\(\int \frac {(d+e x)^5}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [107]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 35, antiderivative size = 131 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {e \left (c d^2-a e^2\right )^3 x}{c^4 d^4}+\frac {\left (c d^2-a e^2\right )^2 (d+e x)^2}{2 c^3 d^3}+\frac {\left (c d^2-a e^2\right ) (d+e x)^3}{3 c^2 d^2}+\frac {(d+e x)^4}{4 c d}+\frac {\left (c d^2-a e^2\right )^4 \log (a e+c d x)}{c^5 d^5} \] Output:

e*(-a*e^2+c*d^2)^3*x/c^4/d^4+1/2*(-a*e^2+c*d^2)^2*(e*x+d)^2/c^3/d^3+1/3*(- 
a*e^2+c*d^2)*(e*x+d)^3/c^2/d^2+1/4*(e*x+d)^4/c/d+(-a*e^2+c*d^2)^4*ln(c*d*x 
+a*e)/c^5/d^5
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {c d e x \left (-12 a^3 e^6+6 a^2 c d e^4 (8 d+e x)-4 a c^2 d^2 e^2 \left (18 d^2+6 d e x+e^2 x^2\right )+c^3 d^3 \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )+12 \left (c d^2-a e^2\right )^4 \log (a e+c d x)}{12 c^5 d^5} \] Input:

Integrate[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2),x]
 

Output:

(c*d*e*x*(-12*a^3*e^6 + 6*a^2*c*d*e^4*(8*d + e*x) - 4*a*c^2*d^2*e^2*(18*d^ 
2 + 6*d*e*x + e^2*x^2) + c^3*d^3*(48*d^3 + 36*d^2*e*x + 16*d*e^2*x^2 + 3*e 
^3*x^3)) + 12*(c*d^2 - a*e^2)^4*Log[a*e + c*d*x])/(12*c^5*d^5)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {1121, 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.

\(\displaystyle \int \frac {(d+e x)^5}{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \, dx\)

\(\Big \downarrow \) 1121

\(\displaystyle \int \left (\frac {\left (c d^2-a e^2\right )^4}{c^4 d^4 (a e+c d x)}+\frac {e \left (c d^2-a e^2\right )^3}{c^4 d^4}+\frac {e (d+e x) \left (c d^2-a e^2\right )^2}{c^3 d^3}+\frac {e (d+e x)^2 \left (c d^2-a e^2\right )}{c^2 d^2}+\frac {e (d+e x)^3}{c d}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (c d^2-a e^2\right )^4 \log (a e+c d x)}{c^5 d^5}+\frac {e x \left (c d^2-a e^2\right )^3}{c^4 d^4}+\frac {(d+e x)^2 \left (c d^2-a e^2\right )^2}{2 c^3 d^3}+\frac {(d+e x)^3 \left (c d^2-a e^2\right )}{3 c^2 d^2}+\frac {(d+e x)^4}{4 c d}\)

Input:

Int[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2),x]
 

Output:

(e*(c*d^2 - a*e^2)^3*x)/(c^4*d^4) + ((c*d^2 - a*e^2)^2*(d + e*x)^2)/(2*c^3 
*d^3) + ((c*d^2 - a*e^2)*(d + e*x)^3)/(3*c^2*d^2) + (d + e*x)^4/(4*c*d) + 
((c*d^2 - a*e^2)^4*Log[a*e + c*d*x])/(c^5*d^5)
 

Defintions of rubi rules used

rule 1121
Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int 
egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
Maple [A] (verified)

Time = 1.73 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.53

method result size
norman \(\frac {e^{4} x^{4}}{4 d c}+\frac {e^{2} \left (a^{2} e^{4}-4 a c \,d^{2} e^{2}+6 c^{2} d^{4}\right ) x^{2}}{2 c^{3} d^{3}}-\frac {e^{3} \left (a \,e^{2}-4 c \,d^{2}\right ) x^{3}}{3 c^{2} d^{2}}-\frac {e \left (e^{6} a^{3}-4 d^{2} e^{4} a^{2} c +6 d^{4} e^{2} a \,c^{2}-4 d^{6} c^{3}\right ) x}{d^{4} c^{4}}+\frac {\left (a^{4} e^{8}-4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}-4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right ) \ln \left (c d x +a e \right )}{d^{5} c^{5}}\) \(201\)
default \(-\frac {e \left (-\frac {x^{4} c^{3} d^{3} e^{3}}{4}+\frac {\left (\left (a \,e^{2}-2 c \,d^{2}\right ) e^{2} c^{2} d^{2}-2 c^{3} d^{4} e^{2}\right ) x^{3}}{3}+\frac {\left (2 \left (a \,e^{2}-2 c \,d^{2}\right ) d^{3} e \,c^{2}-d e c \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+2 c^{2} d^{4}\right )\right ) x^{2}}{2}+\left (a \,e^{2}-2 c \,d^{2}\right ) \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+2 c^{2} d^{4}\right ) x \right )}{d^{4} c^{4}}+\frac {\left (a^{4} e^{8}-4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}-4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right ) \ln \left (c d x +a e \right )}{d^{5} c^{5}}\) \(232\)
risch \(\frac {e^{4} x^{4}}{4 d c}-\frac {e^{5} x^{3} a}{3 d^{2} c^{2}}+\frac {4 e^{3} x^{3}}{3 c}-\frac {2 e^{4} x^{2} a}{d \,c^{2}}+\frac {3 e^{2} d \,x^{2}}{c}+\frac {e^{6} x^{2} a^{2}}{2 d^{3} c^{3}}-\frac {e^{7} a^{3} x}{d^{4} c^{4}}+\frac {4 e^{5} a^{2} x}{d^{2} c^{3}}-\frac {6 e^{3} a x}{c^{2}}+\frac {4 e \,d^{2} x}{c}+\frac {\ln \left (c d x +a e \right ) a^{4} e^{8}}{d^{5} c^{5}}-\frac {4 \ln \left (c d x +a e \right ) a^{3} e^{6}}{d^{3} c^{4}}+\frac {6 \ln \left (c d x +a e \right ) a^{2} e^{4}}{d \,c^{3}}-\frac {4 d \ln \left (c d x +a e \right ) a \,e^{2}}{c^{2}}+\frac {d^{3} \ln \left (c d x +a e \right )}{c}\) \(239\)
parallelrisch \(\frac {3 c^{4} d^{4} e^{4} x^{4}-4 a \,c^{3} d^{3} e^{5} x^{3}+16 c^{4} d^{5} e^{3} x^{3}+6 a^{2} c^{2} d^{2} e^{6} x^{2}-24 a \,c^{3} d^{4} e^{4} x^{2}+36 c^{4} d^{6} e^{2} x^{2}+12 \ln \left (c d x +a e \right ) a^{4} e^{8}-48 \ln \left (c d x +a e \right ) a^{3} c \,d^{2} e^{6}+72 \ln \left (c d x +a e \right ) a^{2} c^{2} d^{4} e^{4}-48 \ln \left (c d x +a e \right ) a \,c^{3} d^{6} e^{2}+12 \ln \left (c d x +a e \right ) c^{4} d^{8}-12 a^{3} c d \,e^{7} x +48 a^{2} c^{2} d^{3} e^{5} x -72 a \,c^{3} d^{5} e^{3} x +48 c^{4} d^{7} e x}{12 d^{5} c^{5}}\) \(247\)

Input:

int((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e),x,method=_RETURNVERBOSE)
 

Output:

1/4/d/c*e^4*x^4+1/2/c^3/d^3*e^2*(a^2*e^4-4*a*c*d^2*e^2+6*c^2*d^4)*x^2-1/3/ 
c^2/d^2*e^3*(a*e^2-4*c*d^2)*x^3-e*(a^3*e^6-4*a^2*c*d^2*e^4+6*a*c^2*d^4*e^2 
-4*c^3*d^6)/d^4/c^4*x+(a^4*e^8-4*a^3*c*d^2*e^6+6*a^2*c^2*d^4*e^4-4*a*c^3*d 
^6*e^2+c^4*d^8)/d^5/c^5*ln(c*d*x+a*e)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.58 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {3 \, c^{4} d^{4} e^{4} x^{4} + 4 \, {\left (4 \, c^{4} d^{5} e^{3} - a c^{3} d^{3} e^{5}\right )} x^{3} + 6 \, {\left (6 \, c^{4} d^{6} e^{2} - 4 \, a c^{3} d^{4} e^{4} + a^{2} c^{2} d^{2} e^{6}\right )} x^{2} + 12 \, {\left (4 \, c^{4} d^{7} e - 6 \, a c^{3} d^{5} e^{3} + 4 \, a^{2} c^{2} d^{3} e^{5} - a^{3} c d e^{7}\right )} x + 12 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \log \left (c d x + a e\right )}{12 \, c^{5} d^{5}} \] Input:

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas 
")
 

Output:

1/12*(3*c^4*d^4*e^4*x^4 + 4*(4*c^4*d^5*e^3 - a*c^3*d^3*e^5)*x^3 + 6*(6*c^4 
*d^6*e^2 - 4*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^2 + 12*(4*c^4*d^7*e - 6*a* 
c^3*d^5*e^3 + 4*a^2*c^2*d^3*e^5 - a^3*c*d*e^7)*x + 12*(c^4*d^8 - 4*a*c^3*d 
^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*log(c*d*x + a*e))/ 
(c^5*d^5)
 

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=x^{3} \left (- \frac {a e^{5}}{3 c^{2} d^{2}} + \frac {4 e^{3}}{3 c}\right ) + x^{2} \left (\frac {a^{2} e^{6}}{2 c^{3} d^{3}} - \frac {2 a e^{4}}{c^{2} d} + \frac {3 d e^{2}}{c}\right ) + x \left (- \frac {a^{3} e^{7}}{c^{4} d^{4}} + \frac {4 a^{2} e^{5}}{c^{3} d^{2}} - \frac {6 a e^{3}}{c^{2}} + \frac {4 d^{2} e}{c}\right ) + \frac {e^{4} x^{4}}{4 c d} + \frac {\left (a e^{2} - c d^{2}\right )^{4} \log {\left (a e + c d x \right )}}{c^{5} d^{5}} \] Input:

integrate((e*x+d)**5/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2),x)
                                                                                    
                                                                                    
 

Output:

x**3*(-a*e**5/(3*c**2*d**2) + 4*e**3/(3*c)) + x**2*(a**2*e**6/(2*c**3*d**3 
) - 2*a*e**4/(c**2*d) + 3*d*e**2/c) + x*(-a**3*e**7/(c**4*d**4) + 4*a**2*e 
**5/(c**3*d**2) - 6*a*e**3/c**2 + 4*d**2*e/c) + e**4*x**4/(4*c*d) + (a*e** 
2 - c*d**2)**4*log(a*e + c*d*x)/(c**5*d**5)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.56 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {3 \, c^{3} d^{3} e^{4} x^{4} + 4 \, {\left (4 \, c^{3} d^{4} e^{3} - a c^{2} d^{2} e^{5}\right )} x^{3} + 6 \, {\left (6 \, c^{3} d^{5} e^{2} - 4 \, a c^{2} d^{3} e^{4} + a^{2} c d e^{6}\right )} x^{2} + 12 \, {\left (4 \, c^{3} d^{6} e - 6 \, a c^{2} d^{4} e^{3} + 4 \, a^{2} c d^{2} e^{5} - a^{3} e^{7}\right )} x}{12 \, c^{4} d^{4}} + \frac {{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \log \left (c d x + a e\right )}{c^{5} d^{5}} \] Input:

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima 
")
 

Output:

1/12*(3*c^3*d^3*e^4*x^4 + 4*(4*c^3*d^4*e^3 - a*c^2*d^2*e^5)*x^3 + 6*(6*c^3 
*d^5*e^2 - 4*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x^2 + 12*(4*c^3*d^6*e - 6*a*c^2* 
d^4*e^3 + 4*a^2*c*d^2*e^5 - a^3*e^7)*x)/(c^4*d^4) + (c^4*d^8 - 4*a*c^3*d^6 
*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*log(c*d*x + a*e)/(c^ 
5*d^5)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.60 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {3 \, c^{3} d^{3} e^{4} x^{4} + 16 \, c^{3} d^{4} e^{3} x^{3} - 4 \, a c^{2} d^{2} e^{5} x^{3} + 36 \, c^{3} d^{5} e^{2} x^{2} - 24 \, a c^{2} d^{3} e^{4} x^{2} + 6 \, a^{2} c d e^{6} x^{2} + 48 \, c^{3} d^{6} e x - 72 \, a c^{2} d^{4} e^{3} x + 48 \, a^{2} c d^{2} e^{5} x - 12 \, a^{3} e^{7} x}{12 \, c^{4} d^{4}} + \frac {{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \log \left ({\left | c d x + a e \right |}\right )}{c^{5} d^{5}} \] Input:

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="giac")
 

Output:

1/12*(3*c^3*d^3*e^4*x^4 + 16*c^3*d^4*e^3*x^3 - 4*a*c^2*d^2*e^5*x^3 + 36*c^ 
3*d^5*e^2*x^2 - 24*a*c^2*d^3*e^4*x^2 + 6*a^2*c*d*e^6*x^2 + 48*c^3*d^6*e*x 
- 72*a*c^2*d^4*e^3*x + 48*a^2*c*d^2*e^5*x - 12*a^3*e^7*x)/(c^4*d^4) + (c^4 
*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*lo 
g(abs(c*d*x + a*e))/(c^5*d^5)
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.66 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=x^3\,\left (\frac {4\,e^3}{3\,c}-\frac {a\,e^5}{3\,c^2\,d^2}\right )+x\,\left (\frac {4\,d^2\,e}{c}-\frac {a\,e\,\left (\frac {6\,d\,e^2}{c}-\frac {a\,e\,\left (\frac {4\,e^3}{c}-\frac {a\,e^5}{c^2\,d^2}\right )}{c\,d}\right )}{c\,d}\right )+x^2\,\left (\frac {3\,d\,e^2}{c}-\frac {a\,e\,\left (\frac {4\,e^3}{c}-\frac {a\,e^5}{c^2\,d^2}\right )}{2\,c\,d}\right )+\frac {\ln \left (a\,e+c\,d\,x\right )\,\left (a^4\,e^8-4\,a^3\,c\,d^2\,e^6+6\,a^2\,c^2\,d^4\,e^4-4\,a\,c^3\,d^6\,e^2+c^4\,d^8\right )}{c^5\,d^5}+\frac {e^4\,x^4}{4\,c\,d} \] Input:

int((d + e*x)^5/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2),x)
 

Output:

x^3*((4*e^3)/(3*c) - (a*e^5)/(3*c^2*d^2)) + x*((4*d^2*e)/c - (a*e*((6*d*e^ 
2)/c - (a*e*((4*e^3)/c - (a*e^5)/(c^2*d^2)))/(c*d)))/(c*d)) + x^2*((3*d*e^ 
2)/c - (a*e*((4*e^3)/c - (a*e^5)/(c^2*d^2)))/(2*c*d)) + (log(a*e + c*d*x)* 
(a^4*e^8 + c^4*d^8 - 4*a*c^3*d^6*e^2 - 4*a^3*c*d^2*e^6 + 6*a^2*c^2*d^4*e^4 
))/(c^5*d^5) + (e^4*x^4)/(4*c*d)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.88 \[ \int \frac {(d+e x)^5}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {12 \,\mathrm {log}\left (c d x +a e \right ) a^{4} e^{8}-48 \,\mathrm {log}\left (c d x +a e \right ) a^{3} c \,d^{2} e^{6}+72 \,\mathrm {log}\left (c d x +a e \right ) a^{2} c^{2} d^{4} e^{4}-48 \,\mathrm {log}\left (c d x +a e \right ) a \,c^{3} d^{6} e^{2}+12 \,\mathrm {log}\left (c d x +a e \right ) c^{4} d^{8}-12 a^{3} c d \,e^{7} x +48 a^{2} c^{2} d^{3} e^{5} x +6 a^{2} c^{2} d^{2} e^{6} x^{2}-72 a \,c^{3} d^{5} e^{3} x -24 a \,c^{3} d^{4} e^{4} x^{2}-4 a \,c^{3} d^{3} e^{5} x^{3}+48 c^{4} d^{7} e x +36 c^{4} d^{6} e^{2} x^{2}+16 c^{4} d^{5} e^{3} x^{3}+3 c^{4} d^{4} e^{4} x^{4}}{12 c^{5} d^{5}} \] Input:

int((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x)
 

Output:

(12*log(a*e + c*d*x)*a**4*e**8 - 48*log(a*e + c*d*x)*a**3*c*d**2*e**6 + 72 
*log(a*e + c*d*x)*a**2*c**2*d**4*e**4 - 48*log(a*e + c*d*x)*a*c**3*d**6*e* 
*2 + 12*log(a*e + c*d*x)*c**4*d**8 - 12*a**3*c*d*e**7*x + 48*a**2*c**2*d** 
3*e**5*x + 6*a**2*c**2*d**2*e**6*x**2 - 72*a*c**3*d**5*e**3*x - 24*a*c**3* 
d**4*e**4*x**2 - 4*a*c**3*d**3*e**5*x**3 + 48*c**4*d**7*e*x + 36*c**4*d**6 
*e**2*x**2 + 16*c**4*d**5*e**3*x**3 + 3*c**4*d**4*e**4*x**4)/(12*c**5*d**5 
)