3.17.53 $\int \frac{e^{\frac{256e^{x}x}{{\log }^4(2)}}(e^x(256x-4096x^2-4352x^3)+(1-34x){\log }^4(2))}{5{\log }^4(2)}dx$

Optimal. Leaf size=23 $$\frac{1}{5}{e}^{\frac{256e^{x}x}{{\log }^4(2)}}(x-17x^2)$$

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Rubi [A]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 2, number of rules used = 2, integrand size = 49, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.041, Rules used = {12, 2288} $$\begin{array}{c}\frac{(-17x^3-16x^2+x)e^{x+\frac{256e^{x}x}{{\log }^4(2)}}}{5(e^{x}x+e^x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^((256*E^x*x)/Log[2]^4)*(E^x*(256*x - 4096*x^2 - 4352*x^3) + (1 - 34*x)*Log[2]^4))/(5*Log[2]^4),x]

[Out]

(E^(x + (256*E^x*x)/Log[2]^4)*(x - 16*x^2 - 17*x^3))/(5*(E^x + E^x*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int e^{\frac{256e^{x}x}{{\log }^4(2)}}(e^x(256x-4096x^2-4352x^3)+(1-34x){\log }^4(2))dx}{5{\log }^4(2)}\\ & =\frac{e^{x+\frac{256e^{x}x}{{\log }^4(2)}}(x-16x^2-17x^3)}{5(e^x+e^{x}x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.96 $$\begin{array}{c}-\frac{1}{5}{e}^{\frac{256e^{x}x}{{\log }^4(2)}}x(-1+17x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^((256*E^x*x)/Log[2]^4)*(E^x*(256*x - 4096*x^2 - 4352*x^3) + (1 - 34*x)*Log[2]^4))/(5*Log[2]^4),x]

[Out]

-1/5*(E^((256*E^x*x)/Log[2]^4)*x*(-1 + 17*x))

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fricas [A]  time = 0.72, size = 21, normalized size = 0.91 $$\begin{array}{c}-\frac{1}{5}(17x^2-x)e^{\left(\frac{256x{e}^x}{\log (2{)}^4}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*log(2)^4)*exp(256*x*exp(x)/log(2)^4)/log(2)^4,x, al
gorithm="fricas")

[Out]

-1/5*(17*x^2 - x)*e^(256*x*e^x/log(2)^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int -\frac{((34x-1)\log (2{)}^4+256(17x^3+16x^2-x)e^x)e^{\left(\frac{256x{e}^x}{\log (2{)}^4}\right)}}{5\log (2{)}^4}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*log(2)^4)*exp(256*x*exp(x)/log(2)^4)/log(2)^4,x, al
gorithm="giac")

[Out]

integrate(-1/5*((34*x - 1)*log(2)^4 + 256*(17*x^3 + 16*x^2 - x)*e^x)*e^(256*x*e^x/log(2)^4)/log(2)^4, x)

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maple [A]  time = 0.08, size = 19, normalized size = 0.83

method	result	size
risch	$-\frac{(17x-1)x{\mathrm{e}}^{\frac{256x{\mathrm{e}}^x}{\ln (2{)}^4}}}{5}$	$19$
norman	$\frac{\frac{x\ln (2{)}^3{\mathrm{e}}^{\frac{256x{\mathrm{e}}^x}{\ln (2{)}^4}}}{5}-\frac{17x^2\ln (2{)}^3{\mathrm{e}}^{\frac{256x{\mathrm{e}}^x}{\ln (2{)}^4}}}{5}}{\ln (2{)}^3}$	$43$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*ln(2)^4)*exp(256*x*exp(x)/ln(2)^4)/ln(2)^4,x,method=_RETU
RNVERBOSE)

[Out]

-1/5*(17*x-1)*x*exp(256*x*exp(x)/ln(2)^4)

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maxima [A]  time = 1.04, size = 33, normalized size = 1.43 $$\begin{array}{c}-\frac{(17x^2\log (2{)}^4-x\log (2{)}^4)e^{\left(\frac{256x{e}^x}{\log (2{)}^4}\right)}}{5\log (2{)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x^3-4096*x^2+256*x)*exp(x)+(-34*x+1)*log(2)^4)*exp(256*x*exp(x)/log(2)^4)/log(2)^4,x, al
gorithm="maxima")

[Out]

-1/5*(17*x^2*log(2)^4 - x*log(2)^4)*e^(256*x*e^x/log(2)^4)/log(2)^4

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mupad [B]  time = 0.11, size = 18, normalized size = 0.78 $$\begin{array}{c}-\frac{x{\mathrm{e}}^{\frac{256x{\mathrm{e}}^x}{{\ln (2)}^4}}(17x-1)}{5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((256*x*exp(x))/log(2)^4)*(log(2)^4*(34*x - 1) + exp(x)*(4096*x^2 - 256*x + 4352*x^3)))/(5*log(2)^4),
x)

[Out]

-(x*exp((256*x*exp(x))/log(2)^4)*(17*x - 1))/5

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sympy [A]  time = 0.22, size = 20, normalized size = 0.87 $$\begin{array}{c}\frac{(-17x^2+x)e^{\frac{256x{e}^x}{\log {(2)}^4}}}{5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4352*x**3-4096*x**2+256*x)*exp(x)+(-34*x+1)*ln(2)**4)*exp(256*x*exp(x)/ln(2)**4)/ln(2)**4,x)

[Out]

(-17*x**2 + x)*exp(256*x*exp(x)/log(2)**4)/5

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