Integrand size = 58, antiderivative size = 26 \[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=e^5-\frac {4 e^3 x}{3 (7-i \pi +x-\log (2))} \] Output:
exp(5)-4*exp(3)*x/(21+3*x-3*ln(2)-3*I*Pi)
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=\frac {4 i e^3 (7-i \pi -\log (2))}{3 (\pi +i (7+x-\log (2)))} \] Input:
Integrate[(-28*E^3 + 4*E^3*(I*Pi + Log[2]))/(147 + 42*x + 3*x^2 + (-42 - 6 *x)*(I*Pi + Log[2]) + 3*(I*Pi + Log[2])^2),x]
Output:
(((4*I)/3)*E^3*(7 - I*Pi - Log[2]))/(Pi + I*(7 + x - Log[2]))
Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 2080, 1077, 17}
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.
\(\displaystyle \int \frac {-28 e^3+4 e^3 (\log (2)+i \pi )}{3 x^2+42 x+(-6 x-42) (\log (2)+i \pi )+147+3 (\log (2)+i \pi )^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 e^3 (7-i \pi -\log (2)) \int \frac {1}{3 x^2+42 x+3 \left (49+(i \pi +\log (2))^2\right )-6 (x+7) (i \pi +\log (2))}dx\) |
\(\Big \downarrow \) 2080 |
\(\displaystyle -4 e^3 (7-i \pi -\log (2)) \int \frac {1}{3 x^2+6 (7-i \pi -\log (2)) x-3 (\pi +i (7-\log (2)))^2}dx\) |
\(\Big \downarrow \) 1077 |
\(\displaystyle -12 e^3 (7-i \pi -\log (2)) \int \frac {1}{(3 x-\log (8)-3 i \pi +21)^2}dx\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {4 e^3 (7-i \pi -\log (2))}{3 x-3 i \pi +21-\log (8)}\) |
Input:
Int[(-28*E^3 + 4*E^3*(I*Pi + Log[2]))/(147 + 42*x + 3*x^2 + (-42 - 6*x)*(I *Pi + Log[2]) + 3*(I*Pi + Log[2])^2),x]
Output:
(4*E^3*(7 - I*Pi - Log[2]))/(21 - (3*I)*Pi + 3*x - Log[8])
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int [(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && Q uadraticQ[u, x] && !QuadraticMatchQ[u, x]
Time = 0.80 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {4 \,{\mathrm e}^{3} \left (\ln \left (2\right )+i \pi \right )-28 \,{\mathrm e}^{3}}{3 \left (-i \pi -\ln \left (2\right )+x +7\right )}\) | \(32\) |
parallelrisch | \(-\frac {i \left (4 \,{\mathrm e}^{3} \left (\ln \left (2\right )+i \pi \right )-28 \,{\mathrm e}^{3}\right )}{3 \left (-i \ln \left (2\right )+i x +\pi +7 i\right )}\) | \(35\) |
risch | \(\frac {4 \pi \,{\mathrm e}^{3}}{3 \left (-i \ln \left (2\right )+i x +\pi +7 i\right )}-\frac {4 i {\mathrm e}^{3} \ln \left (2\right )}{3 \left (-i \ln \left (2\right )+i x +\pi +7 i\right )}+\frac {28 i {\mathrm e}^{3}}{3 \left (-i \ln \left (2\right )+i x +\pi +7 i\right )}\) | \(64\) |
orering | \(\frac {i \left (-i \ln \left (2\right )+i x +\pi +7 i\right ) \left (4 \,{\mathrm e}^{3} \left (\ln \left (2\right )+i \pi \right )-28 \,{\mathrm e}^{3}\right )}{3 \left (\ln \left (2\right )+i \pi \right )^{2}+\left (-6 x -42\right ) \left (\ln \left (2\right )+i \pi \right )+3 x^{2}+42 x +147}\) | \(69\) |
gosper | \(-\frac {4 i \left (-i \ln \left (2\right )+i x +\pi +7 i\right ) {\mathrm e}^{3} \left (i \pi +\ln \left (2\right )-7\right )}{3 \left (-2 i \pi \ln \left (2\right )+2 i x \pi +\pi ^{2}+14 i \pi -\ln \left (2\right )^{2}+2 x \ln \left (2\right )-x^{2}+14 \ln \left (2\right )-14 x -49\right )}\) | \(72\) |
norman | \(\frac {\left (-\frac {4 i {\mathrm e}^{3} \pi }{3}-\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )}{3}+\frac {28 \,{\mathrm e}^{3}}{3}\right ) x +\frac {4 \,{\mathrm e}^{3} \pi ^{2}}{3}+\frac {4 \,{\mathrm e}^{3} \ln \left (2\right )^{2}}{3}-\frac {56 \,{\mathrm e}^{3} \ln \left (2\right )}{3}+\frac {196 \,{\mathrm e}^{3}}{3}}{\pi ^{2}+\ln \left (2\right )^{2}-2 x \ln \left (2\right )+x^{2}-14 \ln \left (2\right )+14 x +49}\) | \(73\) |
Input:
int((4*exp(3)*(ln(2)+I*Pi)-28*exp(3))/(3*(ln(2)+I*Pi)^2+(-6*x-42)*(ln(2)+I *Pi)+3*x^2+42*x+147),x,method=_RETURNVERBOSE)
Output:
-1/3*(4*exp(3)*(ln(2)+I*Pi)-28*exp(3))/(-I*Pi-ln(2)+x+7)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=-\frac {4 \, {\left ({\left (-i \, \pi + 7\right )} e^{3} - e^{3} \log \left (2\right )\right )}}{3 i \, \pi - 3 \, x + 3 \, \log \left (2\right ) - 21} \] Input:
integrate((4*exp(3)*(log(2)+I*pi)-28*exp(3))/(3*(log(2)+I*pi)^2+(-6*x-42)* (log(2)+I*pi)+3*x^2+42*x+147),x, algorithm="fricas")
Output:
-4*((-I*pi + 7)*e^3 - e^3*log(2))/(3*I*pi - 3*x + 3*log(2) - 21)
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=- \frac {- 28 e^{3} + 4 e^{3} \log {\left (2 \right )} + 4 i \pi e^{3}}{3 x - 3 \log {\left (2 \right )} + 21 - 3 i \pi } \] Input:
integrate((4*exp(3)*(ln(2)+I*pi)-28*exp(3))/(3*(ln(2)+I*pi)**2+(-6*x-42)*( ln(2)+I*pi)+3*x**2+42*x+147),x)
Output:
-(-28*exp(3) + 4*exp(3)*log(2) + 4*I*pi*exp(3))/(3*x - 3*log(2) + 21 - 3*I *pi)
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=-\frac {4 \, {\left ({\left (i \, \pi + \log \left (2\right )\right )} e^{3} - 7 \, e^{3}\right )}}{3 \, {\left (-i \, \pi + x - \log \left (2\right ) + 7\right )}} \] Input:
integrate((4*exp(3)*(log(2)+I*pi)-28*exp(3))/(3*(log(2)+I*pi)^2+(-6*x-42)* (log(2)+I*pi)+3*x^2+42*x+147),x, algorithm="maxima")
Output:
-4/3*((I*pi + log(2))*e^3 - 7*e^3)/(-I*pi + x - log(2) + 7)
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=-\frac {4 \, {\left ({\left (i \, \pi + \log \left (2\right )\right )} e^{3} - 7 \, e^{3}\right )}}{3 \, {\left (-i \, \pi + x - \log \left (2\right ) + 7\right )}} \] Input:
integrate((4*exp(3)*(log(2)+I*pi)-28*exp(3))/(3*(log(2)+I*pi)^2+(-6*x-42)* (log(2)+I*pi)+3*x^2+42*x+147),x, algorithm="giac")
Output:
-4/3*((I*pi + log(2))*e^3 - 7*e^3)/(-I*pi + x - log(2) + 7)
Time = 3.62 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=\frac {\frac {4\,{\mathrm {e}}^3\,\ln \left (2\right )}{3}-\frac {28\,{\mathrm {e}}^3}{3}+\frac {\Pi \,{\mathrm {e}}^3\,4{}\mathrm {i}}{3}}{\ln \left (2\right )-x-7+\Pi \,1{}\mathrm {i}} \] Input:
int(-(28*exp(3) - 4*exp(3)*(Pi*1i + log(2)))/(42*x - (6*x + 42)*(Pi*1i + l og(2)) + 3*x^2 + 3*(Pi*1i + log(2))^2 + 147),x)
Output:
((Pi*exp(3)*4i)/3 - (28*exp(3))/3 + (4*exp(3)*log(2))/3)/(Pi*1i - x + log( 2) - 7)
\[ \int \frac {-28 e^3+4 e^3 (i \pi +\log (2))}{147+42 x+3 x^2+(-42-6 x) (i \pi +\log (2))+3 (i \pi +\log (2))^2} \, dx=\frac {4 \left (\int \frac {1}{\mathrm {log}\left (2\right )^{2}+2 \,\mathrm {log}\left (2\right ) i \pi -2 \,\mathrm {log}\left (2\right ) x -14 \,\mathrm {log}\left (2\right )-2 i \pi x -14 i \pi -\pi ^{2}+x^{2}+14 x +49}d x \right ) e^{3} \left (\mathrm {log}\left (2\right )+i \pi -7\right )}{3} \] Input:
int((4*exp(3)*(log(2)+I*Pi)-28*exp(3))/(3*(log(2)+I*Pi)^2+(-6*x-42)*(log(2 )+I*Pi)+3*x^2+42*x+147),x)
Output:
(4*int(1/(log(2)**2 + 2*log(2)*i*pi - 2*log(2)*x - 14*log(2) - 2*i*pi*x - 14*i*pi - pi**2 + x**2 + 14*x + 49),x)*e**3*(log(2) + i*pi - 7))/3