3.44.100 \(\int \frac {-100-10 x^3+x^5+2 x^6-400 e^{50 x^2} x^6}{x^5 \log (3)} \, dx\)

Optimal. Leaf size=27 \[ \frac {-4 e^{50 x^2}+\left (-\frac {5}{x^2}-x\right )^2+x}{\log (3)} \]

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Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.70, number of steps used = 6, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 14, 2209} \begin {gather*} \frac {25}{x^4 \log (3)}+\frac {x^2}{\log (3)}-\frac {4 e^{50 x^2}}{\log (3)}+\frac {x}{\log (3)}+\frac {10}{x \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-100 - 10*x^3 + x^5 + 2*x^6 - 400*E^(50*x^2)*x^6)/(x^5*Log[3]),x]

[Out]

(-4*E^(50*x^2))/Log[3] + 25/(x^4*Log[3]) + 10/(x*Log[3]) + x/Log[3] + x^2/Log[3]

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-100-10 x^3+x^5+2 x^6-400 e^{50 x^2} x^6}{x^5} \, dx}{\log (3)}\\ &=\frac {\int \left (-400 e^{50 x^2} x+\frac {-100-10 x^3+x^5+2 x^6}{x^5}\right ) \, dx}{\log (3)}\\ &=\frac {\int \frac {-100-10 x^3+x^5+2 x^6}{x^5} \, dx}{\log (3)}-\frac {400 \int e^{50 x^2} x \, dx}{\log (3)}\\ &=-\frac {4 e^{50 x^2}}{\log (3)}+\frac {\int \left (1-\frac {100}{x^5}-\frac {10}{x^2}+2 x\right ) \, dx}{\log (3)}\\ &=-\frac {4 e^{50 x^2}}{\log (3)}+\frac {25}{x^4 \log (3)}+\frac {10}{x \log (3)}+\frac {x}{\log (3)}+\frac {x^2}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.07 \begin {gather*} \frac {-4 e^{50 x^2}+\frac {25}{x^4}+\frac {10}{x}+x+x^2}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100 - 10*x^3 + x^5 + 2*x^6 - 400*E^(50*x^2)*x^6)/(x^5*Log[3]),x]

[Out]

(-4*E^(50*x^2) + 25/x^4 + 10/x + x + x^2)/Log[3]

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fricas [A]  time = 0.67, size = 32, normalized size = 1.19 \begin {gather*} \frac {x^{6} + x^{5} - 4 \, x^{4} e^{\left (50 \, x^{2}\right )} + 10 \, x^{3} + 25}{x^{4} \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x^6*exp(25*x^2)^2+2*x^6+x^5-10*x^3-100)/x^5/log(3),x, algorithm="fricas")

[Out]

(x^6 + x^5 - 4*x^4*e^(50*x^2) + 10*x^3 + 25)/(x^4*log(3))

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giac [A]  time = 0.13, size = 32, normalized size = 1.19 \begin {gather*} \frac {x^{6} + x^{5} - 4 \, x^{4} e^{\left (50 \, x^{2}\right )} + 10 \, x^{3} + 25}{x^{4} \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x^6*exp(25*x^2)^2+2*x^6+x^5-10*x^3-100)/x^5/log(3),x, algorithm="giac")

[Out]

(x^6 + x^5 - 4*x^4*e^(50*x^2) + 10*x^3 + 25)/(x^4*log(3))

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maple [A]  time = 0.06, size = 29, normalized size = 1.07

method	result	size
default	\(\frac {x^{2}+x +\frac {25}{x^{4}}+\frac {10}{x}-4 \,{\mathrm e}^{50 x^{2}}}{\ln \relax (3)}\)	\(29\)
risch	\(\frac {x^{2}}{\ln \relax (3)}+\frac {x}{\ln \relax (3)}+\frac {10 x^{3}+25}{\ln \relax (3) x^{4}}-\frac {4 \,{\mathrm e}^{50 x^{2}}}{\ln \relax (3)}\)	\(43\)
norman	\(\frac {\frac {x^{5}}{\ln \relax (3)}+\frac {x^{6}}{\ln \relax (3)}+\frac {25}{\ln \relax (3)}+\frac {10 x^{3}}{\ln \relax (3)}-\frac {4 x^{4} {\mathrm e}^{50 x^{2}}}{\ln \relax (3)}}{x^{4}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-400*x^6*exp(25*x^2)^2+2*x^6+x^5-10*x^3-100)/x^5/ln(3),x,method=_RETURNVERBOSE)

[Out]

1/ln(3)*(x^2+x+25/x^4+10/x-4*exp(x^2)^50)

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maxima [A]  time = 0.35, size = 28, normalized size = 1.04 \begin {gather*} \frac {x^{2} + x + \frac {10}{x} + \frac {25}{x^{4}} - 4 \, e^{\left (50 \, x^{2}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x^6*exp(25*x^2)^2+2*x^6+x^5-10*x^3-100)/x^5/log(3),x, algorithm="maxima")

[Out]

(x^2 + x + 10/x + 25/x^4 - 4*e^(50*x^2))/log(3)

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mupad [B]  time = 0.13, size = 32, normalized size = 1.19 \begin {gather*} \frac {10\,x^3-4\,x^4\,{\mathrm {e}}^{50\,x^2}+x^5+x^6+25}{x^4\,\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(400*x^6*exp(50*x^2) + 10*x^3 - x^5 - 2*x^6 + 100)/(x^5*log(3)),x)

[Out]

(10*x^3 - 4*x^4*exp(50*x^2) + x^5 + x^6 + 25)/(x^4*log(3))

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sympy [A]  time = 0.13, size = 29, normalized size = 1.07 \begin {gather*} \frac {x^{2} + x + \frac {10 x^{3} + 25}{x^{4}}}{\log {\relax (3 )}} - \frac {4 e^{50 x^{2}}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x**6*exp(25*x**2)**2+2*x**6+x**5-10*x**3-100)/x**5/ln(3),x)

[Out]

(x**2 + x + (10*x**3 + 25)/x**4)/log(3) - 4*exp(50*x**2)/log(3)

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