Integrand size = 21, antiderivative size = 143 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {2 d^2 (c d-b e)^2}{5 e^5 (d+e x)^{5/2}}+\frac {4 d (c d-b e) (2 c d-b e)}{3 e^5 (d+e x)^{3/2}}-\frac {2 \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )}{e^5 \sqrt {d+e x}}-\frac {4 c (2 c d-b e) \sqrt {d+e x}}{e^5}+\frac {2 c^2 (d+e x)^{3/2}}{3 e^5} \]
-2/5*d^2*(-b*e+c*d)^2/e^5/(e*x+d)^(5/2)+4/3*d*(-b*e+c*d)*(-b*e+2*c*d)/e^5/ (e*x+d)^(3/2)+2/3*c^2*(e*x+d)^(3/2)/e^5-2*(b^2*e^2-6*b*c*d*e+6*c^2*d^2)/e^ 5/(e*x+d)^(1/2)-4*c*(-b*e+2*c*d)*(e*x+d)^(1/2)/e^5
Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.86 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (b^2 e^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )-6 b c e \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )+c^2 \left (128 d^4+320 d^3 e x+240 d^2 e^2 x^2+40 d e^3 x^3-5 e^4 x^4\right )\right )}{15 e^5 (d+e x)^{5/2}} \]
(-2*(b^2*e^2*(8*d^2 + 20*d*e*x + 15*e^2*x^2) - 6*b*c*e*(16*d^3 + 40*d^2*e* x + 30*d*e^2*x^2 + 5*e^3*x^3) + c^2*(128*d^4 + 320*d^3*e*x + 240*d^2*e^2*x ^2 + 40*d*e^3*x^3 - 5*e^4*x^4)))/(15*e^5*(d + e*x)^(5/2))
Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1140, 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.
| \(\displaystyle \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle \int \left (\frac {b^2 e^2-6 b c d e+6 c^2 d^2}{e^4 (d+e x)^{3/2}}+\frac {d^2 (c d-b e)^2}{e^4 (d+e x)^{7/2}}-\frac {2 c (2 c d-b e)}{e^4 \sqrt {d+e x}}+\frac {2 d (c d-b e) (b e-2 c d)}{e^4 (d+e x)^{5/2}}+\frac {c^2 \sqrt {d+e x}}{e^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )}{e^5 \sqrt {d+e x}}-\frac {2 d^2 (c d-b e)^2}{5 e^5 (d+e x)^{5/2}}-\frac {4 c \sqrt {d+e x} (2 c d-b e)}{e^5}+\frac {4 d (c d-b e) (2 c d-b e)}{3 e^5 (d+e x)^{3/2}}+\frac {2 c^2 (d+e x)^{3/2}}{3 e^5}\) |
(-2*d^2*(c*d - b*e)^2)/(5*e^5*(d + e*x)^(5/2)) + (4*d*(c*d - b*e)*(2*c*d - b*e))/(3*e^5*(d + e*x)^(3/2)) - (2*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2))/(e^ 5*Sqrt[d + e*x]) - (4*c*(2*c*d - b*e)*Sqrt[d + e*x])/e^5 + (2*c^2*(d + e*x )^(3/2))/(3*e^5)
3.4.52.3.1 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*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Time = 1.97 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.76
| method | result | size |
| pseudoelliptic | \(-\frac {16 \left (\frac {15 x^{2} \left (-\frac {1}{3} c^{2} x^{2}-2 b c x +b^{2}\right ) e^{4}}{8}+\frac {5 d x \left (2 c^{2} x^{2}-9 b c x +b^{2}\right ) e^{3}}{2}+d^{2} \left (30 c^{2} x^{2}-30 b c x +b^{2}\right ) e^{2}-12 \left (-\frac {10 c x}{3}+b \right ) c \,d^{3} e +16 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{5}}\) | \(108\) |
| derivativedivides | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+4 b e c \sqrt {e x +d}-8 c^{2} d \sqrt {e x +d}-\frac {2 d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}\right )}{\sqrt {e x +d}}+\frac {4 d \left (b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(140\) |
| default | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+4 b e c \sqrt {e x +d}-8 c^{2} d \sqrt {e x +d}-\frac {2 d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}\right )}{\sqrt {e x +d}}+\frac {4 d \left (b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(140\) |
| gosper | \(-\frac {2 \left (-5 c^{2} x^{4} e^{4}-30 x^{3} b c \,e^{4}+40 x^{3} c^{2} d \,e^{3}+15 x^{2} b^{2} e^{4}-180 x^{2} b c d \,e^{3}+240 x^{2} c^{2} d^{2} e^{2}+20 x \,b^{2} d \,e^{3}-240 x b c \,d^{2} e^{2}+320 x \,c^{2} d^{3} e +8 b^{2} d^{2} e^{2}-96 d^{3} e b c +128 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{5}}\) | \(141\) |
| trager | \(-\frac {2 \left (-5 c^{2} x^{4} e^{4}-30 x^{3} b c \,e^{4}+40 x^{3} c^{2} d \,e^{3}+15 x^{2} b^{2} e^{4}-180 x^{2} b c d \,e^{3}+240 x^{2} c^{2} d^{2} e^{2}+20 x \,b^{2} d \,e^{3}-240 x b c \,d^{2} e^{2}+320 x \,c^{2} d^{3} e +8 b^{2} d^{2} e^{2}-96 d^{3} e b c +128 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{5}}\) | \(141\) |
| risch | \(\frac {2 c \left (c e x +6 b e -11 c d \right ) \sqrt {e x +d}}{3 e^{5}}-\frac {2 \left (15 x^{2} b^{2} e^{4}-90 x^{2} b c d \,e^{3}+90 x^{2} c^{2} d^{2} e^{2}+20 x \,b^{2} d \,e^{3}-150 x b c \,d^{2} e^{2}+160 x \,c^{2} d^{3} e +8 b^{2} d^{2} e^{2}-66 d^{3} e b c +73 c^{2} d^{4}\right )}{15 e^{5} \sqrt {e x +d}\, \left (x^{2} e^{2}+2 d e x +d^{2}\right )}\) | \(153\) |
-16/15/(e*x+d)^(5/2)*(15/8*x^2*(-1/3*c^2*x^2-2*b*c*x+b^2)*e^4+5/2*d*x*(2*c ^2*x^2-9*b*c*x+b^2)*e^3+d^2*(30*c^2*x^2-30*b*c*x+b^2)*e^2-12*(-10/3*c*x+b) *c*d^3*e+16*c^2*d^4)/e^5
Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.17 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, c^{2} e^{4} x^{4} - 128 \, c^{2} d^{4} + 96 \, b c d^{3} e - 8 \, b^{2} d^{2} e^{2} - 10 \, {\left (4 \, c^{2} d e^{3} - 3 \, b c e^{4}\right )} x^{3} - 15 \, {\left (16 \, c^{2} d^{2} e^{2} - 12 \, b c d e^{3} + b^{2} e^{4}\right )} x^{2} - 20 \, {\left (16 \, c^{2} d^{3} e - 12 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \]
2/15*(5*c^2*e^4*x^4 - 128*c^2*d^4 + 96*b*c*d^3*e - 8*b^2*d^2*e^2 - 10*(4*c ^2*d*e^3 - 3*b*c*e^4)*x^3 - 15*(16*c^2*d^2*e^2 - 12*b*c*d*e^3 + b^2*e^4)*x ^2 - 20*(16*c^2*d^3*e - 12*b*c*d^2*e^2 + b^2*d*e^3)*x)*sqrt(e*x + d)/(e^8* x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)
Leaf count of result is larger than twice the leaf count of optimal. 787 vs. \(2 (138) = 276\).
Time = 0.52 (sec) , antiderivative size = 787, normalized size of antiderivative = 5.50 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx=\begin {cases} - \frac {16 b^{2} d^{2} e^{2}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} - \frac {40 b^{2} d e^{3} x}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} - \frac {30 b^{2} e^{4} x^{2}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} + \frac {192 b c d^{3} e}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} + \frac {480 b c d^{2} e^{2} x}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} + \frac {360 b c d e^{3} x^{2}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} + \frac {60 b c e^{4} x^{3}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} - \frac {256 c^{2} d^{4}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} - \frac {640 c^{2} d^{3} e x}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} - \frac {480 c^{2} d^{2} e^{2} x^{2}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} - \frac {80 c^{2} d e^{3} x^{3}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} + \frac {10 c^{2} e^{4} x^{4}}{15 d^{2} e^{5} \sqrt {d + e x} + 30 d e^{6} x \sqrt {d + e x} + 15 e^{7} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {\frac {b^{2} x^{3}}{3} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-16*b**2*d**2*e**2/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sq rt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 40*b**2*d*e**3*x/(15*d**2*e**5 *sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 30*b**2*e**4*x**2/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 192*b*c*d**3*e/(15*d**2*e**5*sqrt(d + e*x ) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 480*b*c*d**2 *e**2*x/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7* x**2*sqrt(d + e*x)) + 360*b*c*d*e**3*x**2/(15*d**2*e**5*sqrt(d + e*x) + 30 *d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 60*b*c*e**4*x**3/( 15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt (d + e*x)) - 256*c**2*d**4/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt( d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 640*c**2*d**3*e*x/(15*d**2*e**5*s qrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 4 80*c**2*d**2*e**2*x**2/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 80*c**2*d*e**3*x**3/(15*d**2*e**5*sqr t(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 10* c**2*e**4*x**4/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 1 5*e**7*x**2*sqrt(d + e*x)), Ne(e, 0)), ((b**2*x**3/3 + b*c*x**4/2 + c**2*x **5/5)/d**(7/2), True))
Time = 0.22 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.03 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} c^{2} - 6 \, {\left (2 \, c^{2} d - b c e\right )} \sqrt {e x + d}\right )}}{e^{4}} - \frac {3 \, c^{2} d^{4} - 6 \, b c d^{3} e + 3 \, b^{2} d^{2} e^{2} + 15 \, {\left (6 \, c^{2} d^{2} - 6 \, b c d e + b^{2} e^{2}\right )} {\left (e x + d\right )}^{2} - 10 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2}\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {5}{2}} e^{4}}\right )}}{15 \, e} \]
2/15*(5*((e*x + d)^(3/2)*c^2 - 6*(2*c^2*d - b*c*e)*sqrt(e*x + d))/e^4 - (3 *c^2*d^4 - 6*b*c*d^3*e + 3*b^2*d^2*e^2 + 15*(6*c^2*d^2 - 6*b*c*d*e + b^2*e ^2)*(e*x + d)^2 - 10*(2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2)*(e*x + d))/((e* x + d)^(5/2)*e^4))/e
Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.22 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (90 \, {\left (e x + d\right )}^{2} c^{2} d^{2} - 20 \, {\left (e x + d\right )} c^{2} d^{3} + 3 \, c^{2} d^{4} - 90 \, {\left (e x + d\right )}^{2} b c d e + 30 \, {\left (e x + d\right )} b c d^{2} e - 6 \, b c d^{3} e + 15 \, {\left (e x + d\right )}^{2} b^{2} e^{2} - 10 \, {\left (e x + d\right )} b^{2} d e^{2} + 3 \, b^{2} d^{2} e^{2}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{5}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} c^{2} e^{10} - 12 \, \sqrt {e x + d} c^{2} d e^{10} + 6 \, \sqrt {e x + d} b c e^{11}\right )}}{3 \, e^{15}} \]
-2/15*(90*(e*x + d)^2*c^2*d^2 - 20*(e*x + d)*c^2*d^3 + 3*c^2*d^4 - 90*(e*x + d)^2*b*c*d*e + 30*(e*x + d)*b*c*d^2*e - 6*b*c*d^3*e + 15*(e*x + d)^2*b^ 2*e^2 - 10*(e*x + d)*b^2*d*e^2 + 3*b^2*d^2*e^2)/((e*x + d)^(5/2)*e^5) + 2/ 3*((e*x + d)^(3/2)*c^2*e^10 - 12*sqrt(e*x + d)*c^2*d*e^10 + 6*sqrt(e*x + d )*b*c*e^11)/e^15
Time = 9.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.98 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{7/2}} \, dx=-\frac {2\,\left (8\,b^2\,d^2\,e^2+20\,b^2\,d\,e^3\,x+15\,b^2\,e^4\,x^2-96\,b\,c\,d^3\,e-240\,b\,c\,d^2\,e^2\,x-180\,b\,c\,d\,e^3\,x^2-30\,b\,c\,e^4\,x^3+128\,c^2\,d^4+320\,c^2\,d^3\,e\,x+240\,c^2\,d^2\,e^2\,x^2+40\,c^2\,d\,e^3\,x^3-5\,c^2\,e^4\,x^4\right )}{15\,e^5\,{\left (d+e\,x\right )}^{5/2}} \]