3.33.2 $\int \frac{16+8x+e^{5x^2}(4+4x-40x^2-40x^3)+(4+4x)\log (\frac{1+2x+x^2}{9x^4})}{e^{10x^2}(1+x)+e^{5x^2}(2+2x)\log (\frac{1+2x+x^2}{9x^4})+(1+x){\log }^2(\frac{1+2x+x^2}{9x^4})}dx$

Optimal. Leaf size=26 $$\frac{4x}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}$$

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Rubi [F]  time = 6.30, 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 = {} $$\begin{array}{c}\int \frac{16+8x+e^{5x^2}(4+4x-40x^2-40x^3)+(4+4x)\log \left(\frac{1+2x+x^2}{9x^4}\right)}{e^{10x^2}(1+x)+e^{5x^2}(2+2x)\log \left(\frac{1+2x+x^2}{9x^4}\right)+(1+x){\log }^2\left(\frac{1+2x+x^2}{9x^4}\right)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(16 + 8*x + E^(5*x^2)*(4 + 4*x - 40*x^2 - 40*x^3) + (4 + 4*x)*Log[(1 + 2*x + x^2)/(9*x^4)])/(E^(10*x^2)*(1
+ x) + E^(5*x^2)*(2 + 2*x)*Log[(1 + 2*x + x^2)/(9*x^4)] + (1 + x)*Log[(1 + 2*x + x^2)/(9*x^4)]^2),x]

[Out]

8*Defer[Int][(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])^(-2), x] + 8*Defer[Int][1/((1 + x)*(E^(5*x^2) + Log[(1 + x)^
2/(9*x^4)])^2), x] + 40*Defer[Int][(x^2*Log[(1 + x)^2/(9*x^4)])/(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])^2, x] + 4
*Defer[Int][(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])^(-1), x] - 40*Defer[Int][x^2/(E^(5*x^2) + Log[(1 + x)^2/(9*x^
4)]), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4(2(2+x)+e^{5x^2}(1+x-10x^2-10x^3)+(1+x)\log \left(\frac{(1+x{)}^2}{9x^4}\right))}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx\\ & =4\int \frac{2(2+x)+e^{5x^2}(1+x-10x^2-10x^3)+(1+x)\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx\\ & =4\int (-\frac{-1+10x^2}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}+\frac{2(2+x+5x^2\log \left(\frac{(1+x{)}^2}{9x^4}\right)+5x^3\log \left(\frac{(1+x{)}^2}{9x^4}\right))}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2})dx\\ & =-(4\int \frac{-1+10x^2}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx)+8\int \frac{2+x+5x^2\log \left(\frac{(1+x{)}^2}{9x^4}\right)+5x^3\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx\\ & =-(4\int (-\frac{1}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}+\frac{10x^2}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)})dx)+8\int \frac{2+x+5x^2(1+x)\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx\\ & =4\int \frac{1}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx+8\int (\frac{2}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}+\frac{x}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}+\frac{5x^2\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}+\frac{5x^3\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2})dx-40\int \frac{x^2}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx\\ & =4\int \frac{1}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx+8\int \frac{x}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx+16\int \frac{1}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx+40\int \frac{x^2\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx+40\int \frac{x^3\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx-40\int \frac{x^2}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx\\ & =4\int \frac{1}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx+8\int (\frac{1}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}-\frac{1}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2})dx+16\int \frac{1}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx-40\int \frac{x^2}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx+40\int (\frac{\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}-\frac{x\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}+\frac{x^2\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}-\frac{\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2})dx+40\int (-\frac{\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}+\frac{x\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}+\frac{\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2})dx\\ & =4\int \frac{1}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx+8\int \frac{1}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx-8\int \frac{1}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx+16\int \frac{1}{(1+x){(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx+40\int \frac{x^2\log \left(\frac{(1+x{)}^2}{9x^4}\right)}{{(e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right))}^2}dx-40\int \frac{x^2}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.60, size = 26, normalized size = 1.00 $$\begin{array}{c}\frac{4x}{e^{5x^2}+\log \left(\frac{(1+x{)}^2}{9x^4}\right)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(16 + 8*x + E^(5*x^2)*(4 + 4*x - 40*x^2 - 40*x^3) + (4 + 4*x)*Log[(1 + 2*x + x^2)/(9*x^4)])/(E^(10*x
^2)*(1 + x) + E^(5*x^2)*(2 + 2*x)*Log[(1 + 2*x + x^2)/(9*x^4)] + (1 + x)*Log[(1 + 2*x + x^2)/(9*x^4)]^2),x]

[Out]

(4*x)/(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])

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fricas [A]  time = 0.57, size = 26, normalized size = 1.00 $$\begin{array}{c}\frac{4x}{e^{\left(5x^2\right)}+\log \left(\frac{x^2+2x+1}{9x^4}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((x+1)*log(1/9*(x^2+2*x+
1)/x^4)^2+(2*x+2)*exp(5*x^2)*log(1/9*(x^2+2*x+1)/x^4)+(x+1)*exp(5*x^2)^2),x, algorithm="fricas")

[Out]

4*x/(e^(5*x^2) + log(1/9*(x^2 + 2*x + 1)/x^4))

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giac [A]  time = 0.83, size = 26, normalized size = 1.00 $$\begin{array}{c}\frac{4x}{e^{\left(5x^2\right)}+\log \left(\frac{x^2+2x+1}{9x^4}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((x+1)*log(1/9*(x^2+2*x+
1)/x^4)^2+(2*x+2)*exp(5*x^2)*log(1/9*(x^2+2*x+1)/x^4)+(x+1)*exp(5*x^2)^2),x, algorithm="giac")

[Out]

4*x/(e^(5*x^2) + log(1/9*(x^2 + 2*x + 1)/x^4))

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maple [C]  time = 0.26, size = 366, normalized size = 14.08

method	result	size
risch	$-\frac{8ix}{-\pi \mathrm{csgn}{\left(\frac{i{(x+1)}^2}{x^4}\right)}^3+\pi \mathrm{csgn}{\left(i{x}^4\right)}^3+\pi \mathrm{csgn}{\left(i{x}^3\right)}^3-\pi \mathrm{csgn}{\left(i{(x+1)}^2\right)}^3-\pi \mathrm{csgn}\left(\frac{i}{x^4}\right)\mathrm{csgn}\left(i{(x+1)}^2\right)\mathrm{csgn}\left(\frac{i{(x+1)}^2}{x^4}\right)+\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i{x}^3\right)\mathrm{csgn}\left(i{x}^4\right)+\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i{x}^2\right)\mathrm{csgn}\left(i{x}^3\right)+8i\ln (x)+4i\ln (3)-2i{\mathrm{e}}^{5x^2}-4i\ln (x+1)-\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^3\right)}^2-\pi \mathrm{csgn}\left(i{x}^2\right)\mathrm{csgn}{\left(i{x}^3\right)}^2-\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^4\right)}^2-\pi \mathrm{csgn}\left(i{x}^3\right)\mathrm{csgn}{\left(i{x}^4\right)}^2+\pi \mathrm{csgn}\left(\frac{i}{x^4}\right)\mathrm{csgn}{\left(\frac{i{(x+1)}^2}{x^4}\right)}^2+\pi \mathrm{csgn}\left(i{(x+1)}^2\right)\mathrm{csgn}{\left(\frac{i{(x+1)}^2}{x^4}\right)}^2-\pi \mathrm{csgn}{\left(i(x+1)\right)}^2\mathrm{csgn}\left(i{(x+1)}^2\right)+2\pi \mathrm{csgn}\left(i(x+1)\right)\mathrm{csgn}{\left(i{(x+1)}^2\right)}^2+\pi \mathrm{csgn}{\left(i{x}^2\right)}^3+\pi \mathrm{csgn}{\left(ix\right)}^2\mathrm{csgn}\left(i{x}^2\right)-2\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^2\right)}^2}$	$366$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+4)*ln(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((x+1)*ln(1/9*(x^2+2*x+1)/x^4)^
2+(2*x+2)*exp(5*x^2)*ln(1/9*(x^2+2*x+1)/x^4)+(x+1)*exp(5*x^2)^2),x,method=_RETURNVERBOSE)

[Out]

-8*I*x/(-Pi*csgn(I/x^4*(x+1)^2)^3+Pi*csgn(I*x^4)^3+Pi*csgn(I*x^3)^3-Pi*csgn(I*(x+1)^2)^3-Pi*csgn(I/x^4)*csgn(I
*(x+1)^2)*csgn(I/x^4*(x+1)^2)+Pi*csgn(I*x)*csgn(I*x^3)*csgn(I*x^4)+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)+8*I*ln
(x)+4*I*ln(3)-2*I*exp(5*x^2)-4*I*ln(x+1)-Pi*csgn(I*x)*csgn(I*x^3)^2-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I*x)*
csgn(I*x^4)^2-Pi*csgn(I*x^3)*csgn(I*x^4)^2+Pi*csgn(I/x^4)*csgn(I/x^4*(x+1)^2)^2+Pi*csgn(I*(x+1)^2)*csgn(I/x^4*
(x+1)^2)^2-Pi*csgn(I*(x+1))^2*csgn(I*(x+1)^2)+2*Pi*csgn(I*(x+1))*csgn(I*(x+1)^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*
x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2)

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maxima [A]  time = 0.60, size = 26, normalized size = 1.00 $$\begin{array}{c}\frac{4x}{e^{\left(5x^2\right)}-2\log (3)+2\log (x+1)-4\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((x+1)*log(1/9*(x^2+2*x+
1)/x^4)^2+(2*x+2)*exp(5*x^2)*log(1/9*(x^2+2*x+1)/x^4)+(x+1)*exp(5*x^2)^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int \frac{8x+{\mathrm{e}}^{5x^2}(-40x^3-40x^2+4x+4)+\ln \left(\frac{\frac{x^2}{9}+\frac{2x}{9}+\frac{1}{9}}{x^4}\right)(4x+4)+16}{(x+1){\ln \left(\frac{\frac{x^2}{9}+\frac{2x}{9}+\frac{1}{9}}{x^4}\right)}^2+{\mathrm{e}}^{5x^2}(2x+2)\ln \left(\frac{\frac{x^2}{9}+\frac{2x}{9}+\frac{1}{9}}{x^4}\right)+{\mathrm{e}}^{10x^2}(x+1)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + exp(5*x^2)*(4*x - 40*x^2 - 40*x^3 + 4) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*(4*x + 4) + 16)/(log(((2*
x)/9 + x^2/9 + 1/9)/x^4)^2*(x + 1) + exp(10*x^2)*(x + 1) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*exp(5*x^2)*(2*x +
2)),x)

[Out]

int((8*x + exp(5*x^2)*(4*x - 40*x^2 - 40*x^3 + 4) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*(4*x + 4) + 16)/(log(((2*
x)/9 + x^2/9 + 1/9)/x^4)^2*(x + 1) + exp(10*x^2)*(x + 1) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*exp(5*x^2)*(2*x +
2)), x)

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sympy [A]  time = 0.39, size = 27, normalized size = 1.04 $$\begin{array}{c}\frac{4x}{e^{5x^2}+\log \left(\frac{\frac{x^2}{9}+\frac{2x}{9}+\frac{1}{9}}{x^4}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*ln(1/9*(x**2+2*x+1)/x**4)+(-40*x**3-40*x**2+4*x+4)*exp(5*x**2)+8*x+16)/((x+1)*ln(1/9*(x**2+
2*x+1)/x**4)**2+(2*x+2)*exp(5*x**2)*ln(1/9*(x**2+2*x+1)/x**4)+(x+1)*exp(5*x**2)**2),x)

[Out]

4*x/(exp(5*x**2) + log((x**2/9 + 2*x/9 + 1/9)/x**4))

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