3.33.43 \(\int \frac {x^2-2 x^3+6 e^{3 x} x^3+e^{5 x} (-2+5 x)+e^x (2 x^4+x^5)}{x^3} \, dx\)

Optimal. Leaf size=26 \[ e^{25}-2 x+e^x \left (\frac {e^{2 x}}{x}+x\right )^2+\log (x) \]

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Rubi [A]  time = 0.11, antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 6, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {14, 2194, 43, 2196, 2176, 2197} \begin {gather*} e^x x^2+\frac {e^{5 x}}{x^2}-2 x+2 e^{3 x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 - 2*x^3 + 6*E^(3*x)*x^3 + E^(5*x)*(-2 + 5*x) + E^x*(2*x^4 + x^5))/x^3,x]

[Out]

2*E^(3*x) + E^(5*x)/x^2 - 2*x + E^x*x^2 + Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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} &=\int \left (6 e^{3 x}+\frac {1-2 x}{x}+e^x x (2+x)+\frac {e^{5 x} (-2+5 x)}{x^3}\right ) \, dx\\ &=6 \int e^{3 x} \, dx+\int \frac {1-2 x}{x} \, dx+\int e^x x (2+x) \, dx+\int \frac {e^{5 x} (-2+5 x)}{x^3} \, dx\\ &=2 e^{3 x}+\frac {e^{5 x}}{x^2}+\int \left (-2+\frac {1}{x}\right ) \, dx+\int \left (2 e^x x+e^x x^2\right ) \, dx\\ &=2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+\log (x)+2 \int e^x x \, dx+\int e^x x^2 \, dx\\ &=2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+2 e^x x+e^x x^2+\log (x)-2 \int e^x \, dx-2 \int e^x x \, dx\\ &=-2 e^x+2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+e^x x^2+\log (x)+2 \int e^x \, dx\\ &=2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+e^x x^2+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.12 \begin {gather*} 2 e^{3 x}+\frac {e^{5 x}}{x^2}-2 x+e^x x^2+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 - 2*x^3 + 6*E^(3*x)*x^3 + E^(5*x)*(-2 + 5*x) + E^x*(2*x^4 + x^5))/x^3,x]

[Out]

2*E^(3*x) + E^(5*x)/x^2 - 2*x + E^x*x^2 + Log[x]

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fricas [A]  time = 0.51, size = 35, normalized size = 1.35 \begin {gather*} \frac {x^{4} e^{x} - 2 \, x^{3} + 2 \, x^{2} e^{\left (3 \, x\right )} + x^{2} \log \relax (x) + e^{\left (5 \, x\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x, algorithm="fricas")

[Out]

(x^4*e^x - 2*x^3 + 2*x^2*e^(3*x) + x^2*log(x) + e^(5*x))/x^2

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giac [A]  time = 0.27, size = 35, normalized size = 1.35 \begin {gather*} \frac {x^{4} e^{x} - 2 \, x^{3} + 2 \, x^{2} e^{\left (3 \, x\right )} + x^{2} \log \relax (x) + e^{\left (5 \, x\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x, algorithm="giac")

[Out]

(x^4*e^x - 2*x^3 + 2*x^2*e^(3*x) + x^2*log(x) + e^(5*x))/x^2

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maple [A]  time = 0.06, size = 27, normalized size = 1.04

method	result	size
default	\(\ln \relax (x )-2 x +{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{3 x}+\frac {{\mathrm e}^{5 x}}{x^{2}}\)	\(27\)
risch	\(\ln \relax (x )-2 x +{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{3 x}+\frac {{\mathrm e}^{5 x}}{x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.47, size = 44, normalized size = 1.69 \begin {gather*} {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 2 \, {\left (x - 1\right )} e^{x} - 2 \, x + 2 \, e^{\left (3 \, x\right )} + 25 \, \Gamma \left (-1, -5 \, x\right ) + 50 \, \Gamma \left (-2, -5 \, x\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x, algorithm="maxima")

[Out]

(x^2 - 2*x + 2)*e^x + 2*(x - 1)*e^x - 2*x + 2*e^(3*x) + 25*gamma(-1, -5*x) + 50*gamma(-2, -5*x) + log(x)

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mupad [B]  time = 1.88, size = 32, normalized size = 1.23 \begin {gather*} \ln \relax (x)+\frac {{\mathrm {e}}^{5\,x}+x^4\,{\mathrm {e}}^x+2\,x^2\,{\mathrm {e}}^{3\,x}-2\,x^3}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^3*exp(3*x) + exp(x)*(2*x^4 + x^5) + exp(5*x)*(5*x - 2) + x^2 - 2*x^3)/x^3,x)

[Out]

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

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sympy [A]  time = 0.16, size = 31, normalized size = 1.19 \begin {gather*} - 2 x + \log {\relax (x )} + \frac {x^{4} e^{x} + 2 x^{2} e^{3 x} + e^{5 x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)**5+6*x**3*exp(x)**3+(x**5+2*x**4)*exp(x)-2*x**3+x**2)/x**3,x)

[Out]

-2*x + log(x) + (x**4*exp(x) + 2*x**2*exp(3*x) + exp(5*x))/x**2

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