\(\int \frac {4^{\frac {8}{x+(12 x+4 x^2) \log (4)}} e^{3 x} (3 x^2+(-8+72 x^2+24 x^3) \log (4)+(-96-64 x+432 x^2+288 x^3+48 x^4) \log ^2(4))}{x^2+(24 x^2+8 x^3) \log (4)+(144 x^2+96 x^3+16 x^4) \log ^2(4)} \, dx\) [7992]

Optimal result

Integrand size = 113, antiderivative size = 31 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=e^{3 x+\frac {2}{3 x+x^2 \left (1+\frac {1}{4 x \log (4)}\right )}} \]

[Out]

exp(2/(3*x+(1/8/x/ln(2)+1)*x^2))*exp(3*x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(93\) vs. \(2(31)=62\).

Time = 1.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6873, 27, 2326} \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=\frac {e^{3 x} 4^{\frac {8}{x (4 x \log (4)+1+12 \log (4))}} (8 x \log (4)+1+12 \log (4))}{x^2 (4 x \log (4)+1+12 \log (4))^2 \left (\frac {1}{x^2 (4 x \log (4)+1+12 \log (4))}+\frac {4 \log (4)}{x (4 x \log (4)+1+12 \log (4))^2}\right )} \]

[In]

Int[(4^(8/(x + (12*x + 4*x^2)*Log[4]))*E^(3*x)*(3*x^2 + (-8 + 72*x^2 + 24*x^3)*Log[4] + (-96 - 64*x + 432*x^2
+ 288*x^3 + 48*x^4)*Log[4]^2))/(x^2 + (24*x^2 + 8*x^3)*Log[4] + (144*x^2 + 96*x^3 + 16*x^4)*Log[4]^2),x]

[Out]

(4^(8/(x*(1 + 12*Log[4] + 4*x*Log[4])))*E^(3*x)*(1 + 12*Log[4] + 8*x*Log[4]))/(x^2*(1 + 12*Log[4] + 4*x*Log[4]
)^2*((4*Log[4])/(x*(1 + 12*Log[4] + 4*x*Log[4])^2) + 1/(x^2*(1 + 12*Log[4] + 4*x*Log[4]))))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {4^{\frac {8}{x (1+12 \log (4)+4 x \log (4))}} e^{3 x} \left (-64 x \log ^2(4)+48 x^4 \log ^2(4)-8 \log (4) (1+12 \log (4))+24 x^3 \log (4) (1+12 \log (4))+3 x^2 (1+12 \log (4))^2\right )}{x^2 \left (16 x^2 \log ^2(4)+8 x \log (4) (1+12 \log (4))+(1+12 \log (4))^2\right )} \, dx \\ & = \int \frac {4^{\frac {8}{x (1+12 \log (4)+4 x \log (4))}} e^{3 x} \left (-64 x \log ^2(4)+48 x^4 \log ^2(4)-8 \log (4) (1+12 \log (4))+24 x^3 \log (4) (1+12 \log (4))+3 x^2 (1+12 \log (4))^2\right )}{x^2 (1+12 \log (4)+4 x \log (4))^2} \, dx \\ & = \frac {4^{\frac {8}{x (1+12 \log (4)+4 x \log (4))}} e^{3 x} (1+12 \log (4)+8 x \log (4))}{x^2 (1+12 \log (4)+4 x \log (4))^2 \left (\frac {4 \log (4)}{x (1+12 \log (4)+4 x \log (4))^2}+\frac {1}{x^2 (1+12 \log (4)+4 x \log (4))}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=4^{\frac {8}{x (1+12 \log (4)+4 x \log (4))}} e^{3 x} \]

[In]

Integrate[(4^(8/(x + (12*x + 4*x^2)*Log[4]))*E^(3*x)*(3*x^2 + (-8 + 72*x^2 + 24*x^3)*Log[4] + (-96 - 64*x + 43
2*x^2 + 288*x^3 + 48*x^4)*Log[4]^2))/(x^2 + (24*x^2 + 8*x^3)*Log[4] + (144*x^2 + 96*x^3 + 16*x^4)*Log[4]^2),x]

[Out]

4^(8/(x*(1 + 12*Log[4] + 4*x*Log[4])))*E^(3*x)

Maple [A] (verified)

Time = 3.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81

method	result	size
risch	\({\mathrm e}^{3 x} 65536^{\frac {1}{x \left (8 x \ln \left (2\right )+24 \ln \left (2\right )+1\right )}}\)	\(25\)
gosper	\({\mathrm e}^{3 x +\frac {16 \ln \left (2\right )}{x \left (8 x \ln \left (2\right )+24 \ln \left (2\right )+1\right )}}\)	\(26\)
parallelrisch	\({\mathrm e}^{\frac {16 \ln \left (2\right )}{x \left (8 x \ln \left (2\right )+24 \ln \left (2\right )+1\right )}} {\mathrm e}^{3 x}\)	\(27\)
norman	\(\frac {\left (24 \ln \left (2\right )+1\right ) x \,{\mathrm e}^{3 x} {\mathrm e}^{\frac {16 \ln \left (2\right )}{2 \left (4 x^{2}+12 x \right ) \ln \left (2\right )+x}}+8 x^{2} \ln \left (2\right ) {\mathrm e}^{3 x} {\mathrm e}^{\frac {16 \ln \left (2\right )}{2 \left (4 x^{2}+12 x \right ) \ln \left (2\right )+x}}}{x \left (8 x \ln \left (2\right )+24 \ln \left (2\right )+1\right )}\)	\(86\)

[In]

int((4*(48*x^4+288*x^3+432*x^2-64*x-96)*ln(2)^2+2*(24*x^3+72*x^2-8)*ln(2)+3*x^2)*exp(16*ln(2)/(2*(4*x^2+12*x)*
ln(2)+x))*exp(3*x)/(4*(16*x^4+96*x^3+144*x^2)*ln(2)^2+2*(8*x^3+24*x^2)*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

exp(3*x)*65536^(1/x/(8*x*ln(2)+24*ln(2)+1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=2^{\frac {16}{8 \, {\left (x^{2} + 3 \, x\right )} \log \left (2\right ) + x}} e^{\left (3 \, x\right )} \]

[In]

integrate((4*(48*x^4+288*x^3+432*x^2-64*x-96)*log(2)^2+2*(24*x^3+72*x^2-8)*log(2)+3*x^2)*exp(16*log(2)/(2*(4*x
^2+12*x)*log(2)+x))*exp(3*x)/(4*(16*x^4+96*x^3+144*x^2)*log(2)^2+2*(8*x^3+24*x^2)*log(2)+x^2),x, algorithm="fr
icas")

[Out]

2^(16/(8*(x^2 + 3*x)*log(2) + x))*e^(3*x)

Sympy [A] (verification not implemented)

Time = 11.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=e^{3 x} e^{\frac {16 \log {\left (2 \right )}}{x + \left (8 x^{2} + 24 x\right ) \log {\left (2 \right )}}} \]

[In]

integrate((4*(48*x**4+288*x**3+432*x**2-64*x-96)*ln(2)**2+2*(24*x**3+72*x**2-8)*ln(2)+3*x**2)*exp(16*ln(2)/(2*
(4*x**2+12*x)*ln(2)+x))*exp(3*x)/(4*(16*x**4+96*x**3+144*x**2)*ln(2)**2+2*(8*x**3+24*x**2)*ln(2)+x**2),x)

[Out]

exp(3*x)*exp(16*log(2)/(x + (8*x**2 + 24*x)*log(2)))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=e^{\left (3 \, x - \frac {128 \, \log \left (2\right )^{2}}{8 \, {\left (24 \, \log \left (2\right )^{2} + \log \left (2\right )\right )} x + 576 \, \log \left (2\right )^{2} + 48 \, \log \left (2\right ) + 1} + \frac {16 \, \log \left (2\right )}{x {\left (24 \, \log \left (2\right ) + 1\right )}}\right )} \]

[In]

integrate((4*(48*x^4+288*x^3+432*x^2-64*x-96)*log(2)^2+2*(24*x^3+72*x^2-8)*log(2)+3*x^2)*exp(16*log(2)/(2*(4*x
^2+12*x)*log(2)+x))*exp(3*x)/(4*(16*x^4+96*x^3+144*x^2)*log(2)^2+2*(8*x^3+24*x^2)*log(2)+x^2),x, algorithm="ma
xima")

[Out]

e^(3*x - 128*log(2)^2/(8*(24*log(2)^2 + log(2))*x + 576*log(2)^2 + 48*log(2) + 1) + 16*log(2)/(x*(24*log(2) +
1)))

Giac [F]

\[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=\int { \frac {{\left (64 \, {\left (3 \, x^{4} + 18 \, x^{3} + 27 \, x^{2} - 4 \, x - 6\right )} \log \left (2\right )^{2} + 3 \, x^{2} + 16 \, {\left (3 \, x^{3} + 9 \, x^{2} - 1\right )} \log \left (2\right )\right )} 2^{\frac {16}{8 \, {\left (x^{2} + 3 \, x\right )} \log \left (2\right ) + x}} e^{\left (3 \, x\right )}}{64 \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )} \log \left (2\right )^{2} + x^{2} + 16 \, {\left (x^{3} + 3 \, x^{2}\right )} \log \left (2\right )} \,d x } \]

[In]

integrate((4*(48*x^4+288*x^3+432*x^2-64*x-96)*log(2)^2+2*(24*x^3+72*x^2-8)*log(2)+3*x^2)*exp(16*log(2)/(2*(4*x
^2+12*x)*log(2)+x))*exp(3*x)/(4*(16*x^4+96*x^3+144*x^2)*log(2)^2+2*(8*x^3+24*x^2)*log(2)+x^2),x, algorithm="gi
ac")

[Out]

integrate((64*(3*x^4 + 18*x^3 + 27*x^2 - 4*x - 6)*log(2)^2 + 3*x^2 + 16*(3*x^3 + 9*x^2 - 1)*log(2))*2^(16/(8*(
x^2 + 3*x)*log(2) + x))*e^(3*x)/(64*(x^4 + 6*x^3 + 9*x^2)*log(2)^2 + x^2 + 16*(x^3 + 3*x^2)*log(2)), x)

Mupad [B] (verification not implemented)

Time = 13.55 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=2^{\frac {16}{x+24\,x\,\ln \left (2\right )+8\,x^2\,\ln \left (2\right )}}\,{\mathrm {e}}^{3\,x} \]

[In]

int((exp(3*x)*exp((16*log(2))/(x + 2*log(2)*(12*x + 4*x^2)))*(4*log(2)^2*(432*x^2 - 64*x + 288*x^3 + 48*x^4 -
96) + 2*log(2)*(72*x^2 + 24*x^3 - 8) + 3*x^2))/(4*log(2)^2*(144*x^2 + 96*x^3 + 16*x^4) + 2*log(2)*(24*x^2 + 8*
x^3) + x^2),x)

[Out]

2^(16/(x + 24*x*log(2) + 8*x^2*log(2)))*exp(3*x)