\(\int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+(10 x+3 x^2) \log ^2(3 x+x \log (16))} \, dx\) [5938]

Optimal result

Integrand size = 85, antiderivative size = 21 \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\log \left (10+e^{\frac {9}{\log (3 x+x \log (16))}}+3 x\right ) \]

[Out]

ln(exp(9/ln(4*x*ln(2)+3*x))+10+3*x)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 12, 6816} \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\log \left (3 x+e^{\frac {9}{\log (x (3+\log (16)))}}+10\right ) \]

[In]

Int[(-9*E^(9/Log[3*x + x*Log[16]]) + 3*x*Log[3*x + x*Log[16]]^2)/(E^(9/Log[3*x + x*Log[16]])*x*Log[3*x + x*Log
[16]]^2 + (10*x + 3*x^2)*Log[3*x + x*Log[16]]^2),x]

[Out]

Log[10 + E^(9/Log[x*(3 + Log[16])]) + 3*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 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {3 \left (-3 e^{\frac {9}{\log (x (3+\log (16)))}}+x \log ^2(x (3+\log (16)))\right )}{x \left (10+e^{\frac {9}{\log (x (3+\log (16)))}}+3 x\right ) \log ^2(x (3+\log (16)))} \, dx \\ & = 3 \int \frac {-3 e^{\frac {9}{\log (x (3+\log (16)))}}+x \log ^2(x (3+\log (16)))}{x \left (10+e^{\frac {9}{\log (x (3+\log (16)))}}+3 x\right ) \log ^2(x (3+\log (16)))} \, dx \\ & = \log \left (10+e^{\frac {9}{\log (x (3+\log (16)))}}+3 x\right ) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(45\) vs. \(2(21)=42\).

Time = 0.90 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\log \left (30+3 e^{\frac {9}{\log (x (3+\log (16)))}}+10 \log (16)+e^{\frac {9}{\log (x (3+\log (16)))}} \log (16)+3 x (3+\log (16))\right ) \]

[In]

Integrate[(-9*E^(9/Log[3*x + x*Log[16]]) + 3*x*Log[3*x + x*Log[16]]^2)/(E^(9/Log[3*x + x*Log[16]])*x*Log[3*x +
x*Log[16]]^2 + (10*x + 3*x^2)*Log[3*x + x*Log[16]]^2),x]

[Out]

Log[30 + 3*E^(9/Log[x*(3 + Log[16])]) + 10*Log[16] + E^(9/Log[x*(3 + Log[16])])*Log[16] + 3*x*(3 + Log[16])]

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\ln \left (\frac {{\mathrm e}^{\frac {9}{\ln \left (x \left (4 \ln \left (2\right )+3\right )\right )}}}{3}+x +\frac {10}{3}\right )\)	\(21\)
norman	\(\ln \left ({\mathrm e}^{\frac {9}{\ln \left (4 x \ln \left (2\right )+3 x \right )}}+10+3 x \right )\)	\(22\)
risch	\(\ln \left ({\mathrm e}^{\frac {9}{\ln \left (4 x \ln \left (2\right )+3 x \right )}}+10+3 x \right )\)	\(22\)

[In]

int((-9*exp(9/ln(4*x*ln(2)+3*x))+3*x*ln(4*x*ln(2)+3*x)^2)/(x*ln(4*x*ln(2)+3*x)^2*exp(9/ln(4*x*ln(2)+3*x))+(3*x
^2+10*x)*ln(4*x*ln(2)+3*x)^2),x,method=_RETURNVERBOSE)

[Out]

ln(1/3*exp(9/ln(x*(4*ln(2)+3)))+x+10/3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\log \left (3 \, x + e^{\left (\frac {9}{\log \left (4 \, x \log \left (2\right ) + 3 \, x\right )}\right )} + 10\right ) \]

[In]

integrate((-9*exp(9/log(4*x*log(2)+3*x))+3*x*log(4*x*log(2)+3*x)^2)/(x*log(4*x*log(2)+3*x)^2*exp(9/log(4*x*log
(2)+3*x))+(3*x^2+10*x)*log(4*x*log(2)+3*x)^2),x, algorithm="fricas")

[Out]

log(3*x + e^(9/log(4*x*log(2) + 3*x)) + 10)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\log {\left (3 x + e^{\frac {9}{\log {\left (4 x \log {\left (2 \right )} + 3 x \right )}}} + 10 \right )} \]

[In]

integrate((-9*exp(9/ln(4*x*ln(2)+3*x))+3*x*ln(4*x*ln(2)+3*x)**2)/(x*ln(4*x*ln(2)+3*x)**2*exp(9/ln(4*x*ln(2)+3*
x))+(3*x**2+10*x)*ln(4*x*ln(2)+3*x)**2),x)

[Out]

log(3*x + exp(9/log(4*x*log(2) + 3*x)) + 10)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\log \left (3 \, x + e^{\left (\frac {9}{\log \left (x\right ) + \log \left (4 \, \log \left (2\right ) + 3\right )}\right )} + 10\right ) \]

[In]

integrate((-9*exp(9/log(4*x*log(2)+3*x))+3*x*log(4*x*log(2)+3*x)^2)/(x*log(4*x*log(2)+3*x)^2*exp(9/log(4*x*log
(2)+3*x))+(3*x^2+10*x)*log(4*x*log(2)+3*x)^2),x, algorithm="maxima")

[Out]

log(3*x + e^(9/(log(x) + log(4*log(2) + 3))) + 10)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\log \left (3 \, x + e^{\left (\frac {9}{\log \left (4 \, x \log \left (2\right ) + 3 \, x\right )}\right )} + 10\right ) \]

[In]

integrate((-9*exp(9/log(4*x*log(2)+3*x))+3*x*log(4*x*log(2)+3*x)^2)/(x*log(4*x*log(2)+3*x)^2*exp(9/log(4*x*log
(2)+3*x))+(3*x^2+10*x)*log(4*x*log(2)+3*x)^2),x, algorithm="giac")

[Out]

log(3*x + e^(9/log(4*x*log(2) + 3*x)) + 10)

Mupad [B] (verification not implemented)

Time = 12.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-9 e^{\frac {9}{\log (3 x+x \log (16))}}+3 x \log ^2(3 x+x \log (16))}{e^{\frac {9}{\log (3 x+x \log (16))}} x \log ^2(3 x+x \log (16))+\left (10 x+3 x^2\right ) \log ^2(3 x+x \log (16))} \, dx=\ln \left (x+\frac {{\mathrm {e}}^{\frac {9}{\ln \left (x\,\left (\ln \left (16\right )+3\right )\right )}}}{3}+\frac {10}{3}\right ) \]

[In]

int(-(9*exp(9/log(3*x + 4*x*log(2))) - 3*x*log(3*x + 4*x*log(2))^2)/(log(3*x + 4*x*log(2))^2*(10*x + 3*x^2) +
x*exp(9/log(3*x + 4*x*log(2)))*log(3*x + 4*x*log(2))^2),x)

[Out]

log(x + exp(9/log(x*(log(16) + 3)))/3 + 10/3)