3.43.66 \(\int \frac {5 x^2-3 e^3 \log (5)}{x^4} \, dx\)

Optimal. Leaf size=22 \[ \frac {5 x (-1+5 x)+\frac {e^3 \log (5)}{x}}{x^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.68, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \begin {gather*} \frac {e^3 \log (5)}{x^3}-\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x^2 - 3*E^3*Log[5])/x^4,x]

[Out]

-5/x + (E^3*Log[5])/x^3

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]]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.68 \begin {gather*} -\frac {5}{x}+\frac {e^3 \log (5)}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5/x + (E^3*Log[5])/x^3

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fricas [A]  time = 0.57, size = 17, normalized size = 0.77 \begin {gather*} -\frac {5 \, x^{2} - e^{3} \log \relax (5)}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*x^2 - e^3*log(5))/x^3

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giac [A]  time = 0.14, size = 17, normalized size = 0.77 \begin {gather*} -\frac {5 \, x^{2} - e^{3} \log \relax (5)}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*x^2 - e^3*log(5))/x^3

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maple [A]  time = 0.03, size = 15, normalized size = 0.68

method	result	size
default	\(\frac {{\mathrm e}^{3} \ln \relax (5)}{x^{3}}-\frac {5}{x}\)	\(15\)
gosper	\(\frac {{\mathrm e}^{3} \ln \relax (5)-5 x^{2}}{x^{3}}\)	\(16\)
norman	\(\frac {{\mathrm e}^{3} \ln \relax (5)-5 x^{2}}{x^{3}}\)	\(16\)
risch	\(\frac {{\mathrm e}^{3} \ln \relax (5)-5 x^{2}}{x^{3}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(3)*ln(5)/x^3-5/x

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maxima [A]  time = 0.34, size = 17, normalized size = 0.77 \begin {gather*} -\frac {5 \, x^{2} - e^{3} \log \relax (5)}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*x^2 - e^3*log(5))/x^3

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mupad [B]  time = 3.07, size = 15, normalized size = 0.68 \begin {gather*} \frac {{\mathrm {e}}^3\,\ln \relax (5)-5\,x^2}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*exp(3)*log(5) - 5*x^2)/x^4,x)

[Out]

(exp(3)*log(5) - 5*x^2)/x^3

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sympy [A]  time = 0.10, size = 14, normalized size = 0.64 \begin {gather*} \frac {- 5 x^{2} + e^{3} \log {\relax (5 )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-5*x**2 + exp(3)*log(5))/x**3

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