3.11.42 \(\int \frac {50 x^5+e^{6-4 x} (3+4 x)}{25 x^4} \, dx\) [1042]

Optimal. Leaf size=28 \[ x \left (-\frac {e^{6-4 x}}{25 x^4}+x\right )-\log \left (\frac {4}{\log (4)}\right ) \]

[Out]

x*(x-1/25*exp(3-2*x)^2/x^4)-ln(2/ln(2))

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Rubi [A]
time = 0.03, antiderivative size = 18, normalized size of antiderivative = 0.64, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 14, 2228} \begin {gather*} x^2-\frac {e^{6-4 x}}{25 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50*x^5 + E^(6 - 4*x)*(3 + 4*x))/(25*x^4),x]

[Out]

-1/25*E^(6 - 4*x)/x^3 + 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 2228

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}{25} \int \frac {50 x^5+e^{6-4 x} (3+4 x)}{x^4} \, dx\\ &=\frac {1}{25} \int \left (50 x+\frac {e^{6-4 x} (3+4 x)}{x^4}\right ) \, dx\\ &=x^2+\frac {1}{25} \int \frac {e^{6-4 x} (3+4 x)}{x^4} \, dx\\ &=-\frac {e^{6-4 x}}{25 x^3}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.11, size = 18, normalized size = 0.64 \begin {gather*} -\frac {e^{6-4 x}}{25 x^3}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50*x^5 + E^(6 - 4*x)*(3 + 4*x))/(25*x^4),x]

[Out]

-1/25*E^(6 - 4*x)/x^3 + x^2

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Maple [A]
time = 0.11, size = 28, normalized size = 1.00

method	result	size
risch	\(x^{2}-\frac {{\mathrm e}^{6-4 x}}{25 x^{3}}\)	\(16\)
norman	\(\frac {x^{5}-\frac {{\mathrm e}^{6-4 x}}{25}}{x^{3}}\)	\(19\)
derivativedivides	\(-\frac {9}{2}+3 x +\frac {\left (3-2 x \right )^{2}}{4}-\frac {{\mathrm e}^{6-4 x}}{25 x^{3}}\)	\(28\)
default	\(-\frac {9}{2}+3 x +\frac {\left (3-2 x \right )^{2}}{4}-\frac {{\mathrm e}^{6-4 x}}{25 x^{3}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*((3+4*x)*exp(3-2*x)^2+50*x^5)/x^4,x,method=_RETURNVERBOSE)

[Out]

-9/2+3*x+1/4*(3-2*x)^2-1/25*exp(3-2*x)^2/x^3

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.32, size = 22, normalized size = 0.79 \begin {gather*} x^{2} - \frac {64}{25} \, e^{6} \Gamma \left (-2, 4 \, x\right ) - \frac {192}{25} \, e^{6} \Gamma \left (-3, 4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((3+4*x)*exp(3-2*x)^2+50*x^5)/x^4,x, algorithm="maxima")

[Out]

x^2 - 64/25*e^6*gamma(-2, 4*x) - 192/25*e^6*gamma(-3, 4*x)

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Fricas [A]
time = 0.39, size = 19, normalized size = 0.68 \begin {gather*} \frac {25 \, x^{5} - e^{\left (-4 \, x + 6\right )}}{25 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((3+4*x)*exp(3-2*x)^2+50*x^5)/x^4,x, algorithm="fricas")

[Out]

1/25*(25*x^5 - e^(-4*x + 6))/x^3

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Sympy [A]
time = 0.03, size = 14, normalized size = 0.50 \begin {gather*} x^{2} - \frac {e^{6 - 4 x}}{25 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((3+4*x)*exp(3-2*x)**2+50*x**5)/x**4,x)

[Out]

x**2 - exp(6 - 4*x)/(25*x**3)

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Giac [A]
time = 0.39, size = 19, normalized size = 0.68 \begin {gather*} \frac {25 \, x^{5} - e^{\left (-4 \, x + 6\right )}}{25 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((3+4*x)*exp(3-2*x)^2+50*x^5)/x^4,x, algorithm="giac")

[Out]

1/25*(25*x^5 - e^(-4*x + 6))/x^3

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Mupad [B]
time = 0.83, size = 15, normalized size = 0.54 \begin {gather*} x^2-\frac {{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^6}{25\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(6 - 4*x)*(4*x + 3))/25 + 2*x^5)/x^4,x)

[Out]

x^2 - (exp(-4*x)*exp(6))/(25*x^3)

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