3.28.59 \(\int \frac {-2 x^2+10 x^6+e^{9 x} (-4+9 x)}{5 x^5} \, dx\)

Optimal. Leaf size=22 \[ x \left (x+\frac {\frac {e^{9 x}}{x}+x}{5 x^4}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2197} \begin {gather*} \frac {e^{9 x}}{5 x^4}+x^2+\frac {1}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^2 + 10*x^6 + E^(9*x)*(-4 + 9*x))/(5*x^5),x]

[Out]

E^(9*x)/(5*x^4) + 1/(5*x^2) + 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 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 {-2 x^2+10 x^6+e^{9 x} (-4+9 x)}{x^5} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{9 x} (-4+9 x)}{x^5}+\frac {2 \left (-1+5 x^4\right )}{x^3}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{9 x} (-4+9 x)}{x^5} \, dx+\frac {2}{5} \int \frac {-1+5 x^4}{x^3} \, dx\\ &=\frac {e^{9 x}}{5 x^4}+\frac {2}{5} \int \left (-\frac {1}{x^3}+5 x\right ) \, dx\\ &=\frac {e^{9 x}}{5 x^4}+\frac {1}{5 x^2}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.05 \begin {gather*} \frac {e^{9 x}}{5 x^4}+\frac {1}{5 x^2}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^2 + 10*x^6 + E^(9*x)*(-4 + 9*x))/(5*x^5),x]

[Out]

E^(9*x)/(5*x^4) + 1/(5*x^2) + x^2

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fricas [A]  time = 0.56, size = 18, normalized size = 0.82 \begin {gather*} \frac {5 \, x^{6} + x^{2} + e^{\left (9 \, x\right )}}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((9*x-4)*exp(9*x)+10*x^6-2*x^2)/x^5,x, algorithm="fricas")

[Out]

1/5*(5*x^6 + x^2 + e^(9*x))/x^4

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giac [A]  time = 0.27, size = 18, normalized size = 0.82 \begin {gather*} \frac {5 \, x^{6} + x^{2} + e^{\left (9 \, x\right )}}{5 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((9*x-4)*exp(9*x)+10*x^6-2*x^2)/x^5,x, algorithm="giac")

[Out]

1/5*(5*x^6 + x^2 + e^(9*x))/x^4

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maple [A]  time = 0.05, size = 19, normalized size = 0.86

method	result	size
derivativedivides	\(x^{2}+\frac {1}{5 x^{2}}+\frac {{\mathrm e}^{9 x}}{5 x^{4}}\)	\(19\)
default	\(x^{2}+\frac {1}{5 x^{2}}+\frac {{\mathrm e}^{9 x}}{5 x^{4}}\)	\(19\)
risch	\(x^{2}+\frac {1}{5 x^{2}}+\frac {{\mathrm e}^{9 x}}{5 x^{4}}\)	\(19\)
norman	\(\frac {x^{6}+\frac {x^{2}}{5}+\frac {{\mathrm e}^{9 x}}{5}}{x^{4}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((9*x-4)*exp(9*x)+10*x^6-2*x^2)/x^5,x,method=_RETURNVERBOSE)

[Out]

x^2+1/5/x^2+1/5/x^4*exp(9*x)

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maxima [C]  time = 0.39, size = 23, normalized size = 1.05 \begin {gather*} x^{2} + \frac {1}{5 \, x^{2}} + \frac {6561}{5} \, \Gamma \left (-3, -9 \, x\right ) + \frac {26244}{5} \, \Gamma \left (-4, -9 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((9*x-4)*exp(9*x)+10*x^6-2*x^2)/x^5,x, algorithm="maxima")

[Out]

x^2 + 1/5/x^2 + 6561/5*gamma(-3, -9*x) + 26244/5*gamma(-4, -9*x)

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mupad [B]  time = 0.07, size = 20, normalized size = 0.91 \begin {gather*} \frac {\frac {{\mathrm {e}}^{9\,x}}{5}+\frac {x^2}{5}}{x^4}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(9*x)*(9*x - 4))/5 - (2*x^2)/5 + 2*x^6)/x^5,x)

[Out]

(exp(9*x)/5 + x^2/5)/x^4 + x^2

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sympy [A]  time = 0.11, size = 19, normalized size = 0.86 \begin {gather*} x^{2} + \frac {1}{5 x^{2}} + \frac {e^{9 x}}{5 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((9*x-4)*exp(9*x)+10*x**6-2*x**2)/x**5,x)

[Out]

x**2 + 1/(5*x**2) + exp(9*x)/(5*x**4)

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