3.53.77 \(\int \frac {18+11 e^x}{15+5 e^x} \, dx\)

Optimal. Leaf size=13 \[ 23+\frac {6 x}{5}+\log \left (3+e^x\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 12, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2282, 72} \begin {gather*} \frac {6 x}{5}+\log \left (e^x+3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 + 11*E^x)/(15 + 5*E^x),x]

[Out]

(6*x)/5 + Log[3 + E^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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {18+11 x}{x (15+5 x)} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {6}{5 x}+\frac {1}{3+x}\right ) \, dx,x,e^x\right )\\ &=\frac {6 x}{5}+\log \left (3+e^x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.92 \begin {gather*} \frac {6 x}{5}+\log \left (3+e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 + 11*E^x)/(15 + 5*E^x),x]

[Out]

(6*x)/5 + Log[3 + E^x]

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fricas [A]  time = 0.63, size = 9, normalized size = 0.69 \begin {gather*} \frac {6}{5} \, x + \log \left (e^{x} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((11*exp(x)+18)/(5*exp(x)+15),x, algorithm="fricas")

[Out]

6/5*x + log(e^x + 3)

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giac [A]  time = 0.19, size = 9, normalized size = 0.69 \begin {gather*} \frac {6}{5} \, x + \log \left (e^{x} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((11*exp(x)+18)/(5*exp(x)+15),x, algorithm="giac")

[Out]

6/5*x + log(e^x + 3)

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maple [A]  time = 0.03, size = 10, normalized size = 0.77

method	result	size
risch	\(\frac {6 x}{5}+\ln \left (3+{\mathrm e}^{x}\right )\)	\(10\)
derivativedivides	\(\ln \left (3+{\mathrm e}^{x}\right )+\frac {6 \ln \left ({\mathrm e}^{x}\right )}{5}\)	\(12\)
default	\(\ln \left (3+{\mathrm e}^{x}\right )+\frac {6 \ln \left ({\mathrm e}^{x}\right )}{5}\)	\(12\)
norman	\(\frac {6 x}{5}+\ln \left (5 \,{\mathrm e}^{x}+15\right )\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((11*exp(x)+18)/(5*exp(x)+15),x,method=_RETURNVERBOSE)

[Out]

6/5*x+ln(3+exp(x))

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maxima [A]  time = 0.37, size = 9, normalized size = 0.69 \begin {gather*} \frac {6}{5} \, x + \log \left (e^{x} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((11*exp(x)+18)/(5*exp(x)+15),x, algorithm="maxima")

[Out]

6/5*x + log(e^x + 3)

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mupad [B]  time = 0.04, size = 9, normalized size = 0.69 \begin {gather*} \frac {6\,x}{5}+\ln \left ({\mathrm {e}}^x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((11*exp(x) + 18)/(5*exp(x) + 15),x)

[Out]

(6*x)/5 + log(exp(x) + 3)

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sympy [A]  time = 0.07, size = 10, normalized size = 0.77 \begin {gather*} \frac {6 x}{5} + \log {\left (e^{x} + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((11*exp(x)+18)/(5*exp(x)+15),x)

[Out]

6*x/5 + log(exp(x) + 3)

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