x)/4)/(16 + x),
x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{3072+2432x+268x^2+8x^3+e^{e^{x/4}}(-64+e^{x/4}(-16x-x^2))}{4(16+x{)}^2}dx\\ & =\frac{1}{4}\int \frac{3072+2432x+268x^2+8x^3+e^{e^{x/4}}(-64+e^{x/4}(-16x-x^2))}{(16+x{)}^2}dx\\ & =\frac{1}{4}\int (\frac{e^{\frac{1}{4}(4e^{x/4}+x)}x}{-16-x}+\frac{3072}{(16+x{)}^2}-\frac{64e^{e^{x/4}}}{(16+x{)}^2}+\frac{2432x}{(16+x{)}^2}+\frac{268x^2}{(16+x{)}^2}+\frac{8x^3}{(16+x{)}^2})dx\\ & =-\frac{768}{16+x}+\frac{1}{4}\int \frac{e^{\frac{1}{4}(4e^{x/4}+x)}x}{-16-x}dx+2\int \frac{x^3}{(16+x{)}^2}dx-16\int \frac{e^{e^{x/4}}}{(16+x{)}^2}dx+67\int \frac{x^2}{(16+x{)}^2}dx+608\int \frac{x}{(16+x{)}^2}dx\\ & =-\frac{768}{16+x}+\frac{1}{4}\int (-e^{\frac{1}{4}(4e^{x/4}+x)}+\frac{16e^{\frac{1}{4}(4e^{x/4}+x)}}{16+x})dx+2\int (-32+x-\frac{4096}{(16+x{)}^2}+\frac{768}{16+x})dx-16\int \frac{e^{e^{x/4}}}{(16+x{)}^2}dx+67\int (1+\frac{256}{(16+x{)}^2}-\frac{32}{16+x})dx+608\int (-\frac{16}{(16+x{)}^2}+\frac{1}{16+x})dx\\ & =3x+x^2-\frac{1}{4}\int e^{\frac{1}{4}(4e^{x/4}+x)}dx+4\int \frac{e^{\frac{1}{4}(4e^{x/4}+x)}}{16+x}dx-16\int \frac{e^{e^{x/4}}}{(16+x{)}^2}dx\\ & =3x+x^2+4\int \frac{e^{\frac{1}{4}(4e^{x/4}+x)}}{16+x}dx-16\int \frac{e^{e^{x/4}}}{(16+x{)}^2}dx-\mathrm{Subst}(\int e^{e^x+x}dx,x,\frac{x}{4})\\ & =3x+x^2+4\int \frac{e^{\frac{1}{4}(4e^{x/4}+x)}}{16+x}dx-16\int \frac{e^{e^{x/4}}}{(16+x{)}^2}dx-\mathrm{Subst}(\int e^{x}dx,x,e^{x/4})\\ & =-e^{e^{x/4}}+3x+x^2+4\int \frac{e^{\frac{1}{4}(4e^{x/4}+x)}}{16+x}dx-16\int \frac{e^{e^{x/4}}}{(16+x{)}^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.25, size = 26, normalized size = 0.90 $$\begin{array}{c}\frac{x(48-e^{e^{x/4}}+19x+x^2)}{16+x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(3072 + 2432*x + 268*x^2 + 8*x^3 + E^E^(x/4)*(-64 + E^(x/4)*(-16*x - x^2)))/(1024 + 128*x + 4*x^2),x
]

[Out]

(x*(48 - E^E^(x/4) + 19*x + x^2))/(16 + x)

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fricas [A]  time = 0.65, size = 26, normalized size = 0.90 $$\begin{array}{c}\frac{x^3+19x^2-x{e}^{\left(e^{\left(\frac{1}{4}x\right)}\right)}+48x}{x+16}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-16*x)*exp(1/4*x)-64)*exp(exp(1/4*x))+8*x^3+268*x^2+2432*x+3072)/(4*x^2+128*x+1024),x, algori
thm="fricas")

[Out]

(x^3 + 19*x^2 - x*e^(e^(1/4*x)) + 48*x)/(x + 16)

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giac [B]  time = 0.18, size = 53, normalized size = 1.83 $$\begin{array}{c}\frac{x^3{e}^{\left(\frac{1}{4}x\right)}+19x^2{e}^{\left(\frac{1}{4}x\right)}+48x{e}^{\left(\frac{1}{4}x\right)}-x{e}^{(\frac{1}{4}x+e^{\left(\frac{1}{4}x\right)})}}{x{e}^{\left(\frac{1}{4}x\right)}+16e^{\left(\frac{1}{4}x\right)}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-16*x)*exp(1/4*x)-64)*exp(exp(1/4*x))+8*x^3+268*x^2+2432*x+3072)/(4*x^2+128*x+1024),x, algori
thm="giac")

[Out]

(x^3*e^(1/4*x) + 19*x^2*e^(1/4*x) + 48*x*e^(1/4*x) - x*e^(1/4*x + e^(1/4*x)))/(x*e^(1/4*x) + 16*e^(1/4*x))

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maple [A]  time = 0.41, size = 21, normalized size = 0.72

method	result	size
risch	$x^2+3x-\frac{{\mathrm{e}}^{{\mathrm{e}}^{\frac{x}{4}}}x}{x+16}$	$21$
norman	$\frac{x^3+19x^2-x{\mathrm{e}}^{{\mathrm{e}}^{\frac{x}{4}}}-768}{x+16}$	$25$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2-16*x)*exp(1/4*x)-64)*exp(exp(1/4*x))+8*x^3+268*x^2+2432*x+3072)/(4*x^2+128*x+1024),x,method=_RETUR
NVERBOSE)

[Out]

x^2+3*x-exp(exp(1/4*x))*x/(x+16)

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maxima [A]  time = 0.41, size = 20, normalized size = 0.69 $$\begin{array}{c}{x}^2+3x-\frac{x{e}^{\left(e^{\left(\frac{1}{4}x\right)}\right)}}{x+16}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2-16*x)*exp(1/4*x)-64)*exp(exp(1/4*x))+8*x^3+268*x^2+2432*x+3072)/(4*x^2+128*x+1024),x, algori
thm="maxima")

[Out]

x^2 + 3*x - x*e^(e^(1/4*x))/(x + 16)

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mupad [B]  time = 5.22, size = 22, normalized size = 0.76 $$\begin{array}{c}\frac{x(19x-{\mathrm{e}}^{{\mathrm{e}}^{x/4}}+x^2+48)}{x+16}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2432*x - exp(exp(x/4))*(exp(x/4)*(16*x + x^2) + 64) + 268*x^2 + 8*x^3 + 3072)/(128*x + 4*x^2 + 1024),x)

[Out]

(x*(19*x - exp(exp(x/4)) + x^2 + 48))/(x + 16)

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sympy [A]  time = 0.16, size = 17, normalized size = 0.59 $$\begin{array}{c}{x}^2+3x-\frac{x{e}^{e^{\frac{x}{4}}}}{x+16}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2-16*x)*exp(1/4*x)-64)*exp(exp(1/4*x))+8*x**3+268*x**2+2432*x+3072)/(4*x**2+128*x+1024),x)

[Out]

x**2 + 3*x - x*exp(exp(x/4))/(x + 16)

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