3.104.24 \(\int \frac {9 e^7-x^2+e^2 x^2+x^2 \log (5)}{x^2 \log (5)} \, dx\)

Optimal. Leaf size=28 \[ x+\frac {(9+x) \left (-x+e^2 \left (-e^5+x\right )\right )}{x \log (5)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6, 12, 14} \begin {gather*} -\frac {x \left (1-e^2-\log (5)\right )}{\log (5)}-\frac {9 e^7}{x \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*E^7 - x^2 + E^2*x^2 + x^2*Log[5])/(x^2*Log[5]),x]

[Out]

(-9*E^7)/(x*Log[5]) - (x*(1 - E^2 - Log[5]))/Log[5]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 e^7+\left (-1+e^2\right ) x^2+x^2 \log (5)}{x^2 \log (5)} \, dx\\ &=\int \frac {9 e^7+x^2 \left (-1+e^2+\log (5)\right )}{x^2 \log (5)} \, dx\\ &=\frac {\int \frac {9 e^7+x^2 \left (-1+e^2+\log (5)\right )}{x^2} \, dx}{\log (5)}\\ &=\frac {\int \left (-1+e^2+\frac {9 e^7}{x^2}+\log (5)\right ) \, dx}{\log (5)}\\ &=-\frac {9 e^7}{x \log (5)}-\frac {x \left (1-e^2-\log (5)\right )}{\log (5)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 23, normalized size = 0.82 \begin {gather*} \frac {-\frac {9 e^7}{x}+x \left (-1+e^2+\log (5)\right )}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*E^7 - x^2 + E^2*x^2 + x^2*Log[5])/(x^2*Log[5]),x]

[Out]

((-9*E^7)/x + x*(-1 + E^2 + Log[5]))/Log[5]

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 30, normalized size = 1.07 \begin {gather*} \frac {x^{2} e^{2} + x^{2} \log \relax (5) - x^{2} - 9 \, e^{7}}{x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(5)+9*exp(1)^2*exp(5)+x^2*exp(1)^2-x^2)/x^2/log(5),x, algorithm="fricas")

[Out]

(x^2*e^2 + x^2*log(5) - x^2 - 9*e^7)/(x*log(5))

________________________________________________________________________________________

giac [A]  time = 0.27, size = 24, normalized size = 0.86 \begin {gather*} \frac {x e^{2} + x \log \relax (5) - x - \frac {9 \, e^{7}}{x}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(5)+9*exp(1)^2*exp(5)+x^2*exp(1)^2-x^2)/x^2/log(5),x, algorithm="giac")

[Out]

(x*e^2 + x*log(5) - x - 9*e^7/x)/log(5)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 25, normalized size = 0.89

method	result	size
default	\(\frac {{\mathrm e}^{2} x +x \ln \relax (5)-x -\frac {9 \,{\mathrm e}^{7}}{x}}{\ln \relax (5)}\)	\(25\)
risch	\(\frac {x \,{\mathrm e}^{2}}{\ln \relax (5)}+x -\frac {x}{\ln \relax (5)}-\frac {9 \,{\mathrm e}^{7}}{\ln \relax (5) x}\)	\(29\)
norman	\(\frac {\frac {\left ({\mathrm e}^{2}+\ln \relax (5)-1\right ) x^{2}}{\ln \relax (5)}-\frac {9 \,{\mathrm e}^{2} {\mathrm e}^{5}}{\ln \relax (5)}}{x}\)	\(34\)
gosper	\(-\frac {-x^{2} {\mathrm e}^{2}+9 \,{\mathrm e}^{2} {\mathrm e}^{5}-x^{2} \ln \relax (5)+x^{2}}{x \ln \relax (5)}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(5)+9*exp(1)^2*exp(5)+x^2*exp(1)^2-x^2)/x^2/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/ln(5)*(exp(2)*x+x*ln(5)-x-9*exp(7)/x)

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 21, normalized size = 0.75 \begin {gather*} \frac {x {\left (e^{2} + \log \relax (5) - 1\right )} - \frac {9 \, e^{7}}{x}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(5)+9*exp(1)^2*exp(5)+x^2*exp(1)^2-x^2)/x^2/log(5),x, algorithm="maxima")

[Out]

(x*(e^2 + log(5) - 1) - 9*e^7/x)/log(5)

________________________________________________________________________________________

mupad [B]  time = 0.09, size = 24, normalized size = 0.86 \begin {gather*} \frac {x\,\left ({\mathrm {e}}^2+\ln \relax (5)-1\right )}{\ln \relax (5)}-\frac {9\,{\mathrm {e}}^7}{x\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*exp(7) + x^2*exp(2) + x^2*log(5) - x^2)/(x^2*log(5)),x)

[Out]

(x*(exp(2) + log(5) - 1))/log(5) - (9*exp(7))/(x*log(5))

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 19, normalized size = 0.68 \begin {gather*} \frac {x \left (-1 + \log {\relax (5 )} + e^{2}\right ) - \frac {9 e^{7}}{x}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(5)+9*exp(1)**2*exp(5)+x**2*exp(1)**2-x**2)/x**2/ln(5),x)

[Out]

(x*(-1 + log(5) + exp(2)) - 9*exp(7)/x)/log(5)

________________________________________________________________________________________