\(\int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} (8 x+24 x^2) \log (x)+(-\pi ^2 (480 x+32 x^2)+i e^{3 x} \pi (-240 x^2-16 x^3)+e^{6 x} (30 x^3+2 x^4)) \log ^2(x)}{(-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3) \log ^2(x)} \, dx\) [560]

Optimal result

Integrand size = 125, antiderivative size = 30 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=(15+x)^2+\frac {2}{\left (i \pi -\frac {1}{4} e^{3 x} x\right ) \log (x)} \]

[Out]

2/(I*Pi-1/4*x*exp(3*x))/ln(x)+(x+15)^2

Rubi [F]

\[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=\int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx \]

[In]

Int[((-32*I)*Pi + 8*E^(3*x)*x + E^(3*x)*(8*x + 24*x^2)*Log[x] + (-(Pi^2*(480*x + 32*x^2)) + I*E^(3*x)*Pi*(-240
*x^2 - 16*x^3) + E^(6*x)*(30*x^3 + 2*x^4))*Log[x]^2)/((-16*Pi^2*x - (8*I)*E^(3*x)*Pi*x^2 + E^(6*x)*x^3)*Log[x]
^2),x]

[Out]

30*x + x^2 + 8*Defer[Int][1/(x*((-4*I)*Pi + E^(3*x)*x)*Log[x]^2), x] - 8*Defer[Int][E^(3*x)/((4*Pi + I*E^(3*x)
*x)^2*Log[x]), x] + 24*Defer[Int][(E^(3*x)*x)/(((-4*I)*Pi + E^(3*x)*x)^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int 2 \left (15+x+\frac {4}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)}+\frac {4 e^{3 x} (1+3 x)}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)}\right ) \, dx \\ & = 2 \int \left (15+x+\frac {4}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)}+\frac {4 e^{3 x} (1+3 x)}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)}\right ) \, dx \\ & = 30 x+x^2+8 \int \frac {1}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)} \, dx+8 \int \frac {e^{3 x} (1+3 x)}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)} \, dx \\ & = 30 x+x^2+8 \int \left (-\frac {e^{3 x}}{\left (4 \pi +i e^{3 x} x\right )^2 \log (x)}+\frac {3 e^{3 x} x}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)}\right ) \, dx+8 \int \frac {1}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)} \, dx \\ & = 30 x+x^2+8 \int \frac {1}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)} \, dx-8 \int \frac {e^{3 x}}{\left (4 \pi +i e^{3 x} x\right )^2 \log (x)} \, dx+24 \int \frac {e^{3 x} x}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=2 \left (15 x+\frac {x^2}{2}-\frac {4 i}{\left (4 \pi +i e^{3 x} x\right ) \log (x)}\right ) \]

[In]

Integrate[((-32*I)*Pi + 8*E^(3*x)*x + E^(3*x)*(8*x + 24*x^2)*Log[x] + (-(Pi^2*(480*x + 32*x^2)) + I*E^(3*x)*Pi
*(-240*x^2 - 16*x^3) + E^(6*x)*(30*x^3 + 2*x^4))*Log[x]^2)/((-16*Pi^2*x - (8*I)*E^(3*x)*Pi*x^2 + E^(6*x)*x^3)*
Log[x]^2),x]

[Out]

2*(15*x + x^2/2 - (4*I)/((4*Pi + I*E^(3*x)*x)*Log[x]))

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
risch	\(x^{2}+30 x -\frac {8 i}{\left (i x \,{\mathrm e}^{3 x}+4 \pi \right ) \ln \left (x \right )}\)	\(29\)
parallelrisch	\(\frac {i \ln \left (x \right ) {\mathrm e}^{3 x} x^{3}+30 i \ln \left (x \right ) {\mathrm e}^{3 x} x^{2}+4 \pi \ln \left (x \right ) x^{2}+120 \pi \ln \left (x \right ) x -8 i}{\left (i x \,{\mathrm e}^{3 x}+4 \pi \right ) \ln \left (x \right )}\)	\(61\)

[In]

int((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*Pi*exp(3*x)-(32*x^2+480*x)*Pi^2)*ln(x)^2+(24*x^2+8*x)*exp(
3*x)*ln(x)+8*x*exp(3*x)-32*I*Pi)/(x^3*exp(3*x)^2-8*I*x^2*Pi*exp(3*x)-16*x*Pi^2)/ln(x)^2,x,method=_RETURNVERBOS
E)

[Out]

x^2+30*x-8*I/(I*x*exp(3*x)+4*Pi)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=\frac {{\left (-4 i \, \pi x^{2} - 120 i \, \pi x + {\left (x^{3} + 30 \, x^{2}\right )} e^{\left (3 \, x\right )}\right )} \log \left (x\right ) - 8}{{\left (-4 i \, \pi + x e^{\left (3 \, x\right )}\right )} \log \left (x\right )} \]

[In]

integrate((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*pi*exp(3*x)-(32*x^2+480*x)*pi^2)*log(x)^2+(24*x^2+8*
x)*exp(3*x)*log(x)+8*x*exp(3*x)-32*I*pi)/(x^3*exp(3*x)^2-8*I*x^2*pi*exp(3*x)-16*x*pi^2)/log(x)^2,x, algorithm=
"fricas")

[Out]

((-4*I*pi*x^2 - 120*I*pi*x + (x^3 + 30*x^2)*e^(3*x))*log(x) - 8)/((-4*I*pi + x*e^(3*x))*log(x))

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=x^{2} + 30 x - \frac {8}{x e^{3 x} \log {\left (x \right )} - 4 i \pi \log {\left (x \right )}} \]

[In]

integrate((((2*x**4+30*x**3)*exp(3*x)**2+I*(-16*x**3-240*x**2)*pi*exp(3*x)-(32*x**2+480*x)*pi**2)*ln(x)**2+(24
*x**2+8*x)*exp(3*x)*ln(x)+8*x*exp(3*x)-32*I*pi)/(x**3*exp(3*x)**2-8*I*x**2*pi*exp(3*x)-16*x*pi**2)/ln(x)**2,x)

[Out]

x**2 + 30*x - 8/(x*exp(3*x)*log(x) - 4*I*pi*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=\frac {{\left (x^{3} + 30 \, x^{2}\right )} e^{\left (3 \, x\right )} \log \left (x\right ) - 4 \, {\left (i \, \pi x^{2} + 30 i \, \pi x\right )} \log \left (x\right ) - 8}{x e^{\left (3 \, x\right )} \log \left (x\right ) - 4 i \, \pi \log \left (x\right )} \]

[In]

integrate((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*pi*exp(3*x)-(32*x^2+480*x)*pi^2)*log(x)^2+(24*x^2+8*
x)*exp(3*x)*log(x)+8*x*exp(3*x)-32*I*pi)/(x^3*exp(3*x)^2-8*I*x^2*pi*exp(3*x)-16*x*pi^2)/log(x)^2,x, algorithm=
"maxima")

[Out]

((x^3 + 30*x^2)*e^(3*x)*log(x) - 4*(I*pi*x^2 + 30*I*pi*x)*log(x) - 8)/(x*e^(3*x)*log(x) - 4*I*pi*log(x))

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).

Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=-\frac {2 \, {\left (-i \, x^{3} e^{\left (3 \, x\right )} \log \left (x\right ) - 4 \, \pi x^{2} \log \left (x\right ) - 30 i \, x^{2} e^{\left (3 \, x\right )} \log \left (x\right ) - 120 \, \pi x \log \left (x\right ) + 8 i\right )}}{2 i \, x e^{\left (3 \, x\right )} \log \left (x\right ) + 8 \, \pi \log \left (x\right )} \]

[In]

integrate((((2*x^4+30*x^3)*exp(3*x)^2+I*(-16*x^3-240*x^2)*pi*exp(3*x)-(32*x^2+480*x)*pi^2)*log(x)^2+(24*x^2+8*
x)*exp(3*x)*log(x)+8*x*exp(3*x)-32*I*pi)/(x^3*exp(3*x)^2-8*I*x^2*pi*exp(3*x)-16*x*pi^2)/log(x)^2,x, algorithm=
"giac")

[Out]

-2*(-I*x^3*e^(3*x)*log(x) - 4*pi*x^2*log(x) - 30*I*x^2*e^(3*x)*log(x) - 120*pi*x*log(x) + 8*I)/(2*I*x*e^(3*x)*
log(x) + 8*pi*log(x))

Mupad [B] (verification not implemented)

Time = 8.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\right ) \log ^2(x)} \, dx=30\,x+\frac {8}{\ln \left (x\right )\,\left (-x\,{\mathrm {e}}^{3\,x}+\Pi \,4{}\mathrm {i}\right )}+x^2 \]

[In]

int((Pi*32i - 8*x*exp(3*x) + log(x)^2*(Pi^2*(480*x + 32*x^2) - exp(6*x)*(30*x^3 + 2*x^4) + Pi*exp(3*x)*(240*x^
2 + 16*x^3)*1i) - exp(3*x)*log(x)*(8*x + 24*x^2))/(log(x)^2*(16*Pi^2*x - x^3*exp(6*x) + Pi*x^2*exp(3*x)*8i)),x
)

[Out]

30*x + 8/(log(x)*(Pi*4i - x*exp(3*x))) + x^2