Integrand size = 37, antiderivative size = 83 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx=-\frac {2 \left (c d^2-a e^2\right )^2}{7 e^3 (d+e x)^{7/2}}+\frac {4 c d \left (c d^2-a e^2\right )}{5 e^3 (d+e x)^{5/2}}-\frac {2 c^2 d^2}{3 e^3 (d+e x)^{3/2}} \]
-2/7*(-a*e^2+c*d^2)^2/e^3/(e*x+d)^(7/2)+4/5*c*d*(-a*e^2+c*d^2)/e^3/(e*x+d) ^(5/2)-2/3*c^2*d^2/e^3/(e*x+d)^(3/2)
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx=-\frac {2 \left (15 a^2 e^4+6 a c d e^2 (2 d+7 e x)+c^2 d^2 \left (8 d^2+28 d e x+35 e^2 x^2\right )\right )}{105 e^3 (d+e x)^{7/2}} \]
(-2*(15*a^2*e^4 + 6*a*c*d*e^2*(2*d + 7*e*x) + c^2*d^2*(8*d^2 + 28*d*e*x + 35*e^2*x^2)))/(105*e^3*(d + e*x)^(7/2))
Time = 0.22 (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.
\(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \left (-\frac {2 c d \left (c d^2-a e^2\right )}{e^2 (d+e x)^{7/2}}+\frac {\left (a e^2-c d^2\right )^2}{e^2 (d+e x)^{9/2}}+\frac {c^2 d^2}{e^2 (d+e x)^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 c d \left (c d^2-a e^2\right )}{5 e^3 (d+e x)^{5/2}}-\frac {2 \left (c d^2-a e^2\right )^2}{7 e^3 (d+e x)^{7/2}}-\frac {2 c^2 d^2}{3 e^3 (d+e x)^{3/2}}\) |
(-2*(c*d^2 - a*e^2)^2)/(7*e^3*(d + e*x)^(7/2)) + (4*c*d*(c*d^2 - a*e^2))/( 5*e^3*(d + e*x)^(5/2)) - (2*c^2*d^2)/(3*e^3*(d + e*x)^(3/2))
3.20.91.3.1 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 = 2.98 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(-\frac {2 \left (a^{2} e^{4}+\frac {14 x a c d \,e^{3}}{5}+\frac {4 c \,d^{2} \left (\frac {35 c \,x^{2}}{12}+a \right ) e^{2}}{5}+\frac {28 x \,c^{2} d^{3} e}{15}+\frac {8 c^{2} d^{4}}{15}\right )}{7 \left (e x +d \right )^{\frac {7}{2}} e^{3}}\) | \(65\) |
gosper | \(-\frac {2 \left (35 x^{2} c^{2} d^{2} e^{2}+42 x a c d \,e^{3}+28 x \,c^{2} d^{3} e +15 a^{2} e^{4}+12 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} e^{3}}\) | \(73\) |
trager | \(-\frac {2 \left (35 x^{2} c^{2} d^{2} e^{2}+42 x a c d \,e^{3}+28 x \,c^{2} d^{3} e +15 a^{2} e^{4}+12 a c \,d^{2} e^{2}+8 c^{2} d^{4}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} e^{3}}\) | \(73\) |
derivativedivides | \(\frac {-\frac {4 c d \left (e^{2} a -c \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{7 \left (e x +d \right )^{\frac {7}{2}}}}{e^{3}}\) | \(79\) |
default | \(\frac {-\frac {4 c d \left (e^{2} a -c \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{7 \left (e x +d \right )^{\frac {7}{2}}}}{e^{3}}\) | \(79\) |
-2/7*(a^2*e^4+14/5*x*a*c*d*e^3+4/5*c*d^2*(35/12*c*x^2+a)*e^2+28/15*x*c^2*d ^3*e+8/15*c^2*d^4)/(e*x+d)^(7/2)/e^3
Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx=-\frac {2 \, {\left (35 \, c^{2} d^{2} e^{2} x^{2} + 8 \, c^{2} d^{4} + 12 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} + 14 \, {\left (2 \, c^{2} d^{3} e + 3 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]
-2/105*(35*c^2*d^2*e^2*x^2 + 8*c^2*d^4 + 12*a*c*d^2*e^2 + 15*a^2*e^4 + 14* (2*c^2*d^3*e + 3*a*c*d*e^3)*x)*sqrt(e*x + d)/(e^7*x^4 + 4*d*e^6*x^3 + 6*d^ 2*e^5*x^2 + 4*d^3*e^4*x + d^4*e^3)
Leaf count of result is larger than twice the leaf count of optimal. 510 vs. \(2 (76) = 152\).
Time = 1.53 (sec) , antiderivative size = 510, normalized size of antiderivative = 6.14 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx=\begin {cases} - \frac {30 a^{2} e^{4}}{105 d^{3} e^{3} \sqrt {d + e x} + 315 d^{2} e^{4} x \sqrt {d + e x} + 315 d e^{5} x^{2} \sqrt {d + e x} + 105 e^{6} x^{3} \sqrt {d + e x}} - \frac {24 a c d^{2} e^{2}}{105 d^{3} e^{3} \sqrt {d + e x} + 315 d^{2} e^{4} x \sqrt {d + e x} + 315 d e^{5} x^{2} \sqrt {d + e x} + 105 e^{6} x^{3} \sqrt {d + e x}} - \frac {84 a c d e^{3} x}{105 d^{3} e^{3} \sqrt {d + e x} + 315 d^{2} e^{4} x \sqrt {d + e x} + 315 d e^{5} x^{2} \sqrt {d + e x} + 105 e^{6} x^{3} \sqrt {d + e x}} - \frac {16 c^{2} d^{4}}{105 d^{3} e^{3} \sqrt {d + e x} + 315 d^{2} e^{4} x \sqrt {d + e x} + 315 d e^{5} x^{2} \sqrt {d + e x} + 105 e^{6} x^{3} \sqrt {d + e x}} - \frac {56 c^{2} d^{3} e x}{105 d^{3} e^{3} \sqrt {d + e x} + 315 d^{2} e^{4} x \sqrt {d + e x} + 315 d e^{5} x^{2} \sqrt {d + e x} + 105 e^{6} x^{3} \sqrt {d + e x}} - \frac {70 c^{2} d^{2} e^{2} x^{2}}{105 d^{3} e^{3} \sqrt {d + e x} + 315 d^{2} e^{4} x \sqrt {d + e x} + 315 d e^{5} x^{2} \sqrt {d + e x} + 105 e^{6} x^{3} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {c^{2} x^{3}}{3 d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-30*a**2*e**4/(105*d**3*e**3*sqrt(d + e*x) + 315*d**2*e**4*x*sq rt(d + e*x) + 315*d*e**5*x**2*sqrt(d + e*x) + 105*e**6*x**3*sqrt(d + e*x)) - 24*a*c*d**2*e**2/(105*d**3*e**3*sqrt(d + e*x) + 315*d**2*e**4*x*sqrt(d + e*x) + 315*d*e**5*x**2*sqrt(d + e*x) + 105*e**6*x**3*sqrt(d + e*x)) - 84 *a*c*d*e**3*x/(105*d**3*e**3*sqrt(d + e*x) + 315*d**2*e**4*x*sqrt(d + e*x) + 315*d*e**5*x**2*sqrt(d + e*x) + 105*e**6*x**3*sqrt(d + e*x)) - 16*c**2* d**4/(105*d**3*e**3*sqrt(d + e*x) + 315*d**2*e**4*x*sqrt(d + e*x) + 315*d* e**5*x**2*sqrt(d + e*x) + 105*e**6*x**3*sqrt(d + e*x)) - 56*c**2*d**3*e*x/ (105*d**3*e**3*sqrt(d + e*x) + 315*d**2*e**4*x*sqrt(d + e*x) + 315*d*e**5* x**2*sqrt(d + e*x) + 105*e**6*x**3*sqrt(d + e*x)) - 70*c**2*d**2*e**2*x**2 /(105*d**3*e**3*sqrt(d + e*x) + 315*d**2*e**4*x*sqrt(d + e*x) + 315*d*e**5 *x**2*sqrt(d + e*x) + 105*e**6*x**3*sqrt(d + e*x)), Ne(e, 0)), (c**2*x**3/ (3*d**(5/2)), True))
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx=-\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{2} c^{2} d^{2} + 15 \, c^{2} d^{4} - 30 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - 42 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} {\left (e x + d\right )}\right )}}{105 \, {\left (e x + d\right )}^{\frac {7}{2}} e^{3}} \]
-2/105*(35*(e*x + d)^2*c^2*d^2 + 15*c^2*d^4 - 30*a*c*d^2*e^2 + 15*a^2*e^4 - 42*(c^2*d^3 - a*c*d*e^2)*(e*x + d))/((e*x + d)^(7/2)*e^3)
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx=-\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{2} c^{2} d^{2} - 42 \, {\left (e x + d\right )} c^{2} d^{3} + 15 \, c^{2} d^{4} + 42 \, {\left (e x + d\right )} a c d e^{2} - 30 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4}\right )}}{105 \, {\left (e x + d\right )}^{\frac {7}{2}} e^{3}} \]
-2/105*(35*(e*x + d)^2*c^2*d^2 - 42*(e*x + d)*c^2*d^3 + 15*c^2*d^4 + 42*(e *x + d)*a*c*d*e^2 - 30*a*c*d^2*e^2 + 15*a^2*e^4)/((e*x + d)^(7/2)*e^3)
Time = 9.66 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^{13/2}} \, dx=-\frac {\frac {2\,a^2\,e^4}{7}+\frac {2\,c^2\,d^4}{7}-\left (\frac {4\,c^2\,d^3}{5}-\frac {4\,a\,c\,d\,e^2}{5}\right )\,\left (d+e\,x\right )+\frac {2\,c^2\,d^2\,{\left (d+e\,x\right )}^2}{3}-\frac {4\,a\,c\,d^2\,e^2}{7}}{e^3\,{\left (d+e\,x\right )}^{7/2}} \]