Integrand size = 37, antiderivative size = 113 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\frac {2 \left (c d^2-a e^2\right )^3}{e^4 \sqrt {d+e x}}+\frac {6 c d \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{e^4}-\frac {2 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{e^4}+\frac {2 c^3 d^3 (d+e x)^{5/2}}{5 e^4} \] Output:
2*(-a*e^2+c*d^2)^3/e^4/(e*x+d)^(1/2)+6*c*d*(-a*e^2+c*d^2)^2*(e*x+d)^(1/2)/ e^4-2*c^2*d^2*(-a*e^2+c*d^2)*(e*x+d)^(3/2)/e^4+2/5*c^3*d^3*(e*x+d)^(5/2)/e ^4
Time = 0.07 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\frac {2 \left (-5 a^3 e^6+15 a^2 c d e^4 (2 d+e x)-5 a c^2 d^2 e^2 \left (8 d^2+4 d e x-e^2 x^2\right )+c^3 d^3 \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )}{5 e^4 \sqrt {d+e x}} \] Input:
Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^(9/2),x]
Output:
(2*(-5*a^3*e^6 + 15*a^2*c*d*e^4*(2*d + e*x) - 5*a*c^2*d^2*e^2*(8*d^2 + 4*d *e*x - e^2*x^2) + c^3*d^3*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3)))/( 5*e^4*Sqrt[d + e*x])
Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, 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.
\(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \left (-\frac {3 c^2 d^2 \sqrt {d+e x} \left (c d^2-a e^2\right )}{e^3}+\frac {3 c d \left (c d^2-a e^2\right )^2}{e^3 \sqrt {d+e x}}+\frac {\left (a e^2-c d^2\right )^3}{e^3 (d+e x)^{3/2}}+\frac {c^3 d^3 (d+e x)^{3/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 c^2 d^2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}{e^4}+\frac {6 c d \sqrt {d+e x} \left (c d^2-a e^2\right )^2}{e^4}+\frac {2 \left (c d^2-a e^2\right )^3}{e^4 \sqrt {d+e x}}+\frac {2 c^3 d^3 (d+e x)^{5/2}}{5 e^4}\) |
Input:
Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^(9/2),x]
Output:
(2*(c*d^2 - a*e^2)^3)/(e^4*Sqrt[d + e*x]) + (6*c*d*(c*d^2 - a*e^2)^2*Sqrt[ d + e*x])/e^4 - (2*c^2*d^2*(c*d^2 - a*e^2)*(d + e*x)^(3/2))/e^4 + (2*c^3*d ^3*(d + e*x)^(5/2))/(5*e^4)
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]))
Time = 1.53 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.93
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \left (d^{3} e^{3} x^{3}-2 d^{4} e^{2} x^{2}+8 d^{5} e x +16 d^{6}\right ) c^{3}}{5}-16 e^{2} \left (-\frac {1}{8} e^{2} x^{2}+\frac {1}{2} d e x +d^{2}\right ) a \,d^{2} c^{2}+12 e^{4} \left (\frac {e x}{2}+d \right ) a^{2} d c -2 e^{6} a^{3}}{\sqrt {e x +d}\, e^{4}}\) | \(105\) |
risch | \(\frac {2 c d \left (x^{2} c^{2} d^{2} e^{2}+5 x a c d \,e^{3}-3 x \,c^{2} d^{3} e +15 a^{2} e^{4}-25 a c \,d^{2} e^{2}+11 c^{2} d^{4}\right ) \sqrt {e x +d}}{5 e^{4}}-\frac {2 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} a \,c^{2}-d^{6} c^{3}\right )}{e^{4} \sqrt {e x +d}}\) | \(127\) |
gosper | \(-\frac {2 \left (-c^{3} d^{3} e^{3} x^{3}-5 x^{2} a \,c^{2} d^{2} e^{4}+2 c^{3} d^{4} e^{2} x^{2}-15 x \,a^{2} c d \,e^{5}+20 x a \,c^{2} d^{3} e^{3}-8 c^{3} d^{5} e x +5 e^{6} a^{3}-30 d^{2} e^{4} a^{2} c +40 d^{4} e^{2} a \,c^{2}-16 d^{6} c^{3}\right )}{5 \sqrt {e x +d}\, e^{4}}\) | \(131\) |
trager | \(-\frac {2 \left (-c^{3} d^{3} e^{3} x^{3}-5 x^{2} a \,c^{2} d^{2} e^{4}+2 c^{3} d^{4} e^{2} x^{2}-15 x \,a^{2} c d \,e^{5}+20 x a \,c^{2} d^{3} e^{3}-8 c^{3} d^{5} e x +5 e^{6} a^{3}-30 d^{2} e^{4} a^{2} c +40 d^{4} e^{2} a \,c^{2}-16 d^{6} c^{3}\right )}{5 \sqrt {e x +d}\, e^{4}}\) | \(131\) |
derivativedivides | \(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {5}{2}}}{5}+2 a \,c^{2} d^{2} e^{2} \left (e x +d \right )^{\frac {3}{2}}-2 c^{3} d^{4} \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} c d \,e^{4} \sqrt {e x +d}-12 a \,c^{2} d^{3} e^{2} \sqrt {e x +d}+6 c^{3} d^{5} \sqrt {e x +d}-\frac {2 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} a \,c^{2}-d^{6} c^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(155\) |
default | \(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {5}{2}}}{5}+2 a \,c^{2} d^{2} e^{2} \left (e x +d \right )^{\frac {3}{2}}-2 c^{3} d^{4} \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} c d \,e^{4} \sqrt {e x +d}-12 a \,c^{2} d^{3} e^{2} \sqrt {e x +d}+6 c^{3} d^{5} \sqrt {e x +d}-\frac {2 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} a \,c^{2}-d^{6} c^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(155\) |
orering | \(-\frac {2 \left (-c^{3} d^{3} e^{3} x^{3}-5 x^{2} a \,c^{2} d^{2} e^{4}+2 c^{3} d^{4} e^{2} x^{2}-15 x \,a^{2} c d \,e^{5}+20 x a \,c^{2} d^{3} e^{3}-8 c^{3} d^{5} e x +5 e^{6} a^{3}-30 d^{2} e^{4} a^{2} c +40 d^{4} e^{2} a \,c^{2}-16 d^{6} c^{3}\right ) {\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{3}}{5 e^{4} \left (c d x +a e \right )^{3} \left (e x +d \right )^{\frac {7}{2}}}\) | \(168\) |
Input:
int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^3/(e*x+d)^(9/2),x,method=_RETURNVERB OSE)
Output:
2/5*((d^3*e^3*x^3-2*d^4*e^2*x^2+8*d^5*e*x+16*d^6)*c^3-40*e^2*(-1/8*e^2*x^2 +1/2*d*e*x+d^2)*a*d^2*c^2+30*e^4*(1/2*e*x+d)*a^2*d*c-5*e^6*a^3)/(e*x+d)^(1 /2)/e^4
Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\frac {2 \, {\left (c^{3} d^{3} e^{3} x^{3} + 16 \, c^{3} d^{6} - 40 \, a c^{2} d^{4} e^{2} + 30 \, a^{2} c d^{2} e^{4} - 5 \, a^{3} e^{6} - {\left (2 \, c^{3} d^{4} e^{2} - 5 \, a c^{2} d^{2} e^{4}\right )} x^{2} + {\left (8 \, c^{3} d^{5} e - 20 \, a c^{2} d^{3} e^{3} + 15 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{5 \, {\left (e^{5} x + d e^{4}\right )}} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(9/2),x, algorithm=" fricas")
Output:
2/5*(c^3*d^3*e^3*x^3 + 16*c^3*d^6 - 40*a*c^2*d^4*e^2 + 30*a^2*c*d^2*e^4 - 5*a^3*e^6 - (2*c^3*d^4*e^2 - 5*a*c^2*d^2*e^4)*x^2 + (8*c^3*d^5*e - 20*a*c^ 2*d^3*e^3 + 15*a^2*c*d*e^5)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (105) = 210\).
Time = 0.78 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.04 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\begin {cases} - \frac {2 a^{3} e^{2}}{\sqrt {d + e x}} + \frac {12 a^{2} c d^{2}}{\sqrt {d + e x}} + \frac {6 a^{2} c d e x}{\sqrt {d + e x}} - \frac {16 a c^{2} d^{4}}{e^{2} \sqrt {d + e x}} - \frac {8 a c^{2} d^{3} x}{e \sqrt {d + e x}} + \frac {2 a c^{2} d^{2} x^{2}}{\sqrt {d + e x}} + \frac {32 c^{3} d^{6}}{5 e^{4} \sqrt {d + e x}} + \frac {16 c^{3} d^{5} x}{5 e^{3} \sqrt {d + e x}} - \frac {4 c^{3} d^{4} x^{2}}{5 e^{2} \sqrt {d + e x}} + \frac {2 c^{3} d^{3} x^{3}}{5 e \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {c^{3} d^{\frac {3}{2}} x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3/(e*x+d)**(9/2),x)
Output:
Piecewise((-2*a**3*e**2/sqrt(d + e*x) + 12*a**2*c*d**2/sqrt(d + e*x) + 6*a **2*c*d*e*x/sqrt(d + e*x) - 16*a*c**2*d**4/(e**2*sqrt(d + e*x)) - 8*a*c**2 *d**3*x/(e*sqrt(d + e*x)) + 2*a*c**2*d**2*x**2/sqrt(d + e*x) + 32*c**3*d** 6/(5*e**4*sqrt(d + e*x)) + 16*c**3*d**5*x/(5*e**3*sqrt(d + e*x)) - 4*c**3* d**4*x**2/(5*e**2*sqrt(d + e*x)) + 2*c**3*d**3*x**3/(5*e*sqrt(d + e*x)), N e(e, 0)), (c**3*d**(3/2)*x**4/4, True))
Time = 0.03 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {5}{2}} c^{3} d^{3} - 5 \, {\left (c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} \sqrt {e x + d}}{e^{3}} + \frac {5 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )}}{\sqrt {e x + d} e^{3}}\right )}}{5 \, e} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(9/2),x, algorithm=" maxima")
Output:
2/5*(((e*x + d)^(5/2)*c^3*d^3 - 5*(c^3*d^4 - a*c^2*d^2*e^2)*(e*x + d)^(3/2 ) + 15*(c^3*d^5 - 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*sqrt(e*x + d))/e^3 + 5*(c ^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)/(sqrt(e*x + d)*e^3)) /e
Time = 0.18 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\frac {2 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )}}{\sqrt {e x + d} e^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {5}{2}} c^{3} d^{3} e^{16} - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{3} d^{4} e^{16} + 15 \, \sqrt {e x + d} c^{3} d^{5} e^{16} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} a c^{2} d^{2} e^{18} - 30 \, \sqrt {e x + d} a c^{2} d^{3} e^{18} + 15 \, \sqrt {e x + d} a^{2} c d e^{20}\right )}}{5 \, e^{20}} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(9/2),x, algorithm=" giac")
Output:
2*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)/(sqrt(e*x + d)*e ^4) + 2/5*((e*x + d)^(5/2)*c^3*d^3*e^16 - 5*(e*x + d)^(3/2)*c^3*d^4*e^16 + 15*sqrt(e*x + d)*c^3*d^5*e^16 + 5*(e*x + d)^(3/2)*a*c^2*d^2*e^18 - 30*sqr t(e*x + d)*a*c^2*d^3*e^18 + 15*sqrt(e*x + d)*a^2*c*d*e^20)/e^20
Time = 5.16 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\frac {2\,c^3\,d^3\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}-\frac {\left (6\,c^3\,d^4-6\,a\,c^2\,d^2\,e^2\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4}-\frac {2\,a^3\,e^6-6\,a^2\,c\,d^2\,e^4+6\,a\,c^2\,d^4\,e^2-2\,c^3\,d^6}{e^4\,\sqrt {d+e\,x}}+\frac {6\,c\,d\,{\left (a\,e^2-c\,d^2\right )}^2\,\sqrt {d+e\,x}}{e^4} \] Input:
int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3/(d + e*x)^(9/2),x)
Output:
(2*c^3*d^3*(d + e*x)^(5/2))/(5*e^4) - ((6*c^3*d^4 - 6*a*c^2*d^2*e^2)*(d + e*x)^(3/2))/(3*e^4) - (2*a^3*e^6 - 2*c^3*d^6 + 6*a*c^2*d^4*e^2 - 6*a^2*c*d ^2*e^4)/(e^4*(d + e*x)^(1/2)) + (6*c*d*(a*e^2 - c*d^2)^2*(d + e*x)^(1/2))/ e^4
Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{9/2}} \, dx=\frac {\frac {2}{5} c^{3} d^{3} e^{3} x^{3}+2 a \,c^{2} d^{2} e^{4} x^{2}-\frac {4}{5} c^{3} d^{4} e^{2} x^{2}+6 a^{2} c d \,e^{5} x -8 a \,c^{2} d^{3} e^{3} x +\frac {16}{5} c^{3} d^{5} e x -2 a^{3} e^{6}+12 a^{2} c \,d^{2} e^{4}-16 a \,c^{2} d^{4} e^{2}+\frac {32}{5} c^{3} d^{6}}{\sqrt {e x +d}\, e^{4}} \] Input:
int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^(9/2),x)
Output:
(2*( - 5*a**3*e**6 + 30*a**2*c*d**2*e**4 + 15*a**2*c*d*e**5*x - 40*a*c**2* d**4*e**2 - 20*a*c**2*d**3*e**3*x + 5*a*c**2*d**2*e**4*x**2 + 16*c**3*d**6 + 8*c**3*d**5*e*x - 2*c**3*d**4*e**2*x**2 + c**3*d**3*e**3*x**3))/(5*sqrt (d + e*x)*e**4)