3.19.12 \(\int e^{\frac {2 (2+(450 x+525 x^2) \log (4))}{x}} (-4+2 x+1050 x^2 \log (4)) \, dx\) [1812]

3.19.12.1 Optimal result

Integrand size = 34, antiderivative size = 25 \[ \int e^{\frac {2 \left (2+\left (450 x+525 x^2\right ) \log (4)\right )}{x}} \left (-4+2 x+1050 x^2 \log (4)\right ) \, dx=e^{\frac {2 (2+75 x (x+6 (1+x)) \log (4))}{x}} x^2 \]

output

exp((150*x*(7*x+6)*ln(2)+2)/x)^2*x^2
 

3.19.12.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int e^{\frac {2 \left (2+\left (450 x+525 x^2\right ) \log (4)\right )}{x}} \left (-4+2 x+1050 x^2 \log (4)\right ) \, dx=2^{1800+2100 x} e^{4/x} x^2 \]

input

Integrate[E^((2*(2 + (450*x + 525*x^2)*Log[4]))/x)*(-4 + 2*x + 1050*x^2*Lo 
g[4]),x]
 

output

2^(1800 + 2100*x)*E^(4/x)*x^2
 

3.19.12.3 Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(25)=50\).

Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.76, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2726}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int e^{\frac {2 \left (\left (525 x^2+450 x\right ) \log (4)+2\right )}{x}} \left (1050 x^2 \log (4)+2 x-4\right ) \, dx\)

\(\Big \downarrow \) 2726

\(\displaystyle -\frac {4^{\frac {2 \left (525 x^2+450 x\right )}{x}} e^{4/x} \left (2-525 x^2 \log (4)\right )}{\frac {150 (7 x+3) \log (4)}{x}-\frac {75 \left (7 x^2+6 x\right ) \log (4)+2}{x^2}}\)

input

Int[E^((2*(2 + (450*x + 525*x^2)*Log[4]))/x)*(-4 + 2*x + 1050*x^2*Log[4]), 
x]
 

output

-((4^((2*(450*x + 525*x^2))/x)*E^(4/x)*(2 - 525*x^2*Log[4]))/((150*(3 + 7* 
x)*Log[4])/x - (2 + 75*(6*x + 7*x^2)*Log[4])/x^2))
 

3.19.12.3.1 Defintions of rubi rules used

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

3.19.12.4 Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
risch	\(x^{2} {\mathrm e}^{\frac {2100 x^{2} \ln \left (2\right )+1800 x \ln \left (2\right )+4}{x}}\)	\(25\)
gosper	\(x^{2} {\mathrm e}^{\frac {2100 x^{2} \ln \left (2\right )+1800 x \ln \left (2\right )+4}{x}}\)	\(27\)
norman	\(x^{2} {\mathrm e}^{\frac {4 \left (525 x^{2}+450 x \right ) \ln \left (2\right )+4}{x}}\)	\(27\)
parallelrisch	\(x^{2} {\mathrm e}^{\frac {4 \left (525 x^{2}+450 x \right ) \ln \left (2\right )+4}{x}}\)	\(27\)

input

int((2100*x^2*ln(2)+2*x-4)*exp((2*(525*x^2+450*x)*ln(2)+2)/x)^2,x,method=_ 
RETURNVERBOSE)
 

output

x^2*exp(4*(525*x^2*ln(2)+450*x*ln(2)+1)/x)
 

3.19.12.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int e^{\frac {2 \left (2+\left (450 x+525 x^2\right ) \log (4)\right )}{x}} \left (-4+2 x+1050 x^2 \log (4)\right ) \, dx=x^{2} e^{\left (\frac {4 \, {\left (75 \, {\left (7 \, x^{2} + 6 \, x\right )} \log \left (2\right ) + 1\right )}}{x}\right )} \]

input

integrate((2100*x^2*log(2)+2*x-4)*exp((2*(525*x^2+450*x)*log(2)+2)/x)^2,x, 
 algorithm=\
 

output

x^2*e^(4*(75*(7*x^2 + 6*x)*log(2) + 1)/x)
 

3.19.12.6 Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int e^{\frac {2 \left (2+\left (450 x+525 x^2\right ) \log (4)\right )}{x}} \left (-4+2 x+1050 x^2 \log (4)\right ) \, dx=x^{2} e^{\frac {2 \left (\left (1050 x^{2} + 900 x\right ) \log {\left (2 \right )} + 2\right )}{x}} \]

input

integrate((2100*x**2*ln(2)+2*x-4)*exp((2*(525*x**2+450*x)*ln(2)+2)/x)**2,x 
)
 

output

x**2*exp(2*((1050*x**2 + 900*x)*log(2) + 2)/x)
 

3.19.12.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int e^{\frac {2 \left (2+\left (450 x+525 x^2\right ) \log (4)\right )}{x}} \left (-4+2 x+1050 x^2 \log (4)\right ) \, dx=71448348576730208360402604523024658663907311448489024669693316988935593287322878666163481950176220037593478347105937422686501991894419788796088422137966026262523598150372719976137911322484446114613284904383977643176193557817897027023063420124852033989626806764509137929914787205373413116077254242653423277386226627159120168223623660139965116969572411841665962582988716865792650075294655252525257343163566042824495509307872827973214736884381496689456792434150079470111661811761376161068055664012337698456291039551943299284254570579952324837376 \, x^{2} e^{\left (2100 \, x \log \left (2\right ) + \frac {4}{x}\right )} \]

input

integrate((2100*x^2*log(2)+2*x-4)*exp((2*(525*x^2+450*x)*log(2)+2)/x)^2,x, 
 algorithm=\
 

output

71448348576730208360402604523024658663907311448489024669693316988935593287 
32287866616348195017622003759347834710593742268650199189441978879608842213 
79660262625235981503727199761379113224844461146132849043839776431761935578 
17897027023063420124852033989626806764509137929914787205373413116077254242 
65342327738622662715912016822362366013996511696957241184166596258298871686 
57926500752946552525252573431635660428244955093078728279732147368843814966 
89456792434150079470111661811761376161068055664012337698456291039551943299 
284254570579952324837376*x^2*e^(2100*x*log(2) + 4/x)
 

3.19.12.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{\frac {2 \left (2+\left (450 x+525 x^2\right ) \log (4)\right )}{x}} \left (-4+2 x+1050 x^2 \log (4)\right ) \, dx=x^{2} e^{\left (\frac {4 \, {\left (525 \, x^{2} \log \left (2\right ) + 450 \, x \log \left (2\right ) + 1\right )}}{x}\right )} \]

input

integrate((2100*x^2*log(2)+2*x-4)*exp((2*(525*x^2+450*x)*log(2)+2)/x)^2,x, 
 algorithm=\
 

output

x^2*e^(4*(525*x^2*log(2) + 450*x*log(2) + 1)/x)
 

3.19.12.9 Mupad [B] (verification not implemented)

Time = 12.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int e^{\frac {2 \left (2+\left (450 x+525 x^2\right ) \log (4)\right )}{x}} \left (-4+2 x+1050 x^2 \log (4)\right ) \, dx=2^{2100\,x+1800}\,x^2\,{\mathrm {e}}^{4/x} \]

input

int(exp((2*(2*log(2)*(450*x + 525*x^2) + 2))/x)*(2*x + 2100*x^2*log(2) - 4 
),x)
 

output

2^(2100*x + 1800)*x^2*exp(4/x)