Integrand size = 21, antiderivative size = 228 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2 d^3}{5 b \left (b x+c x^2\right )^{5/2}}+\frac {2 d^2 (2 c d-3 b e) x}{3 b^2 \left (b x+c x^2\right )^{5/2}}-\frac {2 d (4 c d-3 b e)^2 x^2}{3 b^3 \left (b x+c x^2\right )^{5/2}}+\frac {2 \left (b^3 e^3-2 c d (4 c d-3 b e)^2\right ) x^3}{5 b^4 \left (b x+c x^2\right )^{5/2}}+\frac {8 \left (b^3 e^3-2 c d (4 c d-3 b e)^2\right ) x^2}{15 b^5 \left (b x+c x^2\right )^{3/2}}+\frac {16 \left (b^3 e^3-2 c d (4 c d-3 b e)^2\right ) x}{15 b^6 \sqrt {b x+c x^2}} \] Output:
-2/5*d^3/b/(c*x^2+b*x)^(5/2)+2/3*d^2*(-3*b*e+2*c*d)*x/b^2/(c*x^2+b*x)^(5/2 )-2/3*d*(-3*b*e+4*c*d)^2*x^2/b^3/(c*x^2+b*x)^(5/2)+2/5*(b^3*e^3-2*c*d*(-3* b*e+4*c*d)^2)*x^3/b^4/(c*x^2+b*x)^(5/2)+8/15*(b^3*e^3-2*c*d*(-3*b*e+4*c*d) ^2)*x^2/b^5/(c*x^2+b*x)^(3/2)+16/15*(b^3*e^3-2*c*d*(-3*b*e+4*c*d)^2)*x/b^6 /(c*x^2+b*x)^(1/2)
Time = 0.43 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.83 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \left (256 c^5 d^3 x^5+128 b c^4 d^2 x^4 (5 d-3 e x)+48 b^2 c^3 d x^3 \left (10 d^2-20 d e x+3 e^2 x^2\right )+3 b^5 \left (d^3+5 d^2 e x+15 d e^2 x^2-5 e^3 x^3\right )-8 b^3 c^2 x^2 \left (-10 d^3+90 d^2 e x-45 d e^2 x^2+e^3 x^3\right )-10 b^4 c x \left (d^3+12 d^2 e x-27 d e^2 x^2+2 e^3 x^3\right )\right )}{15 b^6 (x (b+c x))^{5/2}} \] Input:
Integrate[(d + e*x)^3/(b*x + c*x^2)^(7/2),x]
Output:
(-2*(256*c^5*d^3*x^5 + 128*b*c^4*d^2*x^4*(5*d - 3*e*x) + 48*b^2*c^3*d*x^3* (10*d^2 - 20*d*e*x + 3*e^2*x^2) + 3*b^5*(d^3 + 5*d^2*e*x + 15*d*e^2*x^2 - 5*e^3*x^3) - 8*b^3*c^2*x^2*(-10*d^3 + 90*d^2*e*x - 45*d*e^2*x^2 + e^3*x^3) - 10*b^4*c*x*(d^3 + 12*d^2*e*x - 27*d*e^2*x^2 + 2*e^3*x^3)))/(15*b^6*(x*( b + c*x))^(5/2))
Time = 0.54 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.64, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1163, 25, 1227, 1158}
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.
| \(\displaystyle \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1163 |
| \(\displaystyle \frac {2 \int -\frac {(d+e x)^2 (8 c d-3 b e+2 c e x)}{\left (c x^2+b x\right )^{5/2}}dx}{5 b^2}-\frac {2 (b+2 c x) (d+e x)^3}{5 b^2 \left (b x+c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 \int \frac {(d+e x)^2 (8 c d-3 b e+2 c e x)}{\left (c x^2+b x\right )^{5/2}}dx}{5 b^2}-\frac {2 (b+2 c x) (d+e x)^3}{5 b^2 \left (b x+c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1227 |
| \(\displaystyle -\frac {2 \left (-\frac {4 (4 c d-3 b e) (4 c d-b e) \int \frac {d+e x}{\left (c x^2+b x\right )^{3/2}}dx}{3 b^2}-\frac {2 (d+e x)^2 (8 c x (2 c d-b e)+b (8 c d-3 b e))}{3 b^2 \left (b x+c x^2\right )^{3/2}}\right )}{5 b^2}-\frac {2 (b+2 c x) (d+e x)^3}{5 b^2 \left (b x+c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1158 |
| \(\displaystyle -\frac {2 (b+2 c x) (d+e x)^3}{5 b^2 \left (b x+c x^2\right )^{5/2}}-\frac {2 \left (\frac {8 (4 c d-3 b e) (4 c d-b e) (x (2 c d-b e)+b d)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (d+e x)^2 (8 c x (2 c d-b e)+b (8 c d-3 b e))}{3 b^2 \left (b x+c x^2\right )^{3/2}}\right )}{5 b^2}\) |
Input:
Int[(d + e*x)^3/(b*x + c*x^2)^(7/2),x]
Output:
(-2*(b + 2*c*x)*(d + e*x)^3)/(5*b^2*(b*x + c*x^2)^(5/2)) - (2*((-2*(d + e* x)^2*(b*(8*c*d - 3*b*e) + 8*c*(2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2) ) + (8*(4*c*d - 3*b*e)*(4*c*d - b*e)*(b*d + (2*c*d - b*e)*x))/(3*b^4*Sqrt[ b*x + c*x^2])))/(5*b^2)
Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbo l] :> Simp[-2*((b*d - 2*a*e + (2*c*d - b*e)*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c, d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^m*(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)* (b^2 - 4*a*c))), x] - Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 1 )*(b*e*m + 2*c*d*(2*p + 3) + 2*c*e*(m + 2*p + 3)*x)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 0] && (LtQ[ m, 1] || (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*( (b*f - 2*a*g + (2*c*f - b*g)*x)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[m*((b*( e*f + d*g) - 2*(c*d*f + a*e*g))/((p + 1)*(b^2 - 4*a*c))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[Simplify[m + 2*p + 3], 0] && LtQ[p, -1]
Time = 0.82 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.84
| method | result | size |
| pseudoelliptic | \(\frac {\left (30 e^{3} x^{3}-90 d \,e^{2} x^{2}-30 d^{2} e x -6 d^{3}\right ) b^{5}+20 c x \left (2 e^{3} x^{3}-27 d \,e^{2} x^{2}+12 d^{2} e x +d^{3}\right ) b^{4}-160 x^{2} \left (-\frac {1}{10} e^{3} x^{3}+\frac {9}{2} d \,e^{2} x^{2}-9 d^{2} e x +d^{3}\right ) c^{2} b^{3}-960 x^{3} c^{3} \left (\frac {3}{10} e^{2} x^{2}-2 d e x +d^{2}\right ) d \,b^{2}-1280 x^{4} c^{4} \left (-\frac {3 e x}{5}+d \right ) d^{2} b -512 c^{5} d^{3} x^{5}}{15 \sqrt {x \left (c x +b \right )}\, x^{2} \left (c x +b \right )^{2} b^{6}}\) | \(192\) |
| risch | \(-\frac {2 d \left (c x +b \right ) \left (45 b^{2} e^{2} x^{2}-165 b c d e \,x^{2}+128 d^{2} c^{2} x^{2}+15 b^{2} d e x -19 x b c \,d^{2}+3 b^{2} d^{2}\right )}{15 b^{6} x^{2} \sqrt {x \left (c x +b \right )}}+\frac {2 x \left (8 b^{2} c^{2} e^{2} x^{2}-91 b \,c^{3} d e \,x^{2}+128 c^{4} d^{2} x^{2}+20 b^{3} c \,e^{2} x -205 b^{2} c^{2} d e x +275 b \,c^{3} d^{2} x +15 b^{4} e^{2}-120 b^{3} c d e +150 b^{2} c^{2} d^{2}\right ) \left (b e -c d \right )}{15 \sqrt {x \left (c x +b \right )}\, \left (c^{2} x^{2}+2 c b x +b^{2}\right ) b^{6}}\) | \(217\) |
| gosper | \(-\frac {2 x \left (c x +b \right ) \left (-8 b^{3} c^{2} e^{3} x^{5}+144 b^{2} c^{3} d \,e^{2} x^{5}-384 b \,c^{4} d^{2} e \,x^{5}+256 c^{5} d^{3} x^{5}-20 b^{4} c \,e^{3} x^{4}+360 b^{3} c^{2} d \,e^{2} x^{4}-960 b^{2} c^{3} d^{2} e \,x^{4}+640 b \,c^{4} d^{3} x^{4}-15 b^{5} e^{3} x^{3}+270 b^{4} c d \,e^{2} x^{3}-720 b^{3} c^{2} d^{2} e \,x^{3}+480 b^{2} c^{3} d^{3} x^{3}+45 b^{5} d \,e^{2} x^{2}-120 b^{4} c \,d^{2} e \,x^{2}+80 b^{3} c^{2} d^{3} x^{2}+15 b^{5} d^{2} e x -10 b^{4} c \,d^{3} x +3 d^{3} b^{5}\right )}{15 b^{6} \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}\) | \(251\) |
| orering | \(-\frac {2 x \left (c x +b \right ) \left (-8 b^{3} c^{2} e^{3} x^{5}+144 b^{2} c^{3} d \,e^{2} x^{5}-384 b \,c^{4} d^{2} e \,x^{5}+256 c^{5} d^{3} x^{5}-20 b^{4} c \,e^{3} x^{4}+360 b^{3} c^{2} d \,e^{2} x^{4}-960 b^{2} c^{3} d^{2} e \,x^{4}+640 b \,c^{4} d^{3} x^{4}-15 b^{5} e^{3} x^{3}+270 b^{4} c d \,e^{2} x^{3}-720 b^{3} c^{2} d^{2} e \,x^{3}+480 b^{2} c^{3} d^{3} x^{3}+45 b^{5} d \,e^{2} x^{2}-120 b^{4} c \,d^{2} e \,x^{2}+80 b^{3} c^{2} d^{3} x^{2}+15 b^{5} d^{2} e x -10 b^{4} c \,d^{3} x +3 d^{3} b^{5}\right )}{15 b^{6} \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}\) | \(251\) |
| trager | \(-\frac {2 \left (-8 b^{3} c^{2} e^{3} x^{5}+144 b^{2} c^{3} d \,e^{2} x^{5}-384 b \,c^{4} d^{2} e \,x^{5}+256 c^{5} d^{3} x^{5}-20 b^{4} c \,e^{3} x^{4}+360 b^{3} c^{2} d \,e^{2} x^{4}-960 b^{2} c^{3} d^{2} e \,x^{4}+640 b \,c^{4} d^{3} x^{4}-15 b^{5} e^{3} x^{3}+270 b^{4} c d \,e^{2} x^{3}-720 b^{3} c^{2} d^{2} e \,x^{3}+480 b^{2} c^{3} d^{3} x^{3}+45 b^{5} d \,e^{2} x^{2}-120 b^{4} c \,d^{2} e \,x^{2}+80 b^{3} c^{2} d^{3} x^{2}+15 b^{5} d^{2} e x -10 b^{4} c \,d^{3} x +3 d^{3} b^{5}\right ) \sqrt {c \,x^{2}+b x}}{15 b^{6} \left (c x +b \right )^{3} x^{3}}\) | \(255\) |
| default | \(d^{3} \left (-\frac {2 \left (2 c x +b \right )}{5 b^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {16 c \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{5 b^{2}}\right )+e^{3} \left (-\frac {x^{2}}{3 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {b \left (-\frac {x}{4 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {3 b \left (-\frac {1}{5 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{5 b^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {16 c \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{5 b^{2}}\right )}{2 c}\right )}{8 c}\right )}{6 c}\right )+3 d \,e^{2} \left (-\frac {x}{4 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {3 b \left (-\frac {1}{5 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{5 b^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {16 c \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{5 b^{2}}\right )}{2 c}\right )}{8 c}\right )+3 d^{2} e \left (-\frac {1}{5 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{5 b^{2} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}-\frac {16 c \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{5 b^{2}}\right )}{2 c}\right )\) | \(465\) |
Input:
int((e*x+d)^3/(c*x^2+b*x)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15*((30*e^3*x^3-90*d*e^2*x^2-30*d^2*e*x-6*d^3)*b^5+20*c*x*(2*e^3*x^3-27* d*e^2*x^2+12*d^2*e*x+d^3)*b^4-160*x^2*(-1/10*e^3*x^3+9/2*d*e^2*x^2-9*d^2*e *x+d^3)*c^2*b^3-960*x^3*c^3*(3/10*e^2*x^2-2*d*e*x+d^2)*d*b^2-1280*x^4*c^4* (-3/5*e*x+d)*d^2*b-512*c^5*d^3*x^5)/(x*(c*x+b))^(1/2)/x^2/(c*x+b)^2/b^6
Time = 0.09 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (3 \, b^{5} d^{3} + 8 \, {\left (32 \, c^{5} d^{3} - 48 \, b c^{4} d^{2} e + 18 \, b^{2} c^{3} d e^{2} - b^{3} c^{2} e^{3}\right )} x^{5} + 20 \, {\left (32 \, b c^{4} d^{3} - 48 \, b^{2} c^{3} d^{2} e + 18 \, b^{3} c^{2} d e^{2} - b^{4} c e^{3}\right )} x^{4} + 15 \, {\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 18 \, b^{4} c d e^{2} - b^{5} e^{3}\right )} x^{3} + 5 \, {\left (16 \, b^{3} c^{2} d^{3} - 24 \, b^{4} c d^{2} e + 9 \, b^{5} d e^{2}\right )} x^{2} - 5 \, {\left (2 \, b^{4} c d^{3} - 3 \, b^{5} d^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, {\left (b^{6} c^{3} x^{6} + 3 \, b^{7} c^{2} x^{5} + 3 \, b^{8} c x^{4} + b^{9} x^{3}\right )}} \] Input:
integrate((e*x+d)^3/(c*x^2+b*x)^(7/2),x, algorithm="fricas")
Output:
-2/15*(3*b^5*d^3 + 8*(32*c^5*d^3 - 48*b*c^4*d^2*e + 18*b^2*c^3*d*e^2 - b^3 *c^2*e^3)*x^5 + 20*(32*b*c^4*d^3 - 48*b^2*c^3*d^2*e + 18*b^3*c^2*d*e^2 - b ^4*c*e^3)*x^4 + 15*(32*b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 18*b^4*c*d*e^2 - b ^5*e^3)*x^3 + 5*(16*b^3*c^2*d^3 - 24*b^4*c*d^2*e + 9*b^5*d*e^2)*x^2 - 5*(2 *b^4*c*d^3 - 3*b^5*d^2*e)*x)*sqrt(c*x^2 + b*x)/(b^6*c^3*x^6 + 3*b^7*c^2*x^ 5 + 3*b^8*c*x^4 + b^9*x^3)
\[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{3}}{\left (x \left (b + c x\right )\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((e*x+d)**3/(c*x**2+b*x)**(7/2),x)
Output:
Integral((d + e*x)**3/(x*(b + c*x))**(7/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (204) = 408\).
Time = 0.04 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.07 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {e^{3} x^{2}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} c} - \frac {4 \, c d^{3} x}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b^{2}} + \frac {64 \, c^{2} d^{3} x}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} - \frac {512 \, c^{3} d^{3} x}{15 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {6 \, d^{2} e x}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b} - \frac {32 \, c d^{2} e x}{5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} + \frac {256 \, c^{2} d^{2} e x}{5 \, \sqrt {c x^{2} + b x} b^{5}} + \frac {12 \, d e^{2} x}{5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} - \frac {6 \, d e^{2} x}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} c} - \frac {96 \, c d e^{2} x}{5 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {16 \, e^{3} x}{15 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {b e^{3} x}{15 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} c^{2}} - \frac {2 \, e^{3} x}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b c} - \frac {2 \, d^{3}}{5 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b} + \frac {32 \, c d^{3}}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} - \frac {256 \, c^{2} d^{3}}{15 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {16 \, d^{2} e}{5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {128 \, c d^{2} e}{5 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {48 \, d e^{2}}{5 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {6 \, d e^{2}}{5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b c} - \frac {e^{3}}{15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {8 \, e^{3}}{15 \, \sqrt {c x^{2} + b x} b^{2} c} \] Input:
integrate((e*x+d)^3/(c*x^2+b*x)^(7/2),x, algorithm="maxima")
Output:
-1/3*e^3*x^2/((c*x^2 + b*x)^(5/2)*c) - 4/5*c*d^3*x/((c*x^2 + b*x)^(5/2)*b^ 2) + 64/15*c^2*d^3*x/((c*x^2 + b*x)^(3/2)*b^4) - 512/15*c^3*d^3*x/(sqrt(c* x^2 + b*x)*b^6) + 6/5*d^2*e*x/((c*x^2 + b*x)^(5/2)*b) - 32/5*c*d^2*e*x/((c *x^2 + b*x)^(3/2)*b^3) + 256/5*c^2*d^2*e*x/(sqrt(c*x^2 + b*x)*b^5) + 12/5* d*e^2*x/((c*x^2 + b*x)^(3/2)*b^2) - 6/5*d*e^2*x/((c*x^2 + b*x)^(5/2)*c) - 96/5*c*d*e^2*x/(sqrt(c*x^2 + b*x)*b^4) + 16/15*e^3*x/(sqrt(c*x^2 + b*x)*b^ 3) + 1/15*b*e^3*x/((c*x^2 + b*x)^(5/2)*c^2) - 2/15*e^3*x/((c*x^2 + b*x)^(3 /2)*b*c) - 2/5*d^3/((c*x^2 + b*x)^(5/2)*b) + 32/15*c*d^3/((c*x^2 + b*x)^(3 /2)*b^3) - 256/15*c^2*d^3/(sqrt(c*x^2 + b*x)*b^5) - 16/5*d^2*e/((c*x^2 + b *x)^(3/2)*b^2) + 128/5*c*d^2*e/(sqrt(c*x^2 + b*x)*b^4) - 48/5*d*e^2/(sqrt( c*x^2 + b*x)*b^3) + 6/5*d*e^2/((c*x^2 + b*x)^(3/2)*b*c) - 1/15*e^3/((c*x^2 + b*x)^(3/2)*c^2) + 8/15*e^3/(sqrt(c*x^2 + b*x)*b^2*c)
Time = 0.16 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (\frac {3 \, d^{3}}{b} + {\left ({\left ({\left (4 \, x {\left (\frac {2 \, {\left (32 \, c^{5} d^{3} - 48 \, b c^{4} d^{2} e + 18 \, b^{2} c^{3} d e^{2} - b^{3} c^{2} e^{3}\right )} x}{b^{6}} + \frac {5 \, {\left (32 \, b c^{4} d^{3} - 48 \, b^{2} c^{3} d^{2} e + 18 \, b^{3} c^{2} d e^{2} - b^{4} c e^{3}\right )}}{b^{6}}\right )} + \frac {15 \, {\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 18 \, b^{4} c d e^{2} - b^{5} e^{3}\right )}}{b^{6}}\right )} x + \frac {5 \, {\left (16 \, b^{3} c^{2} d^{3} - 24 \, b^{4} c d^{2} e + 9 \, b^{5} d e^{2}\right )}}{b^{6}}\right )} x - \frac {5 \, {\left (2 \, b^{4} c d^{3} - 3 \, b^{5} d^{2} e\right )}}{b^{6}}\right )} x\right )}}{15 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \] Input:
integrate((e*x+d)^3/(c*x^2+b*x)^(7/2),x, algorithm="giac")
Output:
-2/15*(3*d^3/b + (((4*x*(2*(32*c^5*d^3 - 48*b*c^4*d^2*e + 18*b^2*c^3*d*e^2 - b^3*c^2*e^3)*x/b^6 + 5*(32*b*c^4*d^3 - 48*b^2*c^3*d^2*e + 18*b^3*c^2*d* e^2 - b^4*c*e^3)/b^6) + 15*(32*b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 18*b^4*c*d *e^2 - b^5*e^3)/b^6)*x + 5*(16*b^3*c^2*d^3 - 24*b^4*c*d^2*e + 9*b^5*d*e^2) /b^6)*x - 5*(2*b^4*c*d^3 - 3*b^5*d^2*e)/b^6)*x)/(c*x^2 + b*x)^(5/2)
Time = 5.22 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=-\frac {2\,\left (3\,b^5\,d^3+15\,b^5\,d^2\,e\,x+45\,b^5\,d\,e^2\,x^2-15\,b^5\,e^3\,x^3-10\,b^4\,c\,d^3\,x-120\,b^4\,c\,d^2\,e\,x^2+270\,b^4\,c\,d\,e^2\,x^3-20\,b^4\,c\,e^3\,x^4+80\,b^3\,c^2\,d^3\,x^2-720\,b^3\,c^2\,d^2\,e\,x^3+360\,b^3\,c^2\,d\,e^2\,x^4-8\,b^3\,c^2\,e^3\,x^5+480\,b^2\,c^3\,d^3\,x^3-960\,b^2\,c^3\,d^2\,e\,x^4+144\,b^2\,c^3\,d\,e^2\,x^5+640\,b\,c^4\,d^3\,x^4-384\,b\,c^4\,d^2\,e\,x^5+256\,c^5\,d^3\,x^5\right )}{15\,b^6\,{\left (c\,x^2+b\,x\right )}^{5/2}} \] Input:
int((d + e*x)^3/(b*x + c*x^2)^(7/2),x)
Output:
-(2*(3*b^5*d^3 - 15*b^5*e^3*x^3 + 256*c^5*d^3*x^5 + 640*b*c^4*d^3*x^4 - 20 *b^4*c*e^3*x^4 + 45*b^5*d*e^2*x^2 + 80*b^3*c^2*d^3*x^2 + 480*b^2*c^3*d^3*x ^3 - 8*b^3*c^2*e^3*x^5 - 10*b^4*c*d^3*x + 15*b^5*d^2*e*x - 120*b^4*c*d^2*e *x^2 + 270*b^4*c*d*e^2*x^3 - 384*b*c^4*d^2*e*x^5 - 720*b^3*c^2*d^2*e*x^3 - 960*b^2*c^3*d^2*e*x^4 + 360*b^3*c^2*d*e^2*x^4 + 144*b^2*c^3*d*e^2*x^5))/( 15*b^6*(b*x + c*x^2)^(5/2))
Time = 0.26 (sec) , antiderivative size = 569, normalized size of antiderivative = 2.50 \[ \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{7/2}} \, dx=\frac {-\frac {2 \sqrt {x}\, b^{5} c \,d^{3}}{5}-\frac {512 \sqrt {x}\, c^{6} d^{3} x^{5}}{15}+\frac {96 \sqrt {c}\, \sqrt {c x +b}\, b^{4} c d \,e^{2} x^{3}}{5}-\frac {256 \sqrt {c}\, \sqrt {c x +b}\, b^{3} c^{2} d^{2} e \,x^{3}}{5}+\frac {192 \sqrt {c}\, \sqrt {c x +b}\, b^{3} c^{2} d \,e^{2} x^{4}}{5}-\frac {512 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{3} d^{2} e \,x^{4}}{5}+\frac {96 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{3} d \,e^{2} x^{5}}{5}-\frac {256 \sqrt {c}\, \sqrt {c x +b}\, b \,c^{4} d^{2} e \,x^{5}}{5}-\frac {16 \sqrt {c}\, \sqrt {c x +b}\, b^{5} e^{3} x^{3}}{15}+\frac {512 \sqrt {c}\, \sqrt {c x +b}\, c^{5} d^{3} x^{5}}{15}+2 \sqrt {x}\, b^{5} c \,e^{3} x^{3}+\frac {4 \sqrt {x}\, b^{4} c^{2} d^{3} x}{3}+\frac {8 \sqrt {x}\, b^{4} c^{2} e^{3} x^{4}}{3}-\frac {32 \sqrt {x}\, b^{3} c^{3} d^{3} x^{2}}{3}+\frac {16 \sqrt {x}\, b^{3} c^{3} e^{3} x^{5}}{15}-\frac {32 \sqrt {c}\, \sqrt {c x +b}\, b^{4} c \,e^{3} x^{4}}{15}-\frac {16 \sqrt {c}\, \sqrt {c x +b}\, b^{3} c^{2} e^{3} x^{5}}{15}+\frac {512 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{3} d^{3} x^{3}}{15}+\frac {1024 \sqrt {c}\, \sqrt {c x +b}\, b \,c^{4} d^{3} x^{4}}{15}-2 \sqrt {x}\, b^{5} c \,d^{2} e x -6 \sqrt {x}\, b^{5} c d \,e^{2} x^{2}+16 \sqrt {x}\, b^{4} c^{2} d^{2} e \,x^{2}-36 \sqrt {x}\, b^{4} c^{2} d \,e^{2} x^{3}+96 \sqrt {x}\, b^{3} c^{3} d^{2} e \,x^{3}-48 \sqrt {x}\, b^{3} c^{3} d \,e^{2} x^{4}+128 \sqrt {x}\, b^{2} c^{4} d^{2} e \,x^{4}-\frac {96 \sqrt {x}\, b^{2} c^{4} d \,e^{2} x^{5}}{5}+\frac {256 \sqrt {x}\, b \,c^{5} d^{2} e \,x^{5}}{5}-64 \sqrt {x}\, b^{2} c^{4} d^{3} x^{3}-\frac {256 \sqrt {x}\, b \,c^{5} d^{3} x^{4}}{3}}{\sqrt {c x +b}\, b^{6} c \,x^{3} \left (c^{2} x^{2}+2 b c x +b^{2}\right )} \] Input:
int((e*x+d)^3/(c*x^2+b*x)^(7/2),x)
Output:
(2*( - 8*sqrt(c)*sqrt(b + c*x)*b**5*e**3*x**3 + 144*sqrt(c)*sqrt(b + c*x)* b**4*c*d*e**2*x**3 - 16*sqrt(c)*sqrt(b + c*x)*b**4*c*e**3*x**4 - 384*sqrt( c)*sqrt(b + c*x)*b**3*c**2*d**2*e*x**3 + 288*sqrt(c)*sqrt(b + c*x)*b**3*c* *2*d*e**2*x**4 - 8*sqrt(c)*sqrt(b + c*x)*b**3*c**2*e**3*x**5 + 256*sqrt(c) *sqrt(b + c*x)*b**2*c**3*d**3*x**3 - 768*sqrt(c)*sqrt(b + c*x)*b**2*c**3*d **2*e*x**4 + 144*sqrt(c)*sqrt(b + c*x)*b**2*c**3*d*e**2*x**5 + 512*sqrt(c) *sqrt(b + c*x)*b*c**4*d**3*x**4 - 384*sqrt(c)*sqrt(b + c*x)*b*c**4*d**2*e* x**5 + 256*sqrt(c)*sqrt(b + c*x)*c**5*d**3*x**5 - 3*sqrt(x)*b**5*c*d**3 - 15*sqrt(x)*b**5*c*d**2*e*x - 45*sqrt(x)*b**5*c*d*e**2*x**2 + 15*sqrt(x)*b* *5*c*e**3*x**3 + 10*sqrt(x)*b**4*c**2*d**3*x + 120*sqrt(x)*b**4*c**2*d**2* e*x**2 - 270*sqrt(x)*b**4*c**2*d*e**2*x**3 + 20*sqrt(x)*b**4*c**2*e**3*x** 4 - 80*sqrt(x)*b**3*c**3*d**3*x**2 + 720*sqrt(x)*b**3*c**3*d**2*e*x**3 - 3 60*sqrt(x)*b**3*c**3*d*e**2*x**4 + 8*sqrt(x)*b**3*c**3*e**3*x**5 - 480*sqr t(x)*b**2*c**4*d**3*x**3 + 960*sqrt(x)*b**2*c**4*d**2*e*x**4 - 144*sqrt(x) *b**2*c**4*d*e**2*x**5 - 640*sqrt(x)*b*c**5*d**3*x**4 + 384*sqrt(x)*b*c**5 *d**2*e*x**5 - 256*sqrt(x)*c**6*d**3*x**5))/(15*sqrt(b + c*x)*b**6*c*x**3* (b**2 + 2*b*c*x + c**2*x**2))