\(\int \frac {2^{5 x/4} (100+(125 x-75 x^2+25 e^3 x^2) \log (2))}{800-960 x+288 x^2+32 e^6 x^2+e^3 (320 x-192 x^2)} \, dx\) [1124]

Optimal result

Integrand size = 63, antiderivative size = 23 \[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\frac {5\ 2^{-3+\frac {5 x}{4}}}{-3+e^3+\frac {5}{x}} \] Output:

5/8*exp(5/4*x*ln(2))/(exp(3)+5/x-3)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\frac {5\ 2^{-3+\frac {5 x}{4}} x \left (\left (-3+e^3\right ) x \log (2)+\log (32)\right )}{\left (5+\left (-3+e^3\right ) x\right )^2 \log (2)} \] Input:

Integrate[(2^((5*x)/4)*(100 + (125*x - 75*x^2 + 25*E^3*x^2)*Log[2]))/(800 
- 960*x + 288*x^2 + 32*E^6*x^2 + E^3*(320*x - 192*x^2)),x]
 

Output:

(5*2^(-3 + (5*x)/4)*x*((-3 + E^3)*x*Log[2] + Log[32]))/((5 + (-3 + E^3)*x) 
^2*Log[2])
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(23)=46\).

Time = 0.45 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 2007, 2629, 2009}

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.

\(\displaystyle \int \frac {2^{5 x/4} \left (\left (25 e^3 x^2-75 x^2+125 x\right ) \log (2)+100\right )}{32 e^6 x^2+288 x^2+e^3 \left (320 x-192 x^2\right )-960 x+800} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {2^{5 x/4} \left (\left (25 e^3 x^2-75 x^2+125 x\right ) \log (2)+100\right )}{\left (288+32 e^6\right ) x^2+e^3 \left (320 x-192 x^2\right )-960 x+800}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {2^{5 x/4} \left (\left (25 e^3 x^2-75 x^2+125 x\right ) \log (2)+100\right )}{\left (4 \sqrt {2} \left (e^3-3\right ) x+20 \sqrt {2}\right )^2}dx\)

\(\Big \downarrow \) 2629

\(\displaystyle \int \left (\frac {25\ 2^{\frac {5 x}{4}-3}}{\left (5-\left (3-e^3\right ) x\right )^2}+\frac {25\ 2^{\frac {5 x}{4}-5} \log (2)}{e^3-3}+\frac {125\ 2^{\frac {5 x}{4}-5} \log (2)}{\left (3-e^3\right ) \left (5-\left (3-e^3\right ) x\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {25\ 2^{\frac {5 x}{4}-3}}{\left (3-e^3\right ) \left (5-\left (3-e^3\right ) x\right )}-\frac {5\ 2^{\frac {5 x}{4}-3}}{3-e^3}\)

Input:

Int[(2^((5*x)/4)*(100 + (125*x - 75*x^2 + 25*E^3*x^2)*Log[2]))/(800 - 960* 
x + 288*x^2 + 32*E^6*x^2 + E^3*(320*x - 192*x^2)),x]
 

Output:

(-5*2^(-3 + (5*x)/4))/(3 - E^3) + (25*2^(-3 + (5*x)/4))/((3 - E^3)*(5 - (3 
 - E^3)*x))
 

Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 2007

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex 
pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol 
yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2629

Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte 
grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ 
Px, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]
 

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {5 x 2^{\frac {5 x}{4}}}{8 \left (x \,{\mathrm e}^{3}-3 x +5\right )}\)	\(20\)
gosper	\(\frac {5 x \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}}}{8 \left (x \,{\mathrm e}^{3}-3 x +5\right )}\)	\(21\)
norman	\(\frac {5 x \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}}}{8 \left (x \,{\mathrm e}^{3}-3 x +5\right )}\)	\(21\)
parallelrisch	\(\frac {5 x \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}}}{8 \left (x \,{\mathrm e}^{3}-3 x +5\right )}\)	\(21\)
derivativedivides	\(\frac {-\frac {125 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{2}}{8 \left ({\mathrm e}^{3}-3\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}-\frac {125 \ln \left (2\right )^{2} {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{32 \left ({\mathrm e}^{3}-3\right )^{2}}+\frac {3125 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{3}}{32 \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}+9\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}+\frac {125 \left (4 \,{\mathrm e}^{3}-25 \ln \left (2\right )-12\right ) \ln \left (2\right )^{2} {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{128 \left (-{\mathrm e}^{6}+6 \,{\mathrm e}^{3}-9\right ) \left ({\mathrm e}^{3}-3\right )}-\frac {30 \left (\frac {{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{2}}{16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144}-\frac {625 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{4}}{64 \left ({\mathrm e}^{3}-3\right ) \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}+9\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}+\frac {25 \ln \left (2\right )^{3} \left (8 \,{\mathrm e}^{3}-24-25 \ln \left (2\right )\right ) {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{16 \left ({\mathrm e}^{3}-3\right )^{2} \left (16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144\right )}\right )}{\ln \left (2\right )}+\frac {10 \,{\mathrm e}^{3} \left (\frac {{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{2}}{16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144}-\frac {625 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{4}}{64 \left ({\mathrm e}^{3}-3\right ) \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}+9\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}+\frac {25 \ln \left (2\right )^{3} \left (8 \,{\mathrm e}^{3}-24-25 \ln \left (2\right )\right ) {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{16 \left ({\mathrm e}^{3}-3\right )^{2} \left (16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144\right )}\right )}{\ln \left (2\right )}}{\ln \left (2\right )}\)	\(486\)
default	\(\frac {-\frac {125 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{2}}{8 \left ({\mathrm e}^{3}-3\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}-\frac {125 \ln \left (2\right )^{2} {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{32 \left ({\mathrm e}^{3}-3\right )^{2}}+\frac {3125 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{3}}{32 \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}+9\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}+\frac {125 \left (4 \,{\mathrm e}^{3}-25 \ln \left (2\right )-12\right ) \ln \left (2\right )^{2} {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{128 \left (-{\mathrm e}^{6}+6 \,{\mathrm e}^{3}-9\right ) \left ({\mathrm e}^{3}-3\right )}-\frac {30 \left (\frac {{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{2}}{16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144}-\frac {625 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{4}}{64 \left ({\mathrm e}^{3}-3\right ) \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}+9\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}+\frac {25 \ln \left (2\right )^{3} \left (8 \,{\mathrm e}^{3}-24-25 \ln \left (2\right )\right ) {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{16 \left ({\mathrm e}^{3}-3\right )^{2} \left (16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144\right )}\right )}{\ln \left (2\right )}+\frac {10 \,{\mathrm e}^{3} \left (\frac {{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{2}}{16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144}-\frac {625 \,{\mathrm e}^{\frac {5 x \ln \left (2\right )}{4}} \ln \left (2\right )^{4}}{64 \left ({\mathrm e}^{3}-3\right ) \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}+9\right ) \left (5 x \,{\mathrm e}^{3} \ln \left (2\right )+25 \ln \left (2\right )-15 x \ln \left (2\right )\right )}+\frac {25 \ln \left (2\right )^{3} \left (8 \,{\mathrm e}^{3}-24-25 \ln \left (2\right )\right ) {\mathrm e}^{-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}} \operatorname {expIntegral}_{1}\left (-\frac {5 x \ln \left (2\right )}{4}-\frac {25 \ln \left (2\right )}{4 \left ({\mathrm e}^{3}-3\right )}\right )}{16 \left ({\mathrm e}^{3}-3\right )^{2} \left (16 \,{\mathrm e}^{6}-96 \,{\mathrm e}^{3}+144\right )}\right )}{\ln \left (2\right )}}{\ln \left (2\right )}\)	\(486\)

Input:

int(((25*x^2*exp(3)-75*x^2+125*x)*ln(2)+100)*exp(5/4*x*ln(2))/(32*x^2*exp( 
3)^2+(-192*x^2+320*x)*exp(3)+288*x^2-960*x+800),x,method=_RETURNVERBOSE)
 

Output:

5/8*x*2^(5/4*x)/(x*exp(3)-3*x+5)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\frac {5 \cdot 2^{\frac {5}{4} \, x} x}{8 \, {\left (x e^{3} - 3 \, x + 5\right )}} \] Input:

integrate(((25*x^2*exp(3)-75*x^2+125*x)*log(2)+100)*exp(5/4*x*log(2))/(32* 
x^2*exp(3)^2+(-192*x^2+320*x)*exp(3)+288*x^2-960*x+800),x, algorithm="fric 
as")
 

Output:

5/8*2^(5/4*x)*x/(x*e^3 - 3*x + 5)
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\frac {5 x e^{\frac {5 x \log {\left (2 \right )}}{4}}}{- 24 x + 8 x e^{3} + 40} \] Input:

integrate(((25*x**2*exp(3)-75*x**2+125*x)*ln(2)+100)*exp(5/4*x*ln(2))/(32* 
x**2*exp(3)**2+(-192*x**2+320*x)*exp(3)+288*x**2-960*x+800),x)
 

Output:

5*x*exp(5*x*log(2)/4)/(-24*x + 8*x*exp(3) + 40)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\frac {5 \cdot 2^{\frac {5}{4} \, x} x}{8 \, {\left (x {\left (e^{3} - 3\right )} + 5\right )}} \] Input:

integrate(((25*x^2*exp(3)-75*x^2+125*x)*log(2)+100)*exp(5/4*x*log(2))/(32* 
x^2*exp(3)^2+(-192*x^2+320*x)*exp(3)+288*x^2-960*x+800),x, algorithm="maxi 
ma")
 

Output:

5/8*2^(5/4*x)*x/(x*(e^3 - 3) + 5)
 

Giac [F]

\[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\int { \frac {25 \, {\left ({\left (x^{2} e^{3} - 3 \, x^{2} + 5 \, x\right )} \log \left (2\right ) + 4\right )} 2^{\frac {5}{4} \, x}}{32 \, {\left (x^{2} e^{6} + 9 \, x^{2} - 2 \, {\left (3 \, x^{2} - 5 \, x\right )} e^{3} - 30 \, x + 25\right )}} \,d x } \] Input:

integrate(((25*x^2*exp(3)-75*x^2+125*x)*log(2)+100)*exp(5/4*x*log(2))/(32* 
x^2*exp(3)^2+(-192*x^2+320*x)*exp(3)+288*x^2-960*x+800),x, algorithm="giac 
")
 

Output:

integrate(25/32*((x^2*e^3 - 3*x^2 + 5*x)*log(2) + 4)*2^(5/4*x)/(x^2*e^6 + 
9*x^2 - 2*(3*x^2 - 5*x)*e^3 - 30*x + 25), x)
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\frac {20\,2^{\frac {5\,x}{4}}\,x}{32\,x\,{\mathrm {e}}^3-96\,x+160} \] Input:

int((exp((5*x*log(2))/4)*(log(2)*(125*x + 25*x^2*exp(3) - 75*x^2) + 100))/ 
(exp(3)*(320*x - 192*x^2) - 960*x + 32*x^2*exp(6) + 288*x^2 + 800),x)
 

Output:

(20*2^((5*x)/4)*x)/(32*x*exp(3) - 96*x + 160)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {2^{5 x/4} \left (100+\left (125 x-75 x^2+25 e^3 x^2\right ) \log (2)\right )}{800-960 x+288 x^2+32 e^6 x^2+e^3 \left (320 x-192 x^2\right )} \, dx=\frac {5 \,2^{\frac {5 x}{4}} x}{8 e^{3} x -24 x +40} \] Input:

int(((25*x^2*exp(3)-75*x^2+125*x)*log(2)+100)*exp(5/4*x*log(2))/(32*x^2*ex 
p(3)^2+(-192*x^2+320*x)*exp(3)+288*x^2-960*x+800),x)
 

Output:

(5*2**((5*x)/4)*x)/(8*(e**3*x - 3*x + 5))