3.90.17 \(\int \frac {e^2 (2-3 x)+10 x+(-3 e^2 x-15 x^2) \log (\frac {15 x}{e^2+5 x})}{2 e^2 x+10 x^2} \, dx\)

Optimal. Leaf size=25 \[ \log (3 x)-\frac {3}{2} x \log \left (\frac {3 x}{\frac {e^2}{5}+x}\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 7, number of rules used = 5, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {1593, 6742, 72, 2486, 31} \begin {gather*} \log (x)-\frac {3}{2} x \log \left (\frac {15 x}{5 x+e^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(2 - 3*x) + 10*x + (-3*E^2*x - 15*x^2)*Log[(15*x)/(E^2 + 5*x)])/(2*E^2*x + 10*x^2),x]

[Out]

Log[x] - (3*x*Log[(15*x)/(E^2 + 5*x)])/2

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{x \left (2 e^2+10 x\right )} \, dx\\ &=\int \left (\frac {2 e^2+\left (10-3 e^2\right ) x}{2 x \left (e^2+5 x\right )}-\frac {3}{2} \log \left (\frac {15 x}{e^2+5 x}\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {2 e^2+\left (10-3 e^2\right ) x}{x \left (e^2+5 x\right )} \, dx-\frac {3}{2} \int \log \left (\frac {15 x}{e^2+5 x}\right ) \, dx\\ &=-\frac {3}{2} x \log \left (\frac {15 x}{e^2+5 x}\right )+\frac {1}{2} \int \left (\frac {2}{x}-\frac {3 e^2}{e^2+5 x}\right ) \, dx+\frac {1}{2} \left (3 e^2\right ) \int \frac {1}{e^2+5 x} \, dx\\ &=\log (x)-\frac {3}{2} x \log \left (\frac {15 x}{e^2+5 x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (2 \log (x)-3 x \log \left (\frac {15 x}{e^2+5 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(2 - 3*x) + 10*x + (-3*E^2*x - 15*x^2)*Log[(15*x)/(E^2 + 5*x)])/(2*E^2*x + 10*x^2),x]

[Out]

(2*Log[x] - 3*x*Log[(15*x)/(E^2 + 5*x)])/2

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fricas [A]  time = 0.60, size = 18, normalized size = 0.72 \begin {gather*} -\frac {3}{2} \, x \log \left (\frac {15 \, x}{5 \, x + e^{2}}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(2)*x-15*x^2)*log(15*x/(exp(2)+5*x))+(-3*x+2)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x, algorithm=
"fricas")

[Out]

-3/2*x*log(15*x/(5*x + e^2)) + log(x)

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giac [A]  time = 0.20, size = 18, normalized size = 0.72 \begin {gather*} -\frac {3}{2} \, x \log \left (\frac {15 \, x}{5 \, x + e^{2}}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(2)*x-15*x^2)*log(15*x/(exp(2)+5*x))+(-3*x+2)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x, algorithm=
"giac")

[Out]

-3/2*x*log(15*x/(5*x + e^2)) + log(x)

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maple [A]  time = 0.46, size = 19, normalized size = 0.76

method	result	size
norman	\(-\frac {3 x \ln \left (\frac {15 x}{{\mathrm e}^{2}+5 x}\right )}{2}+\ln \relax (x )\)	\(19\)
risch	\(-\frac {3 x \ln \left (\frac {15 x}{{\mathrm e}^{2}+5 x}\right )}{2}+\ln \relax (x )\)	\(19\)
derivativedivides	\(\frac {3 \,{\mathrm e}^{2} \left (\frac {10 \,{\mathrm e}^{-2} \ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )}{3}-\frac {10 \,{\mathrm e}^{-2} \ln \left (-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )}{3}-\frac {\ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) {\mathrm e}^{-2} \left ({\mathrm e}^{2}+5 x \right )}{3}\right )}{10}\)	\(87\)
default	\(\frac {3 \,{\mathrm e}^{2} \left (\frac {10 \,{\mathrm e}^{-2} \ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )}{3}-\frac {10 \,{\mathrm e}^{-2} \ln \left (-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )}{3}-\frac {\ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) {\mathrm e}^{-2} \left ({\mathrm e}^{2}+5 x \right )}{3}\right )}{10}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*exp(2)*x-15*x^2)*ln(15*x/(exp(2)+5*x))+(-3*x+2)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x,method=_RETURNVERB
OSE)

[Out]

-3/2*x*ln(15*x/(exp(2)+5*x))+ln(x)

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maxima [B]  time = 0.51, size = 68, normalized size = 2.72 \begin {gather*} -\frac {3}{2} \, x {\left (\log \relax (5) + \log \relax (3)\right )} - {\left (e^{\left (-2\right )} \log \left (5 \, x + e^{2}\right ) - e^{\left (-2\right )} \log \relax (x)\right )} e^{2} + \frac {3}{10} \, {\left (5 \, x + e^{2}\right )} \log \left (5 \, x + e^{2}\right ) - \frac {3}{10} \, e^{2} \log \left (5 \, x + e^{2}\right ) - \frac {3}{2} \, x \log \relax (x) + \log \left (5 \, x + e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(2)*x-15*x^2)*log(15*x/(exp(2)+5*x))+(-3*x+2)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x, algorithm=
"maxima")

[Out]

-3/2*x*(log(5) + log(3)) - (e^(-2)*log(5*x + e^2) - e^(-2)*log(x))*e^2 + 3/10*(5*x + e^2)*log(5*x + e^2) - 3/1
0*e^2*log(5*x + e^2) - 3/2*x*log(x) + log(5*x + e^2)

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mupad [B]  time = 6.91, size = 18, normalized size = 0.72 \begin {gather*} \ln \relax (x)-\frac {3\,x\,\ln \left (\frac {15\,x}{5\,x+{\mathrm {e}}^2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((15*x)/(5*x + exp(2)))*(3*x*exp(2) + 15*x^2) - 10*x + exp(2)*(3*x - 2))/(2*x*exp(2) + 10*x^2),x)

[Out]

log(x) - (3*x*log((15*x)/(5*x + exp(2))))/2

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sympy [A]  time = 0.16, size = 19, normalized size = 0.76 \begin {gather*} - \frac {3 x \log {\left (\frac {15 x}{5 x + e^{2}} \right )}}{2} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(2)*x-15*x**2)*ln(15*x/(exp(2)+5*x))+(-3*x+2)*exp(2)+10*x)/(2*exp(2)*x+10*x**2),x)

[Out]

-3*x*log(15*x/(5*x + exp(2)))/2 + log(x)

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