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

Optimal. Leaf size=30 \[ (15+x)^2+\frac {2}{\left (i \pi -\frac {1}{4} e^{3 x} x\right ) \log (x)} \]

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Rubi [F]  time = 1.68, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 37, normalized size = 1.23 \begin {gather*} 2 \left (15 x+\frac {x^2}{2}-\frac {4 i}{\left (4 \pi +i e^{3 x} x\right ) \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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]))

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fricas [A]  time = 0.82, size = 47, normalized size = 1.57 \begin {gather*} \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 \relax (x) - 8}{{\left (-4 i \, \pi + x e^{\left (3 \, x\right )}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 0.78, size = 57, normalized size = 1.90 \begin {gather*} -\frac {2 \, {\left (-i \, x^{3} e^{\left (3 \, x\right )} \log \relax (x) - 4 \, \pi x^{2} \log \relax (x) - 30 i \, x^{2} e^{\left (3 \, x\right )} \log \relax (x) - 120 \, \pi x \log \relax (x) + 8 i\right )}}{2 i \, x e^{\left (3 \, x\right )} \log \relax (x) + 8 \, \pi \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.23, size = 29, normalized size = 0.97

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.50, size = 50, normalized size = 1.67 \begin {gather*} \frac {{\left (x^{3} + 30 \, x^{2}\right )} e^{\left (3 \, x\right )} \log \relax (x) - 4 \, {\left (i \, \pi x^{2} + 30 i \, \pi x\right )} \log \relax (x) - 8}{x e^{\left (3 \, x\right )} \log \relax (x) - 4 i \, \pi \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 1.30, size = 27, normalized size = 0.90 \begin {gather*} 30\,x+\frac {8}{\ln \relax (x)\,\left (-x\,{\mathrm {e}}^{3\,x}+\Pi \,4{}\mathrm {i}\right )}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.50, size = 26, normalized size = 0.87 \begin {gather*} x^{2} + 30 x + \frac {8}{- x e^{3 x} \log {\relax (x )} + 4 i \pi \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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