Integrand size = 37, antiderivative size = 119 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}{3 e^4}+\frac {6 c d \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}{5 e^4}-\frac {6 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}{7 e^4}+\frac {2 c^3 d^3 (d+e x)^{9/2}}{9 e^4} \]
-2/3*(-a*e^2+c*d^2)^3*(e*x+d)^(3/2)/e^4+6/5*c*d*(-a*e^2+c*d^2)^2*(e*x+d)^( 5/2)/e^4-6/7*c^2*d^2*(-a*e^2+c*d^2)*(e*x+d)^(7/2)/e^4+2/9*c^3*d^3*(e*x+d)^ (9/2)/e^4
Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx=\frac {2 (d+e x)^{3/2} \left (105 a^3 e^6-63 a^2 c d e^4 (2 d-3 e x)+9 a c^2 d^2 e^2 \left (8 d^2-12 d e x+15 e^2 x^2\right )+c^3 d^3 \left (-16 d^3+24 d^2 e x-30 d e^2 x^2+35 e^3 x^3\right )\right )}{315 e^4} \]
(2*(d + e*x)^(3/2)*(105*a^3*e^6 - 63*a^2*c*d*e^4*(2*d - 3*e*x) + 9*a*c^2*d ^2*e^2*(8*d^2 - 12*d*e*x + 15*e^2*x^2) + c^3*d^3*(-16*d^3 + 24*d^2*e*x - 3 0*d*e^2*x^2 + 35*e^3*x^3)))/(315*e^4)
Time = 0.25 (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.
\(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \left (-\frac {3 c^2 d^2 (d+e x)^{5/2} \left (c d^2-a e^2\right )}{e^3}+\frac {3 c d (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}{e^3}+\frac {\sqrt {d+e x} \left (a e^2-c d^2\right )^3}{e^3}+\frac {c^3 d^3 (d+e x)^{7/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 c^2 d^2 (d+e x)^{7/2} \left (c d^2-a e^2\right )}{7 e^4}+\frac {6 c d (d+e x)^{5/2} \left (c d^2-a e^2\right )^2}{5 e^4}-\frac {2 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}{3 e^4}+\frac {2 c^3 d^3 (d+e x)^{9/2}}{9 e^4}\) |
(-2*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2))/(3*e^4) + (6*c*d*(c*d^2 - a*e^2)^2* (d + e*x)^(5/2))/(5*e^4) - (6*c^2*d^2*(c*d^2 - a*e^2)*(d + e*x)^(7/2))/(7* e^4) + (2*c^3*d^3*(d + e*x)^(9/2))/(9*e^4)
3.20.95.3.1 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 = 2.86 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {6 \left (e^{2} a -c \,d^{2}\right ) c^{2} d^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {6 \left (e^{2} a -c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(97\) |
default | \(\frac {\frac {2 c^{3} d^{3} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {6 \left (e^{2} a -c \,d^{2}\right ) c^{2} d^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {6 \left (e^{2} a -c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(97\) |
pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (e^{6} a^{3}+\frac {9 x \,a^{2} c d \,e^{5}}{5}-\frac {6 \left (-\frac {15 c \,x^{2}}{14}+a \right ) c \,d^{2} a \,e^{4}}{5}-\frac {36 x \left (-\frac {35 c \,x^{2}}{108}+a \right ) c^{2} d^{3} e^{3}}{35}+\frac {24 \left (-\frac {5 c \,x^{2}}{12}+a \right ) c^{2} d^{4} e^{2}}{35}+\frac {8 x \,c^{3} d^{5} e}{35}-\frac {16 c^{3} d^{6}}{105}\right )}{3 e^{4}}\) | \(107\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 x^{3} c^{3} d^{3} e^{3}+135 x^{2} a \,c^{2} d^{2} e^{4}-30 x^{2} c^{3} d^{4} e^{2}+189 x \,a^{2} c d \,e^{5}-108 x a \,c^{2} d^{3} e^{3}+24 x \,c^{3} d^{5} e +105 e^{6} a^{3}-126 d^{2} e^{4} a^{2} c +72 d^{4} e^{2} c^{2} a -16 c^{3} d^{6}\right )}{315 e^{4}}\) | \(131\) |
trager | \(\frac {2 \left (35 c^{3} d^{3} e^{4} x^{4}+135 a \,c^{2} d^{2} e^{5} x^{3}+5 c^{3} d^{4} e^{3} x^{3}+189 a^{2} c d \,e^{6} x^{2}+27 a \,c^{2} d^{3} e^{4} x^{2}-6 c^{3} d^{5} e^{2} x^{2}+105 a^{3} e^{7} x +63 a^{2} c \,d^{2} e^{5} x -36 a \,c^{2} d^{4} e^{3} x +8 c^{3} d^{6} e x +105 a^{3} e^{6} d -126 a^{2} c \,d^{3} e^{4}+72 a \,c^{2} d^{5} e^{2}-16 c^{3} d^{7}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(185\) |
risch | \(\frac {2 \left (35 c^{3} d^{3} e^{4} x^{4}+135 a \,c^{2} d^{2} e^{5} x^{3}+5 c^{3} d^{4} e^{3} x^{3}+189 a^{2} c d \,e^{6} x^{2}+27 a \,c^{2} d^{3} e^{4} x^{2}-6 c^{3} d^{5} e^{2} x^{2}+105 a^{3} e^{7} x +63 a^{2} c \,d^{2} e^{5} x -36 a \,c^{2} d^{4} e^{3} x +8 c^{3} d^{6} e x +105 a^{3} e^{6} d -126 a^{2} c \,d^{3} e^{4}+72 a \,c^{2} d^{5} e^{2}-16 c^{3} d^{7}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(185\) |
2/e^4*(1/9*c^3*d^3*(e*x+d)^(9/2)+3/7*(a*e^2-c*d^2)*c^2*d^2*(e*x+d)^(7/2)+3 /5*(a*e^2-c*d^2)^2*c*d*(e*x+d)^(5/2)+1/3*(a*e^2-c*d^2)^3*(e*x+d)^(3/2))
Time = 0.35 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (35 \, c^{3} d^{3} e^{4} x^{4} - 16 \, c^{3} d^{7} + 72 \, a c^{2} d^{5} e^{2} - 126 \, a^{2} c d^{3} e^{4} + 105 \, a^{3} d e^{6} + 5 \, {\left (c^{3} d^{4} e^{3} + 27 \, a c^{2} d^{2} e^{5}\right )} x^{3} - 3 \, {\left (2 \, c^{3} d^{5} e^{2} - 9 \, a c^{2} d^{3} e^{4} - 63 \, a^{2} c d e^{6}\right )} x^{2} + {\left (8 \, c^{3} d^{6} e - 36 \, a c^{2} d^{4} e^{3} + 63 \, a^{2} c d^{2} e^{5} + 105 \, a^{3} e^{7}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{4}} \]
2/315*(35*c^3*d^3*e^4*x^4 - 16*c^3*d^7 + 72*a*c^2*d^5*e^2 - 126*a^2*c*d^3* e^4 + 105*a^3*d*e^6 + 5*(c^3*d^4*e^3 + 27*a*c^2*d^2*e^5)*x^3 - 3*(2*c^3*d^ 5*e^2 - 9*a*c^2*d^3*e^4 - 63*a^2*c*d*e^6)*x^2 + (8*c^3*d^6*e - 36*a*c^2*d^ 4*e^3 + 63*a^2*c*d^2*e^5 + 105*a^3*e^7)*x)*sqrt(e*x + d)/e^4
Time = 1.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{3} d^{3} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (3 a c^{2} d^{2} e^{2} - 3 c^{3} d^{4}\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (3 a^{2} c d e^{4} - 6 a c^{2} d^{3} e^{2} + 3 c^{3} d^{5}\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a^{3} e^{6} - 3 a^{2} c d^{2} e^{4} + 3 a c^{2} d^{4} e^{2} - c^{3} d^{6}\right )}{3 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {c^{3} d^{\frac {7}{2}} x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((2*(c**3*d**3*(d + e*x)**(9/2)/(9*e**3) + (d + e*x)**(7/2)*(3*a* c**2*d**2*e**2 - 3*c**3*d**4)/(7*e**3) + (d + e*x)**(5/2)*(3*a**2*c*d*e**4 - 6*a*c**2*d**3*e**2 + 3*c**3*d**5)/(5*e**3) + (d + e*x)**(3/2)*(a**3*e** 6 - 3*a**2*c*d**2*e**4 + 3*a*c**2*d**4*e**2 - c**3*d**6)/(3*e**3))/e, Ne(e , 0)), (c**3*d**(7/2)*x**4/4, True))
Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} c^{3} d^{3} - 135 \, {\left (c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 189 \, {\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} {\left (e x + d\right )}^{\frac {5}{2}} - 105 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{315 \, e^{4}} \]
2/315*(35*(e*x + d)^(9/2)*c^3*d^3 - 135*(c^3*d^4 - a*c^2*d^2*e^2)*(e*x + d )^(7/2) + 189*(c^3*d^5 - 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*(e*x + d)^(5/2) - 105*(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)^(3/2 ))/e^4
Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (103) = 206\).
Time = 0.28 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.83 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{3} d e^{3} + 315 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} c d^{2} e + 105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{3} e^{3} + \frac {63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c^{2} d^{3}}{e} + 63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} c d e + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c^{3} d^{4}}{e^{3}} + \frac {27 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a c^{2} d^{2}}{e} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c^{3} d^{3}}{e^{3}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*a^3*d*e^3 + 315*((e*x + d)^(3/2) - 3*sqrt(e*x + d )*d)*a^2*c*d^2*e + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^3*e^3 + 63* (3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*c^2*d^ 3/e + 63*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2) *a^2*c*d*e + 9*(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^4/e^3 + 27*(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^ 2/e + (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^3/e^3)/e
Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{5/2}} \, dx=\frac {2\,{\left (a\,e^2-c\,d^2\right )}^3\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4}-\frac {\left (6\,c^3\,d^4-6\,a\,c^2\,d^2\,e^2\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4}+\frac {2\,c^3\,d^3\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}+\frac {6\,c\,d\,{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4} \]