Integrand size = 37, antiderivative size = 83 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{3 e^3}-\frac {4 c d \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 e^3}+\frac {2 c^2 d^2 (d+e x)^{7/2}}{7 e^3} \] Output:
2/3*(-a*e^2+c*d^2)^2*(e*x+d)^(3/2)/e^3-4/5*c*d*(-a*e^2+c*d^2)*(e*x+d)^(5/2 )/e^3+2/7*c^2*d^2*(e*x+d)^(7/2)/e^3
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 (d+e x)^{3/2} \left (35 a^2 e^4+14 a c d e^2 (-2 d+3 e x)+c^2 d^2 \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3} \] Input:
Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(3/2),x]
Output:
(2*(d + e*x)^(3/2)*(35*a^2*e^4 + 14*a*c*d*e^2*(-2*d + 3*e*x) + c^2*d^2*(8* d^2 - 12*d*e*x + 15*e^2*x^2)))/(105*e^3)
Time = 0.36 (sec) , antiderivative size = 83, 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.
\(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \left (-\frac {2 c d (d+e x)^{3/2} \left (c d^2-a e^2\right )}{e^2}+\frac {\sqrt {d+e x} \left (a e^2-c d^2\right )^2}{e^2}+\frac {c^2 d^2 (d+e x)^{5/2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 c d (d+e x)^{5/2} \left (c d^2-a e^2\right )}{5 e^3}+\frac {2 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}{3 e^3}+\frac {2 c^2 d^2 (d+e x)^{7/2}}{7 e^3}\) |
Input:
Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^(3/2),x]
Output:
(2*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2))/(3*e^3) - (4*c*d*(c*d^2 - a*e^2)*(d + e*x)^(5/2))/(5*e^3) + (2*c^2*d^2*(d + e*x)^(7/2))/(7*e^3)
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 = 2.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (a^{2} e^{4}+\frac {6 x a c d \,e^{3}}{5}-\frac {4 c \,d^{2} \left (-\frac {15 c \,x^{2}}{28}+a \right ) e^{2}}{5}-\frac {12 x \,c^{2} d^{3} e}{35}+\frac {8 c^{2} d^{4}}{35}\right )}{3 e^{3}}\) | \(65\) |
derivativedivides | \(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {4 c d \left (a \,e^{2}-c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\) | \(68\) |
default | \(\frac {\frac {2 c^{2} d^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {4 c d \left (a \,e^{2}-c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\) | \(68\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 x^{2} c^{2} d^{2} e^{2}+42 x a c d \,e^{3}-12 x \,c^{2} d^{3} e +35 a^{2} e^{4}-28 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{105 e^{3}}\) | \(73\) |
trager | \(\frac {2 \left (15 c^{2} d^{2} e^{3} x^{3}+42 a c \,e^{4} d \,x^{2}+3 c^{2} d^{3} e^{2} x^{2}+35 a^{2} e^{5} x +14 a c \,d^{2} e^{3} x -4 c^{2} d^{4} e x +35 a^{2} d \,e^{4}-28 a c \,d^{3} e^{2}+8 c^{2} d^{5}\right ) \sqrt {e x +d}}{105 e^{3}}\) | \(110\) |
risch | \(\frac {2 \left (15 c^{2} d^{2} e^{3} x^{3}+42 a c \,e^{4} d \,x^{2}+3 c^{2} d^{3} e^{2} x^{2}+35 a^{2} e^{5} x +14 a c \,d^{2} e^{3} x -4 c^{2} d^{4} e x +35 a^{2} d \,e^{4}-28 a c \,d^{3} e^{2}+8 c^{2} d^{5}\right ) \sqrt {e x +d}}{105 e^{3}}\) | \(110\) |
orering | \(\frac {2 \left (15 x^{2} c^{2} d^{2} e^{2}+42 x a c d \,e^{3}-12 x \,c^{2} d^{3} e +35 a^{2} e^{4}-28 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right ) {\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{2}}{105 e^{3} \left (c d x +a e \right )^{2} \sqrt {e x +d}}\) | \(110\) |
Input:
int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^2/(e*x+d)^(3/2),x,method=_RETURNVERB OSE)
Output:
2/3*(e*x+d)^(3/2)*(a^2*e^4+6/5*x*a*c*d*e^3-4/5*c*d^2*(-15/28*c*x^2+a)*e^2- 12/35*x*c^2*d^3*e+8/35*c^2*d^4)/e^3
Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (15 \, c^{2} d^{2} e^{3} x^{3} + 8 \, c^{2} d^{5} - 28 \, a c d^{3} e^{2} + 35 \, a^{2} d e^{4} + 3 \, {\left (c^{2} d^{3} e^{2} + 14 \, a c d e^{4}\right )} x^{2} - {\left (4 \, c^{2} d^{4} e - 14 \, a c d^{2} e^{3} - 35 \, a^{2} e^{5}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{3}} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(3/2),x, algorithm=" fricas")
Output:
2/105*(15*c^2*d^2*e^3*x^3 + 8*c^2*d^5 - 28*a*c*d^3*e^2 + 35*a^2*d*e^4 + 3* (c^2*d^3*e^2 + 14*a*c*d*e^4)*x^2 - (4*c^2*d^4*e - 14*a*c*d^2*e^3 - 35*a^2* e^5)*x)*sqrt(e*x + d)/e^3
Time = 0.88 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{2} d^{2} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (2 a c d e^{2} - 2 c^{2} d^{3}\right )}{5 e^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a^{2} e^{4} - 2 a c d^{2} e^{2} + c^{2} d^{4}\right )}{3 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {c^{2} d^{\frac {5}{2}} x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2/(e*x+d)**(3/2),x)
Output:
Piecewise((2*(c**2*d**2*(d + e*x)**(7/2)/(7*e**2) + (d + e*x)**(5/2)*(2*a* c*d*e**2 - 2*c**2*d**3)/(5*e**2) + (d + e*x)**(3/2)*(a**2*e**4 - 2*a*c*d** 2*e**2 + c**2*d**4)/(3*e**2))/e, Ne(e, 0)), (c**2*d**(5/2)*x**3/3, True))
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} c^{2} d^{2} - 42 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{105 \, e^{3}} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(3/2),x, algorithm=" maxima")
Output:
2/105*(15*(e*x + d)^(7/2)*c^2*d^2 - 42*(c^2*d^3 - a*c*d*e^2)*(e*x + d)^(5/ 2) + 35*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*(e*x + d)^(3/2))/e^3
Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (71) = 142\).
Time = 0.17 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.51 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} a^{2} d e^{2} + 70 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a c d^{2} + 35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} e^{2} + 14 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c d + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c^{2} d^{3}}{e^{2}} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c^{2} d^{2}}{e^{2}}\right )}}{105 \, e} \] Input:
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(3/2),x, algorithm=" giac")
Output:
2/105*(105*sqrt(e*x + d)*a^2*d*e^2 + 70*((e*x + d)^(3/2) - 3*sqrt(e*x + d) *d)*a*c*d^2 + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2*e^2 + 14*(3*(e* x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*c*d + 7*(3*( e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*c^2*d^3/e^2 + 3*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 3 5*sqrt(e*x + d)*d^3)*c^2*d^2/e^2)/e
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2\,{\left (d+e\,x\right )}^{3/2}\,\left (35\,a^2\,e^4+35\,c^2\,d^4+15\,c^2\,d^2\,{\left (d+e\,x\right )}^2-42\,c^2\,d^3\,\left (d+e\,x\right )-70\,a\,c\,d^2\,e^2+42\,a\,c\,d\,e^2\,\left (d+e\,x\right )\right )}{105\,e^3} \] Input:
int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2/(d + e*x)^(3/2),x)
Output:
(2*(d + e*x)^(3/2)*(35*a^2*e^4 + 35*c^2*d^4 + 15*c^2*d^2*(d + e*x)^2 - 42* c^2*d^3*(d + e*x) - 70*a*c*d^2*e^2 + 42*a*c*d*e^2*(d + e*x)))/(105*e^3)
Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \sqrt {e x +d}\, \left (15 c^{2} d^{2} e^{3} x^{3}+42 a c d \,e^{4} x^{2}+3 c^{2} d^{3} e^{2} x^{2}+35 a^{2} e^{5} x +14 a c \,d^{2} e^{3} x -4 c^{2} d^{4} e x +35 a^{2} d \,e^{4}-28 a c \,d^{3} e^{2}+8 c^{2} d^{5}\right )}{105 e^{3}} \] Input:
int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^(3/2),x)
Output:
(2*sqrt(d + e*x)*(35*a**2*d*e**4 + 35*a**2*e**5*x - 28*a*c*d**3*e**2 + 14* a*c*d**2*e**3*x + 42*a*c*d*e**4*x**2 + 8*c**2*d**5 - 4*c**2*d**4*e*x + 3*c **2*d**3*e**2*x**2 + 15*c**2*d**2*e**3*x**3))/(105*e**3)