3.40.73 \(\int \frac {-9+24 x+32 x^3+16 x^4+e^4 (-12 x-16 x^3)+e^{4 x} (16 x^4+32 x^5)+e^{2 x} (8 x^3+96 x^4+32 x^5+e^4 (-16 x^3-32 x^4))}{8 x^3} \, dx\)

Optimal. Leaf size=25 \[ \left (-2+e^4+\left (-1-e^{2 x}+\frac {3}{4 x^2}\right ) x\right )^2 \]

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Rubi [B]  time = 0.20, antiderivative size = 104, normalized size of antiderivative = 4.16, number of steps used = 19, number of rules used = 6, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 14, 2196, 2176, 2194, 1590} \begin {gather*} 2 e^{2 x} x^2+e^{4 x} x^2+\frac {\left (-4 x^2-4 \left (2-e^4\right ) x+3\right )^2}{16 x^2}-2 e^{2 x} x+2 \left (3-e^4\right ) e^{2 x} x+e^{2 x}-\left (3-e^4\right ) e^{2 x}+\frac {1}{2} \left (1-2 e^4\right ) e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9 + 24*x + 32*x^3 + 16*x^4 + E^4*(-12*x - 16*x^3) + E^(4*x)*(16*x^4 + 32*x^5) + E^(2*x)*(8*x^3 + 96*x^4
+ 32*x^5 + E^4*(-16*x^3 - 32*x^4)))/(8*x^3),x]

[Out]

E^(2*x) + (E^(2*x)*(1 - 2*E^4))/2 - E^(2*x)*(3 - E^4) - 2*E^(2*x)*x + 2*E^(2*x)*(3 - E^4)*x + 2*E^(2*x)*x^2 +
E^(4*x)*x^2 + (3 - 4*(2 - E^4)*x - 4*x^2)^2/(16*x^2)

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 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 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 2196

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {-9+24 x+32 x^3+16 x^4+e^4 \left (-12 x-16 x^3\right )+e^{4 x} \left (16 x^4+32 x^5\right )+e^{2 x} \left (8 x^3+96 x^4+32 x^5+e^4 \left (-16 x^3-32 x^4\right )\right )}{x^3} \, dx\\ &=\frac {1}{8} \int \left (16 e^{4 x} x (1+2 x)+\frac {\left (-3-4 x^2\right ) \left (3-4 \left (2-e^4\right ) x-4 x^2\right )}{x^3}+8 e^{2 x} \left (1-2 e^4+4 \left (3-e^4\right ) x+4 x^2\right )\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (-3-4 x^2\right ) \left (3-4 \left (2-e^4\right ) x-4 x^2\right )}{x^3} \, dx+2 \int e^{4 x} x (1+2 x) \, dx+\int e^{2 x} \left (1-2 e^4+4 \left (3-e^4\right ) x+4 x^2\right ) \, dx\\ &=\frac {\left (3-4 \left (2-e^4\right ) x-4 x^2\right )^2}{16 x^2}+2 \int \left (e^{4 x} x+2 e^{4 x} x^2\right ) \, dx+\int \left (e^{2 x} \left (1-2 e^4\right )-4 e^{2 x} \left (-3+e^4\right ) x+4 e^{2 x} x^2\right ) \, dx\\ &=\frac {\left (3-4 \left (2-e^4\right ) x-4 x^2\right )^2}{16 x^2}+2 \int e^{4 x} x \, dx+4 \int e^{2 x} x^2 \, dx+4 \int e^{4 x} x^2 \, dx+\left (1-2 e^4\right ) \int e^{2 x} \, dx+\left (4 \left (3-e^4\right )\right ) \int e^{2 x} x \, dx\\ &=\frac {1}{2} e^{2 x} \left (1-2 e^4\right )+\frac {1}{2} e^{4 x} x+2 e^{2 x} \left (3-e^4\right ) x+2 e^{2 x} x^2+e^{4 x} x^2+\frac {\left (3-4 \left (2-e^4\right ) x-4 x^2\right )^2}{16 x^2}-\frac {1}{2} \int e^{4 x} \, dx-2 \int e^{4 x} x \, dx-4 \int e^{2 x} x \, dx-\left (2 \left (3-e^4\right )\right ) \int e^{2 x} \, dx\\ &=-\frac {e^{4 x}}{8}+\frac {1}{2} e^{2 x} \left (1-2 e^4\right )-e^{2 x} \left (3-e^4\right )-2 e^{2 x} x+2 e^{2 x} \left (3-e^4\right ) x+2 e^{2 x} x^2+e^{4 x} x^2+\frac {\left (3-4 \left (2-e^4\right ) x-4 x^2\right )^2}{16 x^2}+\frac {1}{2} \int e^{4 x} \, dx+2 \int e^{2 x} \, dx\\ &=e^{2 x}+\frac {1}{2} e^{2 x} \left (1-2 e^4\right )-e^{2 x} \left (3-e^4\right )-2 e^{2 x} x+2 e^{2 x} \left (3-e^4\right ) x+2 e^{2 x} x^2+e^{4 x} x^2+\frac {\left (3-4 \left (2-e^4\right ) x-4 x^2\right )^2}{16 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.08, size = 72, normalized size = 2.88 \begin {gather*} \frac {9}{16 x^2}-\frac {3}{x}+\frac {3 e^4}{2 x}+4 x-2 e^4 x-2 e^{4+2 x} x+x^2+e^{4 x} x^2+e^{2 x} \left (-\frac {3}{2}+4 x+2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 + 24*x + 32*x^3 + 16*x^4 + E^4*(-12*x - 16*x^3) + E^(4*x)*(16*x^4 + 32*x^5) + E^(2*x)*(8*x^3 + 9
6*x^4 + 32*x^5 + E^4*(-16*x^3 - 32*x^4)))/(8*x^3),x]

[Out]

9/(16*x^2) - 3/x + (3*E^4)/(2*x) + 4*x - 2*E^4*x - 2*E^(4 + 2*x)*x + x^2 + E^(4*x)*x^2 + E^(2*x)*(-3/2 + 4*x +
2*x^2)

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fricas [B]  time = 0.94, size = 71, normalized size = 2.84 \begin {gather*} \frac {16 \, x^{4} e^{\left (4 \, x\right )} + 16 \, x^{4} + 64 \, x^{3} - 8 \, {\left (4 \, x^{3} - 3 \, x\right )} e^{4} + 8 \, {\left (4 \, x^{4} - 4 \, x^{3} e^{4} + 8 \, x^{3} - 3 \, x^{2}\right )} e^{\left (2 \, x\right )} - 48 \, x + 9}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((32*x^5+16*x^4)*exp(x)^4+((-32*x^4-16*x^3)*exp(4)+32*x^5+96*x^4+8*x^3)*exp(x)^2+(-16*x^3-12*x)*
exp(4)+16*x^4+32*x^3+24*x-9)/x^3,x, algorithm="fricas")

[Out]

1/16*(16*x^4*e^(4*x) + 16*x^4 + 64*x^3 - 8*(4*x^3 - 3*x)*e^4 + 8*(4*x^4 - 4*x^3*e^4 + 8*x^3 - 3*x^2)*e^(2*x) -
48*x + 9)/x^2

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giac [B]  time = 0.21, size = 79, normalized size = 3.16 \begin {gather*} \frac {16 \, x^{4} e^{\left (4 \, x\right )} + 32 \, x^{4} e^{\left (2 \, x\right )} + 16 \, x^{4} - 32 \, x^{3} e^{4} + 64 \, x^{3} e^{\left (2 \, x\right )} - 32 \, x^{3} e^{\left (2 \, x + 4\right )} + 64 \, x^{3} - 24 \, x^{2} e^{\left (2 \, x\right )} + 24 \, x e^{4} - 48 \, x + 9}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((32*x^5+16*x^4)*exp(x)^4+((-32*x^4-16*x^3)*exp(4)+32*x^5+96*x^4+8*x^3)*exp(x)^2+(-16*x^3-12*x)*
exp(4)+16*x^4+32*x^3+24*x-9)/x^3,x, algorithm="giac")

[Out]

1/16*(16*x^4*e^(4*x) + 32*x^4*e^(2*x) + 16*x^4 - 32*x^3*e^4 + 64*x^3*e^(2*x) - 32*x^3*e^(2*x + 4) + 64*x^3 - 2
4*x^2*e^(2*x) + 24*x*e^4 - 48*x + 9)/x^2

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maple [B]  time = 0.06, size = 57, normalized size = 2.28

method	result	size
risch	\(-2 x \,{\mathrm e}^{4}+x^{2}+4 x +\frac {\left (12 \,{\mathrm e}^{4}-24\right ) x +\frac {9}{2}}{8 x^{2}}+x^{2} {\mathrm e}^{4 x}+\frac {\left (-16 x \,{\mathrm e}^{4}+16 x^{2}+32 x -12\right ) {\mathrm e}^{2 x}}{8}\)	\(57\)
norman	\(\frac {\frac {9}{16}+x^{4}+x^{4} {\mathrm e}^{4 x}+\left (-2 \,{\mathrm e}^{4}+4\right ) x^{3}+\left (\frac {3 \,{\mathrm e}^{4}}{2}-3\right ) x +\left (-2 \,{\mathrm e}^{4}+4\right ) x^{3} {\mathrm e}^{2 x}-\frac {3 \,{\mathrm e}^{2 x} x^{2}}{2}+2 \,{\mathrm e}^{2 x} x^{4}}{x^{2}}\)	\(68\)
default	\(x^{2}+4 x +\frac {9}{16 x^{2}}-\frac {3}{x}-\frac {3 \,{\mathrm e}^{2 x}}{2}+4 x \,{\mathrm e}^{2 x}+x^{2} {\mathrm e}^{4 x}+\frac {3 \,{\mathrm e}^{4}}{2 x}-{\mathrm e}^{4} {\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x} x^{2}-4 \,{\mathrm e}^{4} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )-2 x \,{\mathrm e}^{4}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*((32*x^5+16*x^4)*exp(x)^4+((-32*x^4-16*x^3)*exp(4)+32*x^5+96*x^4+8*x^3)*exp(x)^2+(-16*x^3-12*x)*exp(4)
+16*x^4+32*x^3+24*x-9)/x^3,x,method=_RETURNVERBOSE)

[Out]

-2*x*exp(4)+x^2+4*x+1/8*((12*exp(4)-24)*x+9/2)/x^2+x^2*exp(4*x)+1/8*(-16*x*exp(4)+16*x^2+32*x-12)*exp(2*x)

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maxima [B]  time = 0.49, size = 112, normalized size = 4.48 \begin {gather*} x^{2} - 2 \, x e^{4} + \frac {1}{8} \, {\left (8 \, x^{2} - 4 \, x + 1\right )} e^{\left (4 \, x\right )} + \frac {1}{8} \, {\left (4 \, x - 1\right )} e^{\left (4 \, x\right )} + {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - {\left (2 \, x e^{4} - e^{4}\right )} e^{\left (2 \, x\right )} + 3 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + 4 \, x + \frac {3 \, e^{4}}{2 \, x} - \frac {3}{x} + \frac {9}{16 \, x^{2}} + \frac {1}{2} \, e^{\left (2 \, x\right )} - e^{\left (2 \, x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((32*x^5+16*x^4)*exp(x)^4+((-32*x^4-16*x^3)*exp(4)+32*x^5+96*x^4+8*x^3)*exp(x)^2+(-16*x^3-12*x)*
exp(4)+16*x^4+32*x^3+24*x-9)/x^3,x, algorithm="maxima")

[Out]

x^2 - 2*x*e^4 + 1/8*(8*x^2 - 4*x + 1)*e^(4*x) + 1/8*(4*x - 1)*e^(4*x) + (2*x^2 - 2*x + 1)*e^(2*x) - (2*x*e^4 -
e^4)*e^(2*x) + 3*(2*x - 1)*e^(2*x) + 4*x + 3/2*e^4/x - 3/x + 9/16/x^2 + 1/2*e^(2*x) - e^(2*x + 4)

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mupad [B]  time = 2.55, size = 58, normalized size = 2.32 \begin {gather*} x^2\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )-\frac {3\,{\mathrm {e}}^{2\,x}}{2}+\frac {x\,\left (\frac {3\,{\mathrm {e}}^4}{2}-3\right )+\frac {9}{16}}{x^2}-x\,\left (2\,{\mathrm {e}}^4+\frac {{\mathrm {e}}^{2\,x}\,\left (16\,{\mathrm {e}}^4-32\right )}{8}-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + (exp(2*x)*(8*x^3 - exp(4)*(16*x^3 + 32*x^4) + 96*x^4 + 32*x^5))/8 - (exp(4)*(12*x + 16*x^3))/8 + (e
xp(4*x)*(16*x^4 + 32*x^5))/8 + 4*x^3 + 2*x^4 - 9/8)/x^3,x)

[Out]

x^2*(2*exp(2*x) + exp(4*x) + 1) - (3*exp(2*x))/2 + (x*((3*exp(4))/2 - 3) + 9/16)/x^2 - x*(2*exp(4) + (exp(2*x)
*(16*exp(4) - 32))/8 - 4)

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sympy [B]  time = 0.25, size = 60, normalized size = 2.40 \begin {gather*} x^{2} e^{4 x} + x^{2} + \frac {x \left (32 - 16 e^{4}\right )}{8} + \frac {\left (4 x^{2} - 4 x e^{4} + 8 x - 3\right ) e^{2 x}}{2} + \frac {x \left (-48 + 24 e^{4}\right ) + 9}{16 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((32*x**5+16*x**4)*exp(x)**4+((-32*x**4-16*x**3)*exp(4)+32*x**5+96*x**4+8*x**3)*exp(x)**2+(-16*x
**3-12*x)*exp(4)+16*x**4+32*x**3+24*x-9)/x**3,x)

[Out]

x**2*exp(4*x) + x**2 + x*(32 - 16*exp(4))/8 + (4*x**2 - 4*x*exp(4) + 8*x - 3)*exp(2*x)/2 + (x*(-48 + 24*exp(4)
) + 9)/(16*x**2)

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