3.73.70 $\int \frac{30x^4+2x^6+e^{\frac{-1+x^2}{x^2}}(30+2x^2)+(2e^{\frac{-1+x^2}{x^2}}{x}^4+2x^6)\log (5e^{\frac{-1+x^2}{x^2}}+5x^2)}{e^{\frac{-1+x^2}{x^2}}{x}^3+x^5}dx$

Optimal. Leaf size=22 $$(15+x^2)\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)$$

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Rubi [F]  time = 1.87, 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 = {} $$\begin{array}{c}\int \frac{30x^4+2x^6+e^{\frac{-1+x^2}{x^2}}(30+2x^2)+(2e^{\frac{-1+x^2}{x^2}}{x}^4+2x^6)\log (5e^{\frac{-1+x^2}{x^2}}+5x^2)}{e^{\frac{-1+x^2}{x^2}}{x}^3+x^5}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(30*x^4 + 2*x^6 + E^((-1 + x^2)/x^2)*(30 + 2*x^2) + (2*E^((-1 + x^2)/x^2)*x^4 + 2*x^6)*Log[5*E^((-1 + x^2)
/x^2) + 5*x^2])/(E^((-1 + x^2)/x^2)*x^3 + x^5),x]

[Out]

30*Log[x] + x^2*Log[5*(E^(1 - x^(-2)) + x^2)] + 15*E*Defer[Subst][Defer[Int][1/(x^2*(E + E^x^(-1)*x)), x], x,
x^2] - 15*E*Defer[Subst][Defer[Int][1/(x*(E + E^x^(-1)*x)), x], x, x^2]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{2(15+x^2)(e+e^{\frac{1}{x^2}}{x}^4)}{x^3(e+e^{\frac{1}{x^2}}{x}^2)}+2x\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right))dx\\ & =2\int \frac{(15+x^2)(e+e^{\frac{1}{x^2}}{x}^4)}{x^3(e+e^{\frac{1}{x^2}}{x}^2)}dx+2\int x\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)dx\\ & =x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)-\int \frac{2(e+e^{\frac{1}{x^2}}{x}^4)}{ex+e^{\frac{1}{x^2}}{x}^3}dx+\mathrm{Subst}(\int \frac{(15+x)(e+e^{\frac{1}{x}}{x}^2)}{x^2(e+e^{\frac{1}{x}}x)}dx,x,x^2)\\ & =x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)-2\int \frac{e+e^{\frac{1}{x^2}}{x}^4}{ex+e^{\frac{1}{x^2}}{x}^3}dx+\mathrm{Subst}(\int (\frac{15+x}{x}-\frac{e(-1+x)(15+x)}{x^2(e+e^{\frac{1}{x}}x)})dx,x,x^2)\\ & =x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)-2\int (x-\frac{e(-1+x^2)}{x(e+e^{\frac{1}{x^2}}{x}^2)})dx-e\mathrm{Subst}(\int \frac{(-1+x)(15+x)}{x^2(e+e^{\frac{1}{x}}x)}dx,x,x^2)+\mathrm{Subst}(\int \frac{15+x}{x}dx,x,x^2)\\ & =-x^2+x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)-e\mathrm{Subst}(\int (\frac{1}{e+e^{\frac{1}{x}}x}-\frac{15}{x^2(e+e^{\frac{1}{x}}x)}+\frac{14}{x(e+e^{\frac{1}{x}}x)})dx,x,x^2)+(2e)\int \frac{-1+x^2}{x(e+e^{\frac{1}{x^2}}{x}^2)}dx+\mathrm{Subst}(\int (1+\frac{15}{x})dx,x,x^2)\\ & =30\log (x)+x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)-e\mathrm{Subst}(\int \frac{1}{e+e^{\frac{1}{x}}x}dx,x,x^2)+e\mathrm{Subst}(\int \frac{-1+x}{x(e+e^{\frac{1}{x}}x)}dx,x,x^2)-(14e)\mathrm{Subst}(\int \frac{1}{x(e+e^{\frac{1}{x}}x)}dx,x,x^2)+(15e)\mathrm{Subst}(\int \frac{1}{x^2(e+e^{\frac{1}{x}}x)}dx,x,x^2)\\ & =30\log (x)+x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)-e\mathrm{Subst}(\int \frac{1}{e+e^{\frac{1}{x}}x}dx,x,x^2)+e\mathrm{Subst}(\int (\frac{1}{e+e^{\frac{1}{x}}x}-\frac{1}{x(e+e^{\frac{1}{x}}x)})dx,x,x^2)-(14e)\mathrm{Subst}(\int \frac{1}{x(e+e^{\frac{1}{x}}x)}dx,x,x^2)+(15e)\mathrm{Subst}(\int \frac{1}{x^2(e+e^{\frac{1}{x}}x)}dx,x,x^2)\\ & =30\log (x)+x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)-e\mathrm{Subst}(\int \frac{1}{x(e+e^{\frac{1}{x}}x)}dx,x,x^2)-(14e)\mathrm{Subst}(\int \frac{1}{x(e+e^{\frac{1}{x}}x)}dx,x,x^2)+(15e)\mathrm{Subst}(\int \frac{1}{x^2(e+e^{\frac{1}{x}}x)}dx,x,x^2)\end{array}\end{array}$$

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Mathematica [A]  time = 0.29, size = 41, normalized size = 1.86 $$\begin{array}{c}1-\frac{15}{x^2}+x^2\log \left(5(e^{1-\frac{1}{x^2}}+x^2)\right)+15\log (e+e^{\frac{1}{x^2}}{x}^2)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(30*x^4 + 2*x^6 + E^((-1 + x^2)/x^2)*(30 + 2*x^2) + (2*E^((-1 + x^2)/x^2)*x^4 + 2*x^6)*Log[5*E^((-1
+ x^2)/x^2) + 5*x^2])/(E^((-1 + x^2)/x^2)*x^3 + x^5),x]

[Out]

1 - 15/x^2 + x^2*Log[5*(E^(1 - x^(-2)) + x^2)] + 15*Log[E + E^x^(-2)*x^2]

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fricas [A]  time = 0.72, size = 25, normalized size = 1.14 $$\begin{array}{c}(x^2+15)\log (5x^2+5e^{\left(\frac{x^2-1}{x^2}\right)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*log(5*exp((x^2-1)/x^2)+5*x^2)+(2*x^2+30)*exp((x^2-1)/x^2)+2*x^6+30*x
^4)/(x^3*exp((x^2-1)/x^2)+x^5),x, algorithm="fricas")

[Out]

(x^2 + 15)*log(5*x^2 + 5*e^((x^2 - 1)/x^2))

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giac [A]  time = 0.17, size = 41, normalized size = 1.86 $$\begin{array}{c}{x}^2\log (5x^2+5e^{\left(\frac{x^2-1}{x^2}\right)})+15\log (x^2+e^{\left(\frac{x^2-1}{x^2}\right)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*log(5*exp((x^2-1)/x^2)+5*x^2)+(2*x^2+30)*exp((x^2-1)/x^2)+2*x^6+30*x
^4)/(x^3*exp((x^2-1)/x^2)+x^5),x, algorithm="giac")

[Out]

x^2*log(5*x^2 + 5*e^((x^2 - 1)/x^2)) + 15*log(x^2 + e^((x^2 - 1)/x^2))

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maple [A]  time = 0.04, size = 45, normalized size = 2.05

method	result	size
risch	$x^2\ln (5{\mathrm{e}}^{\frac{(x-1)(x+1)}{x^2}}+5x^2)+15\ln ({\mathrm{e}}^{\frac{(x-1)(x+1)}{x^2}}+x^2)-15$	$45$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*ln(5*exp((x^2-1)/x^2)+5*x^2)+(2*x^2+30)*exp((x^2-1)/x^2)+2*x^6+30*x^4)/(x^
3*exp((x^2-1)/x^2)+x^5),x,method=_RETURNVERBOSE)

[Out]

x^2*ln(5*exp((x-1)*(x+1)/x^2)+5*x^2)+15*ln(exp((x-1)*(x+1)/x^2)+x^2)-15

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maxima [B]  time = 0.48, size = 51, normalized size = 2.32 $$\begin{array}{c}\frac{x^4\log (5)+x^4\log (x^2{e}^{\left(\frac{1}{x^2}\right)}+e)-15}{x^2}+30\log (x)+15\log \left(\frac{x^2{e}^{\left(\frac{1}{x^2}\right)}+e}{x^2}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4*exp((x^2-1)/x^2)+2*x^6)*log(5*exp((x^2-1)/x^2)+5*x^2)+(2*x^2+30)*exp((x^2-1)/x^2)+2*x^6+30*x
^4)/(x^3*exp((x^2-1)/x^2)+x^5),x, algorithm="maxima")

[Out]

(x^4*log(5) + x^4*log(x^2*e^(x^(-2)) + e) - 15)/x^2 + 30*log(x) + 15*log((x^2*e^(x^(-2)) + e)/x^2)

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mupad [B]  time = 4.46, size = 38, normalized size = 1.73 $$\begin{array}{c}15\ln (\mathrm{e}{\mathrm{e}}^{-\frac{1}{x^2}}+x^2)+x^2\ln (5\mathrm{e}{\mathrm{e}}^{-\frac{1}{x^2}}+5x^2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5*exp((x^2 - 1)/x^2) + 5*x^2)*(2*x^4*exp((x^2 - 1)/x^2) + 2*x^6) + exp((x^2 - 1)/x^2)*(2*x^2 + 30) +
30*x^4 + 2*x^6)/(x^3*exp((x^2 - 1)/x^2) + x^5),x)

[Out]

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

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sympy [A]  time = 0.50, size = 37, normalized size = 1.68 $$\begin{array}{c}{x}^2\log (5x^2+5e^{\frac{x^2-1}{x^2}})+15\log (x^2+e^{\frac{x^2-1}{x^2}})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**4*exp((x**2-1)/x**2)+2*x**6)*ln(5*exp((x**2-1)/x**2)+5*x**2)+(2*x**2+30)*exp((x**2-1)/x**2)+2
*x**6+30*x**4)/(x**3*exp((x**2-1)/x**2)+x**5),x)

[Out]

x**2*log(5*x**2 + 5*exp((x**2 - 1)/x**2)) + 15*log(x**2 + exp((x**2 - 1)/x**2))

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