Integrand size = 19, antiderivative size = 123 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (c d^2+a e^2\right )^2}{3 e^5 (d+e x)^{3/2}}+\frac {8 c d \left (c d^2+a e^2\right )}{e^5 \sqrt {d+e x}}+\frac {4 c \left (3 c d^2+a e^2\right ) \sqrt {d+e x}}{e^5}-\frac {8 c^2 d (d+e x)^{3/2}}{3 e^5}+\frac {2 c^2 (d+e x)^{5/2}}{5 e^5} \]
-2/3*(a*e^2+c*d^2)^2/e^5/(e*x+d)^(3/2)-8/3*c^2*d*(e*x+d)^(3/2)/e^5+2/5*c^2 *(e*x+d)^(5/2)/e^5+8*c*d*(a*e^2+c*d^2)/e^5/(e*x+d)^(1/2)+4*c*(a*e^2+3*c*d^ 2)*(e*x+d)^(1/2)/e^5
Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \left (-5 a^2 e^4+10 a c e^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )+c^2 \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )}{15 e^5 (d+e x)^{3/2}} \]
(2*(-5*a^2*e^4 + 10*a*c*e^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + c^2*(128*d^4 + 192*d^3*e*x + 48*d^2*e^2*x^2 - 8*d*e^3*x^3 + 3*e^4*x^4)))/(15*e^5*(d + e *x)^(3/2))
Time = 0.25 (sec) , antiderivative size = 123, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \int \left (\frac {2 c \left (a e^2+3 c d^2\right )}{e^4 \sqrt {d+e x}}-\frac {4 c d \left (a e^2+c d^2\right )}{e^4 (d+e x)^{3/2}}+\frac {\left (a e^2+c d^2\right )^2}{e^4 (d+e x)^{5/2}}+\frac {c^2 (d+e x)^{3/2}}{e^4}-\frac {4 c^2 d \sqrt {d+e x}}{e^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 c \sqrt {d+e x} \left (a e^2+3 c d^2\right )}{e^5}+\frac {8 c d \left (a e^2+c d^2\right )}{e^5 \sqrt {d+e x}}-\frac {2 \left (a e^2+c d^2\right )^2}{3 e^5 (d+e x)^{3/2}}+\frac {2 c^2 (d+e x)^{5/2}}{5 e^5}-\frac {8 c^2 d (d+e x)^{3/2}}{3 e^5}\) |
(-2*(c*d^2 + a*e^2)^2)/(3*e^5*(d + e*x)^(3/2)) + (8*c*d*(c*d^2 + a*e^2))/( e^5*Sqrt[d + e*x]) + (4*c*(3*c*d^2 + a*e^2)*Sqrt[d + e*x])/e^5 - (8*c^2*d* (d + e*x)^(3/2))/(3*e^5) + (2*c^2*(d + e*x)^(5/2))/(5*e^5)
3.7.3.3.1 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 = 1.97 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\frac {2 c \left (3 c \,x^{2} e^{2}-14 x c d e +30 e^{2} a +73 c \,d^{2}\right ) \sqrt {e x +d}}{15 e^{5}}-\frac {2 \left (-12 x c d e +e^{2} a -11 c \,d^{2}\right ) \left (e^{2} a +c \,d^{2}\right )}{3 e^{5} \left (e x +d \right )^{\frac {3}{2}}}\) | \(84\) |
pseudoelliptic | \(\frac {\frac {2 \left (3 e^{4} x^{4}-8 d \,e^{3} x^{3}+48 d^{2} e^{2} x^{2}+192 d^{3} e x +128 d^{4}\right ) c^{2}}{15}+\frac {32 \left (\frac {3}{8} x^{2} e^{2}+\frac {3}{2} d e x +d^{2}\right ) e^{2} a c}{3}-\frac {2 a^{2} e^{4}}{3}}{\left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(91\) |
gosper | \(-\frac {2 \left (-3 c^{2} x^{4} e^{4}+8 x^{3} c^{2} d \,e^{3}-30 x^{2} a c \,e^{4}-48 x^{2} c^{2} d^{2} e^{2}-120 x a c d \,e^{3}-192 x \,c^{2} d^{3} e +5 a^{2} e^{4}-80 a c \,d^{2} e^{2}-128 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(106\) |
trager | \(-\frac {2 \left (-3 c^{2} x^{4} e^{4}+8 x^{3} c^{2} d \,e^{3}-30 x^{2} a c \,e^{4}-48 x^{2} c^{2} d^{2} e^{2}-120 x a c d \,e^{3}-192 x \,c^{2} d^{3} e +5 a^{2} e^{4}-80 a c \,d^{2} e^{2}-128 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(106\) |
derivativedivides | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+4 a c \,e^{2} \sqrt {e x +d}+12 c^{2} d^{2} \sqrt {e x +d}+\frac {8 d c \left (e^{2} a +c \,d^{2}\right )}{\sqrt {e x +d}}-\frac {2 \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(117\) |
default | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+4 a c \,e^{2} \sqrt {e x +d}+12 c^{2} d^{2} \sqrt {e x +d}+\frac {8 d c \left (e^{2} a +c \,d^{2}\right )}{\sqrt {e x +d}}-\frac {2 \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(117\) |
2/15*c*(3*c*e^2*x^2-14*c*d*e*x+30*a*e^2+73*c*d^2)*(e*x+d)^(1/2)/e^5-2/3*(- 12*c*d*e*x+a*e^2-11*c*d^2)*(a*e^2+c*d^2)/e^5/(e*x+d)^(3/2)
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, c^{2} e^{4} x^{4} - 8 \, c^{2} d e^{3} x^{3} + 128 \, c^{2} d^{4} + 80 \, a c d^{2} e^{2} - 5 \, a^{2} e^{4} + 6 \, {\left (8 \, c^{2} d^{2} e^{2} + 5 \, a c e^{4}\right )} x^{2} + 24 \, {\left (8 \, c^{2} d^{3} e + 5 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]
2/15*(3*c^2*e^4*x^4 - 8*c^2*d*e^3*x^3 + 128*c^2*d^4 + 80*a*c*d^2*e^2 - 5*a ^2*e^4 + 6*(8*c^2*d^2*e^2 + 5*a*c*e^4)*x^2 + 24*(8*c^2*d^3*e + 5*a*c*d*e^3 )*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)
Time = 1.65 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {4 c^{2} d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{4}} + \frac {c^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {4 c d \left (a e^{2} + c d^{2}\right )}{e^{4} \sqrt {d + e x}} + \frac {\sqrt {d + e x} \left (2 a c e^{2} + 6 c^{2} d^{2}\right )}{e^{4}} - \frac {\left (a e^{2} + c d^{2}\right )^{2}}{3 e^{4} \left (d + e x\right )^{\frac {3}{2}}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {a^{2} x + \frac {2 a c x^{3}}{3} + \frac {c^{2} x^{5}}{5}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-4*c**2*d*(d + e*x)**(3/2)/(3*e**4) + c**2*(d + e*x)**(5/2)/ (5*e**4) + 4*c*d*(a*e**2 + c*d**2)/(e**4*sqrt(d + e*x)) + sqrt(d + e*x)*(2 *a*c*e**2 + 6*c**2*d**2)/e**4 - (a*e**2 + c*d**2)**2/(3*e**4*(d + e*x)**(3 /2)))/e, Ne(e, 0)), ((a**2*x + 2*a*c*x**3/3 + c**2*x**5/5)/d**(5/2), True) )
Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{2} - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d + 30 \, {\left (3 \, c^{2} d^{2} + a c e^{2}\right )} \sqrt {e x + d}}{e^{4}} - \frac {5 \, {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4} - 12 \, {\left (c^{2} d^{3} + a c d e^{2}\right )} {\left (e x + d\right )}\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{4}}\right )}}{15 \, e} \]
2/15*((3*(e*x + d)^(5/2)*c^2 - 20*(e*x + d)^(3/2)*c^2*d + 30*(3*c^2*d^2 + a*c*e^2)*sqrt(e*x + d))/e^4 - 5*(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4 - 12*(c ^2*d^3 + a*c*d*e^2)*(e*x + d))/((e*x + d)^(3/2)*e^4))/e
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (e x + d\right )} c^{2} d^{3} - c^{2} d^{4} + 12 \, {\left (e x + d\right )} a c d e^{2} - 2 \, a c d^{2} e^{2} - a^{2} e^{4}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{5}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{2} e^{20} - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d e^{20} + 90 \, \sqrt {e x + d} c^{2} d^{2} e^{20} + 30 \, \sqrt {e x + d} a c e^{22}\right )}}{15 \, e^{25}} \]
2/3*(12*(e*x + d)*c^2*d^3 - c^2*d^4 + 12*(e*x + d)*a*c*d*e^2 - 2*a*c*d^2*e ^2 - a^2*e^4)/((e*x + d)^(3/2)*e^5) + 2/15*(3*(e*x + d)^(5/2)*c^2*e^20 - 2 0*(e*x + d)^(3/2)*c^2*d*e^20 + 90*sqrt(e*x + d)*c^2*d^2*e^20 + 30*sqrt(e*x + d)*a*c*e^22)/e^25
Time = 9.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2\,c^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}-\frac {\frac {2\,a^2\,e^4}{3}+\frac {2\,c^2\,d^4}{3}-\left (8\,c^2\,d^3+8\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )+\frac {4\,a\,c\,d^2\,e^2}{3}}{e^5\,{\left (d+e\,x\right )}^{3/2}}+\frac {\left (12\,c^2\,d^2+4\,a\,c\,e^2\right )\,\sqrt {d+e\,x}}{e^5}-\frac {8\,c^2\,d\,{\left (d+e\,x\right )}^{3/2}}{3\,e^5} \]