3.73.77 $\int \frac{e^{8-4x}(2e^{-8+4x}{x}^4+e^4(-15-20x-24x^4))}{2x^4}dx$

Optimal. Leaf size=21 $$e^{4-4(-2+x)}(3+\frac{5}{2x^3})+x$$

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Rubi [A]  time = 0.40, antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 13, number of rules used = 6, integrand size = 41, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.146, Rules used = {12, 6688, 2199, 2194, 2177, 2178} $$\begin{array}{c}\frac{5e^{12-4x}}{2x^3}+x+3e^{12-4x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(8 - 4*x)*(2*E^(-8 + 4*x)*x^4 + E^4*(-15 - 20*x - 24*x^4)))/(2*x^4),x]

[Out]

3*E^(12 - 4*x) + (5*E^(12 - 4*x))/(2*x^3) + 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{2}\int \frac{e^{8-4x}(2e^{-8+4x}{x}^4+e^4(-15-20x-24x^4))}{x^4}dx\\ & =\frac{1}{2}\int (2-\frac{e^{12-4x}(15+20x+24x^4)}{x^4})dx\\ & =x-\frac{1}{2}\int \frac{e^{12-4x}(15+20x+24x^4)}{x^4}dx\\ & =x-\frac{1}{2}\int (24e^{12-4x}+\frac{15e^{12-4x}}{x^4}+\frac{20e^{12-4x}}{x^3})dx\\ & =x-\frac{15}{2}\int \frac{e^{12-4x}}{x^4}dx-10\int \frac{e^{12-4x}}{x^3}dx-12\int e^{12-4x}dx\\ & =3e^{12-4x}+\frac{5e^{12-4x}}{2x^3}+\frac{5e^{12-4x}}{x^2}+x+10\int \frac{e^{12-4x}}{x^3}dx+20\int \frac{e^{12-4x}}{x^2}dx\\ & =3e^{12-4x}+\frac{5e^{12-4x}}{2x^3}-\frac{20e^{12-4x}}{x}+x-20\int \frac{e^{12-4x}}{x^2}dx-80\int \frac{e^{12-4x}}{x}dx\\ & =3e^{12-4x}+\frac{5e^{12-4x}}{2x^3}+x-80e^{12}\text{Ei}(-4x)+80\int \frac{e^{12-4x}}{x}dx\\ & =3e^{12-4x}+\frac{5e^{12-4x}}{2x^3}+x\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 25, normalized size = 1.19 $$\begin{array}{c}3e^{12-4x}+\frac{5e^{12-4x}}{2x^3}+x\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(8 - 4*x)*(2*E^(-8 + 4*x)*x^4 + E^4*(-15 - 20*x - 24*x^4)))/(2*x^4),x]

[Out]

3*E^(12 - 4*x) + (5*E^(12 - 4*x))/(2*x^3) + x

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fricas [A]  time = 0.57, size = 33, normalized size = 1.57 $$\begin{array}{c}\frac{(2x^4{e}^{(4x-8)}+(6x^3+5)e^4)e^{(-4x+8)}}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x^4*exp(4*x-8)+(-24*x^4-20*x-15)*exp(4))/x^4/exp(4*x-8),x, algorithm="fricas")

[Out]

1/2*(2*x^4*e^(4*x - 8) + (6*x^3 + 5)*e^4)*e^(-4*x + 8)/x^3

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giac [A]  time = 0.20, size = 30, normalized size = 1.43 $$\begin{array}{c}\frac{2x^4+6x^3{e}^{(-4x+12)}+5e^{(-4x+12)}}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x^4*exp(4*x-8)+(-24*x^4-20*x-15)*exp(4))/x^4/exp(4*x-8),x, algorithm="giac")

[Out]

1/2*(2*x^4 + 6*x^3*e^(-4*x + 12) + 5*e^(-4*x + 12))/x^3

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maple [A]  time = 0.12, size = 21, normalized size = 1.00

method	result	size
risch	$x+\frac{(6x^3+5){\mathrm{e}}^{-4x+12}}{2x^3}$	$21$
norman	$\frac{(x^4{\mathrm{e}}^{4x-8}+3x^3{\mathrm{e}}^4+\frac{5{\mathrm{e}}^4}{2}){\mathrm{e}}^{-4x+8}}{x^3}$	$35$
derivativedivides	$x-2-14048{\mathrm{e}}^4(-\frac{{\mathrm{e}}^{-4x+8}({(4x-8)}^2+60x-62)}{384x^3}+\frac{{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{6})-6304{\mathrm{e}}^4(\frac{{\mathrm{e}}^{-4x+8}(11{(4x-8)}^2+660x-688)}{384x^3}-\frac{11{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{6})-1152{\mathrm{e}}^4(-\frac{{\mathrm{e}}^{-4x+8}(59{(4x-8)}^2+3552x-3712)}{192x^3}+\frac{59{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{3})-96{\mathrm{e}}^4(\frac{{\mathrm{e}}^{-4x+8}(77{(4x-8)}^2+4656x-4864)}{24x^3}-\frac{619{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{3})-3{\mathrm{e}}^4(-{\mathrm{e}}^{-4x+8}-\frac{2{\mathrm{e}}^{-4x+8}(49{(4x-8)}^2+2976x-3104)}{3x^3}+\frac{6368{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{3})$	$205$
default	$x-2-14048{\mathrm{e}}^4(-\frac{{\mathrm{e}}^{-4x+8}({(4x-8)}^2+60x-62)}{384x^3}+\frac{{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{6})-6304{\mathrm{e}}^4(\frac{{\mathrm{e}}^{-4x+8}(11{(4x-8)}^2+660x-688)}{384x^3}-\frac{11{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{6})-1152{\mathrm{e}}^4(-\frac{{\mathrm{e}}^{-4x+8}(59{(4x-8)}^2+3552x-3712)}{192x^3}+\frac{59{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{3})-96{\mathrm{e}}^4(\frac{{\mathrm{e}}^{-4x+8}(77{(4x-8)}^2+4656x-4864)}{24x^3}-\frac{619{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{3})-3{\mathrm{e}}^4(-{\mathrm{e}}^{-4x+8}-\frac{2{\mathrm{e}}^{-4x+8}(49{(4x-8)}^2+2976x-3104)}{3x^3}+\frac{6368{\mathrm{e}}^8\text{expIntegralEi}(1,4x)}{3})$	$205$
meijerg	$-\frac{{\mathrm{e}}^{-4x+4x{\mathrm{e}}^8}(1-{\mathrm{e}}^{4x(-{\mathrm{e}}^8+1)})}{4(-{\mathrm{e}}^8+1)}-3{\mathrm{e}}^{-4x+4+4x{\mathrm{e}}^8}(1-{\mathrm{e}}^{-4x{\mathrm{e}}^8})-160{\mathrm{e}}^{-4x+28+4x{\mathrm{e}}^8}(-\frac{{\mathrm{e}}^{-16}}{32x^2}+\frac{{\mathrm{e}}^{-8}}{4x}+\frac{13}{4}+\frac{\ln (x)}{2}+\ln (2)+\frac{{\mathrm{e}}^{-16}(144x^2{\mathrm{e}}^{16}-48x{\mathrm{e}}^8+6)}{192x^2}-\frac{{\mathrm{e}}^{-16-4x{\mathrm{e}}^8}(-12x{\mathrm{e}}^8+3)}{96x^2}-\frac{\ln \left(4x{\mathrm{e}}^8\right)}{2}-\frac{\text{expIntegralEi}(1,4x{\mathrm{e}}^8)}{2})-480{\mathrm{e}}^{4x{\mathrm{e}}^8+36-4x}(-\frac{{\mathrm{e}}^{-24}}{192x^3}+\frac{{\mathrm{e}}^{-16}}{32x^2}-\frac{{\mathrm{e}}^{-8}}{8x}-\frac{37}{36}-\frac{\ln (x)}{6}-\frac{\ln (2)}{3}+\frac{{\mathrm{e}}^{-24}(-1408x^3{\mathrm{e}}^{24}+576x^2{\mathrm{e}}^{16}-144x{\mathrm{e}}^8+24)}{4608x^3}-\frac{{\mathrm{e}}^{-24-4x{\mathrm{e}}^8}(64x^2{\mathrm{e}}^{16}-16x{\mathrm{e}}^8+8)}{1536x^3}+\frac{\ln \left(4x{\mathrm{e}}^8\right)}{6}+\frac{\text{expIntegralEi}(1,4x{\mathrm{e}}^8)}{6})$	$268$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*x^4*exp(4*x-8)+(-24*x^4-20*x-15)*exp(4))/x^4/exp(4*x-8),x,method=_RETURNVERBOSE)

[Out]

x+1/2/x^3*(6*x^3+5)*exp(-4*x+12)

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maxima [C]  time = 0.39, size = 28, normalized size = 1.33 $$\begin{array}{c}160e^{12}\mathrm{\Gamma }(-2,4x)+480e^{12}\mathrm{\Gamma }(-3,4x)+x+3e^{(-4x+12)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x^4*exp(4*x-8)+(-24*x^4-20*x-15)*exp(4))/x^4/exp(4*x-8),x, algorithm="maxima")

[Out]

160*e^12*gamma(-2, 4*x) + 480*e^12*gamma(-3, 4*x) + x + 3*e^(-4*x + 12)

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mupad [B]  time = 4.24, size = 21, normalized size = 1.00 $$\begin{array}{c}x+3{\mathrm{e}}^{12-4x}+\frac{5{\mathrm{e}}^{12-4x}}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8 - 4*x)*((exp(4)*(20*x + 24*x^4 + 15))/2 - x^4*exp(4*x - 8)))/x^4,x)

[Out]

x + 3*exp(12 - 4*x) + (5*exp(12 - 4*x))/(2*x^3)

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sympy [A]  time = 0.15, size = 26, normalized size = 1.24 $$\begin{array}{c}x+\frac{(6x^3{e}^4+5e^4)e^{8-4x}}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x**4*exp(4*x-8)+(-24*x**4-20*x-15)*exp(4))/x**4/exp(4*x-8),x)

[Out]

x + (6*x**3*exp(4) + 5*exp(4))*exp(8 - 4*x)/(2*x**3)

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