Integrand size = 113, antiderivative size = 31 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=e^{3 x+\frac {2}{3 x+x^2 \left (1+\frac {1}{4 x \log (4)}\right )}} \] Output:
exp(2/(3*x+(1/8/x/ln(2)+1)*x^2))*exp(3*x)
Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=4^{\frac {8}{x (1+12 \log (4)+4 x \log (4))}} e^{3 x} \] Input:
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 + 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]
Output:
4^(8/(x*(1 + 12*Log[4] + 4*x*Log[4])))*E^(3*x)
Leaf count is larger than twice the leaf count of optimal. \(93\) vs. \(2(31)=62\).
Time = 2.41 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {2026, 2007, 7292, 2726}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{3 x} 4^{\frac {8}{\left (4 x^2+12 x\right ) \log (4)+x}} \left (3 x^2+\left (24 x^3+72 x^2-8\right ) \log (4)+\left (48 x^4+288 x^3+432 x^2-64 x-96\right ) \log ^2(4)\right )}{x^2+\left (8 x^3+24 x^2\right ) \log (4)+\left (16 x^4+96 x^3+144 x^2\right ) \log ^2(4)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{3 x} 4^{\frac {8}{\left (4 x^2+12 x\right ) \log (4)+x}} \left (3 x^2+\left (24 x^3+72 x^2-8\right ) \log (4)+\left (48 x^4+288 x^3+432 x^2-64 x-96\right ) \log ^2(4)\right )}{x^2 \left (16 x^2 \log ^2(4)+8 x \log (4) (1+12 \log (4))+(1+12 \log (4))^2\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{3 x} 4^{\frac {8}{\left (4 x^2+12 x\right ) \log (4)+x}} \left (3 x^2+\left (24 x^3+72 x^2-8\right ) \log (4)+\left (48 x^4+288 x^3+432 x^2-64 x-96\right ) \log ^2(4)\right )}{x^2 (4 x \log (4)+1+12 \log (4))^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{3 x} 4^{\frac {8}{x (4 x \log (4)+1+12 \log (4))}} \left (48 x^4 \log ^2(4)+24 x^3 \log (4) (1+12 \log (4))+3 x^2 (1+12 \log (4))^2-64 x \log ^2(4)-8 \log (4) (1+12 \log (4))\right )}{x^2 (4 x \log (4)+1+12 \log (4))^2}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {e^{3 x} 4^{\frac {8}{x (4 x \log (4)+1+12 \log (4))}} (8 x \log (4)+1+12 \log (4))}{x^2 (4 x \log (4)+1+12 \log (4))^2 \left (\frac {1}{x^2 (4 x \log (4)+1+12 \log (4))}+\frac {4 \log (4)}{x (4 x \log (4)+1+12 \log (4))^2}\right )}\) |
Input:
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]
Output:
(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]))))
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
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]
Time = 3.79 (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 | \(\frac {\left (24 \ln \left (2\right )+1\right ) x \,{\mathrm e}^{3 x} {\mathrm e}^{\frac {16 \ln \left (2\right )}{2 \left (4 x^{2}+12 x \right ) \ln \left (2\right )+x}}+8 x^{2} \ln \left (2\right ) {\mathrm e}^{3 x} {\mathrm e}^{\frac {16 \ln \left (2\right )}{2 \left (4 x^{2}+12 x \right ) \ln \left (2\right )+x}}}{x \left (8 x \ln \left (2\right )+24 \ln \left (2\right )+1\right )}\) | \(86\) |
Input:
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+1 44*x^2)*ln(2)^2+2*(8*x^3+24*x^2)*ln(2)+x^2),x,method=_RETURNVERBOSE)
Output:
exp(3*x)*65536^(1/x/(8*x*ln(2)+24*ln(2)+1))
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=2^{\frac {16}{8 \, {\left (x^{2} + 3 \, x\right )} \log \left (2\right ) + x}} e^{\left (3 \, x\right )} \] Input:
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="fric as")
Output:
2^(16/(8*(x^2 + 3*x)*log(2) + x))*e^(3*x)
Time = 11.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=e^{3 x} e^{\frac {16 \log {\left (2 \right )}}{x + \left (8 x^{2} + 24 x\right ) \log {\left (2 \right )}}} \] Input:
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*(1 6*x**4+96*x**3+144*x**2)*ln(2)**2+2*(8*x**3+24*x**2)*ln(2)+x**2),x)
Output:
exp(3*x)*exp(16*log(2)/(x + (8*x**2 + 24*x)*log(2)))
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=e^{\left (3 \, x - \frac {128 \, \log \left (2\right )^{2}}{8 \, {\left (24 \, \log \left (2\right )^{2} + \log \left (2\right )\right )} x + 576 \, \log \left (2\right )^{2} + 48 \, \log \left (2\right ) + 1} + \frac {16 \, \log \left (2\right )}{x {\left (24 \, \log \left (2\right ) + 1\right )}}\right )} \] Input:
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="maxi ma")
Output:
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)))
\[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=\int { \frac {{\left (64 \, {\left (3 \, x^{4} + 18 \, x^{3} + 27 \, x^{2} - 4 \, x - 6\right )} \log \left (2\right )^{2} + 3 \, x^{2} + 16 \, {\left (3 \, x^{3} + 9 \, x^{2} - 1\right )} \log \left (2\right )\right )} 2^{\frac {16}{8 \, {\left (x^{2} + 3 \, x\right )} \log \left (2\right ) + x}} e^{\left (3 \, x\right )}}{64 \, {\left (x^{4} + 6 \, x^{3} + 9 \, x^{2}\right )} \log \left (2\right )^{2} + x^{2} + 16 \, {\left (x^{3} + 3 \, x^{2}\right )} \log \left (2\right )} \,d x } \] Input:
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="giac ")
Output:
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)
Time = 3.64 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=2^{\frac {16}{x+24\,x\,\ln \left (2\right )+8\,x^2\,\ln \left (2\right )}}\,{\mathrm {e}}^{3\,x} \] Input:
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)
Output:
2^(16/(x + 24*x*log(2) + 8*x^2*log(2)))*exp(3*x)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {4^{\frac {8}{x+\left (12 x+4 x^2\right ) \log (4)}} e^{3 x} \left (3 x^2+\left (-8+72 x^2+24 x^3\right ) \log (4)+\left (-96-64 x+432 x^2+288 x^3+48 x^4\right ) \log ^2(4)\right )}{x^2+\left (24 x^2+8 x^3\right ) \log (4)+\left (144 x^2+96 x^3+16 x^4\right ) \log ^2(4)} \, dx=e^{\frac {24 \,\mathrm {log}\left (2\right ) x^{3}+72 \,\mathrm {log}\left (2\right ) x^{2}+16 \,\mathrm {log}\left (2\right )+3 x^{2}}{8 \,\mathrm {log}\left (2\right ) x^{2}+24 \,\mathrm {log}\left (2\right ) x +x}} \] Input:
int((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)
Output:
e**((24*log(2)*x**3 + 72*log(2)*x**2 + 16*log(2) + 3*x**2)/(8*log(2)*x**2 + 24*log(2)*x + x))