Integrand size = 21, antiderivative size = 143 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=-\frac {2 d^2 (c d-b e)^2}{3 e^5 (d+e x)^{3/2}}+\frac {4 d (c d-b e) (2 c d-b e)}{e^5 \sqrt {d+e x}}+\frac {2 \left (6 c^2 d^2-6 b c d e+b^2 e^2\right ) \sqrt {d+e x}}{e^5}-\frac {4 c (2 c d-b e) (d+e x)^{3/2}}{3 e^5}+\frac {2 c^2 (d+e x)^{5/2}}{5 e^5} \] Output:
-2/3*d^2*(-b*e+c*d)^2/e^5/(e*x+d)^(3/2)+4*d*(-b*e+c*d)*(-b*e+2*c*d)/e^5/(e *x+d)^(1/2)+2*(b^2*e^2-6*b*c*d*e+6*c^2*d^2)*(e*x+d)^(1/2)/e^5-4/3*c*(-b*e+ 2*c*d)*(e*x+d)^(3/2)/e^5+2/5*c^2*(e*x+d)^(5/2)/e^5
Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.86 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \left (5 b^2 e^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )+10 b c e \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )+c^2 \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )}{15 e^5 (d+e x)^{3/2}} \] Input:
Integrate[(b*x + c*x^2)^2/(d + e*x)^(5/2),x]
Output:
(2*(5*b^2*e^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + 10*b*c*e*(-16*d^3 - 24*d^2* e*x - 6*d*e^2*x^2 + e^3*x^3) + 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 = 0.45 (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.
| \(\displaystyle \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle \int \left (\frac {b^2 e^2-6 b c d e+6 c^2 d^2}{e^4 \sqrt {d+e x}}+\frac {d^2 (c d-b e)^2}{e^4 (d+e x)^{5/2}}-\frac {2 c \sqrt {d+e x} (2 c d-b e)}{e^4}+\frac {2 d (c d-b e) (b e-2 c d)}{e^4 (d+e x)^{3/2}}+\frac {c^2 (d+e x)^{3/2}}{e^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt {d+e x} \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )}{e^5}-\frac {2 d^2 (c d-b e)^2}{3 e^5 (d+e x)^{3/2}}-\frac {4 c (d+e x)^{3/2} (2 c d-b e)}{3 e^5}+\frac {4 d (c d-b e) (2 c d-b e)}{e^5 \sqrt {d+e x}}+\frac {2 c^2 (d+e x)^{5/2}}{5 e^5}\) |
Input:
Int[(b*x + c*x^2)^2/(d + e*x)^(5/2),x]
Output:
(-2*d^2*(c*d - b*e)^2)/(3*e^5*(d + e*x)^(3/2)) + (4*d*(c*d - b*e)*(2*c*d - b*e))/(e^5*Sqrt[d + e*x]) + (2*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2)*Sqrt[d + e*x])/e^5 - (4*c*(2*c*d - b*e)*(d + e*x)^(3/2))/(3*e^5) + (2*c^2*(d + e*x )^(5/2))/(5*e^5)
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 = 0.46 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.77
| method | result | size |
| risch | \(\frac {2 \left (3 c^{2} e^{2} x^{2}+10 e^{2} x b c -14 c^{2} d e x +15 b^{2} e^{2}-80 b c d e +73 c^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{5}}+\frac {2 \left (6 x b \,e^{2}-12 c d x e +5 b d e -11 c \,d^{2}\right ) d \left (b e -c d \right )}{3 e^{5} \left (e x +d \right )^{\frac {3}{2}}}\) | \(110\) |
| pseudoelliptic | \(\frac {\left (6 c^{2} x^{4}+20 b c \,x^{3}+30 b^{2} x^{2}\right ) e^{4}+120 x \left (-\frac {2}{15} c^{2} x^{2}-c b x +b^{2}\right ) d \,e^{3}+80 d^{2} \left (\frac {6}{5} c^{2} x^{2}-6 c b x +b^{2}\right ) e^{2}-320 \left (-\frac {6 c x}{5}+b \right ) c \,d^{3} e +256 c^{2} d^{4}}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(112\) |
| gosper | \(\frac {\frac {2}{5} c^{2} x^{4} e^{4}+\frac {4}{3} x^{3} b c \,e^{4}-\frac {16}{15} d \,c^{2} x^{3} e^{3}+2 x^{2} b^{2} e^{4}-8 x^{2} b c d \,e^{3}+\frac {32}{5} x^{2} c^{2} d^{2} e^{2}+8 x \,b^{2} d \,e^{3}-32 x b c \,d^{2} e^{2}+\frac {128}{5} x \,c^{2} d^{3} e +\frac {16}{3} d^{2} e^{2} b^{2}-\frac {64}{3} d^{3} e b c +\frac {256}{15} c^{2} d^{4}}{\left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(141\) |
| trager | \(\frac {\frac {2}{5} c^{2} x^{4} e^{4}+\frac {4}{3} x^{3} b c \,e^{4}-\frac {16}{15} d \,c^{2} x^{3} e^{3}+2 x^{2} b^{2} e^{4}-8 x^{2} b c d \,e^{3}+\frac {32}{5} x^{2} c^{2} d^{2} e^{2}+8 x \,b^{2} d \,e^{3}-32 x b c \,d^{2} e^{2}+\frac {128}{5} x \,c^{2} d^{3} e +\frac {16}{3} d^{2} e^{2} b^{2}-\frac {64}{3} d^{3} e b c +\frac {256}{15} c^{2} d^{4}}{\left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(141\) |
| derivativedivides | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 b c e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 b^{2} e^{2} \sqrt {e x +d}-12 b c d e \sqrt {e x +d}+12 c^{2} d^{2} \sqrt {e x +d}+\frac {4 d \left (b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right )}{\sqrt {e x +d}}-\frac {2 d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(151\) |
| default | \(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 b c e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 b^{2} e^{2} \sqrt {e x +d}-12 b c d e \sqrt {e x +d}+12 c^{2} d^{2} \sqrt {e x +d}+\frac {4 d \left (b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right )}{\sqrt {e x +d}}-\frac {2 d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(151\) |
| orering | \(\frac {2 \left (3 c^{2} x^{4} e^{4}+10 x^{3} b c \,e^{4}-8 d \,c^{2} x^{3} e^{3}+15 x^{2} b^{2} e^{4}-60 x^{2} b c d \,e^{3}+48 x^{2} c^{2} d^{2} e^{2}+60 x \,b^{2} d \,e^{3}-240 x b c \,d^{2} e^{2}+192 x \,c^{2} d^{3} e +40 d^{2} e^{2} b^{2}-160 d^{3} e b c +128 c^{2} d^{4}\right ) \left (c \,x^{2}+b x \right )^{2}}{15 e^{5} \left (c x +b \right )^{2} \left (e x +d \right )^{\frac {3}{2}} x^{2}}\) | \(162\) |
Input:
int((c*x^2+b*x)^2/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/15*(3*c^2*e^2*x^2+10*b*c*e^2*x-14*c^2*d*e*x+15*b^2*e^2-80*b*c*d*e+73*c^2 *d^2)*(e*x+d)^(1/2)/e^5+2/3*(6*b*e^2*x-12*c*d*e*x+5*b*d*e-11*c*d^2)*d*(b*e -c*d)/e^5/(e*x+d)^(3/2)
Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.11 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, c^{2} e^{4} x^{4} + 128 \, c^{2} d^{4} - 160 \, b c d^{3} e + 40 \, b^{2} d^{2} e^{2} - 2 \, {\left (4 \, c^{2} d e^{3} - 5 \, b c e^{4}\right )} x^{3} + 3 \, {\left (16 \, c^{2} d^{2} e^{2} - 20 \, b c d e^{3} + 5 \, b^{2} e^{4}\right )} x^{2} + 12 \, {\left (16 \, c^{2} d^{3} e - 20 \, b c d^{2} e^{2} + 5 \, b^{2} d e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \] Input:
integrate((c*x^2+b*x)^2/(e*x+d)^(5/2),x, algorithm="fricas")
Output:
2/15*(3*c^2*e^4*x^4 + 128*c^2*d^4 - 160*b*c*d^3*e + 40*b^2*d^2*e^2 - 2*(4* c^2*d*e^3 - 5*b*c*e^4)*x^3 + 3*(16*c^2*d^2*e^2 - 20*b*c*d*e^3 + 5*b^2*e^4) *x^2 + 12*(16*c^2*d^3*e - 20*b*c*d^2*e^2 + 5*b^2*d*e^3)*x)*sqrt(e*x + d)/( e^7*x^2 + 2*d*e^6*x + d^2*e^5)
Time = 2.06 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.21 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} - \frac {d^{2} \left (b e - c d\right )^{2}}{3 e^{4} \left (d + e x\right )^{\frac {3}{2}}} + \frac {2 d \left (b e - 2 c d\right ) \left (b e - c d\right )}{e^{4} \sqrt {d + e x}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (2 b c e - 4 c^{2} d\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right )}{e^{4}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\frac {b^{2} x^{3}}{3} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((c*x**2+b*x)**2/(e*x+d)**(5/2),x)
Output:
Piecewise((2*(c**2*(d + e*x)**(5/2)/(5*e**4) - d**2*(b*e - c*d)**2/(3*e**4 *(d + e*x)**(3/2)) + 2*d*(b*e - 2*c*d)*(b*e - c*d)/(e**4*sqrt(d + e*x)) + (d + e*x)**(3/2)*(2*b*c*e - 4*c**2*d)/(3*e**4) + sqrt(d + e*x)*(b**2*e**2 - 6*b*c*d*e + 6*c**2*d**2)/e**4)/e, Ne(e, 0)), ((b**2*x**3/3 + b*c*x**4/2 + c**2*x**5/5)/d**(5/2), True))
Time = 0.03 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{2} - 10 \, {\left (2 \, c^{2} d - b c e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (6 \, c^{2} d^{2} - 6 \, b c d e + b^{2} e^{2}\right )} \sqrt {e x + d}}{e^{4}} - \frac {5 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} - 6 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2}\right )} {\left (e x + d\right )}\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{4}}\right )}}{15 \, e} \] Input:
integrate((c*x^2+b*x)^2/(e*x+d)^(5/2),x, algorithm="maxima")
Output:
2/15*((3*(e*x + d)^(5/2)*c^2 - 10*(2*c^2*d - b*c*e)*(e*x + d)^(3/2) + 15*( 6*c^2*d^2 - 6*b*c*d*e + b^2*e^2)*sqrt(e*x + d))/e^4 - 5*(c^2*d^4 - 2*b*c*d ^3*e + b^2*d^2*e^2 - 6*(2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2)*(e*x + d))/(( e*x + d)^(3/2)*e^4))/e
Time = 0.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.26 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (e x + d\right )} c^{2} d^{3} - c^{2} d^{4} - 18 \, {\left (e x + d\right )} b c d^{2} e + 2 \, b c d^{3} e + 6 \, {\left (e x + d\right )} b^{2} d e^{2} - b^{2} d^{2} e^{2}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{5}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{2} e^{20} - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d e^{20} + 90 \, \sqrt {e x + d} c^{2} d^{2} e^{20} + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} b c e^{21} - 90 \, \sqrt {e x + d} b c d e^{21} + 15 \, \sqrt {e x + d} b^{2} e^{22}\right )}}{15 \, e^{25}} \] Input:
integrate((c*x^2+b*x)^2/(e*x+d)^(5/2),x, algorithm="giac")
Output:
2/3*(12*(e*x + d)*c^2*d^3 - c^2*d^4 - 18*(e*x + d)*b*c*d^2*e + 2*b*c*d^3*e + 6*(e*x + d)*b^2*d*e^2 - b^2*d^2*e^2)/((e*x + d)^(3/2)*e^5) + 2/15*(3*(e *x + d)^(5/2)*c^2*e^20 - 20*(e*x + d)^(3/2)*c^2*d*e^20 + 90*sqrt(e*x + d)* c^2*d^2*e^20 + 10*(e*x + d)^(3/2)*b*c*e^21 - 90*sqrt(e*x + d)*b*c*d*e^21 + 15*sqrt(e*x + d)*b^2*e^22)/e^25
Time = 0.05 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2\,c^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}+\frac {\sqrt {d+e\,x}\,\left (2\,b^2\,e^2-12\,b\,c\,d\,e+12\,c^2\,d^2\right )}{e^5}-\frac {\left (8\,c^2\,d-4\,b\,c\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^5}+\frac {\left (d+e\,x\right )\,\left (4\,b^2\,d\,e^2-12\,b\,c\,d^2\,e+8\,c^2\,d^3\right )-\frac {2\,c^2\,d^4}{3}-\frac {2\,b^2\,d^2\,e^2}{3}+\frac {4\,b\,c\,d^3\,e}{3}}{e^5\,{\left (d+e\,x\right )}^{3/2}} \] Input:
int((b*x + c*x^2)^2/(d + e*x)^(5/2),x)
Output:
(2*c^2*(d + e*x)^(5/2))/(5*e^5) + ((d + e*x)^(1/2)*(2*b^2*e^2 + 12*c^2*d^2 - 12*b*c*d*e))/e^5 - ((8*c^2*d - 4*b*c*e)*(d + e*x)^(3/2))/(3*e^5) + ((d + e*x)*(8*c^2*d^3 + 4*b^2*d*e^2 - 12*b*c*d^2*e) - (2*c^2*d^4)/3 - (2*b^2*d ^2*e^2)/3 + (4*b*c*d^3*e)/3)/(e^5*(d + e*x)^(3/2))
Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {\frac {2}{5} c^{2} e^{4} x^{4}+\frac {4}{3} b c \,e^{4} x^{3}-\frac {16}{15} c^{2} d \,e^{3} x^{3}+2 b^{2} e^{4} x^{2}-8 b c d \,e^{3} x^{2}+\frac {32}{5} c^{2} d^{2} e^{2} x^{2}+8 b^{2} d \,e^{3} x -32 b c \,d^{2} e^{2} x +\frac {128}{5} c^{2} d^{3} e x +\frac {16}{3} b^{2} d^{2} e^{2}-\frac {64}{3} b c \,d^{3} e +\frac {256}{15} c^{2} d^{4}}{\sqrt {e x +d}\, e^{5} \left (e x +d \right )} \] Input:
int((c*x^2+b*x)^2/(e*x+d)^(5/2),x)
Output:
(2*(40*b**2*d**2*e**2 + 60*b**2*d*e**3*x + 15*b**2*e**4*x**2 - 160*b*c*d** 3*e - 240*b*c*d**2*e**2*x - 60*b*c*d*e**3*x**2 + 10*b*c*e**4*x**3 + 128*c* *2*d**4 + 192*c**2*d**3*e*x + 48*c**2*d**2*e**2*x**2 - 8*c**2*d*e**3*x**3 + 3*c**2*e**4*x**4))/(15*sqrt(d + e*x)*e**5*(d + e*x))