3.4.67 \(\int \frac {80-40 x+(480 e^{25}-40 x) \log (12 e^{25}-x)}{12 e^{25}-x} \, dx\)

Optimal. Leaf size=15 \[ 40 (-2+x) \log \left (12 e^{25}-x\right ) \]

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Rubi [B]  time = 0.09, antiderivative size = 41, normalized size of antiderivative = 2.73, number of steps used = 6, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6742, 43, 2389, 2295} \begin {gather*} -40 \left (12 e^{25}-x\right ) \log \left (12 e^{25}-x\right )-80 \left (1-6 e^{25}\right ) \log \left (12 e^{25}-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(80 - 40*x + (480*E^25 - 40*x)*Log[12*E^25 - x])/(12*E^25 - x),x]

[Out]

-80*(1 - 6*E^25)*Log[12*E^25 - x] - 40*(12*E^25 - x)*Log[12*E^25 - x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

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 \left (-\frac {40 (-2+x)}{12 e^{25}-x}+40 \log \left (12 e^{25}-x\right )\right ) \, dx\\ &=-\left (40 \int \frac {-2+x}{12 e^{25}-x} \, dx\right )+40 \int \log \left (12 e^{25}-x\right ) \, dx\\ &=-\left (40 \int \left (-1+\frac {2 \left (-1+6 e^{25}\right )}{12 e^{25}-x}\right ) \, dx\right )-40 \operatorname {Subst}\left (\int \log (x) \, dx,x,12 e^{25}-x\right )\\ &=-80 \left (1-6 e^{25}\right ) \log \left (12 e^{25}-x\right )-40 \left (12 e^{25}-x\right ) \log \left (12 e^{25}-x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 27, normalized size = 1.80 \begin {gather*} 40 \left (-2 \log \left (12 e^{25}-x\right )+x \log \left (12 e^{25}-x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(80 - 40*x + (480*E^25 - 40*x)*Log[12*E^25 - x])/(12*E^25 - x),x]

[Out]

40*(-2*Log[12*E^25 - x] + x*Log[12*E^25 - x])

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fricas [A]  time = 0.64, size = 14, normalized size = 0.93 \begin {gather*} 40 \, {\left (x - 2\right )} \log \left (-x + 12 \, e^{25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((480*exp(25)-40*x)*log(12*exp(25)-x)-40*x+80)/(12*exp(25)-x),x, algorithm="fricas")

[Out]

40*(x - 2)*log(-x + 12*e^25)

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giac [A]  time = 0.44, size = 22, normalized size = 1.47 \begin {gather*} 40 \, x \log \left (-x + 12 \, e^{25}\right ) - 80 \, \log \left (x - 12 \, e^{25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((480*exp(25)-40*x)*log(12*exp(25)-x)-40*x+80)/(12*exp(25)-x),x, algorithm="giac")

[Out]

40*x*log(-x + 12*e^25) - 80*log(x - 12*e^25)

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maple [A]  time = 0.26, size = 23, normalized size = 1.53

method	result	size
risch	\(40 \ln \left (12 \,{\mathrm e}^{25}-x \right ) x -80 \ln \left (x -12 \,{\mathrm e}^{25}\right )\)	\(23\)
norman	\(-80 \ln \left (12 \,{\mathrm e}^{25}-x \right )+40 \ln \left (12 \,{\mathrm e}^{25}-x \right ) x\)	\(25\)
derivativedivides	\(-40 \left (12 \,{\mathrm e}^{25}-x \right ) \ln \left (12 \,{\mathrm e}^{25}-x \right )+480 \ln \left (12 \,{\mathrm e}^{25}-x \right ) {\mathrm e}^{25}-80 \ln \left (12 \,{\mathrm e}^{25}-x \right )\)	\(45\)
default	\(-40 \left (12 \,{\mathrm e}^{25}-x \right ) \ln \left (12 \,{\mathrm e}^{25}-x \right )+480 \ln \left (12 \,{\mathrm e}^{25}-x \right ) {\mathrm e}^{25}-80 \ln \left (12 \,{\mathrm e}^{25}-x \right )\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((480*exp(25)-40*x)*ln(12*exp(25)-x)-40*x+80)/(12*exp(25)-x),x,method=_RETURNVERBOSE)

[Out]

40*ln(12*exp(25)-x)*x-80*ln(x-12*exp(25))

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maxima [B]  time = 0.66, size = 103, normalized size = 6.87 \begin {gather*} -240 \, e^{25} \log \left (x - 12 \, e^{25}\right )^{2} - 480 \, e^{25} \log \left (x - 12 \, e^{25}\right ) \log \left (-x + 12 \, e^{25}\right ) + 240 \, {\left (2 \, \log \left (x - 12 \, e^{25}\right ) \log \left (-x + 12 \, e^{25}\right ) - \log \left (-x + 12 \, e^{25}\right )^{2}\right )} e^{25} + 40 \, {\left (12 \, e^{25} \log \left (x - 12 \, e^{25}\right ) + x\right )} \log \left (-x + 12 \, e^{25}\right ) - 80 \, \log \left (x - 12 \, e^{25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((480*exp(25)-40*x)*log(12*exp(25)-x)-40*x+80)/(12*exp(25)-x),x, algorithm="maxima")

[Out]

-240*e^25*log(x - 12*e^25)^2 - 480*e^25*log(x - 12*e^25)*log(-x + 12*e^25) + 240*(2*log(x - 12*e^25)*log(-x +
12*e^25) - log(-x + 12*e^25)^2)*e^25 + 40*(12*e^25*log(x - 12*e^25) + x)*log(-x + 12*e^25) - 80*log(x - 12*e^2
5)

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mupad [B]  time = 0.73, size = 14, normalized size = 0.93 \begin {gather*} 40\,\ln \left (12\,{\mathrm {e}}^{25}-x\right )\,\left (x-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((40*x + log(12*exp(25) - x)*(40*x - 480*exp(25)) - 80)/(x - 12*exp(25)),x)

[Out]

40*log(12*exp(25) - x)*(x - 2)

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sympy [A]  time = 0.12, size = 20, normalized size = 1.33 \begin {gather*} 40 x \log {\left (- x + 12 e^{25} \right )} - 80 \log {\left (x - 12 e^{25} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((480*exp(25)-40*x)*ln(12*exp(25)-x)-40*x+80)/(12*exp(25)-x),x)

[Out]

40*x*log(-x + 12*exp(25)) - 80*log(x - 12*exp(25))

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