3.19.50 \(\int \frac {-25 x^4+e^{2 x} (-20+8 x)+e^x (-60 x^2+20 x^3)}{4 x^6} \, dx\)

Optimal. Leaf size=19 \[ \frac {\left (\frac {e^x}{x}+\frac {5 x}{2}\right )^2}{x^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 14, 2197} \begin {gather*} \frac {e^{2 x}}{x^5}+\frac {5 e^x}{x^3}+\frac {25}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25*x^4 + E^(2*x)*(-20 + 8*x) + E^x*(-60*x^2 + 20*x^3))/(4*x^6),x]

[Out]

E^(2*x)/x^5 + (5*E^x)/x^3 + 25/(4*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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {-25 x^4+e^{2 x} (-20+8 x)+e^x \left (-60 x^2+20 x^3\right )}{x^6} \, dx\\ &=\frac {1}{4} \int \left (\frac {20 e^x (-3+x)}{x^4}-\frac {25}{x^2}+\frac {4 e^{2 x} (-5+2 x)}{x^6}\right ) \, dx\\ &=\frac {25}{4 x}+5 \int \frac {e^x (-3+x)}{x^4} \, dx+\int \frac {e^{2 x} (-5+2 x)}{x^6} \, dx\\ &=\frac {e^{2 x}}{x^5}+\frac {5 e^x}{x^3}+\frac {25}{4 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.32 \begin {gather*} \frac {e^{2 x}}{x^5}+\frac {5 e^x}{x^3}+\frac {25}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x^4 + E^(2*x)*(-20 + 8*x) + E^x*(-60*x^2 + 20*x^3))/(4*x^6),x]

[Out]

E^(2*x)/x^5 + (5*E^x)/x^3 + 25/(4*x)

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fricas [A]  time = 0.72, size = 24, normalized size = 1.26 \begin {gather*} \frac {25 \, x^{4} + 20 \, x^{2} e^{x} + 4 \, e^{\left (2 \, x\right )}}{4 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((8*x-20)*exp(x)^2+(20*x^3-60*x^2)*exp(x)-25*x^4)/x^6,x, algorithm="fricas")

[Out]

1/4*(25*x^4 + 20*x^2*e^x + 4*e^(2*x))/x^5

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giac [A]  time = 0.18, size = 24, normalized size = 1.26 \begin {gather*} \frac {25 \, x^{4} + 20 \, x^{2} e^{x} + 4 \, e^{\left (2 \, x\right )}}{4 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((8*x-20)*exp(x)^2+(20*x^3-60*x^2)*exp(x)-25*x^4)/x^6,x, algorithm="giac")

[Out]

1/4*(25*x^4 + 20*x^2*e^x + 4*e^(2*x))/x^5

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maple [A]  time = 0.13, size = 22, normalized size = 1.16

method	result	size
default	\(\frac {25}{4 x}+\frac {5 \,{\mathrm e}^{x}}{x^{3}}+\frac {{\mathrm e}^{2 x}}{x^{5}}\)	\(22\)
norman	\(\frac {{\mathrm e}^{2 x}+\frac {25 x^{4}}{4}+5 \,{\mathrm e}^{x} x^{2}}{x^{5}}\)	\(22\)
risch	\(\frac {25}{4 x}+\frac {5 \,{\mathrm e}^{x}}{x^{3}}+\frac {{\mathrm e}^{2 x}}{x^{5}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((8*x-20)*exp(x)^2+(20*x^3-60*x^2)*exp(x)-25*x^4)/x^6,x,method=_RETURNVERBOSE)

[Out]

25/4/x+5*exp(x)/x^3+1/x^5*exp(x)^2

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maxima [C]  time = 0.45, size = 34, normalized size = 1.79 \begin {gather*} \frac {25}{4 \, x} - 5 \, \Gamma \left (-2, -x\right ) - 15 \, \Gamma \left (-3, -x\right ) - 32 \, \Gamma \left (-4, -2 \, x\right ) - 160 \, \Gamma \left (-5, -2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((8*x-20)*exp(x)^2+(20*x^3-60*x^2)*exp(x)-25*x^4)/x^6,x, algorithm="maxima")

[Out]

25/4/x - 5*gamma(-2, -x) - 15*gamma(-3, -x) - 32*gamma(-4, -2*x) - 160*gamma(-5, -2*x)

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mupad [B]  time = 1.14, size = 21, normalized size = 1.11 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}+5\,x^2\,{\mathrm {e}}^x+\frac {25\,x^4}{4}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x)*(60*x^2 - 20*x^3))/4 - (exp(2*x)*(8*x - 20))/4 + (25*x^4)/4)/x^6,x)

[Out]

(exp(2*x) + 5*x^2*exp(x) + (25*x^4)/4)/x^5

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sympy [A]  time = 0.12, size = 24, normalized size = 1.26 \begin {gather*} \frac {25}{4 x} + \frac {5 x^{5} e^{x} + x^{3} e^{2 x}}{x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((8*x-20)*exp(x)**2+(20*x**3-60*x**2)*exp(x)-25*x**4)/x**6,x)

[Out]

25/(4*x) + (5*x**5*exp(x) + x**3*exp(2*x))/x**8

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