3.21.87 \(\int \frac {5+e^{5/4} (-240+15 x^2)}{e^{17/4} x^2 \log (5)} \, dx\)

Optimal. Leaf size=31 \[ -2+\frac {5 \left (-\frac {1}{e^{5/4}}+3 (4-x)^2\right )}{e^3 x \log (5)} \]

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Rubi [A]  time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 14} \begin {gather*} \frac {15 x}{e^3 \log (5)}-\frac {5 \left (1-48 e^{5/4}\right )}{e^{17/4} x \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E^(5/4)*(-240 + 15*x^2))/(E^(17/4)*x^2*Log[5]),x]

[Out]

(-5*(1 - 48*E^(5/4)))/(E^(17/4)*x*Log[5]) + (15*x)/(E^3*Log[5])

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {5+e^{5/4} \left (-240+15 x^2\right )}{x^2} \, dx}{e^{17/4} \log (5)}\\ &=\frac {\int \left (15 e^{5/4}-\frac {5 \left (-1+48 e^{5/4}\right )}{x^2}\right ) \, dx}{e^{17/4} \log (5)}\\ &=-\frac {5 \left (1-48 e^{5/4}\right )}{e^{17/4} x \log (5)}+\frac {15 x}{e^3 \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.87 \begin {gather*} \frac {-5+15 e^{5/4} \left (16+x^2\right )}{e^{17/4} x \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E^(5/4)*(-240 + 15*x^2))/(E^(17/4)*x^2*Log[5]),x]

[Out]

(-5 + 15*E^(5/4)*(16 + x^2))/(E^(17/4)*x*Log[5])

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fricas [A]  time = 0.59, size = 22, normalized size = 0.71 \begin {gather*} \frac {5 \, {\left (3 \, {\left (x^{2} + 16\right )} e^{\frac {5}{4}} - 1\right )} e^{\left (-\frac {17}{4}\right )}}{x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2-240)*exp(5/4)+5)/x^2/exp(5/4)/exp(3)/log(5),x, algorithm="fricas")

[Out]

5*(3*(x^2 + 16)*e^(5/4) - 1)*e^(-17/4)/(x*log(5))

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giac [A]  time = 0.15, size = 24, normalized size = 0.77 \begin {gather*} \frac {5 \, {\left (3 \, x e^{\frac {5}{4}} + \frac {48 \, e^{\frac {5}{4}} - 1}{x}\right )} e^{\left (-\frac {17}{4}\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2-240)*exp(5/4)+5)/x^2/exp(5/4)/exp(3)/log(5),x, algorithm="giac")

[Out]

5*(3*x*e^(5/4) + (48*e^(5/4) - 1)/x)*e^(-17/4)/log(5)

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maple [A]  time = 0.05, size = 31, normalized size = 1.00

method	result	size
gosper	\(\frac {5 \left (3 x^{2} {\mathrm e}^{\frac {5}{4}}-1+48 \,{\mathrm e}^{\frac {5}{4}}\right ) {\mathrm e}^{-3} {\mathrm e}^{-\frac {5}{4}}}{\ln \relax (5) x}\)	\(31\)
default	\(\frac {{\mathrm e}^{-\frac {5}{4}} {\mathrm e}^{-3} \left (15 x \,{\mathrm e}^{\frac {5}{4}}-\frac {5 \left (-48 \,{\mathrm e}^{\frac {5}{4}}+1\right )}{x}\right )}{\ln \relax (5)}\)	\(31\)
risch	\(\frac {15 \,{\mathrm e}^{-3} x}{\ln \relax (5)}+\frac {240 \,{\mathrm e}^{-\frac {17}{4}} {\mathrm e}^{\frac {5}{4}}}{\ln \relax (5) x}-\frac {5 \,{\mathrm e}^{-\frac {17}{4}}}{\ln \relax (5) x}\)	\(35\)
norman	\(\frac {\frac {15 \,{\mathrm e}^{-3} x^{2}}{\ln \relax (5)}+\frac {5 \left (48 \,{\mathrm e}^{\frac {5}{4}}-1\right ) {\mathrm e}^{-\frac {5}{4}} {\mathrm e}^{-3}}{\ln \relax (5)}}{x}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x^2-240)*exp(5/4)+5)/x^2/exp(5/4)/exp(3)/ln(5),x,method=_RETURNVERBOSE)

[Out]

5*(3*x^2*exp(5/4)-1+48*exp(5/4))/ln(5)/exp(3)/exp(5/4)/x

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maxima [A]  time = 0.54, size = 24, normalized size = 0.77 \begin {gather*} \frac {5 \, {\left (3 \, x e^{\frac {5}{4}} + \frac {48 \, e^{\frac {5}{4}} - 1}{x}\right )} e^{\left (-\frac {17}{4}\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2-240)*exp(5/4)+5)/x^2/exp(5/4)/exp(3)/log(5),x, algorithm="maxima")

[Out]

5*(3*x*e^(5/4) + (48*e^(5/4) - 1)/x)*e^(-17/4)/log(5)

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mupad [B]  time = 1.17, size = 24, normalized size = 0.77 \begin {gather*} \frac {5\,{\mathrm {e}}^{-\frac {17}{4}}\,\left (3\,{\mathrm {e}}^{5/4}\,x^2+48\,{\mathrm {e}}^{5/4}-1\right )}{x\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-17/4)*(exp(5/4)*(15*x^2 - 240) + 5))/(x^2*log(5)),x)

[Out]

(5*exp(-17/4)*(48*exp(5/4) + 3*x^2*exp(5/4) - 1))/(x*log(5))

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sympy [A]  time = 0.10, size = 26, normalized size = 0.84 \begin {gather*} \frac {15 x e^{\frac {5}{4}} + \frac {-5 + 240 e^{\frac {5}{4}}}{x}}{e^{\frac {17}{4}} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x**2-240)*exp(5/4)+5)/x**2/exp(5/4)/exp(3)/ln(5),x)

[Out]

(15*x*exp(5/4) + (-5 + 240*exp(5/4))/x)*exp(-17/4)/log(5)

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