3.7.81 \(\int \frac {-45-15 e x+10 e \log (3+e x)}{3+e x} \, dx\)

Optimal. Leaf size=15 \[ 5 \left (-20-3 x+\log ^2(3+e x)\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6741, 12, 6742, 2390, 2301} \begin {gather*} 5 \log ^2(e x+3)-15 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 - 15*E*x + 10*E*Log[3 + E*x])/(3 + E*x),x]

[Out]

-15*x + 5*Log[3 + E*x]^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 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 6741

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

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 {5 (-9-3 e x+2 e \log (3+e x))}{3+e x} \, dx\\ &=5 \int \frac {-9-3 e x+2 e \log (3+e x)}{3+e x} \, dx\\ &=5 \int \left (-3+\frac {2 e \log (3+e x)}{3+e x}\right ) \, dx\\ &=-15 x+(10 e) \int \frac {\log (3+e x)}{3+e x} \, dx\\ &=-15 x+10 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,3+e x\right )\\ &=-15 x+5 \log ^2(3+e x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.07 \begin {gather*} -5 \left (3 x-\log ^2(3+e x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 - 15*E*x + 10*E*Log[3 + E*x])/(3 + E*x),x]

[Out]

-5*(3*x - Log[3 + E*x]^2)

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fricas [A]  time = 0.47, size = 15, normalized size = 1.00 \begin {gather*} 5 \, \log \left (x e + 3\right )^{2} - 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(1)*log(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x, algorithm="fricas")

[Out]

5*log(x*e + 3)^2 - 15*x

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giac [A]  time = 0.44, size = 15, normalized size = 1.00 \begin {gather*} 5 \, \log \left (x e + 3\right )^{2} - 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(1)*log(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x, algorithm="giac")

[Out]

5*log(x*e + 3)^2 - 15*x

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maple [A]  time = 0.16, size = 16, normalized size = 1.07

method	result	size
norman	\(-15 x +5 \ln \left (x \,{\mathrm e}+3\right )^{2}\)	\(16\)
risch	\(-15 x +5 \ln \left (x \,{\mathrm e}+3\right )^{2}\)	\(16\)
derivativedivides	\(5 \,{\mathrm e}^{-1} \left ({\mathrm e} \ln \left (x \,{\mathrm e}+3\right )^{2}-3 x \,{\mathrm e}-9\right )\)	\(26\)
default	\(5 \,{\mathrm e}^{-1} \left ({\mathrm e} \ln \left (x \,{\mathrm e}+3\right )^{2}-3 x \,{\mathrm e}-9\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(1)*ln(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x,method=_RETURNVERBOSE)

[Out]

-15*x+5*ln(x*exp(1)+3)^2

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maxima [B]  time = 0.48, size = 43, normalized size = 2.87 \begin {gather*} -15 \, {\left (x e^{\left (-1\right )} - 3 \, e^{\left (-2\right )} \log \left (x e + 3\right )\right )} e - 45 \, e^{\left (-1\right )} \log \left (x e + 3\right ) + 5 \, \log \left (x e + 3\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(1)*log(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x, algorithm="maxima")

[Out]

-15*(x*e^(-1) - 3*e^(-2)*log(x*e + 3))*e - 45*e^(-1)*log(x*e + 3) + 5*log(x*e + 3)^2

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mupad [B]  time = 0.25, size = 15, normalized size = 1.00 \begin {gather*} 5\,{\ln \left (x\,\mathrm {e}+3\right )}^2-15\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x*exp(1) - 10*log(x*exp(1) + 3)*exp(1) + 45)/(x*exp(1) + 3),x)

[Out]

5*log(x*exp(1) + 3)^2 - 15*x

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sympy [A]  time = 0.11, size = 14, normalized size = 0.93 \begin {gather*} - 15 x + 5 \log {\left (e x + 3 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(1)*ln(x*exp(1)+3)-15*x*exp(1)-45)/(x*exp(1)+3),x)

[Out]

-15*x + 5*log(E*x + 3)**2

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