Integrand size = 26, antiderivative size = 132 \[ \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=-\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right ) (d+e x)^{5/2}}{5 e^4}+\frac {2 \left (6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)\right ) (d+e x)^{7/2}}{7 e^4}-\frac {2 c (2 c d-b e) (d+e x)^{9/2}}{3 e^4}+\frac {4 c^2 (d+e x)^{11/2}}{11 e^4} \]
-2/5*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)*(e*x+d)^(5/2)/e^4+2/7*(6*c^2*d^2+b^2 *e^2-2*c*e*(-a*e+3*b*d))*(e*x+d)^(7/2)/e^4-2/3*c*(-b*e+2*c*d)*(e*x+d)^(9/2 )/e^4+4/11*c^2*(e*x+d)^(11/2)/e^4
Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.83 \[ \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 (d+e x)^{5/2} \left (33 b e^2 (-2 b d+7 a e+5 b e x)+c^2 \left (-32 d^3+80 d^2 e x-140 d e^2 x^2+210 e^3 x^3\right )+11 c e \left (6 a e (-2 d+5 e x)+b \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )\right )}{1155 e^4} \]
(2*(d + e*x)^(5/2)*(33*b*e^2*(-2*b*d + 7*a*e + 5*b*e*x) + c^2*(-32*d^3 + 8 0*d^2*e*x - 140*d*e^2*x^2 + 210*e^3*x^3) + 11*c*e*(6*a*e*(-2*d + 5*e*x) + b*(8*d^2 - 20*d*e*x + 35*e^2*x^2))))/(1155*e^4)
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1195, 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.
| \(\displaystyle \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (\frac {(d+e x)^{5/2} \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^3}+\frac {(d+e x)^{3/2} (b e-2 c d) \left (a e^2-b d e+c d^2\right )}{e^3}-\frac {3 c (d+e x)^{7/2} (2 c d-b e)}{e^3}+\frac {2 c^2 (d+e x)^{9/2}}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 (d+e x)^{7/2} \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{7 e^4}-\frac {2 (d+e x)^{5/2} (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{5 e^4}-\frac {2 c (d+e x)^{9/2} (2 c d-b e)}{3 e^4}+\frac {4 c^2 (d+e x)^{11/2}}{11 e^4}\) |
(-2*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(5/2))/(5*e^4) + (2*(6 *c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))*(d + e*x)^(7/2))/(7*e^4) - (2*c* (2*c*d - b*e)*(d + e*x)^(9/2))/(3*e^4) + (4*c^2*(d + e*x)^(11/2))/(11*e^4)
3.16.96.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.66 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (\left (\frac {10 c^{2} x^{3}}{11}+\frac {10 \left (\frac {7 b x}{6}+a \right ) x c}{7}+b \left (\frac {5 b x}{7}+a \right )\right ) e^{3}-\frac {4 d \left (\frac {35 c^{2} x^{2}}{33}+\left (\frac {5 b x}{3}+a \right ) c +\frac {b^{2}}{2}\right ) e^{2}}{7}+\frac {8 d^{2} \left (\frac {10 c x}{11}+b \right ) c e}{21}-\frac {32 c^{2} d^{3}}{231}\right )}{5 e^{4}}\) | \(94\) |
| derivativedivides | \(\frac {\frac {4 c^{2} \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 c \left (b e -2 c d \right ) \left (e x +d \right )^{\frac {9}{2}}}{3}+\frac {2 \left (\left (b e -2 c d \right )^{2}+2 c \left (e^{2} a -b d e +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (b e -2 c d \right ) \left (e^{2} a -b d e +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{4}}\) | \(109\) |
| default | \(\frac {\frac {4 c^{2} \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 c \left (b e -2 c d \right ) \left (e x +d \right )^{\frac {9}{2}}}{3}+\frac {2 \left (\left (b e -2 c d \right )^{2}+2 c \left (e^{2} a -b d e +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (b e -2 c d \right ) \left (e^{2} a -b d e +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{4}}\) | \(109\) |
| gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (210 c^{2} x^{3} e^{3}+385 x^{2} b c \,e^{3}-140 x^{2} c^{2} d \,e^{2}+330 x a c \,e^{3}+165 x \,b^{2} e^{3}-220 x b c d \,e^{2}+80 x \,c^{2} d^{2} e +231 a b \,e^{3}-132 a c d \,e^{2}-66 b^{2} d \,e^{2}+88 b c \,d^{2} e -32 c^{2} d^{3}\right )}{1155 e^{4}}\) | \(123\) |
| trager | \(\frac {2 \left (210 c^{2} e^{5} x^{5}+385 b c \,e^{5} x^{4}+280 c^{2} d \,e^{4} x^{4}+330 a c \,e^{5} x^{3}+165 b^{2} e^{5} x^{3}+550 b c d \,e^{4} x^{3}+10 c^{2} d^{2} e^{3} x^{3}+231 a b \,e^{5} x^{2}+528 a c d \,e^{4} x^{2}+264 b^{2} d \,e^{4} x^{2}+33 b c \,d^{2} e^{3} x^{2}-12 c^{2} d^{3} e^{2} x^{2}+462 a b d \,e^{4} x +66 a c \,d^{2} e^{3} x +33 b^{2} d^{2} e^{3} x -44 b c \,d^{3} e^{2} x +16 c^{2} d^{4} e x +231 a b \,d^{2} e^{3}-132 a c \,d^{3} e^{2}-66 b^{2} d^{3} e^{2}+88 b c \,d^{4} e -32 c^{2} d^{5}\right ) \sqrt {e x +d}}{1155 e^{4}}\) | \(253\) |
| risch | \(\frac {2 \left (210 c^{2} e^{5} x^{5}+385 b c \,e^{5} x^{4}+280 c^{2} d \,e^{4} x^{4}+330 a c \,e^{5} x^{3}+165 b^{2} e^{5} x^{3}+550 b c d \,e^{4} x^{3}+10 c^{2} d^{2} e^{3} x^{3}+231 a b \,e^{5} x^{2}+528 a c d \,e^{4} x^{2}+264 b^{2} d \,e^{4} x^{2}+33 b c \,d^{2} e^{3} x^{2}-12 c^{2} d^{3} e^{2} x^{2}+462 a b d \,e^{4} x +66 a c \,d^{2} e^{3} x +33 b^{2} d^{2} e^{3} x -44 b c \,d^{3} e^{2} x +16 c^{2} d^{4} e x +231 a b \,d^{2} e^{3}-132 a c \,d^{3} e^{2}-66 b^{2} d^{3} e^{2}+88 b c \,d^{4} e -32 c^{2} d^{5}\right ) \sqrt {e x +d}}{1155 e^{4}}\) | \(253\) |
2/5*(e*x+d)^(5/2)*((10/11*c^2*x^3+10/7*(7/6*b*x+a)*x*c+b*(5/7*b*x+a))*e^3- 4/7*d*(35/33*c^2*x^2+(5/3*b*x+a)*c+1/2*b^2)*e^2+8/21*d^2*(10/11*c*x+b)*c*e -32/231*c^2*d^3)/e^4
Time = 0.28 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.67 \[ \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \, {\left (210 \, c^{2} e^{5} x^{5} - 32 \, c^{2} d^{5} + 88 \, b c d^{4} e + 231 \, a b d^{2} e^{3} - 66 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + 35 \, {\left (8 \, c^{2} d e^{4} + 11 \, b c e^{5}\right )} x^{4} + 5 \, {\left (2 \, c^{2} d^{2} e^{3} + 110 \, b c d e^{4} + 33 \, {\left (b^{2} + 2 \, a c\right )} e^{5}\right )} x^{3} - 3 \, {\left (4 \, c^{2} d^{3} e^{2} - 11 \, b c d^{2} e^{3} - 77 \, a b e^{5} - 88 \, {\left (b^{2} + 2 \, a c\right )} d e^{4}\right )} x^{2} + {\left (16 \, c^{2} d^{4} e - 44 \, b c d^{3} e^{2} + 462 \, a b d e^{4} + 33 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{1155 \, e^{4}} \]
2/1155*(210*c^2*e^5*x^5 - 32*c^2*d^5 + 88*b*c*d^4*e + 231*a*b*d^2*e^3 - 66 *(b^2 + 2*a*c)*d^3*e^2 + 35*(8*c^2*d*e^4 + 11*b*c*e^5)*x^4 + 5*(2*c^2*d^2* e^3 + 110*b*c*d*e^4 + 33*(b^2 + 2*a*c)*e^5)*x^3 - 3*(4*c^2*d^3*e^2 - 11*b* c*d^2*e^3 - 77*a*b*e^5 - 88*(b^2 + 2*a*c)*d*e^4)*x^2 + (16*c^2*d^4*e - 44* b*c*d^3*e^2 + 462*a*b*d*e^4 + 33*(b^2 + 2*a*c)*d^2*e^3)*x)*sqrt(e*x + d)/e ^4
Time = 1.08 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.31 \[ \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\begin {cases} \frac {2 \cdot \left (\frac {2 c^{2} \left (d + e x\right )^{\frac {11}{2}}}{11 e^{3}} + \frac {\left (d + e x\right )^{\frac {9}{2}} \cdot \left (3 b c e - 6 c^{2} d\right )}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (2 a c e^{2} + b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (a b e^{3} - 2 a c d e^{2} - b^{2} d e^{2} + 3 b c d^{2} e - 2 c^{2} d^{3}\right )}{5 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {d^{\frac {3}{2}} \left (a + b x + c x^{2}\right )^{2}}{2} & \text {otherwise} \end {cases} \]
Piecewise((2*(2*c**2*(d + e*x)**(11/2)/(11*e**3) + (d + e*x)**(9/2)*(3*b*c *e - 6*c**2*d)/(9*e**3) + (d + e*x)**(7/2)*(2*a*c*e**2 + b**2*e**2 - 6*b*c *d*e + 6*c**2*d**2)/(7*e**3) + (d + e*x)**(5/2)*(a*b*e**3 - 2*a*c*d*e**2 - b**2*d*e**2 + 3*b*c*d**2*e - 2*c**2*d**3)/(5*e**3))/e, Ne(e, 0)), (d**(3/ 2)*(a + b*x + c*x**2)**2/2, True))
Time = 0.19 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \, {\left (210 \, {\left (e x + d\right )}^{\frac {11}{2}} c^{2} - 385 \, {\left (2 \, c^{2} d - b c e\right )} {\left (e x + d\right )}^{\frac {9}{2}} + 165 \, {\left (6 \, c^{2} d^{2} - 6 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} {\left (e x + d\right )}^{\frac {7}{2}} - 231 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{1155 \, e^{4}} \]
2/1155*(210*(e*x + d)^(11/2)*c^2 - 385*(2*c^2*d - b*c*e)*(e*x + d)^(9/2) + 165*(6*c^2*d^2 - 6*b*c*d*e + (b^2 + 2*a*c)*e^2)*(e*x + d)^(7/2) - 231*(2* c^2*d^3 - 3*b*c*d^2*e - a*b*e^3 + (b^2 + 2*a*c)*d*e^2)*(e*x + d)^(5/2))/e^ 4
Leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (116) = 232\).
Time = 0.27 (sec) , antiderivative size = 673, normalized size of antiderivative = 5.10 \[ \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \, {\left (3465 \, \sqrt {e x + d} a b d^{2} + 2310 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b d + \frac {1155 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} b^{2} d^{2}}{e} + \frac {2310 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a c d^{2}}{e} + 231 \, {\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 )} a b + \frac {693 \, {\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 c d^{2}}{e^{2}} + \frac {462 \, {\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} d}{e} + \frac {924 \, {\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 )} a c d}{e} + \frac {198 \, {\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 )} c^{2} d^{2}}{e^{3}} + \frac {594 \, {\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 d}{e^{2}} + \frac {99 \, {\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^{2}}{e} + \frac {198 \, {\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 )} a c}{e} + \frac {44 \, {\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} d}{e^{3}} + \frac {33 \, {\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 )} b c}{e^{2}} + \frac {10 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} - 385 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 990 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 1386 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} + 1155 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 693 \, \sqrt {e x + d} d^{5}\right )} c^{2}}{e^{3}}\right )}}{3465 \, e} \]
2/3465*(3465*sqrt(e*x + d)*a*b*d^2 + 2310*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*b*d + 1155*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*b^2*d^2/e + 2310* ((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*c*d^2/e + 231*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*b + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*b*c*d^2/e^2 + 462*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*b^2*d/e + 924*(3 *(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*c*d/e + 198*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 3 5*sqrt(e*x + d)*d^3)*c^2*d^2/e^3 + 594*(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*d/e^2 + 99*(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^2/e + 198*(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)*a*c/e + 44*(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*d/e^3 + 33*(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)*b*c/e^2 + 10*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d)^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d)^(3/2) *d^4 - 693*sqrt(e*x + d)*d^5)*c^2/e^3)/e
Time = 10.82 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89 \[ \int (b+2 c x) (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {4\,c^2\,{\left (d+e\,x\right )}^{11/2}}{11\,e^4}+\frac {{\left (d+e\,x\right )}^{7/2}\,\left (2\,b^2\,e^2-12\,b\,c\,d\,e+12\,c^2\,d^2+4\,a\,c\,e^2\right )}{7\,e^4}-\frac {\left (12\,c^2\,d-6\,b\,c\,e\right )\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}+\frac {2\,\left (b\,e-2\,c\,d\right )\,{\left (d+e\,x\right )}^{5/2}\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{5\,e^4} \]