3.50.85 \(\int \frac {1200+e^x (-1200-25 x)+25 x-50 \log (\frac {48+x}{12})}{48+x} \, dx\)

Optimal. Leaf size=25 \[ 25 \left (1+e^3-e^x+x-\log ^2\left (4+\frac {x}{12}\right )\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6688, 2194, 2390, 12, 2301} \begin {gather*} -25 e^x+25 x-25 \log ^2\left (\frac {x}{12}+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1200 + E^x*(-1200 - 25*x) + 25*x - 50*Log[(48 + x)/12])/(48 + x),x]

[Out]

-25*E^x + 25*x - 25*Log[4 + x/12]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (25-25 e^x-\frac {50 \log \left (4+\frac {x}{12}\right )}{48+x}\right ) \, dx\\ &=25 x-25 \int e^x \, dx-50 \int \frac {\log \left (4+\frac {x}{12}\right )}{48+x} \, dx\\ &=-25 e^x+25 x-600 \operatorname {Subst}\left (\int \frac {\log (x)}{12 x} \, dx,x,4+\frac {x}{12}\right )\\ &=-25 e^x+25 x-50 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,4+\frac {x}{12}\right )\\ &=-25 e^x+25 x-25 \log ^2\left (4+\frac {x}{12}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.84 \begin {gather*} -25 e^x+25 x-25 \log ^2\left (4+\frac {x}{12}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1200 + E^x*(-1200 - 25*x) + 25*x - 50*Log[(48 + x)/12])/(48 + x),x]

[Out]

-25*E^x + 25*x - 25*Log[4 + x/12]^2

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fricas [A]  time = 1.05, size = 18, normalized size = 0.72 \begin {gather*} -25 \, \log \left (\frac {1}{12} \, x + 4\right )^{2} + 25 \, x - 25 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*log(1/12*x+4)+(-25*x-1200)*exp(x)+25*x+1200)/(x+48),x, algorithm="fricas")

[Out]

-25*log(1/12*x + 4)^2 + 25*x - 25*e^x

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giac [A]  time = 0.14, size = 18, normalized size = 0.72 \begin {gather*} -25 \, \log \left (\frac {1}{12} \, x + 4\right )^{2} + 25 \, x - 25 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*log(1/12*x+4)+(-25*x-1200)*exp(x)+25*x+1200)/(x+48),x, algorithm="giac")

[Out]

-25*log(1/12*x + 4)^2 + 25*x - 25*e^x

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maple [A]  time = 0.13, size = 19, normalized size = 0.76

method	result	size
default	\(25 x -25 \ln \left (\frac {x}{12}+4\right )^{2}-25 \,{\mathrm e}^{x}\)	\(19\)
norman	\(25 x -25 \ln \left (\frac {x}{12}+4\right )^{2}-25 \,{\mathrm e}^{x}\)	\(19\)
risch	\(25 x -25 \ln \left (\frac {x}{12}+4\right )^{2}-25 \,{\mathrm e}^{x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-50*ln(1/12*x+4)+(-25*x-1200)*exp(x)+25*x+1200)/(x+48),x,method=_RETURNVERBOSE)

[Out]

25*x-25*ln(1/12*x+4)^2-25*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 1200 \, e^{\left (-48\right )} E_{1}\left (-x - 48\right ) + 50 \, {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \left (x + 48\right ) - 25 \, \log \left (x + 48\right )^{2} + 25 \, x - \frac {25 \, x e^{x}}{x + 48} + 1200 \, \int \frac {e^{x}}{x^{2} + 96 \, x + 2304}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*log(1/12*x+4)+(-25*x-1200)*exp(x)+25*x+1200)/(x+48),x, algorithm="maxima")

[Out]

1200*e^(-48)*exp_integral_e(1, -x - 48) + 50*(log(3) + 2*log(2))*log(x + 48) - 25*log(x + 48)^2 + 25*x - 25*x*
e^x/(x + 48) + 1200*integrate(e^x/(x^2 + 96*x + 2304), x)

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mupad [B]  time = 0.22, size = 18, normalized size = 0.72 \begin {gather*} -25\,{\ln \left (\frac {x}{12}+4\right )}^2+25\,x-25\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x - 50*log(x/12 + 4) - exp(x)*(25*x + 1200) + 1200)/(x + 48),x)

[Out]

25*x - 25*exp(x) - 25*log(x/12 + 4)^2

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sympy [A]  time = 0.27, size = 17, normalized size = 0.68 \begin {gather*} 25 x - 25 e^{x} - 25 \log {\left (\frac {x}{12} + 4 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*ln(1/12*x+4)+(-25*x-1200)*exp(x)+25*x+1200)/(x+48),x)

[Out]

25*x - 25*exp(x) - 25*log(x/12 + 4)**2

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