3.93.79 \(\int \frac {32 e^{3+2 x} x^3+e^3 (-18 e^2+18 e x+6 x^3+2 x^4)+e^{3+x} (32 x^3+8 x^4+e (24 x-24 x^2))}{x^3} \, dx\)

Optimal. Leaf size=20 \[ e^3 \left (3+4 e^x-\frac {3 e}{x}+x\right )^2 \]

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Rubi [B]  time = 0.16, antiderivative size = 56, normalized size of antiderivative = 2.80, number of steps used = 12, number of rules used = 7, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.101, Rules used = {14, 2194, 1590, 2199, 2177, 2178, 2176} \begin {gather*} \frac {e^3 \left (-x^2-3 x+3 e\right )^2}{x^2}+24 e^{x+3}+16 e^{2 x+3}+8 e^{x+3} x-\frac {24 e^{x+4}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*E^(3 + 2*x)*x^3 + E^3*(-18*E^2 + 18*E*x + 6*x^3 + 2*x^4) + E^(3 + x)*(32*x^3 + 8*x^4 + E*(24*x - 24*x^
2)))/x^3,x]

[Out]

24*E^(3 + x) + 16*E^(3 + 2*x) - (24*E^(4 + x))/x + 8*E^(3 + x)*x + (E^3*(3*E - 3*x - x^2)^2)/x^2

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 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x
, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

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 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 \left (32 e^{3+2 x}-\frac {2 e^3 \left (3 e-3 x-x^2\right ) \left (3 e+x^2\right )}{x^3}-\frac {8 e^{3+x} \left (-3 e+3 e x-4 x^2-x^3\right )}{x^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{3+x} \left (-3 e+3 e x-4 x^2-x^3\right )}{x^2} \, dx\right )+32 \int e^{3+2 x} \, dx-\left (2 e^3\right ) \int \frac {\left (3 e-3 x-x^2\right ) \left (3 e+x^2\right )}{x^3} \, dx\\ &=16 e^{3+2 x}+\frac {e^3 \left (3 e-3 x-x^2\right )^2}{x^2}-8 \int \left (-4 e^{3+x}-\frac {3 e^{4+x}}{x^2}+\frac {3 e^{4+x}}{x}-e^{3+x} x\right ) \, dx\\ &=16 e^{3+2 x}+\frac {e^3 \left (3 e-3 x-x^2\right )^2}{x^2}+8 \int e^{3+x} x \, dx+24 \int \frac {e^{4+x}}{x^2} \, dx-24 \int \frac {e^{4+x}}{x} \, dx+32 \int e^{3+x} \, dx\\ &=32 e^{3+x}+16 e^{3+2 x}-\frac {24 e^{4+x}}{x}+8 e^{3+x} x+\frac {e^3 \left (3 e-3 x-x^2\right )^2}{x^2}-24 e^4 \text {Ei}(x)-8 \int e^{3+x} \, dx+24 \int \frac {e^{4+x}}{x} \, dx\\ &=24 e^{3+x}+16 e^{3+2 x}-\frac {24 e^{4+x}}{x}+8 e^{3+x} x+\frac {e^3 \left (3 e-3 x-x^2\right )^2}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 54, normalized size = 2.70 \begin {gather*} -2 e^3 \left (-8 e^{2 x}+e^x \left (-12+\frac {12 e}{x}-4 x\right )-\frac {9 e^2}{2 x^2}+\frac {9 e}{x}-3 x-\frac {x^2}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*E^(3 + 2*x)*x^3 + E^3*(-18*E^2 + 18*E*x + 6*x^3 + 2*x^4) + E^(3 + x)*(32*x^3 + 8*x^4 + E*(24*x -
24*x^2)))/x^3,x]

[Out]

-2*E^3*(-8*E^(2*x) + E^x*(-12 + (12*E)/x - 4*x) - (9*E^2)/(2*x^2) + (9*E)/x - 3*x - x^2/2)

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fricas [B]  time = 0.77, size = 64, normalized size = 3.20 \begin {gather*} \frac {{\left (16 \, x^{2} e^{\left (2 \, x + 6\right )} - 18 \, x e^{7} + {\left (x^{4} + 6 \, x^{3}\right )} e^{6} - 8 \, {\left (3 \, x e^{4} - {\left (x^{3} + 3 \, x^{2}\right )} e^{3}\right )} e^{\left (x + 3\right )} + 9 \, e^{8}\right )} e^{\left (-3\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^3*exp(3)*exp(x)^2+((-24*x^2+24*x)*exp(1)+8*x^4+32*x^3)*exp(3)*exp(x)+(-18*exp(1)^2+18*x*exp(1)
+2*x^4+6*x^3)*exp(3))/x^3,x, algorithm="fricas")

[Out]

(16*x^2*e^(2*x + 6) - 18*x*e^7 + (x^4 + 6*x^3)*e^6 - 8*(3*x*e^4 - (x^3 + 3*x^2)*e^3)*e^(x + 3) + 9*e^8)*e^(-3)
/x^2

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giac [B]  time = 0.14, size = 63, normalized size = 3.15 \begin {gather*} \frac {x^{4} e^{3} + 6 \, x^{3} e^{3} + 8 \, x^{3} e^{\left (x + 3\right )} + 16 \, x^{2} e^{\left (2 \, x + 3\right )} + 24 \, x^{2} e^{\left (x + 3\right )} - 18 \, x e^{4} - 24 \, x e^{\left (x + 4\right )} + 9 \, e^{5}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^3*exp(3)*exp(x)^2+((-24*x^2+24*x)*exp(1)+8*x^4+32*x^3)*exp(3)*exp(x)+(-18*exp(1)^2+18*x*exp(1)
+2*x^4+6*x^3)*exp(3))/x^3,x, algorithm="giac")

[Out]

(x^4*e^3 + 6*x^3*e^3 + 8*x^3*e^(x + 3) + 16*x^2*e^(2*x + 3) + 24*x^2*e^(x + 3) - 18*x*e^4 - 24*x*e^(x + 4) + 9
*e^5)/x^2

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maple [B]  time = 0.11, size = 57, normalized size = 2.85

method	result	size
risch	\(x^{2} {\mathrm e}^{3}+6 x \,{\mathrm e}^{3}+\frac {9 \,{\mathrm e}^{5}-18 x \,{\mathrm e}^{4}}{x^{2}}+16 \,{\mathrm e}^{2 x +3}-\frac {8 \left (-x^{2}+3 \,{\mathrm e}-3 x \right ) {\mathrm e}^{3+x}}{x}\)	\(57\)
norman	\(\frac {x^{4} {\mathrm e}^{3}+6 x^{3} {\mathrm e}^{3}+9 \,{\mathrm e}^{2} {\mathrm e}^{3}-18 x \,{\mathrm e} \,{\mathrm e}^{3}+24 x^{2} {\mathrm e}^{3} {\mathrm e}^{x}+16 x^{2} {\mathrm e}^{3} {\mathrm e}^{2 x}+8 \,{\mathrm e}^{x} {\mathrm e}^{3} x^{3}-24 \,{\mathrm e}^{x} {\mathrm e}^{3} x \,{\mathrm e}}{x^{2}}\)	\(72\)
default	\(x^{2} {\mathrm e}^{3}+16 \,{\mathrm e}^{3} {\mathrm e}^{2 x}+32 \,{\mathrm e}^{x} {\mathrm e}^{3}+\frac {9 \,{\mathrm e}^{2} {\mathrm e}^{3}}{x^{2}}-\frac {18 \,{\mathrm e} \,{\mathrm e}^{3}}{x}+8 \,{\mathrm e}^{3} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+24 \,{\mathrm e} \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{x}-\expIntegralEi \left (1, -x \right )\right )+24 \,{\mathrm e} \,{\mathrm e}^{3} \expIntegralEi \left (1, -x \right )+6 x \,{\mathrm e}^{3}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x^3*exp(3)*exp(x)^2+((-24*x^2+24*x)*exp(1)+8*x^4+32*x^3)*exp(3)*exp(x)+(-18*exp(1)^2+18*x*exp(1)+2*x^4
+6*x^3)*exp(3))/x^3,x,method=_RETURNVERBOSE)

[Out]

x^2*exp(3)+6*x*exp(3)+(9*exp(5)-18*x*exp(4))/x^2+16*exp(2*x+3)-8/x*(-x^2+3*exp(1)-3*x)*exp(3+x)

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maxima [C]  time = 0.41, size = 68, normalized size = 3.40 \begin {gather*} x^{2} e^{3} - 24 \, {\rm Ei}\relax (x) e^{4} + 6 \, x e^{3} + 8 \, {\left (x e^{3} - e^{3}\right )} e^{x} + 24 \, e^{4} \Gamma \left (-1, -x\right ) - \frac {18 \, e^{4}}{x} + \frac {9 \, e^{5}}{x^{2}} + 16 \, e^{\left (2 \, x + 3\right )} + 32 \, e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^3*exp(3)*exp(x)^2+((-24*x^2+24*x)*exp(1)+8*x^4+32*x^3)*exp(3)*exp(x)+(-18*exp(1)^2+18*x*exp(1)
+2*x^4+6*x^3)*exp(3))/x^3,x, algorithm="maxima")

[Out]

x^2*e^3 - 24*Ei(x)*e^4 + 6*x*e^3 + 8*(x*e^3 - e^3)*e^x + 24*e^4*gamma(-1, -x) - 18*e^4/x + 9*e^5/x^2 + 16*e^(2
*x + 3) + 32*e^(x + 3)

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mupad [B]  time = 0.17, size = 56, normalized size = 2.80 \begin {gather*} \frac {9\,{\mathrm {e}}^5-x\,{\mathrm {e}}^3\,\left (24\,{\mathrm {e}}^{x+1}+18\,\mathrm {e}\right )}{x^2}+x^2\,{\mathrm {e}}^3+{\mathrm {e}}^3\,\left (16\,{\mathrm {e}}^{2\,x}+24\,{\mathrm {e}}^x\right )+x\,{\mathrm {e}}^3\,\left (8\,{\mathrm {e}}^x+6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(18*x*exp(1) - 18*exp(2) + 6*x^3 + 2*x^4) + exp(3)*exp(x)*(exp(1)*(24*x - 24*x^2) + 32*x^3 + 8*x^4
) + 32*x^3*exp(2*x)*exp(3))/x^3,x)

[Out]

(9*exp(5) - x*exp(3)*(24*exp(x + 1) + 18*exp(1)))/x^2 + x^2*exp(3) + exp(3)*(16*exp(2*x) + 24*exp(x)) + x*exp(
3)*(8*exp(x) + 6)

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sympy [B]  time = 0.23, size = 65, normalized size = 3.25 \begin {gather*} x^{2} e^{3} + 6 x e^{3} + \frac {16 x e^{3} e^{2 x} + \left (8 x^{2} e^{3} + 24 x e^{3} - 24 e^{4}\right ) e^{x}}{x} + \frac {- 18 x e^{4} + 9 e^{5}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x**3*exp(3)*exp(x)**2+((-24*x**2+24*x)*exp(1)+8*x**4+32*x**3)*exp(3)*exp(x)+(-18*exp(1)**2+18*x*
exp(1)+2*x**4+6*x**3)*exp(3))/x**3,x)

[Out]

x**2*exp(3) + 6*x*exp(3) + (16*x*exp(3)*exp(2*x) + (8*x**2*exp(3) + 24*x*exp(3) - 24*exp(4))*exp(x))/x + (-18*
x*exp(4) + 9*exp(5))/x**2

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