3.11.34 \(\int \frac {12 e^x+15 e^{6 x}+(3 e^{6 x}+e^x (48+12 x)) \log (\frac {1}{4} (16+e^{5 x}+4 x))}{16+e^{5 x}+4 x} \, dx\)

Optimal. Leaf size=18 \[ 3 e^x \log \left (4+\frac {e^{5 x}}{4}+x\right ) \]

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Rubi [A]  time = 1.00, antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6742, 2194, 2554} \begin {gather*} 3 e^x \log \left (\frac {1}{4} \left (4 x+e^{5 x}+16\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*E^x + 15*E^(6*x) + (3*E^(6*x) + E^x*(48 + 12*x))*Log[(16 + E^(5*x) + 4*x)/4])/(16 + E^(5*x) + 4*x),x]

[Out]

3*E^x*Log[(16 + E^(5*x) + 4*x)/4]

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, 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 {12 e^x (19+5 x)}{16+e^{5 x}+4 x}+3 e^x \left (5+\log \left (4+\frac {e^{5 x}}{4}+x\right )\right )\right ) \, dx\\ &=3 \int e^x \left (5+\log \left (4+\frac {e^{5 x}}{4}+x\right )\right ) \, dx-12 \int \frac {e^x (19+5 x)}{16+e^{5 x}+4 x} \, dx\\ &=3 \int \left (5 e^x+e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )\right ) \, dx-12 \int \left (\frac {19 e^x}{16+e^{5 x}+4 x}+\frac {5 e^x x}{16+e^{5 x}+4 x}\right ) \, dx\\ &=3 \int e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right ) \, dx+15 \int e^x \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx\\ &=15 e^x+3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )-3 \int \frac {e^x \left (4+5 e^{5 x}\right )}{e^{5 x}+4 (4+x)} \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx\\ &=15 e^x+3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )-3 \int \left (5 e^x-\frac {4 e^x (19+5 x)}{16+e^{5 x}+4 x}\right ) \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx\\ &=15 e^x+3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )+12 \int \frac {e^x (19+5 x)}{16+e^{5 x}+4 x} \, dx-15 \int e^x \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx\\ &=3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )+12 \int \left (\frac {19 e^x}{16+e^{5 x}+4 x}+\frac {5 e^x x}{16+e^{5 x}+4 x}\right ) \, dx-60 \int \frac {e^x x}{16+e^{5 x}+4 x} \, dx-228 \int \frac {e^x}{16+e^{5 x}+4 x} \, dx\\ &=3 e^x \log \left (\frac {1}{4} \left (16+e^{5 x}+4 x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 1.00 \begin {gather*} 3 e^x \log \left (4+\frac {e^{5 x}}{4}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*E^x + 15*E^(6*x) + (3*E^(6*x) + E^x*(48 + 12*x))*Log[(16 + E^(5*x) + 4*x)/4])/(16 + E^(5*x) + 4*
x),x]

[Out]

3*E^x*Log[4 + E^(5*x)/4 + x]

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fricas [A]  time = 0.63, size = 14, normalized size = 0.78 \begin {gather*} 3 \, e^{x} \log \left (x + \frac {1}{4} \, e^{\left (5 \, x\right )} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*exp(5*x)+(12*x+48)*exp(x))*log(1/4*exp(5*x)+x+4)+15*exp(x)*exp(5*x)+12*exp(x))/(exp(5*x)+
4*x+16),x, algorithm="fricas")

[Out]

3*e^x*log(x + 1/4*e^(5*x) + 4)

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giac [A]  time = 0.55, size = 21, normalized size = 1.17 \begin {gather*} -6 \, e^{x} \log \relax (2) + 3 \, e^{x} \log \left (4 \, x + e^{\left (5 \, x\right )} + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*exp(5*x)+(12*x+48)*exp(x))*log(1/4*exp(5*x)+x+4)+15*exp(x)*exp(5*x)+12*exp(x))/(exp(5*x)+
4*x+16),x, algorithm="giac")

[Out]

-6*e^x*log(2) + 3*e^x*log(4*x + e^(5*x) + 16)

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maple [A]  time = 0.05, size = 15, normalized size = 0.83

method	result	size
risch	\(3 \ln \left (\frac {{\mathrm e}^{5 x}}{4}+x +4\right ) {\mathrm e}^{x}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*exp(x)*exp(5*x)+(12*x+48)*exp(x))*ln(1/4*exp(5*x)+x+4)+15*exp(x)*exp(5*x)+12*exp(x))/(exp(5*x)+4*x+16)
,x,method=_RETURNVERBOSE)

[Out]

3*ln(1/4*exp(5*x)+x+4)*exp(x)

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maxima [A]  time = 0.90, size = 21, normalized size = 1.17 \begin {gather*} -6 \, e^{x} \log \relax (2) + 3 \, e^{x} \log \left (4 \, x + e^{\left (5 \, x\right )} + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*exp(5*x)+(12*x+48)*exp(x))*log(1/4*exp(5*x)+x+4)+15*exp(x)*exp(5*x)+12*exp(x))/(exp(5*x)+
4*x+16),x, algorithm="maxima")

[Out]

-6*e^x*log(2) + 3*e^x*log(4*x + e^(5*x) + 16)

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mupad [B]  time = 0.80, size = 14, normalized size = 0.78 \begin {gather*} 3\,{\mathrm {e}}^x\,\ln \left (x+\frac {{\mathrm {e}}^{5\,x}}{4}+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*exp(6*x) + 12*exp(x) + log(x + exp(5*x)/4 + 4)*(3*exp(6*x) + exp(x)*(12*x + 48)))/(4*x + exp(5*x) + 16
),x)

[Out]

3*exp(x)*log(x + exp(5*x)/4 + 4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*exp(5*x)+(12*x+48)*exp(x))*ln(1/4*exp(5*x)+x+4)+15*exp(x)*exp(5*x)+12*exp(x))/(exp(5*x)+4
*x+16),x)

[Out]

Timed out

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