\(\int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx\) [6404]

Optimal result

Integrand size = 51, antiderivative size = 18 \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=4+\log \left (2+e^{4+\frac {16}{5} \log (4) \log (x)}+x\right ) \]

[Out]

4+ln(2+exp(16/5*ln(2)*ln(x)+2)^2+x)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6873, 12, 6816} \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=\log \left (x+e^4 2^{\frac {32 \log (x)}{5}}+2\right ) \]

[In]

Int[(5*x + 16*E^((2*(10 + 8*Log[4]*Log[x]))/5)*Log[4])/(10*x + 5*E^((2*(10 + 8*Log[4]*Log[x]))/5)*x + 5*x^2),x
]

[Out]

Log[2 + 2^((32*Log[x])/5)*E^4 + x]

Rule 12

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

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{5 x \left (2+e^{4+\frac {16}{5} \log (4) \log (x)}+x\right )} \, dx \\ & = \frac {1}{5} \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{x \left (2+e^{4+\frac {16}{5} \log (4) \log (x)}+x\right )} \, dx \\ & = \log \left (2+2^{\frac {32 \log (x)}{5}} e^4+x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=\log \left (2+e^{4+\frac {16}{5} \log (4) \log (x)}+x\right ) \]

[In]

Integrate[(5*x + 16*E^((2*(10 + 8*Log[4]*Log[x]))/5)*Log[4])/(10*x + 5*E^((2*(10 + 8*Log[4]*Log[x]))/5)*x + 5*
x^2),x]

[Out]

Log[2 + E^(4 + (16*Log[4]*Log[x])/5) + x]

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
norman	\(\ln \left ({\mathrm e}^{\frac {32 \ln \left (2\right ) \ln \left (x \right )}{5}+4}+x +2\right )\)	\(16\)
parallelrisch	\(\ln \left ({\mathrm e}^{\frac {32 \ln \left (2\right ) \ln \left (x \right )}{5}+4}+x +2\right )\)	\(16\)
risch	\(-4+\ln \left (x^{\frac {32 \ln \left (2\right )}{5}} {\mathrm e}^{4}+x +2\right )\)	\(18\)

[In]

int((32*ln(2)*exp(16/5*ln(2)*ln(x)+2)^2+5*x)/(5*x*exp(16/5*ln(2)*ln(x)+2)^2+5*x^2+10*x),x,method=_RETURNVERBOS
E)

[Out]

ln(2+exp(16/5*ln(2)*ln(x)+2)^2+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=\log \left (x + e^{\left (\frac {32}{5} \, \log \left (2\right ) \log \left (x\right ) + 4\right )} + 2\right ) \]

[In]

integrate((32*log(2)*exp(16/5*log(2)*log(x)+2)^2+5*x)/(5*x*exp(16/5*log(2)*log(x)+2)^2+5*x^2+10*x),x, algorith
m="fricas")

[Out]

log(x + e^(32/5*log(2)*log(x) + 4) + 2)

Sympy [A] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=\log {\left (x + e^{4} e^{\frac {32 \log {\left (2 \right )} \log {\left (x \right )}}{5}} + 2 \right )} \]

[In]

integrate((32*ln(2)*exp(16/5*ln(2)*ln(x)+2)**2+5*x)/(5*x*exp(16/5*ln(2)*ln(x)+2)**2+5*x**2+10*x),x)

[Out]

log(x + exp(4)*exp(32*log(2)*log(x)/5) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=\log \left ({\left (x + e^{\left (\frac {32}{5} \, \log \left (2\right ) \log \left (x\right ) + 4\right )} + 2\right )} e^{\left (-4\right )}\right ) \]

[In]

integrate((32*log(2)*exp(16/5*log(2)*log(x)+2)^2+5*x)/(5*x*exp(16/5*log(2)*log(x)+2)^2+5*x^2+10*x),x, algorith
m="maxima")

[Out]

log((x + e^(32/5*log(2)*log(x) + 4) + 2)*e^(-4))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=\log \left (x + e^{\left (\frac {32}{5} \, \log \left (2\right ) \log \left (x\right ) + 4\right )} + 2\right ) \]

[In]

integrate((32*log(2)*exp(16/5*log(2)*log(x)+2)^2+5*x)/(5*x*exp(16/5*log(2)*log(x)+2)^2+5*x^2+10*x),x, algorith
m="giac")

[Out]

log(x + e^(32/5*log(2)*log(x) + 4) + 2)

Mupad [B] (verification not implemented)

Time = 12.81 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {5 x+16 e^{\frac {2}{5} (10+8 \log (4) \log (x))} \log (4)}{10 x+5 e^{\frac {2}{5} (10+8 \log (4) \log (x))} x+5 x^2} \, dx=\ln \left (x+{\mathrm {e}}^{\frac {32\,\ln \left (2\right )\,\ln \left (x\right )}{5}+4}+2\right ) \]

[In]

int((5*x + 32*exp((32*log(2)*log(x))/5 + 4)*log(2))/(10*x + 5*x*exp((32*log(2)*log(x))/5 + 4) + 5*x^2),x)

[Out]

log(x + exp((32*log(2)*log(x))/5 + 4) + 2)