Integrand size = 37, antiderivative size = 119 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx=-\frac {2 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}{5 e^4}+\frac {6 c d \left (c d^2-a e^2\right )^2 (d+e x)^{7/2}}{7 e^4}-\frac {2 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{9/2}}{3 e^4}+\frac {2 c^3 d^3 (d+e x)^{11/2}}{11 e^4} \]
-2/5*(-a*e^2+c*d^2)^3*(e*x+d)^(5/2)/e^4+6/7*c*d*(-a*e^2+c*d^2)^2*(e*x+d)^( 7/2)/e^4-2/3*c^2*d^2*(-a*e^2+c*d^2)*(e*x+d)^(9/2)/e^4+2/11*c^3*d^3*(e*x+d) ^(11/2)/e^4
Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx=\frac {2 (d+e x)^{5/2} \left (231 a^3 e^6-99 a^2 c d e^4 (2 d-5 e x)+11 a c^2 d^2 e^2 \left (8 d^2-20 d e x+35 e^2 x^2\right )+c^3 d^3 \left (-16 d^3+40 d^2 e x-70 d e^2 x^2+105 e^3 x^3\right )\right )}{1155 e^4} \]
(2*(d + e*x)^(5/2)*(231*a^3*e^6 - 99*a^2*c*d*e^4*(2*d - 5*e*x) + 11*a*c^2* d^2*e^2*(8*d^2 - 20*d*e*x + 35*e^2*x^2) + c^3*d^3*(-16*d^3 + 40*d^2*e*x - 70*d*e^2*x^2 + 105*e^3*x^3)))/(1155*e^4)
Time = 0.26 (sec) , antiderivative size = 119, 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.
| \(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1121 |
| \(\displaystyle \int \left (-\frac {3 c^2 d^2 (d+e x)^{7/2} \left (c d^2-a e^2\right )}{e^3}+\frac {3 c d (d+e x)^{5/2} \left (c d^2-a e^2\right )^2}{e^3}+\frac {(d+e x)^{3/2} \left (a e^2-c d^2\right )^3}{e^3}+\frac {c^3 d^3 (d+e x)^{9/2}}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 c^2 d^2 (d+e x)^{9/2} \left (c d^2-a e^2\right )}{3 e^4}+\frac {6 c d (d+e x)^{7/2} \left (c d^2-a e^2\right )^2}{7 e^4}-\frac {2 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3}{5 e^4}+\frac {2 c^3 d^3 (d+e x)^{11/2}}{11 e^4}\) |
(-2*(c*d^2 - a*e^2)^3*(d + e*x)^(5/2))/(5*e^4) + (6*c*d*(c*d^2 - a*e^2)^2* (d + e*x)^(7/2))/(7*e^4) - (2*c^2*d^2*(c*d^2 - a*e^2)*(d + e*x)^(9/2))/(3* e^4) + (2*c^3*d^3*(d + e*x)^(11/2))/(11*e^4)
3.20.94.3.1 Defintions of rubi rules used
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 = 2.73 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (e^{2} a -c \,d^{2}\right ) c^{2} d^{2} \left (e x +d \right )^{\frac {9}{2}}}{3}+\frac {6 \left (e^{2} a -c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{4}}\) | \(97\) |
| default | \(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (e^{2} a -c \,d^{2}\right ) c^{2} d^{2} \left (e x +d \right )^{\frac {9}{2}}}{3}+\frac {6 \left (e^{2} a -c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{4}}\) | \(97\) |
| pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (e^{6} a^{3}+\frac {15 x \,a^{2} c d \,e^{5}}{7}-\frac {6 \left (-\frac {35 c \,x^{2}}{18}+a \right ) c \,d^{2} a \,e^{4}}{7}-\frac {20 x \,c^{2} \left (-\frac {21 c \,x^{2}}{44}+a \right ) d^{3} e^{3}}{21}+\frac {8 c^{2} \left (-\frac {35 c \,x^{2}}{44}+a \right ) d^{4} e^{2}}{21}+\frac {40 x \,c^{3} d^{5} e}{231}-\frac {16 c^{3} d^{6}}{231}\right )}{5 e^{4}}\) | \(107\) |
| gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (105 x^{3} c^{3} d^{3} e^{3}+385 x^{2} a \,c^{2} d^{2} e^{4}-70 x^{2} c^{3} d^{4} e^{2}+495 x \,a^{2} c d \,e^{5}-220 x a \,c^{2} d^{3} e^{3}+40 x \,c^{3} d^{5} e +231 e^{6} a^{3}-198 d^{2} e^{4} a^{2} c +88 d^{4} e^{2} c^{2} a -16 c^{3} d^{6}\right )}{1155 e^{4}}\) | \(131\) |
| trager | \(\frac {2 \left (105 c^{3} d^{3} e^{5} x^{5}+385 a \,c^{2} d^{2} e^{6} x^{4}+140 c^{3} d^{4} e^{4} x^{4}+495 d \,e^{7} a^{2} c \,x^{3}+550 a \,c^{2} d^{3} e^{5} x^{3}+5 c^{3} d^{5} e^{3} x^{3}+231 a^{3} e^{8} x^{2}+792 a^{2} c \,d^{2} e^{6} x^{2}+33 a \,c^{2} d^{4} e^{4} x^{2}-6 c^{3} d^{6} e^{2} x^{2}+462 a^{3} d \,e^{7} x +99 a^{2} c \,d^{3} e^{5} x -44 a \,c^{2} d^{5} e^{3} x +8 c^{3} d^{7} e x +231 a^{3} e^{6} d^{2}-198 d^{4} e^{4} a^{2} c +88 a \,c^{2} d^{6} e^{2}-16 c^{3} d^{8}\right ) \sqrt {e x +d}}{1155 e^{4}}\) | \(243\) |
| risch | \(\frac {2 \left (105 c^{3} d^{3} e^{5} x^{5}+385 a \,c^{2} d^{2} e^{6} x^{4}+140 c^{3} d^{4} e^{4} x^{4}+495 d \,e^{7} a^{2} c \,x^{3}+550 a \,c^{2} d^{3} e^{5} x^{3}+5 c^{3} d^{5} e^{3} x^{3}+231 a^{3} e^{8} x^{2}+792 a^{2} c \,d^{2} e^{6} x^{2}+33 a \,c^{2} d^{4} e^{4} x^{2}-6 c^{3} d^{6} e^{2} x^{2}+462 a^{3} d \,e^{7} x +99 a^{2} c \,d^{3} e^{5} x -44 a \,c^{2} d^{5} e^{3} x +8 c^{3} d^{7} e x +231 a^{3} e^{6} d^{2}-198 d^{4} e^{4} a^{2} c +88 a \,c^{2} d^{6} e^{2}-16 c^{3} d^{8}\right ) \sqrt {e x +d}}{1155 e^{4}}\) | \(243\) |
2/e^4*(1/11*c^3*d^3*(e*x+d)^(11/2)+1/3*(a*e^2-c*d^2)*c^2*d^2*(e*x+d)^(9/2) +3/7*(a*e^2-c*d^2)^2*c*d*(e*x+d)^(7/2)+1/5*(a*e^2-c*d^2)^3*(e*x+d)^(5/2))
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (103) = 206\).
Time = 0.40 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (105 \, c^{3} d^{3} e^{5} x^{5} - 16 \, c^{3} d^{8} + 88 \, a c^{2} d^{6} e^{2} - 198 \, a^{2} c d^{4} e^{4} + 231 \, a^{3} d^{2} e^{6} + 35 \, {\left (4 \, c^{3} d^{4} e^{4} + 11 \, a c^{2} d^{2} e^{6}\right )} x^{4} + 5 \, {\left (c^{3} d^{5} e^{3} + 110 \, a c^{2} d^{3} e^{5} + 99 \, a^{2} c d e^{7}\right )} x^{3} - 3 \, {\left (2 \, c^{3} d^{6} e^{2} - 11 \, a c^{2} d^{4} e^{4} - 264 \, a^{2} c d^{2} e^{6} - 77 \, a^{3} e^{8}\right )} x^{2} + {\left (8 \, c^{3} d^{7} e - 44 \, a c^{2} d^{5} e^{3} + 99 \, a^{2} c d^{3} e^{5} + 462 \, a^{3} d e^{7}\right )} x\right )} \sqrt {e x + d}}{1155 \, e^{4}} \]
2/1155*(105*c^3*d^3*e^5*x^5 - 16*c^3*d^8 + 88*a*c^2*d^6*e^2 - 198*a^2*c*d^ 4*e^4 + 231*a^3*d^2*e^6 + 35*(4*c^3*d^4*e^4 + 11*a*c^2*d^2*e^6)*x^4 + 5*(c ^3*d^5*e^3 + 110*a*c^2*d^3*e^5 + 99*a^2*c*d*e^7)*x^3 - 3*(2*c^3*d^6*e^2 - 11*a*c^2*d^4*e^4 - 264*a^2*c*d^2*e^6 - 77*a^3*e^8)*x^2 + (8*c^3*d^7*e - 44 *a*c^2*d^5*e^3 + 99*a^2*c*d^3*e^5 + 462*a^3*d*e^7)*x)*sqrt(e*x + d)/e^4
Time = 1.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{3} d^{3} \left (d + e x\right )^{\frac {11}{2}}}{11 e^{3}} + \frac {\left (d + e x\right )^{\frac {9}{2}} \cdot \left (3 a c^{2} d^{2} e^{2} - 3 c^{3} d^{4}\right )}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (3 a^{2} c d e^{4} - 6 a c^{2} d^{3} e^{2} + 3 c^{3} d^{5}\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (a^{3} e^{6} - 3 a^{2} c d^{2} e^{4} + 3 a c^{2} d^{4} e^{2} - c^{3} d^{6}\right )}{5 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {c^{3} d^{\frac {9}{2}} x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((2*(c**3*d**3*(d + e*x)**(11/2)/(11*e**3) + (d + e*x)**(9/2)*(3* a*c**2*d**2*e**2 - 3*c**3*d**4)/(9*e**3) + (d + e*x)**(7/2)*(3*a**2*c*d*e* *4 - 6*a*c**2*d**3*e**2 + 3*c**3*d**5)/(7*e**3) + (d + e*x)**(5/2)*(a**3*e **6 - 3*a**2*c*d**2*e**4 + 3*a*c**2*d**4*e**2 - c**3*d**6)/(5*e**3))/e, Ne (e, 0)), (c**3*d**(9/2)*x**4/4, True))
Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (105 \, {\left (e x + d\right )}^{\frac {11}{2}} c^{3} d^{3} - 385 \, {\left (c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {9}{2}} + 495 \, {\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} {\left (e x + d\right )}^{\frac {7}{2}} - 231 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{1155 \, e^{4}} \]
2/1155*(105*(e*x + d)^(11/2)*c^3*d^3 - 385*(c^3*d^4 - a*c^2*d^2*e^2)*(e*x + d)^(9/2) + 495*(c^3*d^5 - 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*(e*x + d)^(7/2) - 231*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*(e*x + d)^( 5/2))/e^4
Leaf count of result is larger than twice the leaf count of optimal. 582 vs. \(2 (103) = 206\).
Time = 0.27 (sec) , antiderivative size = 582, normalized size of antiderivative = 4.89 \[ \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)^{3/2}} \, dx=\frac {2 \, {\left (3465 \, \sqrt {e x + d} a^{3} d^{2} e^{3} + 3465 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} c d^{3} e + 2310 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{3} d e^{3} + \frac {693 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c^{2} d^{4}}{e} + 1386 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} c d^{2} e + 231 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{3} e^{3} + \frac {99 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c^{3} d^{5}}{e^{3}} + \frac {594 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a c^{2} d^{3}}{e} + 297 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a^{2} c d e + \frac {22 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c^{3} d^{4}}{e^{3}} + \frac {33 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} a c^{2} d^{2}}{e} + \frac {5 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} - 385 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 990 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 1386 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} + 1155 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 693 \, \sqrt {e x + d} d^{5}\right )} c^{3} d^{3}}{e^{3}}\right )}}{3465 \, e} \]
2/3465*(3465*sqrt(e*x + d)*a^3*d^2*e^3 + 3465*((e*x + d)^(3/2) - 3*sqrt(e* x + d)*d)*a^2*c*d^3*e + 2310*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^3*d*e ^3 + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2) *a*c^2*d^4/e + 1386*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e* x + d)*d^2)*a^2*c*d^2*e + 231*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a^3*e^3 + 99*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2) *d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*c^3*d^5/e^3 + 594*(5*( e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e *x + d)*d^3)*a*c^2*d^3/e + 297*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a^2*c*d*e + 22*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d) ^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*c^3*d^4/e^3 + 33*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*a*c^2*d^2/e + 5*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d)^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*( e*x + d)^(3/2)*d^4 - 693*sqrt(e*x + d)*d^5)*c^3*d^3/e^3)/e
Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx=\frac {2\,{\left (a\,e^2-c\,d^2\right )}^3\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}-\frac {\left (6\,c^3\,d^4-6\,a\,c^2\,d^2\,e^2\right )\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}+\frac {2\,c^3\,d^3\,{\left (d+e\,x\right )}^{11/2}}{11\,e^4}+\frac {6\,c\,d\,{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4} \]