Integrand size = 19, antiderivative size = 125 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \left (c d^2+a e^2\right )^2 \sqrt {d+e x}}{e^5}-\frac {8 c d \left (c d^2+a e^2\right ) (d+e x)^{3/2}}{3 e^5}+\frac {4 c \left (3 c d^2+a e^2\right ) (d+e x)^{5/2}}{5 e^5}-\frac {8 c^2 d (d+e x)^{7/2}}{7 e^5}+\frac {2 c^2 (d+e x)^{9/2}}{9 e^5} \]
-8/3*c*d*(a*e^2+c*d^2)*(e*x+d)^(3/2)/e^5+4/5*c*(a*e^2+3*c*d^2)*(e*x+d)^(5/ 2)/e^5-8/7*c^2*d*(e*x+d)^(7/2)/e^5+2/9*c^2*(e*x+d)^(9/2)/e^5+2*(a*e^2+c*d^ 2)^2*(e*x+d)^(1/2)/e^5
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (315 a^2 e^4+42 a c e^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )+c^2 \left (128 d^4-64 d^3 e x+48 d^2 e^2 x^2-40 d e^3 x^3+35 e^4 x^4\right )\right )}{315 e^5} \]
(2*Sqrt[d + e*x]*(315*a^2*e^4 + 42*a*c*e^2*(8*d^2 - 4*d*e*x + 3*e^2*x^2) + c^2*(128*d^4 - 64*d^3*e*x + 48*d^2*e^2*x^2 - 40*d*e^3*x^3 + 35*e^4*x^4))) /(315*e^5)
Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {476, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \int \left (\frac {2 c (d+e x)^{3/2} \left (a e^2+3 c d^2\right )}{e^4}-\frac {4 c d \sqrt {d+e x} \left (a e^2+c d^2\right )}{e^4}+\frac {\left (a e^2+c d^2\right )^2}{e^4 \sqrt {d+e x}}+\frac {c^2 (d+e x)^{7/2}}{e^4}-\frac {4 c^2 d (d+e x)^{5/2}}{e^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 c (d+e x)^{5/2} \left (a e^2+3 c d^2\right )}{5 e^5}-\frac {8 c d (d+e x)^{3/2} \left (a e^2+c d^2\right )}{3 e^5}+\frac {2 \sqrt {d+e x} \left (a e^2+c d^2\right )^2}{e^5}+\frac {2 c^2 (d+e x)^{9/2}}{9 e^5}-\frac {8 c^2 d (d+e x)^{7/2}}{7 e^5}\) |
(2*(c*d^2 + a*e^2)^2*Sqrt[d + e*x])/e^5 - (8*c*d*(c*d^2 + a*e^2)*(d + e*x) ^(3/2))/(3*e^5) + (4*c*(3*c*d^2 + a*e^2)*(d + e*x)^(5/2))/(5*e^5) - (8*c^2 *d*(d + e*x)^(7/2))/(7*e^5) + (2*c^2*(d + e*x)^(9/2))/(9*e^5)
3.7.1.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Time = 2.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {e x +d}\, \left (\left (\frac {1}{9} x^{4} c^{2}+\frac {2}{5} a c \,x^{2}+a^{2}\right ) e^{4}-\frac {8 x c \left (\frac {5 c \,x^{2}}{21}+a \right ) d \,e^{3}}{15}+\frac {16 c \left (\frac {c \,x^{2}}{7}+a \right ) d^{2} e^{2}}{15}-\frac {64 x \,c^{2} d^{3} e}{315}+\frac {128 c^{2} d^{4}}{315}\right )}{e^{5}}\) | \(88\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (35 c^{2} x^{4} e^{4}-40 x^{3} c^{2} d \,e^{3}+126 x^{2} a c \,e^{4}+48 x^{2} c^{2} d^{2} e^{2}-168 x a c d \,e^{3}-64 x \,c^{2} d^{3} e +315 a^{2} e^{4}+336 a c \,d^{2} e^{2}+128 c^{2} d^{4}\right )}{315 e^{5}}\) | \(106\) |
trager | \(\frac {2 \sqrt {e x +d}\, \left (35 c^{2} x^{4} e^{4}-40 x^{3} c^{2} d \,e^{3}+126 x^{2} a c \,e^{4}+48 x^{2} c^{2} d^{2} e^{2}-168 x a c d \,e^{3}-64 x \,c^{2} d^{3} e +315 a^{2} e^{4}+336 a c \,d^{2} e^{2}+128 c^{2} d^{4}\right )}{315 e^{5}}\) | \(106\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (35 c^{2} x^{4} e^{4}-40 x^{3} c^{2} d \,e^{3}+126 x^{2} a c \,e^{4}+48 x^{2} c^{2} d^{2} e^{2}-168 x a c d \,e^{3}-64 x \,c^{2} d^{3} e +315 a^{2} e^{4}+336 a c \,d^{2} e^{2}+128 c^{2} d^{4}\right )}{315 e^{5}}\) | \(106\) |
derivativedivides | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {8 \left (e^{2} a +c \,d^{2}\right ) c d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (e^{2} a +c \,d^{2}\right )^{2} \sqrt {e x +d}}{e^{5}}\) | \(107\) |
default | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {8 \left (e^{2} a +c \,d^{2}\right ) c d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (e^{2} a +c \,d^{2}\right )^{2} \sqrt {e x +d}}{e^{5}}\) | \(107\) |
2*(e*x+d)^(1/2)*((1/9*x^4*c^2+2/5*a*c*x^2+a^2)*e^4-8/15*x*c*(5/21*c*x^2+a) *d*e^3+16/15*c*(1/7*c*x^2+a)*d^2*e^2-64/315*x*c^2*d^3*e+128/315*c^2*d^4)/e ^5
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (35 \, c^{2} e^{4} x^{4} - 40 \, c^{2} d e^{3} x^{3} + 128 \, c^{2} d^{4} + 336 \, a c d^{2} e^{2} + 315 \, a^{2} e^{4} + 6 \, {\left (8 \, c^{2} d^{2} e^{2} + 21 \, a c e^{4}\right )} x^{2} - 8 \, {\left (8 \, c^{2} d^{3} e + 21 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{5}} \]
2/315*(35*c^2*e^4*x^4 - 40*c^2*d*e^3*x^3 + 128*c^2*d^4 + 336*a*c*d^2*e^2 + 315*a^2*e^4 + 6*(8*c^2*d^2*e^2 + 21*a*c*e^4)*x^2 - 8*(8*c^2*d^3*e + 21*a* c*d*e^3)*x)*sqrt(e*x + d)/e^5
Time = 0.58 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (- \frac {4 c^{2} d \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} + \frac {c^{2} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (2 a c e^{2} + 6 c^{2} d^{2}\right )}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 4 a c d e^{2} - 4 c^{2} d^{3}\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (a^{2} e^{4} + 2 a c d^{2} e^{2} + c^{2} d^{4}\right )}{e^{4}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {a^{2} x + \frac {2 a c x^{3}}{3} + \frac {c^{2} x^{5}}{5}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-4*c**2*d*(d + e*x)**(7/2)/(7*e**4) + c**2*(d + e*x)**(9/2)/ (9*e**4) + (d + e*x)**(5/2)*(2*a*c*e**2 + 6*c**2*d**2)/(5*e**4) + (d + e*x )**(3/2)*(-4*a*c*d*e**2 - 4*c**2*d**3)/(3*e**4) + sqrt(d + e*x)*(a**2*e**4 + 2*a*c*d**2*e**2 + c**2*d**4)/e**4)/e, Ne(e, 0)), ((a**2*x + 2*a*c*x**3/ 3 + c**2*x**5/5)/sqrt(d), True))
Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{2} + \frac {42 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c}{e^{2}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c^{2}}{e^{4}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*a^2 + 42*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)* d + 15*sqrt(e*x + d)*d^2)*a*c/e^2 + (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^2/e^4)/e
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{2} + \frac {42 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a c}{e^{2}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c^{2}}{e^{4}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*a^2 + 42*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)* d + 15*sqrt(e*x + d)*d^2)*a*c/e^2 + (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^2/e^4)/e
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2\,c^2\,{\left (d+e\,x\right )}^{9/2}}{9\,e^5}-\frac {\left (8\,c^2\,d^3+8\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^5}+\frac {2\,{\left (c\,d^2+a\,e^2\right )}^2\,\sqrt {d+e\,x}}{e^5}+\frac {\left (12\,c^2\,d^2+4\,a\,c\,e^2\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}-\frac {8\,c^2\,d\,{\left (d+e\,x\right )}^{7/2}}{7\,e^5} \]