3.104.9 \(\int \frac {e^{13} (5+3 e^3-e^{3+3 x} x) (10+6 e^3+e^{3+3 x} (-x+3 x^2))}{x^2 (-5 x-3 e^3 x+e^{3+3 x} x^2)} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{16} \left (-e^{3 x}+\frac {3}{x}+\frac {5}{e^3 x}\right )}{x} \]

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Rubi [A]  time = 0.45, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 12, 6688, 14, 2197} \begin {gather*} \frac {e^{13} \left (5+3 e^3\right )}{x^2}-\frac {e^{3 x+16}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^13*(5 + 3*E^3 - E^(3 + 3*x)*x)*(10 + 6*E^3 + E^(3 + 3*x)*(-x + 3*x^2)))/(x^2*(-5*x - 3*E^3*x + E^(3 + 3
*x)*x^2)),x]

[Out]

(E^13*(5 + 3*E^3))/x^2 - E^(16 + 3*x)/x

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 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
]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{13} \left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (\left (-5-3 e^3\right ) x+e^{3+3 x} x^2\right )} \, dx\\ &=e^{13} \int \frac {\left (5+3 e^3-e^{3+3 x} x\right ) \left (10+6 e^3+e^{3+3 x} \left (-x+3 x^2\right )\right )}{x^2 \left (\left (-5-3 e^3\right ) x+e^{3+3 x} x^2\right )} \, dx\\ &=e^{13} \int \frac {-10 \left (1+\frac {3 e^3}{5}\right )+e^{3+3 x} \left (x-3 x^2\right )}{x^3} \, dx\\ &=e^{13} \int \left (-\frac {2 \left (5+3 e^3\right )}{x^3}-\frac {e^{3+3 x} (-1+3 x)}{x^2}\right ) \, dx\\ &=\frac {e^{13} \left (5+3 e^3\right )}{x^2}-e^{13} \int \frac {e^{3+3 x} (-1+3 x)}{x^2} \, dx\\ &=\frac {e^{13} \left (5+3 e^3\right )}{x^2}-\frac {e^{16+3 x}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 0.86 \begin {gather*} -\frac {e^{13} \left (-5-3 e^3+e^{3+3 x} x\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^13*(5 + 3*E^3 - E^(3 + 3*x)*x)*(10 + 6*E^3 + E^(3 + 3*x)*(-x + 3*x^2)))/(x^2*(-5*x - 3*E^3*x + E^
(3 + 3*x)*x^2)),x]

[Out]

-((E^13*(-5 - 3*E^3 + E^(3 + 3*x)*x))/x^2)

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fricas [A]  time = 1.05, size = 22, normalized size = 0.79 \begin {gather*} -\frac {x e^{\left (3 \, x + 16\right )} - 3 \, e^{16} - 5 \, e^{13}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(log((-x*exp(3)*exp(3*x)+3*exp(3)+5)/x^2/exp(3))+16)/(x^2
*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x, algorithm="fricas")

[Out]

-(x*e^(3*x + 16) - 3*e^16 - 5*e^13)/x^2

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giac [A]  time = 0.13, size = 22, normalized size = 0.79 \begin {gather*} -\frac {x e^{\left (3 \, x + 16\right )} - 3 \, e^{16} - 5 \, e^{13}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(log((-x*exp(3)*exp(3*x)+3*exp(3)+5)/x^2/exp(3))+16)/(x^2
*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x, algorithm="giac")

[Out]

-(x*e^(3*x + 16) - 3*e^16 - 5*e^13)/x^2

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maple [A]  time = 0.32, size = 28, normalized size = 1.00

method	result	size
norman	\(\frac {{\mathrm e}^{-3} {\mathrm e}^{16} \left (3 \,{\mathrm e}^{3}+5\right )-x \,{\mathrm e}^{16} {\mathrm e}^{3 x}}{x^{2}}\)	\(28\)
risch	\(-\frac {\left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right ) {\mathrm e}^{13+\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right )}{x^{2}}\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right )}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right )}{x^{2}}\right )^{2} \mathrm {csgn}\left (i \left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right )}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right )\right )}{2}-i \mathrm {csgn}\left (\frac {i \left (-5+\left (x \,{\mathrm e}^{3 x}-3\right ) {\mathrm e}^{3}\right )}{x^{2}}\right )^{2} \pi }}{x^{2}}\)	\(247\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(ln((-x*exp(3)*exp(3*x)+3*exp(3)+5)/x^2/exp(3))+16)/(x^2*exp(3)
*exp(3*x)-3*x*exp(3)-5*x),x,method=_RETURNVERBOSE)

[Out]

(1/exp(3)*exp(16)*(3*exp(3)+5)-x*exp(16)*exp(3*x))/x^2

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maxima [A]  time = 0.39, size = 21, normalized size = 0.75 \begin {gather*} -\frac {{\left (x e^{\left (3 \, x + 3\right )} - 3 \, e^{3} - 5\right )} e^{13}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(log((-x*exp(3)*exp(3*x)+3*exp(3)+5)/x^2/exp(3))+16)/(x^2
*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x, algorithm="maxima")

[Out]

-(x*e^(3*x + 3) - 3*e^3 - 5)*e^13/x^2

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mupad [B]  time = 9.36, size = 24, normalized size = 0.86 \begin {gather*} -\frac {x\,{\mathrm {e}}^{3\,x+16}-{\mathrm {e}}^{13}\,\left (3\,{\mathrm {e}}^3+5\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log((exp(-3)*(3*exp(3) - x*exp(3*x)*exp(3) + 5))/x^2) + 16)*(6*exp(3) - exp(3*x)*exp(3)*(x - 3*x^2)
+ 10))/(5*x + 3*x*exp(3) - x^2*exp(3*x)*exp(3)),x)

[Out]

-(x*exp(3*x + 16) - exp(13)*(3*exp(3) + 5))/x^2

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sympy [A]  time = 0.18, size = 27, normalized size = 0.96 \begin {gather*} - \frac {e^{16} e^{3 x}}{x} - \frac {- 6 e^{16} - 10 e^{13}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2-x)*exp(3)*exp(3*x)+6*exp(3)+10)*exp(ln((-x*exp(3)*exp(3*x)+3*exp(3)+5)/x**2/exp(3))+16)/(x*
*2*exp(3)*exp(3*x)-3*x*exp(3)-5*x),x)

[Out]

-exp(16)*exp(3*x)/x - (-6*exp(16) - 10*exp(13))/(2*x**2)

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