\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^2} \, dx\) [1855]

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

Optimal result

Integrand size = 35, antiderivative size = 54 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {\left (c d^2-a e^2\right ) (a e+c d x)^4}{4 c^2 d^2}+\frac {e (a e+c d x)^5}{5 c^2 d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 45} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {\left (c d^2-a e^2\right ) (a e+c d x)^4}{4 c^2 d^2}+\frac {e (a e+c d x)^5}{5 c^2 d^2} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^2,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
 IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int (a e+c d x)^3 (d+e x) \, dx \\ & = \int \left (\frac {\left (c d^2-a e^2\right ) (a e+c d x)^3}{c d}+\frac {e (a e+c d x)^4}{c d}\right ) \, dx \\ & = \frac {\left (c d^2-a e^2\right ) (a e+c d x)^4}{4 c^2 d^2}+\frac {e (a e+c d x)^5}{5 c^2 d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {1}{20} x \left (10 a^3 e^3 (2 d+e x)+10 a^2 c d e^2 x (3 d+2 e x)+5 a c^2 d^2 e x^2 (4 d+3 e x)+c^3 d^3 x^3 (5 d+4 e x)\right ) \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^2,x]

[Out]

(x*(10*a^3*e^3*(2*d + e*x) + 10*a^2*c*d*e^2*x*(3*d + 2*e*x) + 5*a*c^2*d^2*e*x^2*(4*d + 3*e*x) + c^3*d^3*x^3*(5
*d + 4*e*x)))/20

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.83

method result size
risch \(\frac {1}{5} d^{3} e \,c^{3} x^{5}+\frac {3}{4} x^{4} d^{2} e^{2} c^{2} a +\frac {1}{4} c^{3} d^{4} x^{4}+a^{2} c d \,e^{3} x^{3}+a \,c^{2} d^{3} e \,x^{3}+\frac {1}{2} x^{2} e^{4} a^{3}+\frac {3}{2} x^{2} d^{2} e^{2} a^{2} c +a^{3} d \,e^{3} x\) \(99\)
parallelrisch \(\frac {1}{5} d^{3} e \,c^{3} x^{5}+\frac {3}{4} x^{4} d^{2} e^{2} c^{2} a +\frac {1}{4} c^{3} d^{4} x^{4}+a^{2} c d \,e^{3} x^{3}+a \,c^{2} d^{3} e \,x^{3}+\frac {1}{2} x^{2} e^{4} a^{3}+\frac {3}{2} x^{2} d^{2} e^{2} a^{2} c +a^{3} d \,e^{3} x\) \(99\)
gosper \(\frac {x \left (4 d^{3} e \,c^{3} x^{4}+15 d^{2} e^{2} c^{2} a \,x^{3}+5 d^{4} c^{3} x^{3}+20 a^{2} c d \,e^{3} x^{2}+20 d^{3} e \,c^{2} a \,x^{2}+10 e^{4} a^{3} x +30 d^{2} e^{2} a^{2} c x +20 a^{3} d \,e^{3}\right )}{20}\) \(100\)
default \(\frac {d^{3} e \,c^{3} x^{5}}{5}+\frac {\left (2 d^{2} e^{2} c^{2} a +c^{2} d^{2} \left (e^{2} a +c \,d^{2}\right )\right ) x^{4}}{4}+\frac {\left (a^{2} c d \,e^{3}+2 a c d e \left (e^{2} a +c \,d^{2}\right )+d^{3} e \,c^{2} a \right ) x^{3}}{3}+\frac {\left (a^{2} e^{2} \left (e^{2} a +c \,d^{2}\right )+2 d^{2} e^{2} a^{2} c \right ) x^{2}}{2}+a^{3} d \,e^{3} x\) \(136\)
norman \(\frac {\left (\frac {3}{2} d \,e^{4} a^{3}+\frac {3}{2} d^{3} e^{2} a^{2} c \right ) x^{2}+\left (\frac {3}{4} e^{3} a \,c^{2} d^{2}+\frac {9}{20} d^{4} e \,c^{3}\right ) x^{5}+\left (\frac {1}{2} a^{3} e^{5}+\frac {5}{2} d^{2} e^{3} a^{2} c +d^{4} a \,c^{2} e \right ) x^{3}+\left (d \,e^{4} a^{2} c +\frac {7}{4} d^{3} e^{2} c^{2} a +\frac {1}{4} d^{5} c^{3}\right ) x^{4}+d^{2} a^{3} e^{3} x +\frac {e^{2} c^{3} d^{3} x^{6}}{5}}{e x +d}\) \(155\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/5*d^3*e*c^3*x^5+3/4*x^4*d^2*e^2*c^2*a+1/4*c^3*d^4*x^4+a^2*c*d*e^3*x^3+a*c^2*d^3*e*x^3+1/2*x^2*e^4*a^3+3/2*x^
2*d^2*e^2*a^2*c+a^3*d*e^3*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {1}{5} \, c^{3} d^{3} e x^{5} + a^{3} d e^{3} x + \frac {1}{4} \, {\left (c^{3} d^{4} + 3 \, a c^{2} d^{2} e^{2}\right )} x^{4} + {\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} x^{2} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^2,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).

Time = 0.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=a^{3} d e^{3} x + \frac {c^{3} d^{3} e x^{5}}{5} + x^{4} \cdot \left (\frac {3 a c^{2} d^{2} e^{2}}{4} + \frac {c^{3} d^{4}}{4}\right ) + x^{3} \left (a^{2} c d e^{3} + a c^{2} d^{3} e\right ) + x^{2} \left (\frac {a^{3} e^{4}}{2} + \frac {3 a^{2} c d^{2} e^{2}}{2}\right ) \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3/(e*x+d)**2,x)

[Out]

a**3*d*e**3*x + c**3*d**3*e*x**5/5 + x**4*(3*a*c**2*d**2*e**2/4 + c**3*d**4/4) + x**3*(a**2*c*d*e**3 + a*c**2*
d**3*e) + x**2*(a**3*e**4/2 + 3*a**2*c*d**2*e**2/2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {1}{5} \, c^{3} d^{3} e x^{5} + a^{3} d e^{3} x + \frac {1}{4} \, {\left (c^{3} d^{4} + 3 \, a c^{2} d^{2} e^{2}\right )} x^{4} + {\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} x^{2} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (50) = 100\).

Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 3.22 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {{\left (4 \, c^{3} d^{3} - \frac {15 \, c^{3} d^{4}}{e x + d} + \frac {20 \, c^{3} d^{5}}{{\left (e x + d\right )}^{2}} - \frac {10 \, c^{3} d^{6}}{{\left (e x + d\right )}^{3}} + \frac {15 \, a c^{2} d^{2} e^{2}}{e x + d} - \frac {40 \, a c^{2} d^{3} e^{2}}{{\left (e x + d\right )}^{2}} + \frac {30 \, a c^{2} d^{4} e^{2}}{{\left (e x + d\right )}^{3}} + \frac {20 \, a^{2} c d e^{4}}{{\left (e x + d\right )}^{2}} - \frac {30 \, a^{2} c d^{2} e^{4}}{{\left (e x + d\right )}^{3}} + \frac {10 \, a^{3} e^{6}}{{\left (e x + d\right )}^{3}}\right )} {\left (e x + d\right )}^{5}}{20 \, e^{4}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^2,x, algorithm="giac")

[Out]

1/20*(4*c^3*d^3 - 15*c^3*d^4/(e*x + d) + 20*c^3*d^5/(e*x + d)^2 - 10*c^3*d^6/(e*x + d)^3 + 15*a*c^2*d^2*e^2/(e
*x + d) - 40*a*c^2*d^3*e^2/(e*x + d)^2 + 30*a*c^2*d^4*e^2/(e*x + d)^3 + 20*a^2*c*d*e^4/(e*x + d)^2 - 30*a^2*c*
d^2*e^4/(e*x + d)^3 + 10*a^3*e^6/(e*x + d)^3)*(e*x + d)^5/e^4

Mupad [B] (verification not implemented)

Time = 9.95 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=x^2\,\left (\frac {a^3\,e^4}{2}+\frac {3\,c\,a^2\,d^2\,e^2}{2}\right )+x^4\,\left (\frac {c^3\,d^4}{4}+\frac {3\,a\,c^2\,d^2\,e^2}{4}\right )+\frac {c^3\,d^3\,e\,x^5}{5}+a^3\,d\,e^3\,x+a\,c\,d\,e\,x^3\,\left (c\,d^2+a\,e^2\right ) \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3/(d + e*x)^2,x)

[Out]

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