3.93.22 \(\int \frac {48 (i \pi +\log (3))+48 (i \pi +\log (3)) \log (5 x)-15 x^2 (i \pi +\log (3))^2 \log ^2(5 x)}{512+320 x^2 (i \pi +\log (3)) \log (5 x)+50 x^4 (i \pi +\log (3))^2 \log ^2(5 x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {3}{2 \left (5 x+\frac {16}{x (i \pi +\log (3)) \log (5 x)}\right )} \]

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Rubi [F]  time = 1.21, 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 {48 (i \pi +\log (3))+48 (i \pi +\log (3)) \log (5 x)-15 x^2 (i \pi +\log (3))^2 \log ^2(5 x)}{512+320 x^2 (i \pi +\log (3)) \log (5 x)+50 x^4 (i \pi +\log (3))^2 \log ^2(5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(48*(I*Pi + Log[3]) + 48*(I*Pi + Log[3])*Log[5*x] - 15*x^2*(I*Pi + Log[3])^2*Log[5*x]^2)/(512 + 320*x^2*(I
*Pi + Log[3])*Log[5*x] + 50*x^4*(I*Pi + Log[3])^2*Log[5*x]^2),x]

[Out]

3/(10*x) - 24*(I*Pi + Log[3])*Defer[Int][(16*I - 5*Pi*x^2*(1 - (I*Log[3])/Pi)*Log[5*x])^(-2), x] + (768*Defer[
Int][1/(x^2*(16*I - 5*Pi*x^2*(1 - (I*Log[3])/Pi)*Log[5*x])^2), x])/5 + ((72*I)/5)*Defer[Int][1/(x^2*(16*I - 5*
Pi*x^2*(1 - (I*Log[3])/Pi)*Log[5*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 (\pi -i \log (3)) \left (-16 i-16 i \log (5 x)-5 x^2 (\pi -i \log (3)) \log ^2(5 x)\right )}{2 \left (16 i-5 x^2 (\pi -i \log (3)) \log (5 x)\right )^2} \, dx\\ &=\frac {1}{2} (3 (\pi -i \log (3))) \int \frac {-16 i-16 i \log (5 x)-5 x^2 (\pi -i \log (3)) \log ^2(5 x)}{\left (16 i-5 x^2 (\pi -i \log (3)) \log (5 x)\right )^2} \, dx\\ &=\frac {1}{2} (3 (\pi -i \log (3))) \int \left (-\frac {1}{5 x^2 (\pi -i \log (3))}+\frac {16 \left (32-5 x^2 (i \pi +\log (3))\right )}{5 x^2 (\pi -i \log (3)) \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2}-\frac {48 i}{5 x^2 (-\pi +i \log (3)) \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )}\right ) \, dx\\ &=\frac {3}{10 x}+\frac {72}{5} i \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )} \, dx+\frac {24}{5} \int \frac {32-5 x^2 (i \pi +\log (3))}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2} \, dx\\ &=\frac {3}{10 x}+\frac {72}{5} i \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )} \, dx+\frac {24}{5} \int \left (\frac {32}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2}-\frac {5 i (\pi -i \log (3))}{\left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2}\right ) \, dx\\ &=\frac {3}{10 x}+\frac {72}{5} i \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )} \, dx+\frac {768}{5} \int \frac {1}{x^2 \left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2} \, dx-(24 (i \pi +\log (3))) \int \frac {1}{\left (16 i-5 \pi x^2 \left (1-\frac {i \log (3)}{\pi }\right ) \log (5 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 40, normalized size = 1.29 \begin {gather*} \frac {3 x (\pi -i \log (3)) \log (5 x)}{2 \left (-16 i+5 x^2 (\pi -i \log (3)) \log (5 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48*(I*Pi + Log[3]) + 48*(I*Pi + Log[3])*Log[5*x] - 15*x^2*(I*Pi + Log[3])^2*Log[5*x]^2)/(512 + 320*
x^2*(I*Pi + Log[3])*Log[5*x] + 50*x^4*(I*Pi + Log[3])^2*Log[5*x]^2),x]

[Out]

(3*x*(Pi - I*Log[3])*Log[5*x])/(2*(-16*I + 5*x^2*(Pi - I*Log[3])*Log[5*x]))

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fricas [A]  time = 0.71, size = 40, normalized size = 1.29 \begin {gather*} \frac {3 \, {\left (-i \, \pi x - x \log \relax (3)\right )} \log \left (5 \, x\right )}{2 \, {\left (5 \, {\left (-i \, \pi x^{2} - x^{2} \log \relax (3)\right )} \log \left (5 \, x\right ) - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^2*(log(3)+I*pi)^2*log(5*x)^2+48*(log(3)+I*pi)*log(5*x)+48*log(3)+48*I*pi)/(50*x^4*(log(3)+I*p
i)^2*log(5*x)^2+320*x^2*(log(3)+I*pi)*log(5*x)+512),x, algorithm="fricas")

[Out]

3/2*(-I*pi*x - x*log(3))*log(5*x)/(5*(-I*pi*x^2 - x^2*log(3))*log(5*x) - 16)

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giac [A]  time = 0.24, size = 35, normalized size = 1.13 \begin {gather*} -\frac {24}{25 i \, \pi x^{3} \log \left (5 \, x\right ) + 25 \, x^{3} \log \relax (3) \log \left (5 \, x\right ) + 80 \, x} + \frac {3}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^2*(log(3)+I*pi)^2*log(5*x)^2+48*(log(3)+I*pi)*log(5*x)+48*log(3)+48*I*pi)/(50*x^4*(log(3)+I*p
i)^2*log(5*x)^2+320*x^2*(log(3)+I*pi)*log(5*x)+512),x, algorithm="giac")

[Out]

-24/(25*I*pi*x^3*log(5*x) + 25*x^3*log(3)*log(5*x) + 80*x) + 3/10/x

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maple [A]  time = 0.25, size = 40, normalized size = 1.29

method	result	size
risch	\(\frac {3}{10 x}+\frac {24 i}{5 x \left (-5 i \ln \relax (3) x^{2} \ln \left (5 x \right )+5 \pi \ln \left (5 x \right ) x^{2}-16 i\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-15*x^2*(ln(3)+I*Pi)^2*ln(5*x)^2+48*(ln(3)+I*Pi)*ln(5*x)+48*ln(3)+48*I*Pi)/(50*x^4*(ln(3)+I*Pi)^2*ln(5*x)
^2+320*x^2*(ln(3)+I*Pi)*ln(5*x)+512),x,method=_RETURNVERBOSE)

[Out]

3/10/x+24/5*I/x/(-5*I*ln(3)*x^2*ln(5*x)+5*Pi*ln(5*x)*x^2-16*I)

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maxima [B]  time = 0.60, size = 59, normalized size = 1.90 \begin {gather*} \frac {3 \, {\left ({\left (\pi - i \, \log \relax (3)\right )} x \log \relax (x) + {\left (\pi \log \relax (5) - i \, \log \relax (5) \log \relax (3)\right )} x\right )}}{2 \, {\left (5 \, {\left (\pi - i \, \log \relax (3)\right )} x^{2} \log \relax (x) + 5 \, {\left (\pi \log \relax (5) - i \, \log \relax (5) \log \relax (3)\right )} x^{2} - 16 i\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x^2*(log(3)+I*pi)^2*log(5*x)^2+48*(log(3)+I*pi)*log(5*x)+48*log(3)+48*I*pi)/(50*x^4*(log(3)+I*p
i)^2*log(5*x)^2+320*x^2*(log(3)+I*pi)*log(5*x)+512),x, algorithm="maxima")

[Out]

3/2*((pi - I*log(3))*x*log(x) + (pi*log(5) - I*log(5)*log(3))*x)/(5*(pi - I*log(3))*x^2*log(x) + 5*(pi*log(5)
- I*log(5)*log(3))*x^2 - 16*I)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\Pi \,48{}\mathrm {i}+48\,\ln \relax (3)+48\,\ln \left (5\,x\right )\,\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )-15\,x^2\,{\ln \left (5\,x\right )}^2\,{\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )}^2}{50\,x^4\,{\ln \left (5\,x\right )}^2\,{\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )}^2+320\,x^2\,\ln \left (5\,x\right )\,\left (\ln \relax (3)+\Pi \,1{}\mathrm {i}\right )+512} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((Pi*48i + 48*log(3) + 48*log(5*x)*(Pi*1i + log(3)) - 15*x^2*log(5*x)^2*(Pi*1i + log(3))^2)/(50*x^4*log(5*x
)^2*(Pi*1i + log(3))^2 + 320*x^2*log(5*x)*(Pi*1i + log(3)) + 512),x)

[Out]

int((Pi*48i + 48*log(3) + 48*log(5*x)*(Pi*1i + log(3)) - 15*x^2*log(5*x)^2*(Pi*1i + log(3))^2)/(50*x^4*log(5*x
)^2*(Pi*1i + log(3))^2 + 320*x^2*log(5*x)*(Pi*1i + log(3)) + 512), x)

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sympy [B]  time = 40.22, size = 54, normalized size = 1.74 \begin {gather*} \frac {24}{- 25 x^{3} \log {\relax (3 )} \log {\relax (5 )} - 25 i \pi x^{3} \log {\relax (5 )} - 80 x + \left (- 25 x^{3} \log {\relax (3 )} - 25 i \pi x^{3}\right ) \log {\relax (x )}} + \frac {3}{10 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x**2*(ln(3)+I*pi)**2*ln(5*x)**2+48*(ln(3)+I*pi)*ln(5*x)+48*ln(3)+48*I*pi)/(50*x**4*(ln(3)+I*pi)
**2*ln(5*x)**2+320*x**2*(ln(3)+I*pi)*ln(5*x)+512),x)

[Out]

24/(-25*x**3*log(3)*log(5) - 25*I*pi*x**3*log(5) - 80*x + (-25*x**3*log(3) - 25*I*pi*x**3)*log(x)) + 3/(10*x)

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