3.89.29 $\int \frac{(-256x^5+e^{\frac{-1+x-2x^2+2\log (5)}{2x}}(2-4x^2-4\log (5))){\log }^3(e^{\frac{-1+x-2x^2+2\log (5)}{2x}}-16x^4)}{e^{\frac{-1+x-2x^2+2\log (5)}{2x}}{x}^2-16x^6}dx$

Optimal. Leaf size=29 $${\log }^4(e^{-x+\frac{-1+x+2\log (5)}{2x}}-16x^4)$$

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Rubi [A]  time = 0.54, antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, integrand size = 103, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.010, Rules used = {6686} $$\begin{array}{c}{\log }^4(5^{\frac{1}{x}}{e}^{-\frac{2x^2-x+1}{2x}}-16x^4)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[((-256*x^5 + E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*(2 - 4*x^2 - 4*Log[5]))*Log[E^((-1 + x - 2*x^2 + 2*Log[
5])/(2*x)) - 16*x^4]^3)/(E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*x^2 - 16*x^6),x]

[Out]

Log[5^x^(-1)/E^((1 - x + 2*x^2)/(2*x)) - 16*x^4]^4

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& ={\log }^4(5^{\frac{1}{x}}{e}^{-\frac{1-x+2x^2}{2x}}-16x^4)\end{array}\end{array}$$

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Mathematica [A]  time = 0.10, size = 28, normalized size = 0.97 $$\begin{array}{c}{\log }^4(e^{\frac{-1+x-2x^2+\log (25)}{2x}}-16x^4)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[((-256*x^5 + E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*(2 - 4*x^2 - 4*Log[5]))*Log[E^((-1 + x - 2*x^2 +
2*Log[5])/(2*x)) - 16*x^4]^3)/(E^((-1 + x - 2*x^2 + 2*Log[5])/(2*x))*x^2 - 16*x^6),x]

[Out]

Log[E^((-1 + x - 2*x^2 + Log[25])/(2*x)) - 16*x^4]^4

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fricas [A]  time = 0.56, size = 29, normalized size = 1.00 $$\begin{array}{c}\log {(-16x^4+e^{(-\frac{2x^2-x-2\log (5)+1}{2x})})}^4\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(5)-4*x^2+2)*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-256*x^5)*log(exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16
*x^4)^3/(x^2*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16*x^6),x, algorithm="fricas")

[Out]

log(-16*x^4 + e^(-1/2*(2*x^2 - x - 2*log(5) + 1)/x))^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{2(128x^5+(2x^2+2\log (5)-1)e^{(-\frac{2x^2-x-2\log (5)+1}{2x})})\log {(-16x^4+e^{(-\frac{2x^2-x-2\log (5)+1}{2x})})}^3}{16x^6-x^2{e}^{(-\frac{2x^2-x-2\log (5)+1}{2x})}}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(5)-4*x^2+2)*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-256*x^5)*log(exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16
*x^4)^3/(x^2*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16*x^6),x, algorithm="giac")

[Out]

integrate(2*(128*x^5 + (2*x^2 + 2*log(5) - 1)*e^(-1/2*(2*x^2 - x - 2*log(5) + 1)/x))*log(-16*x^4 + e^(-1/2*(2*
x^2 - x - 2*log(5) + 1)/x))^3/(16*x^6 - x^2*e^(-1/2*(2*x^2 - x - 2*log(5) + 1)/x)), x)

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maple [A]  time = 0.06, size = 30, normalized size = 1.03

method	result	size
risch	$\ln {({\mathrm{e}}^{\frac{2\ln (5)-2x^2+x-1}{2x}}-16x^4)}^4-\frac{3}{2}$	$30$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*ln(5)-4*x^2+2)*exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-256*x^5)*ln(exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-16*x^4)^3/(x
^2*exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-16*x^6),x,method=_RETURNVERBOSE)

[Out]

ln(exp(1/2*(2*ln(5)-2*x^2+x-1)/x)-16*x^4)^4-3/2

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maxima [B]  time = 0.77, size = 574, normalized size = 19.79 $$\begin{array}{c}-2(\frac{2x^2+1}{x}-2\log (-(16x^4{e}^{(x+\frac{1}{2x})}-e^{(\frac{\log (5)}{x}+\frac{1}{2})})e^{(-\frac{1}{2})}))\log {(-16x^4+e^{(-\frac{2x^2-x-2\log (5)+1}{2x})})}^3-\frac{3(4x^4+4x^2\log {(-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})}^2+4x^3-4(2x^3+x^2+x)\log (-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})+2x+1)\log {(-16x^4+e^{(-\frac{2x^2-x-2\log (5)+1}{2x})})}^2}{2x^2}-\frac{(8x^6-8x^3\log {(-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})}^3+12x^5-12x^4+12(2x^4+x^3+x^2)\log {(-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})}^2-6x^2-6(4x^5+4x^4+2x^2+x)\log (-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})+3x+1)\log (-16x^4+e^{(-\frac{2x^2-x-2\log (5)+1}{2x})})}{2x^3}-\frac{16x^8+16x^4\log {(-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})}^4+32x^7-64x^6-48x^5-32(2x^5+x^4+x^3)\log {(-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})}^3-24x^3+24(4x^6+4x^5+2x^3+x^2)\log {(-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})}^2-16x^2-8(8x^7+12x^6-12x^5-6x^3+3x^2+x)\log (-16x^4{e}^{(x+\frac{1}{2x})}+e^{(\frac{\log (5)}{x}+\frac{1}{2})})+4x+1}{16x^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(5)-4*x^2+2)*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-256*x^5)*log(exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16
*x^4)^3/(x^2*exp(1/2*(2*log(5)-2*x^2+x-1)/x)-16*x^6),x, algorithm="maxima")

[Out]

-2*((2*x^2 + 1)/x - 2*log(-(16*x^4*e^(x + 1/2/x) - e^(log(5)/x + 1/2))*e^(-1/2)))*log(-16*x^4 + e^(-1/2*(2*x^2
- x - 2*log(5) + 1)/x))^3 - 3/2*(4*x^4 + 4*x^2*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^2 + 4*x^3 - 4*
(2*x^3 + x^2 + x)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2)) + 2*x + 1)*log(-16*x^4 + e^(-1/2*(2*x^2 - x
- 2*log(5) + 1)/x))^2/x^2 - 1/2*(8*x^6 - 8*x^3*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^3 + 12*x^5 - 12
*x^4 + 12*(2*x^4 + x^3 + x^2)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^2 - 6*x^2 - 6*(4*x^5 + 4*x^4 + 2
*x^2 + x)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2)) + 3*x + 1)*log(-16*x^4 + e^(-1/2*(2*x^2 - x - 2*log(
5) + 1)/x))/x^3 - 1/16*(16*x^8 + 16*x^4*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^4 + 32*x^7 - 64*x^6 -
48*x^5 - 32*(2*x^5 + x^4 + x^3)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^3 - 24*x^3 + 24*(4*x^6 + 4*x^5
+ 2*x^3 + x^2)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2))^2 - 16*x^2 - 8*(8*x^7 + 12*x^6 - 12*x^5 - 6*x^
3 + 3*x^2 + x)*log(-16*x^4*e^(x + 1/2/x) + e^(log(5)/x + 1/2)) + 4*x + 1)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}\int -\frac{{\ln ({\mathrm{e}}^{\frac{-x^2+\frac{x}{2}+\ln (5)-\frac{1}{2}}{x}}-16x^4)}^3({\mathrm{e}}^{\frac{-x^2+\frac{x}{2}+\ln (5)-\frac{1}{2}}{x}}(4x^2+4\ln (5)-2)+256x^5)}{x^2{\mathrm{e}}^{\frac{-x^2+\frac{x}{2}+\ln (5)-\frac{1}{2}}{x}}-16x^6}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^4)^3*(exp((x/2 + log(5) - x^2 - 1/2)/x)*(4*log(5) + 4*x^2 -
2) + 256*x^5))/(x^2*exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^6),x)

[Out]

int(-(log(exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^4)^3*(exp((x/2 + log(5) - x^2 - 1/2)/x)*(4*log(5) + 4*x^2 -
2) + 256*x^5))/(x^2*exp((x/2 + log(5) - x^2 - 1/2)/x) - 16*x^6), x)

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sympy [A]  time = 0.63, size = 24, normalized size = 0.83 $$\begin{array}{c}\log {(-16x^4+e^{\frac{-x^2+\frac{x}{2}-\frac{1}{2}+\log (5)}{x}})}^4\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*ln(5)-4*x**2+2)*exp(1/2*(2*ln(5)-2*x**2+x-1)/x)-256*x**5)*ln(exp(1/2*(2*ln(5)-2*x**2+x-1)/x)-16
*x**4)**3/(x**2*exp(1/2*(2*ln(5)-2*x**2+x-1)/x)-16*x**6),x)

[Out]

log(-16*x**4 + exp((-x**2 + x/2 - 1/2 + log(5))/x))**4

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