3.85.55 $\int \frac{-1-x+(-x^2+x^3)\log (4)+e^{1+x}(-2+2x+(-2x^2+2x^3)\log (4))+e^{2+2x}(-1+2x+(-x^2+2x^3)\log (4))}{x^2+x^4\log (4)}dx$

Optimal. Leaf size=23 $$\frac{{(1+e^{1+x})}^2}{x}+\log (\frac{1}{x}+x\log (4))$$

________________________________________________________________________________________

Rubi [B]  time = 0.92, antiderivative size = 48, normalized size of antiderivative = 2.09, number of steps used = 8, number of rules used = 5, integrand size = 82, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.061, Rules used = {1593, 6725, 2197, 1802, 260} $$\begin{array}{c}\frac{\log (16)\log (x^2\log (4)+1)}{2\log (4)}+\frac{2e^{x+1}}{x}+\frac{e^{2x+2}}{x}+\frac{1}{x}-\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-1 - x + (-x^2 + x^3)*Log[4] + E^(1 + x)*(-2 + 2*x + (-2*x^2 + 2*x^3)*Log[4]) + E^(2 + 2*x)*(-1 + 2*x + (
-x^2 + 2*x^3)*Log[4]))/(x^2 + x^4*Log[4]),x]

[Out]

x^(-1) + (2*E^(1 + x))/x + E^(2 + 2*x)/x - Log[x] + (Log[16]*Log[1 + x^2*Log[4]])/(2*Log[4])

Rule 260

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-1-x+(-x^2+x^3)\log (4)+e^{1+x}(-2+2x+(-2x^2+2x^3)\log (4))+e^{2+2x}(-1+2x+(-x^2+2x^3)\log (4))}{x^2(1+x^2\log (4))}dx\\ & =\int (\frac{2e^{1+x}(-1+x)}{x^2}+\frac{e^{2+2x}(-1+2x)}{x^2}+\frac{-1-x-x^2\log (4)+x^3\log (4)}{x^2(1+x^2\log (4))})dx\\ & =2\int \frac{e^{1+x}(-1+x)}{x^2}dx+\int \frac{e^{2+2x}(-1+2x)}{x^2}dx+\int \frac{-1-x-x^2\log (4)+x^3\log (4)}{x^2(1+x^2\log (4))}dx\\ & =\frac{2e^{1+x}}{x}+\frac{e^{2+2x}}{x}+\int (-\frac{1}{x^2}-\frac{1}{x}+\frac{x\log (16)}{1+x^2\log (4)})dx\\ & =\frac{1}{x}+\frac{2e^{1+x}}{x}+\frac{e^{2+2x}}{x}-\log (x)+\log (16)\int \frac{x}{1+x^2\log (4)}dx\\ & =\frac{1}{x}+\frac{2e^{1+x}}{x}+\frac{e^{2+2x}}{x}-\log (x)+\frac{\log (16)\log (1+x^2\log (4))}{2\log (4)}\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [B]  time = 0.18, size = 50, normalized size = 2.17 $$\begin{array}{c}\frac{4e^{1+x}\log (4)+\log (16)+e^{2+2x}\log (16)-2x\log (4)\log (x)+x\log (16)\log (1+x^2\log (4))}{x\log (16)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-1 - x + (-x^2 + x^3)*Log[4] + E^(1 + x)*(-2 + 2*x + (-2*x^2 + 2*x^3)*Log[4]) + E^(2 + 2*x)*(-1 + 2
*x + (-x^2 + 2*x^3)*Log[4]))/(x^2 + x^4*Log[4]),x]

[Out]

(4*E^(1 + x)*Log[4] + Log[16] + E^(2 + 2*x)*Log[16] - 2*x*Log[4]*Log[x] + x*Log[16]*Log[1 + x^2*Log[4]])/(x*Lo
g[16])

________________________________________________________________________________________

fricas [A]  time = 0.81, size = 35, normalized size = 1.52 $$\begin{array}{c}\frac{x\log (2x^2\log (2)+1)-x\log (x)+e^{(2x+2)}+2e^{(x+1)}+1}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^3-x^2)*log(2)+2*x-1)*exp(x+1)^2+(2*(2*x^3-2*x^2)*log(2)+2*x-2)*exp(x+1)+2*(x^3-x^2)*log(2)-
x-1)/(2*x^4*log(2)+x^2),x, algorithm="fricas")

[Out]

(x*log(2*x^2*log(2) + 1) - x*log(x) + e^(2*x + 2) + 2*e^(x + 1) + 1)/x

________________________________________________________________________________________

giac [A]  time = 0.12, size = 35, normalized size = 1.52 $$\begin{array}{c}\frac{x\log (2x^2\log (2)+1)-x\log (x)+e^{(2x+2)}+2e^{(x+1)}+1}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^3-x^2)*log(2)+2*x-1)*exp(x+1)^2+(2*(2*x^3-2*x^2)*log(2)+2*x-2)*exp(x+1)+2*(x^3-x^2)*log(2)-
x-1)/(2*x^4*log(2)+x^2),x, algorithm="giac")

[Out]

(x*log(2*x^2*log(2) + 1) - x*log(x) + e^(2*x + 2) + 2*e^(x + 1) + 1)/x

________________________________________________________________________________________

maple [A]  time = 0.38, size = 34, normalized size = 1.48

method	result	size
norman	$\frac{1+{\mathrm{e}}^{2x+2}+2{\mathrm{e}}^{x+1}}{x}-\ln (x)+\ln (2x^2\ln (2)+1)$	$34$
risch	$\frac{1}{x}-\ln (x)+\ln (-2x^2\ln (2)-1)+\frac{{\mathrm{e}}^{2x+2}}{x}+\frac{2{\mathrm{e}}^{x+1}}{x}$	$38$
derivativedivides	$\frac{{\mathrm{e}}^{2x+2}}{x}-\ln (x)+\frac{2{\mathrm{e}}^{x+1}}{x}+\frac{1}{x}+\ln (2\ln (2){(x+1)}^2-4\ln (2)(x+1)+2\ln (2)+1)$	$51$
default	$\frac{{\mathrm{e}}^{2x+2}}{x}-\ln (x)+\frac{2{\mathrm{e}}^{x+1}}{x}+\frac{1}{x}+\ln (2\ln (2){(x+1)}^2-4\ln (2)(x+1)+2\ln (2)+1)$	$51$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(2*x^3-x^2)*ln(2)+2*x-1)*exp(x+1)^2+(2*(2*x^3-2*x^2)*ln(2)+2*x-2)*exp(x+1)+2*(x^3-x^2)*ln(2)-x-1)/(2*x
^4*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 37, normalized size = 1.61 $$\begin{array}{c}\frac{e^{(2x+2)}+2e^{(x+1)}}{x}+\frac{1}{x}+\log (2x^2\log (2)+1)-\frac{1}{2}\log \left(x^2\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x^3-x^2)*log(2)+2*x-1)*exp(x+1)^2+(2*(2*x^3-2*x^2)*log(2)+2*x-2)*exp(x+1)+2*(x^3-x^2)*log(2)-
x-1)/(2*x^4*log(2)+x^2),x, algorithm="maxima")

[Out]

(e^(2*x + 2) + 2*e^(x + 1))/x + 1/x + log(2*x^2*log(2) + 1) - 1/2*log(x^2)

________________________________________________________________________________________

mupad [B]  time = 5.25, size = 33, normalized size = 1.43 $$\begin{array}{c}\ln (2\ln (2)x^2+1)-\ln (x)+\frac{2{\mathrm{e}}^{x+1}+{\mathrm{e}}^{2x+2}+1}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + 2*log(2)*(x^2 - x^3) + exp(x + 1)*(2*log(2)*(2*x^2 - 2*x^3) - 2*x + 2) + exp(2*x + 2)*(2*log(2)*(x^2
- 2*x^3) - 2*x + 1) + 1)/(2*x^4*log(2) + x^2),x)

[Out]

log(2*x^2*log(2) + 1) - log(x) + (2*exp(x + 1) + exp(2*x + 2) + 1)/x

________________________________________________________________________________________

sympy [A]  time = 0.36, size = 37, normalized size = 1.61 $$\begin{array}{c}-\log (x)+\log (x^2+\frac{1}{2\log (2)})+\frac{1}{x}+\frac{2x{e}^{x+1}+x{e}^{2x+2}}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x**3-x**2)*ln(2)+2*x-1)*exp(x+1)**2+(2*(2*x**3-2*x**2)*ln(2)+2*x-2)*exp(x+1)+2*(x**3-x**2)*ln
(2)-x-1)/(2*x**4*ln(2)+x**2),x)

[Out]

-log(x) + log(x**2 + 1/(2*log(2))) + 1/x + (2*x*exp(x + 1) + x*exp(2*x + 2))/x**2

________________________________________________________________________________________