3.17.76 $\int \frac{-e^{8}x+(x^2{)}^{\frac{25e^8-10x^2}{e^8}}(50e^8-20x^2-20x^2\log (x^2))}{e^{8}x}dx$

Optimal. Leaf size=20 $$-x+{\left(x^2\right)}^{5(5-\frac{2x^2}{e^8})}$$

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Rubi [F]  time = 1.29, 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 = {} $$\begin{array}{c}\int \frac{-e^{8}x+{\left(x^2\right)}^{\frac{25e^8-10x^2}{e^8}}(50e^8-20x^2-20x^2\log \left(x^2\right))}{e^{8}x}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-(E^8*x) + (x^2)^((25*E^8 - 10*x^2)/E^8)*(50*E^8 - 20*x^2 - 20*x^2*Log[x^2]))/(E^8*x),x]

[Out]

-x + 25*Defer[Subst][Defer[Int][x^(24 - (10*x)/E^8), x], x, x^2] - (10*Defer[Subst][Defer[Int][x^(25 - (10*x)/
E^8), x], x, x^2])/E^8 - (10*Log[x^2]*Defer[Subst][Defer[Int][x^(25 - (10*x)/E^8), x], x, x^2])/E^8 + (10*Defe
r[Subst][Defer[Int][Defer[Int][x^(25 - (10*x)/E^8), x]/x, x], x, x^2])/E^8

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int \frac{-e^{8}x+{\left(x^2\right)}^{\frac{25e^8-10x^2}{e^8}}(50e^8-20x^2-20x^2\log \left(x^2\right))}{x}dx}{e^8}\\ & =\frac{\int (-e^8-10x^{49}{\left(x^2\right)}^{-\frac{10x^2}{e^8}}(-5e^8+2x^2+2x^2\log \left(x^2\right)))dx}{e^8}\\ & =-x-\frac{10\int x^{49}{\left(x^2\right)}^{-\frac{10x^2}{e^8}}(-5e^8+2x^2+2x^2\log \left(x^2\right))dx}{e^8}\\ & =-x-\frac{5\mathrm{Subst}(\int x^{24-\frac{10x}{e^8}}(-5e^8+2x+2x\log (x))dx,x,x^2)}{e^8}\\ & =-x-\frac{5\mathrm{Subst}(\int (-5e^8{x}^{24-\frac{10x}{e^8}}+2x^{25-\frac{10x}{e^8}}+2x^{25-\frac{10x}{e^8}}\log (x))dx,x,x^2)}{e^8}\\ & =-x+25\mathrm{Subst}(\int x^{24-\frac{10x}{e^8}}dx,x,x^2)-\frac{10\mathrm{Subst}(\int x^{25-\frac{10x}{e^8}}dx,x,x^2)}{e^8}-\frac{10\mathrm{Subst}(\int x^{25-\frac{10x}{e^8}}\log (x)dx,x,x^2)}{e^8}\\ & =-x+25\mathrm{Subst}(\int x^{24-\frac{10x}{e^8}}dx,x,x^2)-\frac{10\mathrm{Subst}(\int x^{25-\frac{10x}{e^8}}dx,x,x^2)}{e^8}+\frac{10\mathrm{Subst}(\int \frac{\int x^{25-\frac{10x}{e^8}}dx}{x}dx,x,x^2)}{e^8}-\frac{(10\log \left(x^2\right))\mathrm{Subst}(\int x^{25-\frac{10x}{e^8}}dx,x,x^2)}{e^8}\end{array}\end{array}$$

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Mathematica [A]  time = 0.11, size = 18, normalized size = 0.90 $$\begin{array}{c}-x+{\left(x^2\right)}^{25-\frac{10x^2}{e^8}}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-(E^8*x) + (x^2)^((25*E^8 - 10*x^2)/E^8)*(50*E^8 - 20*x^2 - 20*x^2*Log[x^2]))/(E^8*x),x]

[Out]

-x + (x^2)^(25 - (10*x^2)/E^8)

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fricas [B]  time = 0.68, size = 44, normalized size = 2.20 $$\begin{array}{c}-\frac{{\left(x^2\right)}^{5(2x^2-5e^8)e^{(-8)}}x-1}{{\left(x^2\right)}^{5(2x^2-5e^8)e^{(-8)}}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2*log(x^2)+50*exp(4)^2-20*x^2)*exp((25*exp(4)^2-10*x^2)*log(x^2)/exp(4)^2)-x*exp(4)^2)/x/exp
(4)^2,x, algorithm="fricas")

[Out]

-((x^2)^(5*(2*x^2 - 5*e^8)*e^(-8))*x - 1)/(x^2)^(5*(2*x^2 - 5*e^8)*e^(-8))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Timed out}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2*log(x^2)+50*exp(4)^2-20*x^2)*exp((25*exp(4)^2-10*x^2)*log(x^2)/exp(4)^2)-x*exp(4)^2)/x/exp
(4)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.14, size = 18, normalized size = 0.90

method	result	size
risch	${\left(x^2\right)}^{25-10x^2{\mathrm{e}}^{-8}}-x$	$18$
norman	$({\mathrm{e}}^4{\mathrm{e}}^{(25{\mathrm{e}}^8-10x^2)\ln \left(x^2\right){\mathrm{e}}^{-8}}-x{\mathrm{e}}^4){\mathrm{e}}^{-4}$	$37$
default	${\mathrm{e}}^{-8}({\mathrm{e}}^8{\mathrm{e}}^{(25{\mathrm{e}}^8-10x^2)\ln \left(x^2\right){\mathrm{e}}^{-8}}-x{\mathrm{e}}^8)$	$41$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x^2*ln(x^2)+50*exp(4)^2-20*x^2)*exp((25*exp(4)^2-10*x^2)*ln(x^2)/exp(4)^2)-x*exp(4)^2)/x/exp(4)^2,x,
method=_RETURNVERBOSE)

[Out]

(x^2)^(25-10*x^2*exp(-8))-x

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maxima [A]  time = 0.49, size = 25, normalized size = 1.25 $$\begin{array}{c}(x^{50}{e}^{(-20x^2{e}^{(-8)}\log (x)+8)}-x{e}^8)e^{(-8)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2*log(x^2)+50*exp(4)^2-20*x^2)*exp((25*exp(4)^2-10*x^2)*log(x^2)/exp(4)^2)-x*exp(4)^2)/x/exp
(4)^2,x, algorithm="maxima")

[Out]

(x^50*e^(-20*x^2*e^(-8)*log(x) + 8) - x*e^8)*e^(-8)

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mupad [B]  time = 1.28, size = 21, normalized size = 1.05 $$\begin{array}{c}\frac{x^{50}}{{\left(x^2\right)}^{10x^2{\mathrm{e}}^{-8}}}-x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-8)*(x*exp(8) + exp(log(x^2)*exp(-8)*(25*exp(8) - 10*x^2))*(20*x^2*log(x^2) - 50*exp(8) + 20*x^2)))/
x,x)

[Out]

x^50/(x^2)^(10*x^2*exp(-8)) - x

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sympy [A]  time = 0.39, size = 20, normalized size = 1.00 $$\begin{array}{c}-x+e^{\frac{(-10x^2+25e^8)\log \left(x^2\right)}{e^8}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x**2*ln(x**2)+50*exp(4)**2-20*x**2)*exp((25*exp(4)**2-10*x**2)*ln(x**2)/exp(4)**2)-x*exp(4)**2
)/x/exp(4)**2,x)

[Out]

-x + exp((-10*x**2 + 25*exp(8))*exp(-8)*log(x**2))

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