3.84.55 \(\int \frac {\frac {e^{2 x} (16+e^{3 x} (8-40 x)-32 x)}{x}+10 x+\frac {e^{4 x} (-4+8 x+e^{3 x} (-2+7 x))}{x^2}+e^{3 x} (53 x+15 x^2+15 x \log (4))}{5 x} \, dx\)

Optimal. Leaf size=29 \[ \left (2+e^{3 x}\right ) \left (\frac {1}{5} \left (-4+\frac {e^{2 x}}{x}\right )^2+x+\log (4)\right ) \]

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Rubi [B]  time = 0.19, antiderivative size = 79, normalized size of antiderivative = 2.72, number of steps used = 9, number of rules used = 5, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 14, 2197, 2176, 2194} \begin {gather*} \frac {2 e^{4 x}}{5 x^2}+\frac {e^{7 x}}{5 x^2}+2 x-\frac {e^{3 x}}{3}-\frac {16 e^{2 x}}{5 x}-\frac {8 e^{5 x}}{5 x}+\frac {1}{15} e^{3 x} (15 x+53+15 \log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((E^(2*x)*(16 + E^(3*x)*(8 - 40*x) - 32*x))/x + 10*x + (E^(4*x)*(-4 + 8*x + E^(3*x)*(-2 + 7*x)))/x^2 + E^(
3*x)*(53*x + 15*x^2 + 15*x*Log[4]))/(5*x),x]

[Out]

-1/3*E^(3*x) + (2*E^(4*x))/(5*x^2) + E^(7*x)/(5*x^2) - (16*E^(2*x))/(5*x) - (8*E^(5*x))/(5*x) + 2*x + (E^(3*x)
*(53 + 15*x + 15*Log[4]))/15

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (10+\frac {4 e^{4 x} (-1+2 x)}{x^3}-\frac {16 e^{2 x} (-1+2 x)}{x^2}-\frac {8 e^{5 x} (-1+5 x)}{x^2}+\frac {e^{7 x} (-2+7 x)}{x^3}+e^{3 x} (53+15 x+15 \log (4))\right ) \, dx\\ &=2 x+\frac {1}{5} \int \frac {e^{7 x} (-2+7 x)}{x^3} \, dx+\frac {1}{5} \int e^{3 x} (53+15 x+15 \log (4)) \, dx+\frac {4}{5} \int \frac {e^{4 x} (-1+2 x)}{x^3} \, dx-\frac {8}{5} \int \frac {e^{5 x} (-1+5 x)}{x^2} \, dx-\frac {16}{5} \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx\\ &=\frac {2 e^{4 x}}{5 x^2}+\frac {e^{7 x}}{5 x^2}-\frac {16 e^{2 x}}{5 x}-\frac {8 e^{5 x}}{5 x}+2 x+\frac {1}{15} e^{3 x} (53+15 x+15 \log (4))-\int e^{3 x} \, dx\\ &=-\frac {e^{3 x}}{3}+\frac {2 e^{4 x}}{5 x^2}+\frac {e^{7 x}}{5 x^2}-\frac {16 e^{2 x}}{5 x}-\frac {8 e^{5 x}}{5 x}+2 x+\frac {1}{15} e^{3 x} (53+15 x+15 \log (4))\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.11, size = 62, normalized size = 2.14 \begin {gather*} \frac {1}{5} \left (\frac {2 e^{4 x}}{x^2}+\frac {e^{7 x}}{x^2}-\frac {16 e^{2 x}}{x}-\frac {8 e^{5 x}}{x}+10 x+e^{3 x} (16+5 x+5 \log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((E^(2*x)*(16 + E^(3*x)*(8 - 40*x) - 32*x))/x + 10*x + (E^(4*x)*(-4 + 8*x + E^(3*x)*(-2 + 7*x)))/x^2
+ E^(3*x)*(53*x + 15*x^2 + 15*x*Log[4]))/(5*x),x]

[Out]

((2*E^(4*x))/x^2 + E^(7*x)/x^2 - (16*E^(2*x))/x - (8*E^(5*x))/x + 10*x + E^(3*x)*(16 + 5*x + 5*Log[4]))/5

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fricas [A]  time = 0.73, size = 63, normalized size = 2.17 \begin {gather*} \frac {{\left (10 \, x^{3} e^{\left (-7 \, x\right )} - 8 \, x e^{\left (-2 \, x\right )} + {\left (5 \, x^{3} + 10 \, x^{2} \log \relax (2) + 16 \, x^{2}\right )} e^{\left (-4 \, x\right )} - 16 \, x e^{\left (-5 \, x\right )} + 2 \, e^{\left (-3 \, x\right )} + 1\right )} e^{\left (7 \, x\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-log(x/exp(x))+x)^2+((-40*x+8)*exp(3*x)-32*x+16)*exp(-log(x/exp(x)
)+x)+(30*x*log(2)+15*x^2+53*x)*exp(3*x)+10*x)/x,x, algorithm="fricas")

[Out]

1/5*(10*x^3*e^(-7*x) - 8*x*e^(-2*x) + (5*x^3 + 10*x^2*log(2) + 16*x^2)*e^(-4*x) - 16*x*e^(-5*x) + 2*e^(-3*x) +
1)*e^(7*x)/x^2

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giac [A]  time = 0.12, size = 64, normalized size = 2.21 \begin {gather*} \frac {5 \, x^{3} e^{\left (3 \, x\right )} + 10 \, x^{2} e^{\left (3 \, x\right )} \log \relax (2) + 10 \, x^{3} + 16 \, x^{2} e^{\left (3 \, x\right )} - 8 \, x e^{\left (5 \, x\right )} - 16 \, x e^{\left (2 \, x\right )} + e^{\left (7 \, x\right )} + 2 \, e^{\left (4 \, x\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-log(x/exp(x))+x)^2+((-40*x+8)*exp(3*x)-32*x+16)*exp(-log(x/exp(x)
)+x)+(30*x*log(2)+15*x^2+53*x)*exp(3*x)+10*x)/x,x, algorithm="giac")

[Out]

1/5*(5*x^3*e^(3*x) + 10*x^2*e^(3*x)*log(2) + 10*x^3 + 16*x^2*e^(3*x) - 8*x*e^(5*x) - 16*x*e^(2*x) + e^(7*x) +
2*e^(4*x))/x^2

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maple [A]  time = 0.23, size = 61, normalized size = 2.10

method	result	size
default	\(2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}+x \,{\mathrm e}^{3 x}+2 \ln \relax (2) {\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{5 x}}{5 x}+\frac {{\mathrm e}^{7 x}}{5 x^{2}}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}-\frac {16 \,{\mathrm e}^{2 x}}{5 x}\)	\(61\)
risch	\(2 \ln \relax (2) {\mathrm e}^{3 x}+x \,{\mathrm e}^{3 x}+\frac {16 \,{\mathrm e}^{3 x}}{5}+2 x +\frac {\left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{4 x}}{5 x^{2}}-\frac {8 \left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{2 x} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i x \right )}{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )}{2}}}{5 x}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-ln(x/exp(x))+x)^2+((-40*x+8)*exp(3*x)-32*x+16)*exp(-ln(x/exp(x))+x)+(30
*x*ln(2)+15*x^2+53*x)*exp(3*x)+10*x)/x,x,method=_RETURNVERBOSE)

[Out]

2*x+16/5*exp(x)^3+x*exp(x)^3+2*exp(x)^3*ln(2)-8/5*exp(x)^5/x+1/5*exp(x)^7/x^2+2/5/x^2*exp(4*x)-16/5*exp(2*x)/x

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maxima [C]  time = 0.41, size = 83, normalized size = 2.86 \begin {gather*} \frac {1}{3} \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} + 2 \, e^{\left (3 \, x\right )} \log \relax (2) + 2 \, x - 8 \, {\rm Ei}\left (5 \, x\right ) - \frac {32}{5} \, {\rm Ei}\left (2 \, x\right ) + \frac {53}{15} \, e^{\left (3 \, x\right )} + \frac {32}{5} \, \Gamma \left (-1, -2 \, x\right ) + \frac {32}{5} \, \Gamma \left (-1, -4 \, x\right ) + 8 \, \Gamma \left (-1, -5 \, x\right ) + \frac {49}{5} \, \Gamma \left (-1, -7 \, x\right ) + \frac {64}{5} \, \Gamma \left (-2, -4 \, x\right ) + \frac {98}{5} \, \Gamma \left (-2, -7 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-log(x/exp(x))+x)^2+((-40*x+8)*exp(3*x)-32*x+16)*exp(-log(x/exp(x)
)+x)+(30*x*log(2)+15*x^2+53*x)*exp(3*x)+10*x)/x,x, algorithm="maxima")

[Out]

1/3*(3*x - 1)*e^(3*x) + 2*e^(3*x)*log(2) + 2*x - 8*Ei(5*x) - 32/5*Ei(2*x) + 53/15*e^(3*x) + 32/5*gamma(-1, -2*
x) + 32/5*gamma(-1, -4*x) + 8*gamma(-1, -5*x) + 49/5*gamma(-1, -7*x) + 64/5*gamma(-2, -4*x) + 98/5*gamma(-2, -
7*x)

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mupad [B]  time = 5.23, size = 53, normalized size = 1.83 \begin {gather*} \frac {\frac {2\,{\mathrm {e}}^{4\,x}}{5}+\frac {{\mathrm {e}}^{7\,x}}{5}-x\,\left (\frac {16\,{\mathrm {e}}^{2\,x}}{5}+\frac {8\,{\mathrm {e}}^{5\,x}}{5}\right )}{x^2}+x\,\left ({\mathrm {e}}^{3\,x}+2\right )+{\mathrm {e}}^{3\,x}\,\left (2\,\ln \relax (2)+\frac {16}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + (exp(3*x)*(53*x + 30*x*log(2) + 15*x^2))/5 - (exp(x - log(x*exp(-x)))*(32*x + exp(3*x)*(40*x - 8) -
16))/5 + (exp(2*x - 2*log(x*exp(-x)))*(8*x + exp(3*x)*(7*x - 2) - 4))/5)/x,x)

[Out]

((2*exp(4*x))/5 + exp(7*x)/5 - x*((16*exp(2*x))/5 + (8*exp(5*x))/5))/x^2 + x*(exp(3*x) + 2) + exp(3*x)*(2*log(
2) + 16/5)

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sympy [B]  time = 0.24, size = 71, normalized size = 2.45 \begin {gather*} 2 x + \frac {- 5000 x^{5} e^{5 x} - 10000 x^{5} e^{2 x} + 625 x^{4} e^{7 x} + 1250 x^{4} e^{4 x} + \left (3125 x^{7} + 6250 x^{6} \log {\relax (2 )} + 10000 x^{6}\right ) e^{3 x}}{3125 x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-ln(x/exp(x))+x)**2+((-40*x+8)*exp(3*x)-32*x+16)*exp(-ln(x/exp(x))
+x)+(30*x*ln(2)+15*x**2+53*x)*exp(3*x)+10*x)/x,x)

[Out]

2*x + (-5000*x**5*exp(5*x) - 10000*x**5*exp(2*x) + 625*x**4*exp(7*x) + 1250*x**4*exp(4*x) + (3125*x**7 + 6250*
x**6*log(2) + 10000*x**6)*exp(3*x))/(3125*x**6)

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