3.102.47 \(\int \frac {-225-300 \log (2)+(-60 x+75 \log (2)) \log (49 x^4)+(-4 x^2+20 x \log (2)) \log ^2(49 x^4)}{225 x^2+60 x^3 \log (49 x^4)+4 x^4 \log ^2(49 x^4)} \, dx\) [10147]

Optimal. Leaf size=28 \[ \frac {1-\frac {\log (2)}{\frac {2 x}{5}+\frac {3}{\log \left (49 x^4\right )}}}{x} \]

[Out]

(1-ln(2)/(2/5*x+3/ln(49*x^4)))/x

________________________________________________________________________________________

Rubi [F]
time = 0.67, 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 {-225-300 \log (2)+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{225 x^2+60 x^3 \log \left (49 x^4\right )+4 x^4 \log ^2\left (49 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-225 - 300*Log[2] + (-60*x + 75*Log[2])*Log[49*x^4] + (-4*x^2 + 20*x*Log[2])*Log[49*x^4]^2)/(225*x^2 + 60
*x^3*Log[49*x^4] + 4*x^4*Log[49*x^4]^2),x]

[Out]

-1/2*(x - Log[32])^2/(x^2*Log[32]) + (1125*Log[2]*Defer[Int][1/(x^3*(15 + 2*x*Log[49*x^4])^2), x])/2 - 300*Log
[2]*Defer[Int][1/(x^2*(15 + 2*x*Log[49*x^4])^2), x] - (225*Log[2]*Defer[Int][1/(x^3*(15 + 2*x*Log[49*x^4])), x
])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-225 \left (1+\frac {4 \log (2)}{3}\right )+(-60 x+75 \log (2)) \log \left (49 x^4\right )+\left (-4 x^2+20 x \log (2)\right ) \log ^2\left (49 x^4\right )}{x^2 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx\\ &=\int \left (\frac {-x+\log (32)}{x^3}-\frac {75 (-15+8 x) \log (2)}{2 x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2}-\frac {225 \log (2)}{2 x^3 \left (15+2 x \log \left (49 x^4\right )\right )}\right ) \, dx\\ &=-\left (\frac {1}{2} (75 \log (2)) \int \frac {-15+8 x}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx\right )-\frac {1}{2} (225 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )} \, dx+\int \frac {-x+\log (32)}{x^3} \, dx\\ &=-\frac {(x-\log (32))^2}{2 x^2 \log (32)}-\frac {1}{2} (75 \log (2)) \int \left (-\frac {15}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2}+\frac {8}{x^2 \left (15+2 x \log \left (49 x^4\right )\right )^2}\right ) \, dx-\frac {1}{2} (225 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )} \, dx\\ &=-\frac {(x-\log (32))^2}{2 x^2 \log (32)}-\frac {1}{2} (225 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )} \, dx-(300 \log (2)) \int \frac {1}{x^2 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx+\frac {1}{2} (1125 \log (2)) \int \frac {1}{x^3 \left (15+2 x \log \left (49 x^4\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 34, normalized size = 1.21 \begin {gather*} \frac {15+(2 x-5 \log (2)) \log \left (49 x^4\right )}{x \left (15+2 x \log \left (49 x^4\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-225 - 300*Log[2] + (-60*x + 75*Log[2])*Log[49*x^4] + (-4*x^2 + 20*x*Log[2])*Log[49*x^4]^2)/(225*x^
2 + 60*x^3*Log[49*x^4] + 4*x^4*Log[49*x^4]^2),x]

[Out]

(15 + (2*x - 5*Log[2])*Log[49*x^4])/(x*(15 + 2*x*Log[49*x^4]))

________________________________________________________________________________________

Maple [A]
time = 0.91, size = 40, normalized size = 1.43

method	result	size
risch	\(-\frac {5 \ln \left (2\right )-2 x}{2 x^{2}}+\frac {75 \ln \left (2\right )}{2 x^{2} \left (2 \ln \left (49 x^{4}\right ) x +15\right )}\)	\(35\)
norman	\(\frac {15+2 \ln \left (49 x^{4}\right ) x -5 \ln \left (49 x^{4}\right ) \ln \left (2\right )}{x \left (2 \ln \left (49 x^{4}\right ) x +15\right )}\)	\(39\)
default	\(\frac {1}{x}+\frac {-10 \ln \left (2\right ) \ln \left (7\right )-5 \ln \left (2\right ) \ln \left (x^{4}\right )}{x \left (4 x \ln \left (7\right )+2 x \ln \left (x^{4}\right )+15\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x*ln(2)-4*x^2)*ln(49*x^4)^2+(75*ln(2)-60*x)*ln(49*x^4)-300*ln(2)-225)/(4*x^4*ln(49*x^4)^2+60*x^3*ln(4
9*x^4)+225*x^2),x,method=_RETURNVERBOSE)

[Out]

1/x+(-10*ln(2)*ln(7)-5*ln(2)*ln(x^4))/x/(4*x*ln(7)+2*x*ln(x^4)+15)

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 46, normalized size = 1.64 \begin {gather*} \frac {4 \, x \log \left (7\right ) - 10 \, \log \left (7\right ) \log \left (2\right ) + 4 \, {\left (2 \, x - 5 \, \log \left (2\right )\right )} \log \left (x\right ) + 15}{4 \, x^{2} \log \left (7\right ) + 8 \, x^{2} \log \left (x\right ) + 15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*log(2)-4*x^2)*log(49*x^4)^2+(75*log(2)-60*x)*log(49*x^4)-300*log(2)-225)/(4*x^4*log(49*x^4)^2
+60*x^3*log(49*x^4)+225*x^2),x, algorithm="maxima")

[Out]

(4*x*log(7) - 10*log(7)*log(2) + 4*(2*x - 5*log(2))*log(x) + 15)/(4*x^2*log(7) + 8*x^2*log(x) + 15*x)

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 35, normalized size = 1.25 \begin {gather*} \frac {{\left (2 \, x - 5 \, \log \left (2\right )\right )} \log \left (49 \, x^{4}\right ) + 15}{2 \, x^{2} \log \left (49 \, x^{4}\right ) + 15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*log(2)-4*x^2)*log(49*x^4)^2+(75*log(2)-60*x)*log(49*x^4)-300*log(2)-225)/(4*x^4*log(49*x^4)^2
+60*x^3*log(49*x^4)+225*x^2),x, algorithm="fricas")

[Out]

((2*x - 5*log(2))*log(49*x^4) + 15)/(2*x^2*log(49*x^4) + 15*x)

________________________________________________________________________________________

Sympy [A]
time = 0.10, size = 34, normalized size = 1.21 \begin {gather*} \frac {75 \log {\left (2 \right )}}{4 x^{3} \log {\left (49 x^{4} \right )} + 30 x^{2}} - \frac {- 2 x + 5 \log {\left (2 \right )}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*ln(2)-4*x**2)*ln(49*x**4)**2+(75*ln(2)-60*x)*ln(49*x**4)-300*ln(2)-225)/(4*x**4*ln(49*x**4)**
2+60*x**3*ln(49*x**4)+225*x**2),x)

[Out]

75*log(2)/(4*x**3*log(49*x**4) + 30*x**2) - (-2*x + 5*log(2))/(2*x**2)

________________________________________________________________________________________

Giac [A]
time = 0.43, size = 37, normalized size = 1.32 \begin {gather*} \frac {75 \, \log \left (2\right )}{2 \, {\left (2 \, x^{3} \log \left (49 \, x^{4}\right ) + 15 \, x^{2}\right )}} + \frac {2 \, x - 5 \, \log \left (2\right )}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x*log(2)-4*x^2)*log(49*x^4)^2+(75*log(2)-60*x)*log(49*x^4)-300*log(2)-225)/(4*x^4*log(49*x^4)^2
+60*x^3*log(49*x^4)+225*x^2),x, algorithm="giac")

[Out]

75/2*log(2)/(2*x^3*log(49*x^4) + 15*x^2) + 1/2*(2*x - 5*log(2))/x^2

________________________________________________________________________________________

Mupad [B]
time = 7.62, size = 38, normalized size = 1.36 \begin {gather*} \frac {2\,x\,\ln \left (49\,x^4\right )-\ln \left (32\right )\,\ln \left (49\,x^4\right )+15}{x\,\left (2\,x\,\ln \left (49\,x^4\right )+15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(300*log(2) + log(49*x^4)*(60*x - 75*log(2)) - log(49*x^4)^2*(20*x*log(2) - 4*x^2) + 225)/(4*x^4*log(49*x
^4)^2 + 225*x^2 + 60*x^3*log(49*x^4)),x)

[Out]

(2*x*log(49*x^4) - log(32)*log(49*x^4) + 15)/(x*(2*x*log(49*x^4) + 15))

________________________________________________________________________________________