Integrand size = 37, antiderivative size = 81 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^{5/2}}+\frac {4 c d \left (c d^2-a e^2\right )}{3 e^3 (d+e x)^{3/2}}-\frac {2 c^2 d^2}{e^3 \sqrt {d+e x}} \] Output:
-2/5*(-a*e^2+c*d^2)^2/e^3/(e*x+d)^(5/2)+4/3*c*d*(-a*e^2+c*d^2)/e^3/(e*x+d) ^(3/2)-2*c^2*d^2/e^3/(e*x+d)^(1/2)
Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \left (3 a^2 e^4+2 a c d e^2 (2 d+5 e x)+c^2 d^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (d+e x)^{5/2}} \] Input:
Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(11/2),x]
Output:
(-2*(3*a^2*e^4 + 2*a*c*d*e^2*(2*d + 5*e*x) + c^2*d^2*(8*d^2 + 20*d*e*x + 1 5*e^2*x^2)))/(15*e^3*(d + e*x)^(5/2))
Time = 0.36 (sec) , antiderivative size = 81, 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.
\(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \left (-\frac {2 c d \left (c d^2-a e^2\right )}{e^2 (d+e x)^{5/2}}+\frac {\left (a e^2-c d^2\right )^2}{e^2 (d+e x)^{7/2}}+\frac {c^2 d^2}{e^2 (d+e x)^{3/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 c d \left (c d^2-a e^2\right )}{3 e^3 (d+e x)^{3/2}}-\frac {2 \left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^{5/2}}-\frac {2 c^2 d^2}{e^3 \sqrt {d+e x}}\) |
Input:
Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(11/2),x]
Output:
(-2*(c*d^2 - a*e^2)^2)/(5*e^3*(d + e*x)^(5/2)) + (4*c*d*(c*d^2 - a*e^2))/( 3*e^3*(d + e*x)^(3/2)) - (2*c^2*d^2)/(e^3*Sqrt[d + e*x])
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 = 1.39 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80
method | result | size |
pseudoelliptic | \(-\frac {2 \left (a^{2} e^{4}+\frac {10 x a c d \,e^{3}}{3}+\frac {4 \left (\frac {15 c \,x^{2}}{4}+a \right ) c \,d^{2} e^{2}}{3}+\frac {20 x \,c^{2} d^{3} e}{3}+\frac {8 c^{2} d^{4}}{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\) | \(65\) |
gosper | \(-\frac {2 \left (15 x^{2} c^{2} d^{2} e^{2}+10 x a c d \,e^{3}+20 x \,c^{2} d^{3} e +3 a^{2} e^{4}+4 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\) | \(73\) |
trager | \(-\frac {2 \left (15 x^{2} c^{2} d^{2} e^{2}+10 x a c d \,e^{3}+20 x \,c^{2} d^{3} e +3 a^{2} e^{4}+4 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\) | \(73\) |
derivativedivides | \(\frac {-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\sqrt {e x +d}}-\frac {4 c d \left (a \,e^{2}-c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\) | \(79\) |
default | \(\frac {-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\sqrt {e x +d}}-\frac {4 c d \left (a \,e^{2}-c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\) | \(79\) |
orering | \(-\frac {2 \left (15 x^{2} c^{2} d^{2} e^{2}+10 x a c d \,e^{3}+20 x \,c^{2} d^{3} e +3 a^{2} e^{4}+4 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right ) {\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{2}}{15 e^{3} \left (c d x +a e \right )^{2} \left (e x +d \right )^{\frac {9}{2}}}\) | \(110\) |
Input:
int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^2/(e*x+d)^(11/2),x,method=_RETURNVER BOSE)
Output:
-2/5/(e*x+d)^(5/2)*(a^2*e^4+10/3*x*a*c*d*e^3+4/3*(15/4*c*x^2+a)*c*d^2*e^2+ 20/3*x*c^2*d^3*e+8/3*c^2*d^4)/e^3
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (15 \, c^{2} d^{2} e^{2} x^{2} + 8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 10 \, {\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x, algorithm= "fricas")
Output:
-2/15*(15*c^2*d^2*e^2*x^2 + 8*c^2*d^4 + 4*a*c*d^2*e^2 + 3*a^2*e^4 + 10*(2* c^2*d^3*e + a*c*d*e^3)*x)*sqrt(e*x + d)/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4 *x + d^3*e^3)
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (75) = 150\).
Time = 0.96 (sec) , antiderivative size = 388, normalized size of antiderivative = 4.79 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=\begin {cases} - \frac {6 a^{2} e^{4}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {8 a c d^{2} e^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {20 a c d e^{3} x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {16 c^{2} d^{4}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {40 c^{2} d^{3} e x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {30 c^{2} d^{2} e^{2} x^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {c^{2} x^{3}}{3 d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2/(e*x+d)**(11/2),x)
Output:
Piecewise((-6*a**2*e**4/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 8*a*c*d**2*e**2/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 20*a*c *d*e**3*x/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e** 5*x**2*sqrt(d + e*x)) - 16*c**2*d**4/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e* *4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 40*c**2*d**3*e*x/(15*d* *2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 30*c**2*d**2*e**2*x**2/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*s qrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)), Ne(e, 0)), (c**2*x**3/(3*d**(3 /2)), True))
Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c^{2} d^{2} + 3 \, c^{2} d^{4} - 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} - 10 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} {\left (e x + d\right )}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x, algorithm= "maxima")
Output:
-2/15*(15*(e*x + d)^2*c^2*d^2 + 3*c^2*d^4 - 6*a*c*d^2*e^2 + 3*a^2*e^4 - 10 *(c^2*d^3 - a*c*d*e^2)*(e*x + d))/((e*x + d)^(5/2)*e^3)
Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c^{2} d^{2} - 10 \, {\left (e x + d\right )} c^{2} d^{3} + 3 \, c^{2} d^{4} + 10 \, {\left (e x + d\right )} a c d e^{2} - 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x, algorithm= "giac")
Output:
-2/15*(15*(e*x + d)^2*c^2*d^2 - 10*(e*x + d)*c^2*d^3 + 3*c^2*d^4 + 10*(e*x + d)*a*c*d*e^2 - 6*a*c*d^2*e^2 + 3*a^2*e^4)/((e*x + d)^(5/2)*e^3)
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=-\frac {\frac {2\,a^2\,e^4}{5}+\frac {2\,c^2\,d^4}{5}-\left (\frac {4\,c^2\,d^3}{3}-\frac {4\,a\,c\,d\,e^2}{3}\right )\,\left (d+e\,x\right )+2\,c^2\,d^2\,{\left (d+e\,x\right )}^2-\frac {4\,a\,c\,d^2\,e^2}{5}}{e^3\,{\left (d+e\,x\right )}^{5/2}} \] Input:
int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2/(d + e*x)^(11/2),x)
Output:
-((2*a^2*e^4)/5 + (2*c^2*d^4)/5 - ((4*c^2*d^3)/3 - (4*a*c*d*e^2)/3)*(d + e *x) + 2*c^2*d^2*(d + e*x)^2 - (4*a*c*d^2*e^2)/5)/(e^3*(d + e*x)^(5/2))
Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{11/2}} \, dx=\frac {-2 c^{2} d^{2} e^{2} x^{2}-\frac {4}{3} a c d \,e^{3} x -\frac {8}{3} c^{2} d^{3} e x -\frac {2}{5} a^{2} e^{4}-\frac {8}{15} a c \,d^{2} e^{2}-\frac {16}{15} c^{2} d^{4}}{\sqrt {e x +d}\, e^{3} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(11/2),x)
Output:
(2*( - 3*a**2*e**4 - 4*a*c*d**2*e**2 - 10*a*c*d*e**3*x - 8*c**2*d**4 - 20* c**2*d**3*e*x - 15*c**2*d**2*e**2*x**2))/(15*sqrt(d + e*x)*e**3*(d**2 + 2* d*e*x + e**2*x**2))