Integrand size = 19, antiderivative size = 136 \[ \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {256 b^4 \left (b x+c x^2\right )^{5/2}}{15015 c^5 x^{5/2}}-\frac {128 b^3 \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac {32 b^2 \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt {x}}-\frac {16 b \sqrt {x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c} \]
256/15015*b^4*(c*x^2+b*x)^(5/2)/c^5/x^(5/2)-128/3003*b^3*(c*x^2+b*x)^(5/2) /c^4/x^(3/2)+2/13*x^(3/2)*(c*x^2+b*x)^(5/2)/c+32/429*b^2*(c*x^2+b*x)^(5/2) /c^3/x^(1/2)-16/143*b*(c*x^2+b*x)^(5/2)*x^(1/2)/c^2
Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.47 \[ \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 (x (b+c x))^{5/2} \left (128 b^4-320 b^3 c x+560 b^2 c^2 x^2-840 b c^3 x^3+1155 c^4 x^4\right )}{15015 c^5 x^{5/2}} \]
(2*(x*(b + c*x))^(5/2)*(128*b^4 - 320*b^3*c*x + 560*b^2*c^2*x^2 - 840*b*c^ 3*x^3 + 1155*c^4*x^4))/(15015*c^5*x^(5/2))
Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1128, 1128, 1128, 1128, 1122}
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.
\(\displaystyle \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac {8 b \int x^{3/2} \left (c x^2+b x\right )^{3/2}dx}{13 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {6 b \int \sqrt {x} \left (c x^2+b x\right )^{3/2}dx}{11 c}\right )}{13 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {4 b \int \frac {\left (c x^2+b x\right )^{3/2}}{\sqrt {x}}dx}{9 c}\right )}{11 c}\right )}{13 c}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {2 b \int \frac {\left (c x^2+b x\right )^{3/2}}{x^{3/2}}dx}{7 c}\right )}{9 c}\right )}{11 c}\right )}{13 c}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}}\right )}{9 c}\right )}{11 c}\right )}{13 c}\) |
(2*x^(3/2)*(b*x + c*x^2)^(5/2))/(13*c) - (8*b*((2*Sqrt[x]*(b*x + c*x^2)^(5 /2))/(11*c) - (6*b*((2*(b*x + c*x^2)^(5/2))/(9*c*Sqrt[x]) - (4*b*((-4*b*(b *x + c*x^2)^(5/2))/(35*c^2*x^(5/2)) + (2*(b*x + c*x^2)^(5/2))/(7*c*x^(3/2) )))/(9*c)))/(11*c)))/(13*c)
3.1.83.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[Simplify[m + p], 0]
Time = 2.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.49
method | result | size |
gosper | \(\frac {2 \left (c x +b \right ) \left (1155 c^{4} x^{4}-840 b \,c^{3} x^{3}+560 b^{2} c^{2} x^{2}-320 b^{3} c x +128 b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 c^{5} x^{\frac {3}{2}}}\) | \(66\) |
default | \(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{2} \left (1155 c^{4} x^{4}-840 b \,c^{3} x^{3}+560 b^{2} c^{2} x^{2}-320 b^{3} c x +128 b^{4}\right )}{15015 \sqrt {x}\, c^{5}}\) | \(66\) |
risch | \(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (1155 x^{6} c^{6}+1470 x^{5} b \,c^{5}+35 b^{2} x^{4} c^{4}-40 x^{3} b^{3} c^{3}+48 b^{4} x^{2} c^{2}-64 b^{5} x c +128 b^{6}\right )}{15015 \sqrt {x \left (c x +b \right )}\, c^{5}}\) | \(86\) |
2/15015*(c*x+b)*(1155*c^4*x^4-840*b*c^3*x^3+560*b^2*c^2*x^2-320*b^3*c*x+12 8*b^4)*(c*x^2+b*x)^(3/2)/c^5/x^(3/2)
Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.60 \[ \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left (1155 \, c^{6} x^{6} + 1470 \, b c^{5} x^{5} + 35 \, b^{2} c^{4} x^{4} - 40 \, b^{3} c^{3} x^{3} + 48 \, b^{4} c^{2} x^{2} - 64 \, b^{5} c x + 128 \, b^{6}\right )} \sqrt {c x^{2} + b x}}{15015 \, c^{5} \sqrt {x}} \]
2/15015*(1155*c^6*x^6 + 1470*b*c^5*x^5 + 35*b^2*c^4*x^4 - 40*b^3*c^3*x^3 + 48*b^4*c^2*x^2 - 64*b^5*c*x + 128*b^6)*sqrt(c*x^2 + b*x)/(c^5*sqrt(x))
\[ \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx=\int x^{\frac {5}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.08 \[ \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left (5 \, {\left (693 \, c^{6} x^{6} + 63 \, b c^{5} x^{5} - 70 \, b^{2} c^{4} x^{4} + 80 \, b^{3} c^{3} x^{3} - 96 \, b^{4} c^{2} x^{2} + 128 \, b^{5} c x - 256 \, b^{6}\right )} x^{5} + 13 \, {\left (315 \, b c^{5} x^{6} + 35 \, b^{2} c^{4} x^{5} - 40 \, b^{3} c^{3} x^{4} + 48 \, b^{4} c^{2} x^{3} - 64 \, b^{5} c x^{2} + 128 \, b^{6} x\right )} x^{4}\right )} \sqrt {c x + b}}{45045 \, c^{5} x^{5}} \]
2/45045*(5*(693*c^6*x^6 + 63*b*c^5*x^5 - 70*b^2*c^4*x^4 + 80*b^3*c^3*x^3 - 96*b^4*c^2*x^2 + 128*b^5*c*x - 256*b^6)*x^5 + 13*(315*b*c^5*x^6 + 35*b^2* c^4*x^5 - 40*b^3*c^3*x^4 + 48*b^4*c^2*x^3 - 64*b^5*c*x^2 + 128*b^6*x)*x^4) *sqrt(c*x + b)/(c^5*x^5)
Time = 0.28 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.16 \[ \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {2}{9009} \, c {\left (\frac {256 \, b^{\frac {13}{2}}}{c^{6}} + \frac {693 \, {\left (c x + b\right )}^{\frac {13}{2}} - 4095 \, {\left (c x + b\right )}^{\frac {11}{2}} b + 10010 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{2} - 12870 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{4} - 3003 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{5}}{c^{6}}\right )} - \frac {2}{3465} \, b {\left (\frac {128 \, b^{\frac {11}{2}}}{c^{5}} - \frac {315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}}{c^{5}}\right )} \]
2/9009*c*(256*b^(13/2)/c^6 + (693*(c*x + b)^(13/2) - 4095*(c*x + b)^(11/2) *b + 10010*(c*x + b)^(9/2)*b^2 - 12870*(c*x + b)^(7/2)*b^3 + 9009*(c*x + b )^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)/c^6) - 2/3465*b*(128*b^(11/2)/c^5 - (315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^ 2 - 2772*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b)^(3/2)*b^4)/c^5)
Timed out. \[ \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx=\int x^{5/2}\,{\left (c\,x^2+b\,x\right )}^{3/2} \,d x \]