3.8.51 \(\int \frac {e^2 (-9600-7680 x)+7680 e x+1920 e^3 x+128 e^{2 x/3} x^3+e^{x/3} (512 x^3+128 e^2 x^3+e (1920 x-640 x^2-512 x^3))}{3 x^3} \, dx\)

Optimal. Leaf size=29 \[ 5+64 \left (-(-2+e)^2-e^{x/3}+\frac {5 e}{x}\right )^2 \]

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Rubi [B]  time = 0.16, antiderivative size = 61, normalized size of antiderivative = 2.10, number of steps used = 12, number of rules used = 8, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6, 12, 14, 2194, 37, 2199, 2177, 2178} \begin {gather*} \frac {64 \left (5 e-(2-e)^2 x\right )^2}{x^2}+128 (2-e)^2 e^{x/3}+64 e^{2 x/3}-\frac {640 e^{\frac {x}{3}+1}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-9600 - 7680*x) + 7680*E*x + 1920*E^3*x + 128*E^((2*x)/3)*x^3 + E^(x/3)*(512*x^3 + 128*E^2*x^3 + E*(
1920*x - 640*x^2 - 512*x^3)))/(3*x^3),x]

[Out]

128*(2 - E)^2*E^(x/3) + 64*E^((2*x)/3) - (640*E^(1 + x/3))/x + (64*(5*E - (2 - E)^2*x)^2)/x^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 (-9600-7680 x)+\left (7680 e+1920 e^3\right ) x+128 e^{2 x/3} x^3+e^{x/3} \left (512 x^3+128 e^2 x^3+e \left (1920 x-640 x^2-512 x^3\right )\right )}{3 x^3} \, dx\\ &=\frac {1}{3} \int \frac {e^2 (-9600-7680 x)+\left (7680 e+1920 e^3\right ) x+128 e^{2 x/3} x^3+e^{x/3} \left (512 x^3+128 e^2 x^3+e \left (1920 x-640 x^2-512 x^3\right )\right )}{x^3} \, dx\\ &=\frac {1}{3} \int \left (128 e^{2 x/3}+\frac {1920 e \left (-5 e+(2-e)^2 x\right )}{x^3}+\frac {128 e^{x/3} \left (15 e-5 e x+(2-e)^2 x^2\right )}{x^2}\right ) \, dx\\ &=\frac {128}{3} \int e^{2 x/3} \, dx+\frac {128}{3} \int \frac {e^{x/3} \left (15 e-5 e x+(2-e)^2 x^2\right )}{x^2} \, dx+(640 e) \int \frac {-5 e+(2-e)^2 x}{x^3} \, dx\\ &=64 e^{2 x/3}+\frac {64 \left (5 e-(2-e)^2 x\right )^2}{x^2}+\frac {128}{3} \int \left ((-2+e)^2 e^{x/3}+\frac {15 e^{1+\frac {x}{3}}}{x^2}-\frac {5 e^{1+\frac {x}{3}}}{x}\right ) \, dx\\ &=64 e^{2 x/3}+\frac {64 \left (5 e-(2-e)^2 x\right )^2}{x^2}-\frac {640}{3} \int \frac {e^{1+\frac {x}{3}}}{x} \, dx+640 \int \frac {e^{1+\frac {x}{3}}}{x^2} \, dx+\frac {1}{3} \left (128 (2-e)^2\right ) \int e^{x/3} \, dx\\ &=128 (2-e)^2 e^{x/3}+64 e^{2 x/3}-\frac {640 e^{1+\frac {x}{3}}}{x}+\frac {64 \left (5 e-(2-e)^2 x\right )^2}{x^2}-\frac {640 e \text {Ei}\left (\frac {x}{3}\right )}{3}+\frac {640}{3} \int \frac {e^{1+\frac {x}{3}}}{x} \, dx\\ &=128 (2-e)^2 e^{x/3}+64 e^{2 x/3}-\frac {640 e^{1+\frac {x}{3}}}{x}+\frac {64 \left (5 e-(2-e)^2 x\right )^2}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 50, normalized size = 1.72 \begin {gather*} 64 \left (e^{2 x/3}+2 e^{x/3} \left ((-2+e)^2-\frac {5 e}{x}\right )+\frac {25 e^2}{x^2}-\frac {10 (-2+e)^2 e}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-9600 - 7680*x) + 7680*E*x + 1920*E^3*x + 128*E^((2*x)/3)*x^3 + E^(x/3)*(512*x^3 + 128*E^2*x^3
+ E*(1920*x - 640*x^2 - 512*x^3)))/(3*x^3),x]

[Out]

64*(E^((2*x)/3) + 2*E^(x/3)*((-2 + E)^2 - (5*E)/x) + (25*E^2)/x^2 - (10*(-2 + E)^2*E)/x)

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fricas [B]  time = 0.56, size = 64, normalized size = 2.21 \begin {gather*} \frac {64 \, {\left (x^{2} e^{\left (\frac {2}{3} \, x\right )} - 10 \, x e^{3} + 5 \, {\left (8 \, x + 5\right )} e^{2} - 40 \, x e + 2 \, {\left (x^{2} e^{2} + 4 \, x^{2} - {\left (4 \, x^{2} + 5 \, x\right )} e\right )} e^{\left (\frac {1}{3} \, x\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(128*x^3*exp(1/3*x)^2+(128*x^3*exp(1)^2+(-512*x^3-640*x^2+1920*x)*exp(1)+512*x^3)*exp(1/3*x)+192
0*x*exp(1)^3+(-7680*x-9600)*exp(1)^2+7680*x*exp(1))/x^3,x, algorithm="fricas")

[Out]

64*(x^2*e^(2/3*x) - 10*x*e^3 + 5*(8*x + 5)*e^2 - 40*x*e + 2*(x^2*e^2 + 4*x^2 - (4*x^2 + 5*x)*e)*e^(1/3*x))/x^2

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giac [B]  time = 0.28, size = 73, normalized size = 2.52 \begin {gather*} \frac {64 \, {\left (x^{2} e^{\left (\frac {2}{3} \, x\right )} + 8 \, x^{2} e^{\left (\frac {1}{3} \, x\right )} + 2 \, x^{2} e^{\left (\frac {1}{3} \, x + 2\right )} - 8 \, x^{2} e^{\left (\frac {1}{3} \, x + 1\right )} - 10 \, x e^{3} + 40 \, x e^{2} - 40 \, x e - 10 \, x e^{\left (\frac {1}{3} \, x + 1\right )} + 25 \, e^{2}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(128*x^3*exp(1/3*x)^2+(128*x^3*exp(1)^2+(-512*x^3-640*x^2+1920*x)*exp(1)+512*x^3)*exp(1/3*x)+192
0*x*exp(1)^3+(-7680*x-9600)*exp(1)^2+7680*x*exp(1))/x^3,x, algorithm="giac")

[Out]

64*(x^2*e^(2/3*x) + 8*x^2*e^(1/3*x) + 2*x^2*e^(1/3*x + 2) - 8*x^2*e^(1/3*x + 1) - 10*x*e^3 + 40*x*e^2 - 40*x*e
- 10*x*e^(1/3*x + 1) + 25*e^2)/x^2

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maple [A]  time = 0.07, size = 59, normalized size = 2.03

method	result	size
risch	\(\frac {\left (-1920 \,{\mathrm e}^{3}+7680 \,{\mathrm e}^{2}-7680 \,{\mathrm e}\right ) x +4800 \,{\mathrm e}^{2}}{3 x^{2}}+64 \,{\mathrm e}^{\frac {2 x}{3}}+\frac {128 \left ({\mathrm e}^{2} x -4 x \,{\mathrm e}-5 \,{\mathrm e}+4 x \right ) {\mathrm e}^{\frac {x}{3}}}{x}\)	\(59\)
norman	\(\frac {\left (-640 \,{\mathrm e}^{3}+2560 \,{\mathrm e}^{2}-2560 \,{\mathrm e}\right ) x +\left (128 \,{\mathrm e}^{2}-512 \,{\mathrm e}+512\right ) x^{2} {\mathrm e}^{\frac {x}{3}}+1600 \,{\mathrm e}^{2}+64 x^{2} {\mathrm e}^{\frac {2 x}{3}}-640 \,{\mathrm e} \,{\mathrm e}^{\frac {x}{3}} x}{x^{2}}\)	\(71\)
derivativedivides	\(64 \,{\mathrm e}^{\frac {2 x}{3}}+\frac {1600 \,{\mathrm e}^{2}}{x^{2}}-\frac {2560 \,{\mathrm e}}{x}+\frac {2560 \,{\mathrm e}^{2}}{x}-\frac {640 \,{\mathrm e}^{3}}{x}-512 \,{\mathrm e} \,{\mathrm e}^{\frac {x}{3}}+128 \,{\mathrm e}^{2} {\mathrm e}^{\frac {x}{3}}+\frac {640 \,{\mathrm e} \left (-\frac {3 \,{\mathrm e}^{\frac {x}{3}}}{x}-\expIntegralEi \left (1, -\frac {x}{3}\right )\right )}{3}+\frac {640 \,{\mathrm e} \expIntegralEi \left (1, -\frac {x}{3}\right )}{3}+512 \,{\mathrm e}^{\frac {x}{3}}\)	\(98\)
default	\(64 \,{\mathrm e}^{\frac {2 x}{3}}+\frac {1600 \,{\mathrm e}^{2}}{x^{2}}-\frac {2560 \,{\mathrm e}}{x}+\frac {2560 \,{\mathrm e}^{2}}{x}-\frac {640 \,{\mathrm e}^{3}}{x}-512 \,{\mathrm e} \,{\mathrm e}^{\frac {x}{3}}+128 \,{\mathrm e}^{2} {\mathrm e}^{\frac {x}{3}}+\frac {640 \,{\mathrm e} \left (-\frac {3 \,{\mathrm e}^{\frac {x}{3}}}{x}-\expIntegralEi \left (1, -\frac {x}{3}\right )\right )}{3}+\frac {640 \,{\mathrm e} \expIntegralEi \left (1, -\frac {x}{3}\right )}{3}+512 \,{\mathrm e}^{\frac {x}{3}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(128*x^3*exp(1/3*x)^2+(128*x^3*exp(1)^2+(-512*x^3-640*x^2+1920*x)*exp(1)+512*x^3)*exp(1/3*x)+1920*x*ex
p(1)^3+(-7680*x-9600)*exp(1)^2+7680*x*exp(1))/x^3,x,method=_RETURNVERBOSE)

[Out]

1/3*((-1920*exp(3)+7680*exp(2)-7680*exp(1))*x+4800*exp(2))/x^2+64*exp(2/3*x)+128*(exp(2)*x-4*x*exp(1)-5*exp(1)
+4*x)/x*exp(1/3*x)

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maxima [C]  time = 0.58, size = 74, normalized size = 2.55 \begin {gather*} -\frac {640}{3} \, {\rm Ei}\left (\frac {1}{3} \, x\right ) e + \frac {640}{3} \, e \Gamma \left (-1, -\frac {1}{3} \, x\right ) - \frac {640 \, e^{3}}{x} + \frac {2560 \, e^{2}}{x} - \frac {2560 \, e}{x} + \frac {1600 \, e^{2}}{x^{2}} + 64 \, e^{\left (\frac {2}{3} \, x\right )} + 512 \, e^{\left (\frac {1}{3} \, x\right )} + 128 \, e^{\left (\frac {1}{3} \, x + 2\right )} - 512 \, e^{\left (\frac {1}{3} \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(128*x^3*exp(1/3*x)^2+(128*x^3*exp(1)^2+(-512*x^3-640*x^2+1920*x)*exp(1)+512*x^3)*exp(1/3*x)+192
0*x*exp(1)^3+(-7680*x-9600)*exp(1)^2+7680*x*exp(1))/x^3,x, algorithm="maxima")

[Out]

-640/3*Ei(1/3*x)*e + 640/3*e*gamma(-1, -1/3*x) - 640*e^3/x + 2560*e^2/x - 2560*e/x + 1600*e^2/x^2 + 64*e^(2/3*
x) + 512*e^(1/3*x) + 128*e^(1/3*x + 2) - 512*e^(1/3*x + 1)

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mupad [B]  time = 0.14, size = 50, normalized size = 1.72 \begin {gather*} 64\,{\mathrm {e}}^{\frac {2\,x}{3}}+\frac {1600\,{\mathrm {e}}^2-x\,\left (640\,{\mathrm {e}}^{\frac {x}{3}+1}+640\,\mathrm {e}\,{\left (\mathrm {e}-2\right )}^2\right )}{x^2}+128\,{\mathrm {e}}^{x/3}\,{\left (\mathrm {e}-2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x/3)*(128*x^3*exp(2) - exp(1)*(640*x^2 - 1920*x + 512*x^3) + 512*x^3))/3 + 2560*x*exp(1) + 640*x*exp
(3) + (128*x^3*exp((2*x)/3))/3 - (exp(2)*(7680*x + 9600))/3)/x^3,x)

[Out]

64*exp((2*x)/3) + (1600*exp(2) - x*(640*exp(x/3 + 1) + 640*exp(1)*(exp(1) - 2)^2))/x^2 + 128*exp(x/3)*(exp(1)
- 2)^2

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sympy [A]  time = 0.31, size = 63, normalized size = 2.17 \begin {gather*} \frac {64 x e^{\frac {2 x}{3}} + \left (- 512 e x + 512 x + 128 x e^{2} - 640 e\right ) e^{\frac {x}{3}}}{x} + \frac {x \left (- 640 e^{3} - 2560 e + 2560 e^{2}\right ) + 1600 e^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(128*x**3*exp(1/3*x)**2+(128*x**3*exp(1)**2+(-512*x**3-640*x**2+1920*x)*exp(1)+512*x**3)*exp(1/3
*x)+1920*x*exp(1)**3+(-7680*x-9600)*exp(1)**2+7680*x*exp(1))/x**3,x)

[Out]

(64*x*exp(2*x/3) + (-512*E*x + 512*x + 128*x*exp(2) - 640*E)*exp(x/3))/x + (x*(-640*exp(3) - 2560*E + 2560*exp
(2)) + 1600*exp(2))/x**2

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