Integrand size = 21, antiderivative size = 145 \[ \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 d^2 (c d-b e)^2 \sqrt {d+e x}}{e^5}-\frac {4 d (c d-b e) (2 c d-b e) (d+e x)^{3/2}}{3 e^5}+\frac {2 \left (6 c^2 d^2-6 b c d e+b^2 e^2\right ) (d+e x)^{5/2}}{5 e^5}-\frac {4 c (2 c d-b e) (d+e x)^{7/2}}{7 e^5}+\frac {2 c^2 (d+e x)^{9/2}}{9 e^5} \]
-4/3*d*(-b*e+c*d)*(-b*e+2*c*d)*(e*x+d)^(3/2)/e^5+2/5*(b^2*e^2-6*b*c*d*e+6* c^2*d^2)*(e*x+d)^(5/2)/e^5-4/7*c*(-b*e+2*c*d)*(e*x+d)^(7/2)/e^5+2/9*c^2*(e *x+d)^(9/2)/e^5+2*d^2*(-b*e+c*d)^2*(e*x+d)^(1/2)/e^5
Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.86 \[ \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (21 b^2 e^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )+18 b c e \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )+c^2 \left (128 d^4-64 d^3 e x+48 d^2 e^2 x^2-40 d e^3 x^3+35 e^4 x^4\right )\right )}{315 e^5} \]
(2*Sqrt[d + e*x]*(21*b^2*e^2*(8*d^2 - 4*d*e*x + 3*e^2*x^2) + 18*b*c*e*(-16 *d^3 + 8*d^2*e*x - 6*d*e^2*x^2 + 5*e^3*x^3) + 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 = 0.26 (sec) , antiderivative size = 145, 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.
\(\displaystyle \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \int \left (\frac {(d+e x)^{3/2} \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )}{e^4}+\frac {d^2 (c d-b e)^2}{e^4 \sqrt {d+e x}}-\frac {2 c (d+e x)^{5/2} (2 c d-b e)}{e^4}+\frac {2 d \sqrt {d+e x} (c d-b e) (b e-2 c d)}{e^4}+\frac {c^2 (d+e x)^{7/2}}{e^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (d+e x)^{5/2} \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )}{5 e^5}+\frac {2 d^2 \sqrt {d+e x} (c d-b e)^2}{e^5}-\frac {4 c (d+e x)^{7/2} (2 c d-b e)}{7 e^5}-\frac {4 d (d+e x)^{3/2} (c d-b e) (2 c d-b e)}{3 e^5}+\frac {2 c^2 (d+e x)^{9/2}}{9 e^5}\) |
(2*d^2*(c*d - b*e)^2*Sqrt[d + e*x])/e^5 - (4*d*(c*d - b*e)*(2*c*d - b*e)*( d + e*x)^(3/2))/(3*e^5) + (2*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2)*(d + e*x)^( 5/2))/(5*e^5) - (4*c*(2*c*d - b*e)*(d + e*x)^(7/2))/(7*e^5) + (2*c^2*(d + e*x)^(9/2))/(9*e^5)
3.4.49.3.1 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 = 2.00 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.74
method | result | size |
pseudoelliptic | \(\frac {16 \left (\frac {3 x^{2} \left (\frac {5}{9} c^{2} x^{2}+\frac {10}{7} b c x +b^{2}\right ) e^{4}}{8}-\frac {x \left (\frac {10}{21} c^{2} x^{2}+\frac {9}{7} b c x +b^{2}\right ) d \,e^{3}}{2}+d^{2} \left (\frac {2}{7} c^{2} x^{2}+\frac {6}{7} b c x +b^{2}\right ) e^{2}-\frac {12 c \,d^{3} \left (\frac {2 c x}{9}+b \right ) e}{7}+\frac {16 c^{2} d^{4}}{21}\right ) \sqrt {e x +d}}{15 e^{5}}\) | \(108\) |
gosper | \(\frac {2 \left (35 c^{2} x^{4} e^{4}+90 x^{3} b c \,e^{4}-40 x^{3} c^{2} d \,e^{3}+63 x^{2} b^{2} e^{4}-108 x^{2} b c d \,e^{3}+48 x^{2} c^{2} d^{2} e^{2}-84 x \,b^{2} d \,e^{3}+144 x b c \,d^{2} e^{2}-64 x \,c^{2} d^{3} e +168 b^{2} d^{2} e^{2}-288 d^{3} e b c +128 c^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{5}}\) | \(141\) |
trager | \(\frac {2 \left (35 c^{2} x^{4} e^{4}+90 x^{3} b c \,e^{4}-40 x^{3} c^{2} d \,e^{3}+63 x^{2} b^{2} e^{4}-108 x^{2} b c d \,e^{3}+48 x^{2} c^{2} d^{2} e^{2}-84 x \,b^{2} d \,e^{3}+144 x b c \,d^{2} e^{2}-64 x \,c^{2} d^{3} e +168 b^{2} d^{2} e^{2}-288 d^{3} e b c +128 c^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{5}}\) | \(141\) |
risch | \(\frac {2 \left (35 c^{2} x^{4} e^{4}+90 x^{3} b c \,e^{4}-40 x^{3} c^{2} d \,e^{3}+63 x^{2} b^{2} e^{4}-108 x^{2} b c d \,e^{3}+48 x^{2} c^{2} d^{2} e^{2}-84 x \,b^{2} d \,e^{3}+144 x b c \,d^{2} e^{2}-64 x \,c^{2} d^{3} e +168 b^{2} d^{2} e^{2}-288 d^{3} e b c +128 c^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{5}}\) | \(141\) |
derivativedivides | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (-2 c^{2} d +2 \left (b e -c d \right ) c \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (c^{2} d^{2}-4 d \left (b e -c d \right ) c +\left (b e -c d \right )^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (2 d^{2} \left (b e -c d \right ) c -2 d \left (b e -c d \right )^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 d^{2} \left (b e -c d \right )^{2} \sqrt {e x +d}}{e^{5}}\) | \(143\) |
default | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (-2 c^{2} d +2 \left (b e -c d \right ) c \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (c^{2} d^{2}-4 d \left (b e -c d \right ) c +\left (b e -c d \right )^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (2 d^{2} \left (b e -c d \right ) c -2 d \left (b e -c d \right )^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 d^{2} \left (b e -c d \right )^{2} \sqrt {e x +d}}{e^{5}}\) | \(143\) |
16/15*(3/8*x^2*(5/9*c^2*x^2+10/7*b*c*x+b^2)*e^4-1/2*x*(10/21*c^2*x^2+9/7*b *c*x+b^2)*d*e^3+d^2*(2/7*c^2*x^2+6/7*b*c*x+b^2)*e^2-12/7*c*d^3*(2/9*c*x+b) *e+16/21*c^2*d^4)*(e*x+d)^(1/2)/e^5
Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (35 \, c^{2} e^{4} x^{4} + 128 \, c^{2} d^{4} - 288 \, b c d^{3} e + 168 \, b^{2} d^{2} e^{2} - 10 \, {\left (4 \, c^{2} d e^{3} - 9 \, b c e^{4}\right )} x^{3} + 3 \, {\left (16 \, c^{2} d^{2} e^{2} - 36 \, b c d e^{3} + 21 \, b^{2} e^{4}\right )} x^{2} - 4 \, {\left (16 \, c^{2} d^{3} e - 36 \, b c d^{2} e^{2} + 21 \, b^{2} d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{5}} \]
2/315*(35*c^2*e^4*x^4 + 128*c^2*d^4 - 288*b*c*d^3*e + 168*b^2*d^2*e^2 - 10 *(4*c^2*d*e^3 - 9*b*c*e^4)*x^3 + 3*(16*c^2*d^2*e^2 - 36*b*c*d*e^3 + 21*b^2 *e^4)*x^2 - 4*(16*c^2*d^3*e - 36*b*c*d^2*e^2 + 21*b^2*d*e^3)*x)*sqrt(e*x + d)/e^5
Time = 0.66 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.39 \[ \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{2} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (2 b c e - 4 c^{2} d\right )}{7 e^{4}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right )}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 2 b^{2} d e^{2} + 6 b c d^{2} e - 4 c^{2} d^{3}\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (b^{2} d^{2} e^{2} - 2 b c d^{3} e + c^{2} d^{4}\right )}{e^{4}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\frac {b^{2} x^{3}}{3} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
Piecewise((2*(c**2*(d + e*x)**(9/2)/(9*e**4) + (d + e*x)**(7/2)*(2*b*c*e - 4*c**2*d)/(7*e**4) + (d + e*x)**(5/2)*(b**2*e**2 - 6*b*c*d*e + 6*c**2*d** 2)/(5*e**4) + (d + e*x)**(3/2)*(-2*b**2*d*e**2 + 6*b*c*d**2*e - 4*c**2*d** 3)/(3*e**4) + sqrt(d + e*x)*(b**2*d**2*e**2 - 2*b*c*d**3*e + c**2*d**4)/e* *4)/e, Ne(e, 0)), ((b**2*x**3/3 + b*c*x**4/2 + c**2*x**5/5)/sqrt(d), True) )
Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10 \[ \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (\frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2}}{e^{2}} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b c}{e^{3}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c^{2}}{e^{4}}\right )}}{315 \, e} \]
2/315*(21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2 )*b^2/e^2 + 18*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3 /2)*d^2 - 35*sqrt(e*x + d)*d^3)*b*c/e^3 + (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 = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10 \[ \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (\frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2}}{e^{2}} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b c}{e^{3}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c^{2}}{e^{4}}\right )}}{315 \, e} \]
2/315*(21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2 )*b^2/e^2 + 18*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3 /2)*d^2 - 35*sqrt(e*x + d)*d^3)*b*c/e^3 + (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 = 138, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x+c x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2\,c^2\,{\left (d+e\,x\right )}^{9/2}}{9\,e^5}-\frac {{\left (d+e\,x\right )}^{3/2}\,\left (4\,b^2\,d\,e^2-12\,b\,c\,d^2\,e+8\,c^2\,d^3\right )}{3\,e^5}+\frac {{\left (d+e\,x\right )}^{5/2}\,\left (2\,b^2\,e^2-12\,b\,c\,d\,e+12\,c^2\,d^2\right )}{5\,e^5}-\frac {\left (8\,c^2\,d-4\,b\,c\,e\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^5}+\frac {2\,d^2\,{\left (b\,e-c\,d\right )}^2\,\sqrt {d+e\,x}}{e^5} \]