\(\int \frac {-3 x^3+e^3 (10 x^2-x^3-x^4)+e^x (-6 x^3+e^3 (-10 x^2-2 x^3-2 x^4))+e^{2 x} (6 x^3+e^3 (2 x^3+2 x^4))+(-30 x+e^3 (40 x-10 x^2)+e^x (30 x-30 x^2+e^3 (10 x-10 x^3))) \log (3+e^3 (1+x))+(-150+e^3 (-50-50 x)) \log ^2(3+e^3 (1+x))}{3 x^3+e^3 (x^3+x^4)} \, dx\) [1712]

Optimal result

Integrand size = 181, antiderivative size = 26 \[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=-x+\left (-1+e^x-\frac {5 \log \left (3+e^3 (1+x)\right )}{x}\right )^2 \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=-2 e^x+e^{2 x}-x-\frac {10 \left (-1+e^x\right ) \log \left (3+e^3 (1+x)\right )}{x}+\frac {25 \log ^2\left (3+e^3 (1+x)\right )}{x^2} \] Input:

Integrate[(-3*x^3 + E^3*(10*x^2 - x^3 - x^4) + E^x*(-6*x^3 + E^3*(-10*x^2 
- 2*x^3 - 2*x^4)) + E^(2*x)*(6*x^3 + E^3*(2*x^3 + 2*x^4)) + (-30*x + E^3*( 
40*x - 10*x^2) + E^x*(30*x - 30*x^2 + E^3*(10*x - 10*x^3)))*Log[3 + E^3*(1 
 + x)] + (-150 + E^3*(-50 - 50*x))*Log[3 + E^3*(1 + x)]^2)/(3*x^3 + E^3*(x 
^3 + x^4)),x]
 

Output:

-2*E^x + E^(2*x) - x - (10*(-1 + E^x)*Log[3 + E^3*(1 + x)])/x + (25*Log[3 
+ E^3*(1 + x)]^2)/x^2
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 3.58 (sec) , antiderivative size = 365, normalized size of antiderivative = 14.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {2026, 7293, 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.

\(\displaystyle \int \frac {-3 x^3+e^{2 x} \left (6 x^3+e^3 \left (2 x^4+2 x^3\right )\right )+\left (e^3 \left (40 x-10 x^2\right )+e^x \left (e^3 \left (10 x-10 x^3\right )-30 x^2+30 x\right )-30 x\right ) \log \left (e^3 (x+1)+3\right )+e^3 \left (-x^4-x^3+10 x^2\right )+e^x \left (e^3 \left (-2 x^4-2 x^3-10 x^2\right )-6 x^3\right )+\left (e^3 (-50 x-50)-150\right ) \log ^2\left (e^3 (x+1)+3\right )}{3 x^3+e^3 \left (x^4+x^3\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {-3 x^3+e^{2 x} \left (6 x^3+e^3 \left (2 x^4+2 x^3\right )\right )+\left (e^3 \left (40 x-10 x^2\right )+e^x \left (e^3 \left (10 x-10 x^3\right )-30 x^2+30 x\right )-30 x\right ) \log \left (e^3 (x+1)+3\right )+e^3 \left (-x^4-x^3+10 x^2\right )+e^x \left (e^3 \left (-2 x^4-2 x^3-10 x^2\right )-6 x^3\right )+\left (e^3 (-50 x-50)-150\right ) \log ^2\left (e^3 (x+1)+3\right )}{x^3 \left (e^3 x+e^3+3\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 e^x \left (-e^3 x^3-3 \left (1+\frac {e^3}{3}\right ) x^2-5 e^3 x^2 \log \left (e^3 (x+1)+3\right )-5 e^3 x-15 x \log \left (e^3 (x+1)+3\right )+15 \left (1+\frac {e^3}{3}\right ) \log \left (e^3 (x+1)+3\right )\right )}{x^2 \left (e^3 x+e^3+3\right )}+\frac {-e^3 x^4-3 \left (1+\frac {e^3}{3}\right ) x^3+10 e^3 x^2-10 e^3 x^2 \log \left (e^3 (x+1)+3\right )-50 e^3 x \log ^2\left (e^3 (x+1)+3\right )-150 \left (1+\frac {e^3}{3}\right ) \log ^2\left (e^3 (x+1)+3\right )-30 \left (1-\frac {4 e^3}{3}\right ) x \log \left (e^3 (x+1)+3\right )}{x^3 \left (e^3 x+e^3+3\right )}+2 e^{2 x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {50 e^6 \operatorname {PolyLog}\left (2,-\frac {e^3 x}{3+e^3}\right )}{\left (3+e^3\right )^2}-\frac {50 e^6 \operatorname {PolyLog}\left (2,\frac {3+e^3}{e^3 x+e^3+3}\right )}{\left (3+e^3\right )^2}+\frac {25 \log ^2\left (e^3 x+e^3+3\right )}{x^2}-2 e^x+e^{2 x}-x+\frac {25 e^6 \log ^2\left (e^3 (x+1)+3\right )}{\left (3+e^3\right )^2}-\frac {10 e^x \log \left (e^3 x+e^3+3\right )}{x}+\frac {10 \left (3-4 e^3\right ) \log \left (e^3 x+e^3+3\right )}{\left (3+e^3\right ) x}-\frac {10 e^3 \log \left (e^3 x+e^3+3\right )}{3+e^3}+\frac {10 e^3 \left (3-4 e^3\right ) \log \left (e^3 x+e^3+3\right )}{\left (3+e^3\right )^2}-\frac {50 e^6 \log \left (3+e^3\right ) \log (x)}{\left (3+e^3\right )^2}+\frac {10 e^3 \log (x)}{3+e^3}-\frac {10 e^3 \left (3-4 e^3\right ) \log (x)}{\left (3+e^3\right )^2}-\frac {50 e^6 \log (x)}{\left (3+e^3\right )^2}+\frac {50 e^3 \left (e^3 x+e^3+3\right ) \log \left (e^3 (x+1)+3\right )}{\left (3+e^3\right )^2 x}+\frac {50 e^6 \log \left (e^3 (x+1)+3\right ) \log \left (1-\frac {3+e^3}{e^3 x+e^3+3}\right )}{\left (3+e^3\right )^2}\)

Input:

Int[(-3*x^3 + E^3*(10*x^2 - x^3 - x^4) + E^x*(-6*x^3 + E^3*(-10*x^2 - 2*x^ 
3 - 2*x^4)) + E^(2*x)*(6*x^3 + E^3*(2*x^3 + 2*x^4)) + (-30*x + E^3*(40*x - 
 10*x^2) + E^x*(30*x - 30*x^2 + E^3*(10*x - 10*x^3)))*Log[3 + E^3*(1 + x)] 
 + (-150 + E^3*(-50 - 50*x))*Log[3 + E^3*(1 + x)]^2)/(3*x^3 + E^3*(x^3 + x 
^4)),x]
 

Output:

-2*E^x + E^(2*x) - x - (50*E^6*Log[x])/(3 + E^3)^2 - (10*E^3*(3 - 4*E^3)*L 
og[x])/(3 + E^3)^2 + (10*E^3*Log[x])/(3 + E^3) - (50*E^6*Log[3 + E^3]*Log[ 
x])/(3 + E^3)^2 + (10*E^3*(3 - 4*E^3)*Log[3 + E^3 + E^3*x])/(3 + E^3)^2 - 
(10*E^3*Log[3 + E^3 + E^3*x])/(3 + E^3) - (10*E^x*Log[3 + E^3 + E^3*x])/x 
+ (10*(3 - 4*E^3)*Log[3 + E^3 + E^3*x])/((3 + E^3)*x) + (25*Log[3 + E^3 + 
E^3*x]^2)/x^2 + (50*E^3*(3 + E^3 + E^3*x)*Log[3 + E^3*(1 + x)])/((3 + E^3) 
^2*x) + (25*E^6*Log[3 + E^3*(1 + x)]^2)/(3 + E^3)^2 + (50*E^6*Log[3 + E^3* 
(1 + x)]*Log[1 - (3 + E^3)/(3 + E^3 + E^3*x)])/(3 + E^3)^2 + (50*E^6*PolyL 
og[2, -((E^3*x)/(3 + E^3))])/(3 + E^3)^2 - (50*E^6*PolyLog[2, (3 + E^3)/(3 
 + E^3 + E^3*x)])/(3 + E^3)^2
 

Defintions of rubi rules used

rule 2009

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

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81

\[\frac {25 \ln \left (\left (1+x \right ) {\mathrm e}^{3}+3\right )^{2}}{x^{2}}-\frac {10 \left ({\mathrm e}^{x}-1\right ) \ln \left (\left (1+x \right ) {\mathrm e}^{3}+3\right )}{x}+{\mathrm e}^{2 x}-x -2 \,{\mathrm e}^{x}\]

Input:

int((((-50*x-50)*exp(3)-150)*ln((1+x)*exp(3)+3)^2+(((-10*x^3+10*x)*exp(3)- 
30*x^2+30*x)*exp(x)+(-10*x^2+40*x)*exp(3)-30*x)*ln((1+x)*exp(3)+3)+((2*x^4 
+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x^3)*exp(x) 
+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x)
                                                                                    
                                                                                    
 

Output:

25/x^2*ln((1+x)*exp(3)+3)^2-10*(exp(x)-1)/x*ln((1+x)*exp(3)+3)+exp(2*x)-x- 
2*exp(x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (26) = 52\).

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.19 \[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=-\frac {x^{3} - x^{2} e^{\left (2 \, x\right )} + 2 \, x^{2} e^{x} + 10 \, {\left (x e^{x} - x\right )} \log \left ({\left (x + 1\right )} e^{3} + 3\right ) - 25 \, \log \left ({\left (x + 1\right )} e^{3} + 3\right )^{2}}{x^{2}} \] Input:

integrate((((-50*x-50)*exp(3)-150)*log((1+x)*exp(3)+3)^2+(((-10*x^3+10*x)* 
exp(3)-30*x^2+30*x)*exp(x)+(-10*x^2+40*x)*exp(3)-30*x)*log((1+x)*exp(3)+3) 
+((2*x^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x^3 
)*exp(x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x, algor 
ithm="fricas")
 

Output:

-(x^3 - x^2*e^(2*x) + 2*x^2*e^x + 10*(x*e^x - x)*log((x + 1)*e^3 + 3) - 25 
*log((x + 1)*e^3 + 3)^2)/x^2
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (20) = 40\).

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=- x + \frac {x e^{2 x} + \left (- 2 x - 10 \log {\left (\left (x + 1\right ) e^{3} + 3 \right )}\right ) e^{x}}{x} + \frac {10 \log {\left (\left (x + 1\right ) e^{3} + 3 \right )}}{x} + \frac {25 \log {\left (\left (x + 1\right ) e^{3} + 3 \right )}^{2}}{x^{2}} \] Input:

integrate((((-50*x-50)*exp(3)-150)*ln((1+x)*exp(3)+3)**2+(((-10*x**3+10*x) 
*exp(3)-30*x**2+30*x)*exp(x)+(-10*x**2+40*x)*exp(3)-30*x)*ln((1+x)*exp(3)+ 
3)+((2*x**4+2*x**3)*exp(3)+6*x**3)*exp(x)**2+((-2*x**4-2*x**3-10*x**2)*exp 
(3)-6*x**3)*exp(x)+(-x**4-x**3+10*x**2)*exp(3)-3*x**3)/((x**4+x**3)*exp(3) 
+3*x**3),x)
 

Output:

-x + (x*exp(2*x) + (-2*x - 10*log((x + 1)*exp(3) + 3))*exp(x))/x + 10*log( 
(x + 1)*exp(3) + 3)/x + 25*log((x + 1)*exp(3) + 3)**2/x**2
 

Maxima [F]

\[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=\int { -\frac {3 \, x^{3} + 50 \, {\left ({\left (x + 1\right )} e^{3} + 3\right )} \log \left ({\left (x + 1\right )} e^{3} + 3\right )^{2} + {\left (x^{4} + x^{3} - 10 \, x^{2}\right )} e^{3} - 2 \, {\left (3 \, x^{3} + {\left (x^{4} + x^{3}\right )} e^{3}\right )} e^{\left (2 \, x\right )} + 2 \, {\left (3 \, x^{3} + {\left (x^{4} + x^{3} + 5 \, x^{2}\right )} e^{3}\right )} e^{x} + 10 \, {\left ({\left (x^{2} - 4 \, x\right )} e^{3} + {\left (3 \, x^{2} + {\left (x^{3} - x\right )} e^{3} - 3 \, x\right )} e^{x} + 3 \, x\right )} \log \left ({\left (x + 1\right )} e^{3} + 3\right )}{3 \, x^{3} + {\left (x^{4} + x^{3}\right )} e^{3}} \,d x } \] Input:

integrate((((-50*x-50)*exp(3)-150)*log((1+x)*exp(3)+3)^2+(((-10*x^3+10*x)* 
exp(3)-30*x^2+30*x)*exp(x)+(-10*x^2+40*x)*exp(3)-30*x)*log((1+x)*exp(3)+3) 
+((2*x^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x^3 
)*exp(x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x, algor 
ithm="maxima")
 

Output:

((e^3 + 3)*e^(-6)*log(x*e^3 + e^3 + 3) - x*e^(-3))*e^3 - 10*(log(x*e^3 + e 
^3 + 3)/(e^3 + 3) - log(x)/(e^3 + 3))*e^3 + 6*e^(-(e^3 + 3)*e^(-3) - 3)*ex 
p_integral_e(1, -(x*e^3 + e^3 + 3)*e^(-3)) - 6*e^(-2*(e^3 + 3)*e^(-3) - 3) 
*exp_integral_e(1, -2*(x*e^3 + e^3 + 3)*e^(-3)) - 3*e^(-3)*log(x*e^3 + e^3 
 + 3) - 10*e^3*log(x)/(e^3 + 3) + (x^3*(e^6 + 3*e^3)*e^(2*x) - 2*x^3*(e^6 
+ 3*e^3)*e^x + 25*(x*(e^6 + 3*e^3) + e^6 + 6*e^3 + 9)*log(x*e^3 + e^3 + 3) 
^2 + 10*(x^3*e^6 + 2*x^2*(e^6 + 3*e^3) + x*(e^6 + 6*e^3 + 9) - (x^2*(e^6 + 
 3*e^3) + x*(e^6 + 6*e^3 + 9))*e^x)*log(x*e^3 + e^3 + 3))/(x^3*(e^6 + 3*e^ 
3) + x^2*(e^6 + 6*e^3 + 9)) + integrate((2*x*e^6 + e^6 + 3*e^3)*e^(2*x)/(x 
^2*e^6 + 2*x*(e^6 + 3*e^3) + e^6 + 6*e^3 + 9), x) - 2*integrate(x*e^(x + 6 
)/(x^2*e^6 + 2*x*(e^6 + 3*e^3) + e^6 + 6*e^3 + 9), x) - log(x*e^3 + e^3 + 
3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (26) = 52\).

Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.04 \[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=-\frac {{\left (x^{3} e^{3} - x^{2} e^{\left (2 \, x + 3\right )} + 2 \, x^{2} e^{\left (x + 3\right )} - 10 \, x e^{3} \log \left (x e^{3} + e^{3} + 3\right ) + 10 \, x e^{\left (x + 3\right )} \log \left (x e^{3} + e^{3} + 3\right ) - 25 \, e^{3} \log \left (x e^{3} + e^{3} + 3\right )^{2}\right )} e^{\left (-3\right )}}{x^{2}} \] Input:

integrate((((-50*x-50)*exp(3)-150)*log((1+x)*exp(3)+3)^2+(((-10*x^3+10*x)* 
exp(3)-30*x^2+30*x)*exp(x)+(-10*x^2+40*x)*exp(3)-30*x)*log((1+x)*exp(3)+3) 
+((2*x^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x^3 
)*exp(x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x, algor 
ithm="giac")
 

Output:

-(x^3*e^3 - x^2*e^(2*x + 3) + 2*x^2*e^(x + 3) - 10*x*e^3*log(x*e^3 + e^3 + 
 3) + 10*x*e^(x + 3)*log(x*e^3 + e^3 + 3) - 25*e^3*log(x*e^3 + e^3 + 3)^2) 
*e^(-3)/x^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=\int -\frac {{\mathrm {e}}^3\,\left (x^4+x^3-10\,x^2\right )-\ln \left ({\mathrm {e}}^3\,\left (x+1\right )+3\right )\,\left ({\mathrm {e}}^3\,\left (40\,x-10\,x^2\right )-30\,x+{\mathrm {e}}^x\,\left (30\,x+{\mathrm {e}}^3\,\left (10\,x-10\,x^3\right )-30\,x^2\right )\right )+{\ln \left ({\mathrm {e}}^3\,\left (x+1\right )+3\right )}^2\,\left ({\mathrm {e}}^3\,\left (50\,x+50\right )+150\right )+{\mathrm {e}}^x\,\left ({\mathrm {e}}^3\,\left (2\,x^4+2\,x^3+10\,x^2\right )+6\,x^3\right )-{\mathrm {e}}^{2\,x}\,\left ({\mathrm {e}}^3\,\left (2\,x^4+2\,x^3\right )+6\,x^3\right )+3\,x^3}{{\mathrm {e}}^3\,\left (x^4+x^3\right )+3\,x^3} \,d x \] Input:

int(-(exp(3)*(x^3 - 10*x^2 + x^4) - log(exp(3)*(x + 1) + 3)*(exp(3)*(40*x 
- 10*x^2) - 30*x + exp(x)*(30*x + exp(3)*(10*x - 10*x^3) - 30*x^2)) + log( 
exp(3)*(x + 1) + 3)^2*(exp(3)*(50*x + 50) + 150) + exp(x)*(exp(3)*(10*x^2 
+ 2*x^3 + 2*x^4) + 6*x^3) - exp(2*x)*(exp(3)*(2*x^3 + 2*x^4) + 6*x^3) + 3* 
x^3)/(exp(3)*(x^3 + x^4) + 3*x^3),x)
 

Output:

int(-(exp(3)*(x^3 - 10*x^2 + x^4) - log(exp(3)*(x + 1) + 3)*(exp(3)*(40*x 
- 10*x^2) - 30*x + exp(x)*(30*x + exp(3)*(10*x - 10*x^3) - 30*x^2)) + log( 
exp(3)*(x + 1) + 3)^2*(exp(3)*(50*x + 50) + 150) + exp(x)*(exp(3)*(10*x^2 
+ 2*x^3 + 2*x^4) + 6*x^3) - exp(2*x)*(exp(3)*(2*x^3 + 2*x^4) + 6*x^3) + 3* 
x^3)/(exp(3)*(x^3 + x^4) + 3*x^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.81 \[ \int \frac {-3 x^3+e^3 \left (10 x^2-x^3-x^4\right )+e^x \left (-6 x^3+e^3 \left (-10 x^2-2 x^3-2 x^4\right )\right )+e^{2 x} \left (6 x^3+e^3 \left (2 x^3+2 x^4\right )\right )+\left (-30 x+e^3 \left (40 x-10 x^2\right )+e^x \left (30 x-30 x^2+e^3 \left (10 x-10 x^3\right )\right )\right ) \log \left (3+e^3 (1+x)\right )+\left (-150+e^3 (-50-50 x)\right ) \log ^2\left (3+e^3 (1+x)\right )}{3 x^3+e^3 \left (x^3+x^4\right )} \, dx=\frac {e^{2 x} x^{2}-10 e^{x} \mathrm {log}\left (e^{3} x +e^{3}+3\right ) x -2 e^{x} x^{2}+25 \mathrm {log}\left (e^{3} x +e^{3}+3\right )^{2}+10 \,\mathrm {log}\left (e^{3} x +e^{3}+3\right ) x -x^{3}}{x^{2}} \] Input:

int((((-50*x-50)*exp(3)-150)*log((1+x)*exp(3)+3)^2+(((-10*x^3+10*x)*exp(3) 
-30*x^2+30*x)*exp(x)+(-10*x^2+40*x)*exp(3)-30*x)*log((1+x)*exp(3)+3)+((2*x 
^4+2*x^3)*exp(3)+6*x^3)*exp(x)^2+((-2*x^4-2*x^3-10*x^2)*exp(3)-6*x^3)*exp( 
x)+(-x^4-x^3+10*x^2)*exp(3)-3*x^3)/((x^4+x^3)*exp(3)+3*x^3),x)
 

Output:

(e**(2*x)*x**2 - 10*e**x*log(e**3*x + e**3 + 3)*x - 2*e**x*x**2 + 25*log(e 
**3*x + e**3 + 3)**2 + 10*log(e**3*x + e**3 + 3)*x - x**3)/x**2