3.3.72 \(\int \frac {-300-480 x+e^{2 x} (50 x^2+100 x^3)+(60+120 x+e^{2 x} (-20 x^2-40 x^3)) \log (2+4 x)+e^{2 x} (2 x^2+4 x^3) \log ^2(2+4 x)}{25 x^2+50 x^3+(-10 x^2-20 x^3) \log (2+4 x)+(x^2+2 x^3) \log ^2(2+4 x)} \, dx\)

Optimal. Leaf size=21 \[ e^{2 x}-\frac {60}{x (-5+\log (2+4 x))} \]

________________________________________________________________________________________

Rubi [F]  time = 3.88, 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 {-300-480 x+e^{2 x} \left (50 x^2+100 x^3\right )+\left (60+120 x+e^{2 x} \left (-20 x^2-40 x^3\right )\right ) \log (2+4 x)+e^{2 x} \left (2 x^2+4 x^3\right ) \log ^2(2+4 x)}{25 x^2+50 x^3+\left (-10 x^2-20 x^3\right ) \log (2+4 x)+\left (x^2+2 x^3\right ) \log ^2(2+4 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-300 - 480*x + E^(2*x)*(50*x^2 + 100*x^3) + (60 + 120*x + E^(2*x)*(-20*x^2 - 40*x^3))*Log[2 + 4*x] + E^(2
*x)*(2*x^2 + 4*x^3)*Log[2 + 4*x]^2)/(25*x^2 + 50*x^3 + (-10*x^2 - 20*x^3)*Log[2 + 4*x] + (x^2 + 2*x^3)*Log[2 +
4*x]^2),x]

[Out]

E^(2*x) - 120/(5 - Log[2 + 4*x]) + 120*Defer[Int][1/(x*(-5 + Log[2 + 4*x])^2), x] + 60*Defer[Int][1/(x^2*(-5 +
Log[2 + 4*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-300-480 x+e^{2 x} \left (50 x^2+100 x^3\right )+\left (60+120 x+e^{2 x} \left (-20 x^2-40 x^3\right )\right ) \log (2+4 x)+e^{2 x} \left (2 x^2+4 x^3\right ) \log ^2(2+4 x)}{x^2 (1+2 x) (5-\log (2+4 x))^2} \, dx\\ &=\int \left (2 e^{2 x}-\frac {300}{x^2 (1+2 x) (-5+\log (2+4 x))^2}-\frac {480}{x (1+2 x) (-5+\log (2+4 x))^2}+\frac {60 \log (2+4 x)}{x^2 (1+2 x) (-5+\log (2+4 x))^2}+\frac {120 \log (2+4 x)}{x (1+2 x) (-5+\log (2+4 x))^2}\right ) \, dx\\ &=2 \int e^{2 x} \, dx+60 \int \frac {\log (2+4 x)}{x^2 (1+2 x) (-5+\log (2+4 x))^2} \, dx+120 \int \frac {\log (2+4 x)}{x (1+2 x) (-5+\log (2+4 x))^2} \, dx-300 \int \frac {1}{x^2 (1+2 x) (-5+\log (2+4 x))^2} \, dx-480 \int \frac {1}{x (1+2 x) (-5+\log (2+4 x))^2} \, dx\\ &=e^{2 x}+60 \int \left (\frac {5}{x^2 (1+2 x) (-5+\log (2+4 x))^2}+\frac {1}{x^2 (1+2 x) (-5+\log (2+4 x))}\right ) \, dx+120 \int \left (\frac {5}{x (1+2 x) (-5+\log (2+4 x))^2}+\frac {1}{x (1+2 x) (-5+\log (2+4 x))}\right ) \, dx-300 \int \left (\frac {1}{x^2 (-5+\log (2+4 x))^2}-\frac {2}{x (-5+\log (2+4 x))^2}+\frac {4}{(1+2 x) (-5+\log (2+4 x))^2}\right ) \, dx-480 \int \left (\frac {1}{x (-5+\log (2+4 x))^2}-\frac {2}{(1+2 x) (-5+\log (2+4 x))^2}\right ) \, dx\\ &=e^{2 x}+60 \int \frac {1}{x^2 (1+2 x) (-5+\log (2+4 x))} \, dx+120 \int \frac {1}{x (1+2 x) (-5+\log (2+4 x))} \, dx-300 \int \frac {1}{x^2 (-5+\log (2+4 x))^2} \, dx+300 \int \frac {1}{x^2 (1+2 x) (-5+\log (2+4 x))^2} \, dx-480 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx+600 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx+600 \int \frac {1}{x (1+2 x) (-5+\log (2+4 x))^2} \, dx+960 \int \frac {1}{(1+2 x) (-5+\log (2+4 x))^2} \, dx-1200 \int \frac {1}{(1+2 x) (-5+\log (2+4 x))^2} \, dx\\ &=e^{2 x}+60 \int \left (\frac {1}{x^2 (-5+\log (2+4 x))}-\frac {2}{x (-5+\log (2+4 x))}+\frac {4}{(1+2 x) (-5+\log (2+4 x))}\right ) \, dx+120 \int \left (\frac {1}{x (-5+\log (2+4 x))}-\frac {2}{(1+2 x) (-5+\log (2+4 x))}\right ) \, dx+240 \operatorname {Subst}\left (\int \frac {2}{x (-5+\log (x))^2} \, dx,x,2+4 x\right )+300 \int \left (\frac {1}{x^2 (-5+\log (2+4 x))^2}-\frac {2}{x (-5+\log (2+4 x))^2}+\frac {4}{(1+2 x) (-5+\log (2+4 x))^2}\right ) \, dx-300 \int \frac {1}{x^2 (-5+\log (2+4 x))^2} \, dx-300 \operatorname {Subst}\left (\int \frac {2}{x (-5+\log (x))^2} \, dx,x,2+4 x\right )-480 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx+600 \int \left (\frac {1}{x (-5+\log (2+4 x))^2}-\frac {2}{(1+2 x) (-5+\log (2+4 x))^2}\right ) \, dx+600 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx\\ &=e^{2 x}+60 \int \frac {1}{x^2 (-5+\log (2+4 x))} \, dx-480 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx+480 \operatorname {Subst}\left (\int \frac {1}{x (-5+\log (x))^2} \, dx,x,2+4 x\right )+600 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx-600 \operatorname {Subst}\left (\int \frac {1}{x (-5+\log (x))^2} \, dx,x,2+4 x\right )\\ &=e^{2 x}+60 \int \frac {1}{x^2 (-5+\log (2+4 x))} \, dx-480 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx+480 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-5+\log (2+4 x)\right )+600 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx-600 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-5+\log (2+4 x)\right )\\ &=e^{2 x}-\frac {120}{5-\log (2+4 x)}+60 \int \frac {1}{x^2 (-5+\log (2+4 x))} \, dx-480 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx+600 \int \frac {1}{x (-5+\log (2+4 x))^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 27, normalized size = 1.29 \begin {gather*} 2 \left (\frac {e^{2 x}}{2}-\frac {30}{x (-5+\log (2+4 x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-300 - 480*x + E^(2*x)*(50*x^2 + 100*x^3) + (60 + 120*x + E^(2*x)*(-20*x^2 - 40*x^3))*Log[2 + 4*x]
+ E^(2*x)*(2*x^2 + 4*x^3)*Log[2 + 4*x]^2)/(25*x^2 + 50*x^3 + (-10*x^2 - 20*x^3)*Log[2 + 4*x] + (x^2 + 2*x^3)*L
og[2 + 4*x]^2),x]

[Out]

2*(E^(2*x)/2 - 30/(x*(-5 + Log[2 + 4*x])))

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 36, normalized size = 1.71 \begin {gather*} \frac {x e^{\left (2 \, x\right )} \log \left (4 \, x + 2\right ) - 5 \, x e^{\left (2 \, x\right )} - 60}{x \log \left (4 \, x + 2\right ) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+2*x^2)*exp(x)^2*log(4*x+2)^2+((-40*x^3-20*x^2)*exp(x)^2+120*x+60)*log(4*x+2)+(100*x^3+50*x^2
)*exp(x)^2-480*x-300)/((2*x^3+x^2)*log(4*x+2)^2+(-20*x^3-10*x^2)*log(4*x+2)+50*x^3+25*x^2),x, algorithm="frica
s")

[Out]

(x*e^(2*x)*log(4*x + 2) - 5*x*e^(2*x) - 60)/(x*log(4*x + 2) - 5*x)

________________________________________________________________________________________

giac [A]  time = 0.85, size = 36, normalized size = 1.71 \begin {gather*} \frac {x e^{\left (2 \, x\right )} \log \left (4 \, x + 2\right ) - 5 \, x e^{\left (2 \, x\right )} - 60}{x \log \left (4 \, x + 2\right ) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+2*x^2)*exp(x)^2*log(4*x+2)^2+((-40*x^3-20*x^2)*exp(x)^2+120*x+60)*log(4*x+2)+(100*x^3+50*x^2
)*exp(x)^2-480*x-300)/((2*x^3+x^2)*log(4*x+2)^2+(-20*x^3-10*x^2)*log(4*x+2)+50*x^3+25*x^2),x, algorithm="giac"
)

[Out]

(x*e^(2*x)*log(4*x + 2) - 5*x*e^(2*x) - 60)/(x*log(4*x + 2) - 5*x)

________________________________________________________________________________________

maple [A]  time = 0.25, size = 21, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{2 x}-\frac {60}{\left (\ln \left (4 x +2\right )-5\right ) x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3+2*x^2)*exp(x)^2*ln(4*x+2)^2+((-40*x^3-20*x^2)*exp(x)^2+120*x+60)*ln(4*x+2)+(100*x^3+50*x^2)*exp(x)
^2-480*x-300)/((2*x^3+x^2)*ln(4*x+2)^2+(-20*x^3-10*x^2)*ln(4*x+2)+50*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

exp(2*x)-60/(ln(4*x+2)-5)/x

________________________________________________________________________________________

maxima [B]  time = 0.64, size = 42, normalized size = 2.00 \begin {gather*} \frac {x {\left (\log \relax (2) - 5\right )} e^{\left (2 \, x\right )} + x e^{\left (2 \, x\right )} \log \left (2 \, x + 1\right ) - 60}{x {\left (\log \relax (2) - 5\right )} + x \log \left (2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+2*x^2)*exp(x)^2*log(4*x+2)^2+((-40*x^3-20*x^2)*exp(x)^2+120*x+60)*log(4*x+2)+(100*x^3+50*x^2
)*exp(x)^2-480*x-300)/((2*x^3+x^2)*log(4*x+2)^2+(-20*x^3-10*x^2)*log(4*x+2)+50*x^3+25*x^2),x, algorithm="maxim
a")

[Out]

(x*(log(2) - 5)*e^(2*x) + x*e^(2*x)*log(2*x + 1) - 60)/(x*(log(2) - 5) + x*log(2*x + 1))

________________________________________________________________________________________

mupad [B]  time = 0.48, size = 37, normalized size = 1.76 \begin {gather*} -\frac {5\,x\,{\mathrm {e}}^{2\,x}-x\,{\mathrm {e}}^{2\,x}\,\ln \left (4\,x+2\right )+60}{x\,\left (\ln \left (4\,x+2\right )-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(50*x^2 + 100*x^3) - 480*x + log(4*x + 2)*(120*x - exp(2*x)*(20*x^2 + 40*x^3) + 60) + exp(2*x)*l
og(4*x + 2)^2*(2*x^2 + 4*x^3) - 300)/(log(4*x + 2)^2*(x^2 + 2*x^3) - log(4*x + 2)*(10*x^2 + 20*x^3) + 25*x^2 +
50*x^3),x)

[Out]

-(5*x*exp(2*x) - x*exp(2*x)*log(4*x + 2) + 60)/(x*(log(4*x + 2) - 5))

________________________________________________________________________________________

sympy [A]  time = 0.39, size = 17, normalized size = 0.81 \begin {gather*} e^{2 x} - \frac {60}{x \log {\left (4 x + 2 \right )} - 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3+2*x**2)*exp(x)**2*ln(4*x+2)**2+((-40*x**3-20*x**2)*exp(x)**2+120*x+60)*ln(4*x+2)+(100*x**3+
50*x**2)*exp(x)**2-480*x-300)/((2*x**3+x**2)*ln(4*x+2)**2+(-20*x**3-10*x**2)*ln(4*x+2)+50*x**3+25*x**2),x)

[Out]

exp(2*x) - 60/(x*log(4*x + 2) - 5*x)

________________________________________________________________________________________