Integrand size = 114, antiderivative size = 24 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\left (-3-x-x \left (-2+x+\frac {1}{4} e^x \log (4)\right )\right )^2} \] Output:
exp((-x-(1/2*exp(x)*ln(2)+x-2)*x-3)^2)
\[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=\int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx \] Input:
Integrate[(E^((144 - 96*x + 112*x^2 - 32*x^3 + 16*x^4 + E^x*(24*x - 8*x^2 + 8*x^3)*Log[4] + E^(2*x)*x^2*Log[4]^2)/16)*(-48 + 112*x - 48*x^2 + 32*x^3 + E^x*(12 + 4*x + 8*x^2 + 4*x^3)*Log[4] + E^(2*x)*(x + x^2)*Log[4]^2))/8, x]
Output:
Integrate[E^((144 - 96*x + 112*x^2 - 32*x^3 + 16*x^4 + E^x*(24*x - 8*x^2 + 8*x^3)*Log[4] + E^(2*x)*x^2*Log[4]^2)/16)*(-48 + 112*x - 48*x^2 + 32*x^3 + E^x*(12 + 4*x + 8*x^2 + 4*x^3)*Log[4] + E^(2*x)*(x + x^2)*Log[4]^2), x]/ 8
Leaf count is larger than twice the leaf count of optimal. \(112\) vs. \(2(24)=48\).
Time = 0.71 (sec) , antiderivative size = 112, normalized size of antiderivative = 4.67, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {1}{8} \left (32 x^3-48 x^2+e^{2 x} \left (x^2+x\right ) \log ^2(4)+e^x \left (4 x^3+8 x^2+4 x+12\right ) \log (4)+112 x-48\right ) \exp \left (\frac {1}{16} \left (16 x^4-32 x^3+112 x^2+e^{2 x} x^2 \log ^2(4)+e^x \left (8 x^3-8 x^2+24 x\right ) \log (4)-96 x+144\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int -2^{\frac {1}{8} e^x \left (8 x^3-8 x^2+24 x\right )} \exp \left (\frac {1}{16} \left (16 x^4-32 x^3+e^{2 x} \log ^2(4) x^2+112 x^2-96 x+144\right )\right ) \left (-32 x^3+48 x^2-112 x-e^{2 x} \left (x^2+x\right ) \log ^2(4)-4 e^x \left (x^3+2 x^2+x+3\right ) \log (4)+48\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{8} \int 2^{e^x \left (x^3-x^2+3 x\right )} \exp \left (\frac {1}{16} \left (16 x^4-32 x^3+e^{2 x} \log ^2(4) x^2+112 x^2-96 x+144\right )\right ) \left (-32 x^3+48 x^2-112 x-e^{2 x} \left (x^2+x\right ) \log ^2(4)-4 e^x \left (x^3+2 x^2+x+3\right ) \log (4)+48\right )dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {2^{e^x \left (x^3-x^2+3 x\right )-1} \left (x^3+2 x^2+x+3\right ) \log (4) \exp \left (\frac {1}{16} \left (16 x^4-32 x^3+112 x^2+e^{2 x} x^2 \log ^2(4)-96 x+144\right )+x\right )}{\left (e^x \left (3 x^2-2 x+3\right )+e^x \left (x^3-x^2+3 x\right )\right ) \log (2)}\) |
Input:
Int[(E^((144 - 96*x + 112*x^2 - 32*x^3 + 16*x^4 + E^x*(24*x - 8*x^2 + 8*x^ 3)*Log[4] + E^(2*x)*x^2*Log[4]^2)/16)*(-48 + 112*x - 48*x^2 + 32*x^3 + E^x *(12 + 4*x + 8*x^2 + 4*x^3)*Log[4] + E^(2*x)*(x + x^2)*Log[4]^2))/8,x]
Output:
(2^(-1 + E^x*(3*x - x^2 + x^3))*E^(x + (144 - 96*x + 112*x^2 - 32*x^3 + 16 *x^4 + E^(2*x)*x^2*Log[4]^2)/16)*(3 + x + 2*x^2 + x^3)*Log[4])/((E^x*(3 - 2*x + 3*x^2) + E^x*(3*x - x^2 + x^3))*Log[2])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(47\) vs. \(2(20)=40\).
Time = 0.55 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00
method | result | size |
risch | \(2^{x \left (x^{2}-x +3\right ) {\mathrm e}^{x}} {\mathrm e}^{\frac {x^{2} \ln \left (2\right )^{2} {\mathrm e}^{2 x}}{4}+9+x^{4}-2 x^{3}+7 x^{2}-6 x}\) | \(48\) |
norman | \({\mathrm e}^{\frac {x^{2} \ln \left (2\right )^{2} {\mathrm e}^{2 x}}{4}+\frac {\left (8 x^{3}-8 x^{2}+24 x \right ) \ln \left (2\right ) {\mathrm e}^{x}}{8}+x^{4}-2 x^{3}+7 x^{2}-6 x +9}\) | \(53\) |
parallelrisch | \({\mathrm e}^{9-6 x +x^{4}-2 x^{3}+7 x^{2}+\frac {x^{2} \ln \left (2\right )^{2} {\mathrm e}^{2 x}}{4}+\frac {\ln \left (2\right ) {\mathrm e}^{x} \left (4 x^{3}-4 x^{2}+12 x \right )}{4}}\) | \(53\) |
Input:
int(1/8*(4*(x^2+x)*ln(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*ln(2)*exp(x)+32 *x^3-48*x^2+112*x-48)*exp(1/4*x^2*ln(2)^2*exp(x)^2+1/8*(8*x^3-8*x^2+24*x)* ln(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x,method=_RETURNVERBOSE)
Output:
2^(x*(x^2-x+3)*exp(x))*exp(1/4*x^2*ln(2)^2*exp(2*x)+9+x^4-2*x^3+7*x^2-6*x)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\left (\frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \left (2\right )^{2} + x^{4} - 2 \, x^{3} + {\left (x^{3} - x^{2} + 3 \, x\right )} e^{x} \log \left (2\right ) + 7 \, x^{2} - 6 \, x + 9\right )} \] Input:
integrate(1/8*(4*(x^2+x)*log(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*log(2)*e xp(x)+32*x^3-48*x^2+112*x-48)*exp(1/4*x^2*log(2)^2*exp(x)^2+1/8*(8*x^3-8*x ^2+24*x)*log(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x, algorithm="fricas")
Output:
e^(1/4*x^2*e^(2*x)*log(2)^2 + x^4 - 2*x^3 + (x^3 - x^2 + 3*x)*e^x*log(2) + 7*x^2 - 6*x + 9)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).
Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{x^{4} - 2 x^{3} + \frac {x^{2} e^{2 x} \log {\left (2 \right )}^{2}}{4} + 7 x^{2} - 6 x + \left (x^{3} - x^{2} + 3 x\right ) e^{x} \log {\left (2 \right )} + 9} \] Input:
integrate(1/8*(4*(x**2+x)*ln(2)**2*exp(x)**2+2*(4*x**3+8*x**2+4*x+12)*ln(2 )*exp(x)+32*x**3-48*x**2+112*x-48)*exp(1/4*x**2*ln(2)**2*exp(x)**2+1/8*(8* x**3-8*x**2+24*x)*ln(2)*exp(x)+x**4-2*x**3+7*x**2-6*x+9),x)
Output:
exp(x**4 - 2*x**3 + x**2*exp(2*x)*log(2)**2/4 + 7*x**2 - 6*x + (x**3 - x** 2 + 3*x)*exp(x)*log(2) + 9)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).
Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=e^{\left (x^{3} e^{x} \log \left (2\right ) + \frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \left (2\right )^{2} + x^{4} - x^{2} e^{x} \log \left (2\right ) - 2 \, x^{3} + 3 \, x e^{x} \log \left (2\right ) + 7 \, x^{2} - 6 \, x + 9\right )} \] Input:
integrate(1/8*(4*(x^2+x)*log(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*log(2)*e xp(x)+32*x^3-48*x^2+112*x-48)*exp(1/4*x^2*log(2)^2*exp(x)^2+1/8*(8*x^3-8*x ^2+24*x)*log(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x, algorithm="maxima")
Output:
e^(x^3*e^x*log(2) + 1/4*x^2*e^(2*x)*log(2)^2 + x^4 - x^2*e^x*log(2) - 2*x^ 3 + 3*x*e^x*log(2) + 7*x^2 - 6*x + 9)
\[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=\int { \frac {1}{2} \, {\left ({\left (x^{2} + x\right )} e^{\left (2 \, x\right )} \log \left (2\right )^{2} + 8 \, x^{3} + 2 \, {\left (x^{3} + 2 \, x^{2} + x + 3\right )} e^{x} \log \left (2\right ) - 12 \, x^{2} + 28 \, x - 12\right )} e^{\left (\frac {1}{4} \, x^{2} e^{\left (2 \, x\right )} \log \left (2\right )^{2} + x^{4} - 2 \, x^{3} + {\left (x^{3} - x^{2} + 3 \, x\right )} e^{x} \log \left (2\right ) + 7 \, x^{2} - 6 \, x + 9\right )} \,d x } \] Input:
integrate(1/8*(4*(x^2+x)*log(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*log(2)*e xp(x)+32*x^3-48*x^2+112*x-48)*exp(1/4*x^2*log(2)^2*exp(x)^2+1/8*(8*x^3-8*x ^2+24*x)*log(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x, algorithm="giac")
Output:
integrate(1/2*((x^2 + x)*e^(2*x)*log(2)^2 + 8*x^3 + 2*(x^3 + 2*x^2 + x + 3 )*e^x*log(2) - 12*x^2 + 28*x - 12)*e^(1/4*x^2*e^(2*x)*log(2)^2 + x^4 - 2*x ^3 + (x^3 - x^2 + 3*x)*e^x*log(2) + 7*x^2 - 6*x + 9), x)
Time = 3.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=2^{x^3\,{\mathrm {e}}^x-x^2\,{\mathrm {e}}^x+3\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-6\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^9\,{\mathrm {e}}^{-2\,x^3}\,{\mathrm {e}}^{7\,x^2}\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{2\,x}\,{\ln \left (2\right )}^2}{4}} \] Input:
int((exp(7*x^2 - 6*x - 2*x^3 + x^4 + (x^2*exp(2*x)*log(2)^2)/4 + (exp(x)*l og(2)*(24*x - 8*x^2 + 8*x^3))/8 + 9)*(112*x - 48*x^2 + 32*x^3 + 4*exp(2*x) *log(2)^2*(x + x^2) + 2*exp(x)*log(2)*(4*x + 8*x^2 + 4*x^3 + 12) - 48))/8, x)
Output:
2^(x^3*exp(x) - x^2*exp(x) + 3*x*exp(x))*exp(-6*x)*exp(x^4)*exp(9)*exp(-2* x^3)*exp(7*x^2)*exp((x^2*exp(2*x)*log(2)^2)/4)
Time = 9.87 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.83 \[ \int \frac {1}{8} e^{\frac {1}{16} \left (144-96 x+112 x^2-32 x^3+16 x^4+e^x \left (24 x-8 x^2+8 x^3\right ) \log (4)+e^{2 x} x^2 \log ^2(4)\right )} \left (-48+112 x-48 x^2+32 x^3+e^x \left (12+4 x+8 x^2+4 x^3\right ) \log (4)+e^{2 x} \left (x+x^2\right ) \log ^2(4)\right ) \, dx=\frac {e^{\frac {e^{2 x} \mathrm {log}\left (2\right )^{2} x^{2}}{4}+e^{x} \mathrm {log}\left (2\right ) x^{3}+3 e^{x} \mathrm {log}\left (2\right ) x +x^{4}+7 x^{2}} e^{9}}{e^{e^{x} \mathrm {log}\left (2\right ) x^{2}+2 x^{3}+6 x}} \] Input:
int(1/8*(4*(x^2+x)*log(2)^2*exp(x)^2+2*(4*x^3+8*x^2+4*x+12)*log(2)*exp(x)+ 32*x^3-48*x^2+112*x-48)*exp(1/4*x^2*log(2)^2*exp(x)^2+1/8*(8*x^3-8*x^2+24* x)*log(2)*exp(x)+x^4-2*x^3+7*x^2-6*x+9),x)
Output:
(e**((e**(2*x)*log(2)**2*x**2 + 4*e**x*log(2)*x**3 + 12*e**x*log(2)*x + 4* x**4 + 28*x**2)/4)*e**9)/e**(e**x*log(2)*x**2 + 2*x**3 + 6*x)