Integrand size = 17, antiderivative size = 126 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx=-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac {16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac {32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac {128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac {256 c^4 \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5} \]
-2/13*(c*x^2+b*x)^(5/2)/b/x^9+16/143*c*(c*x^2+b*x)^(5/2)/b^2/x^8-32/429*c^ 2*(c*x^2+b*x)^(5/2)/b^3/x^7+128/3003*c^3*(c*x^2+b*x)^(5/2)/b^4/x^6-256/150 15*c^4*(c*x^2+b*x)^(5/2)/b^5/x^5
Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.49 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx=-\frac {2 (x (b+c x))^{5/2} \left (1155 b^4-840 b^3 c x+560 b^2 c^2 x^2-320 b c^3 x^3+128 c^4 x^4\right )}{15015 b^5 x^9} \]
(-2*(x*(b + c*x))^(5/2)*(1155*b^4 - 840*b^3*c*x + 560*b^2*c^2*x^2 - 320*b* c^3*x^3 + 128*c^4*x^4))/(15015*b^5*x^9)
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1129, 1129, 1129, 1129, 1123}
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.
\(\displaystyle \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle -\frac {8 c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^8}dx}{13 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^7}dx}{11 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}\right )}{13 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^6}dx}{9 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7}\right )}{11 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}\right )}{13 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^5}dx}{7 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{7 b x^6}\right )}{9 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7}\right )}{11 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}\right )}{13 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \left (\frac {4 c \left (b x+c x^2\right )^{5/2}}{35 b^2 x^5}-\frac {2 \left (b x+c x^2\right )^{5/2}}{7 b x^6}\right )}{9 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{9 b x^7}\right )}{11 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}\right )}{13 b}-\frac {2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}\) |
(-2*(b*x + c*x^2)^(5/2))/(13*b*x^9) - (8*c*((-2*(b*x + c*x^2)^(5/2))/(11*b *x^8) - (6*c*((-2*(b*x + c*x^2)^(5/2))/(9*b*x^7) - (4*c*((-2*(b*x + c*x^2) ^(5/2))/(7*b*x^6) + (4*c*(b*x + c*x^2)^(5/2))/(35*b^2*x^5)))/(9*b)))/(11*b )))/(13*b)
3.1.23.3.1 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*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b *e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) )) 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] && ILtQ[Simplify[m + 2*p + 2], 0]
Time = 2.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(-\frac {2 \left (c x +b \right ) \left (128 c^{4} x^{4}-320 b \,c^{3} x^{3}+560 b^{2} c^{2} x^{2}-840 b^{3} c x +1155 b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 b^{5} x^{8}}\) | \(66\) |
pseudoelliptic | \(-\frac {2 \left (c x +b \right )^{2} \sqrt {x \left (c x +b \right )}\, \left (128 c^{4} x^{4}-320 b \,c^{3} x^{3}+560 b^{2} c^{2} x^{2}-840 b^{3} c x +1155 b^{4}\right )}{15015 x^{7} b^{5}}\) | \(66\) |
trager | \(-\frac {2 \left (128 c^{6} x^{6}-64 x^{5} b \,c^{5}+48 b^{2} c^{4} x^{4}-40 x^{3} b^{3} c^{3}+35 b^{4} c^{2} x^{2}+1470 b^{5} c x +1155 b^{6}\right ) \sqrt {c \,x^{2}+b x}}{15015 b^{5} x^{7}}\) | \(83\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (128 c^{6} x^{6}-64 x^{5} b \,c^{5}+48 b^{2} c^{4} x^{4}-40 x^{3} b^{3} c^{3}+35 b^{4} c^{2} x^{2}+1470 b^{5} c x +1155 b^{6}\right )}{15015 x^{6} \sqrt {x \left (c x +b \right )}\, b^{5}}\) | \(86\) |
default | \(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{13 b \,x^{9}}-\frac {8 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{11 b \,x^{8}}-\frac {6 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{9 b \,x^{7}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{7 b \,x^{6}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{35 b^{2} x^{5}}\right )}{9 b}\right )}{11 b}\right )}{13 b}\) | \(119\) |
-2/15015*(c*x+b)*(128*c^4*x^4-320*b*c^3*x^3+560*b^2*c^2*x^2-840*b^3*c*x+11 55*b^4)*(c*x^2+b*x)^(3/2)/b^5/x^8
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx=-\frac {2 \, {\left (128 \, c^{6} x^{6} - 64 \, b c^{5} x^{5} + 48 \, b^{2} c^{4} x^{4} - 40 \, b^{3} c^{3} x^{3} + 35 \, b^{4} c^{2} x^{2} + 1470 \, b^{5} c x + 1155 \, b^{6}\right )} \sqrt {c x^{2} + b x}}{15015 \, b^{5} x^{7}} \]
-2/15015*(128*c^6*x^6 - 64*b*c^5*x^5 + 48*b^2*c^4*x^4 - 40*b^3*c^3*x^3 + 3 5*b^4*c^2*x^2 + 1470*b^5*c*x + 1155*b^6)*sqrt(c*x^2 + b*x)/(b^5*x^7)
\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{9}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.28 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx=-\frac {256 \, \sqrt {c x^{2} + b x} c^{6}}{15015 \, b^{5} x} + \frac {128 \, \sqrt {c x^{2} + b x} c^{5}}{15015 \, b^{4} x^{2}} - \frac {32 \, \sqrt {c x^{2} + b x} c^{4}}{5005 \, b^{3} x^{3}} + \frac {16 \, \sqrt {c x^{2} + b x} c^{3}}{3003 \, b^{2} x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} c^{2}}{429 \, b x^{5}} + \frac {3 \, \sqrt {c x^{2} + b x} c}{715 \, x^{6}} + \frac {3 \, \sqrt {c x^{2} + b x} b}{65 \, x^{7}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{5 \, x^{8}} \]
-256/15015*sqrt(c*x^2 + b*x)*c^6/(b^5*x) + 128/15015*sqrt(c*x^2 + b*x)*c^5 /(b^4*x^2) - 32/5005*sqrt(c*x^2 + b*x)*c^4/(b^3*x^3) + 16/3003*sqrt(c*x^2 + b*x)*c^3/(b^2*x^4) - 2/429*sqrt(c*x^2 + b*x)*c^2/(b*x^5) + 3/715*sqrt(c* x^2 + b*x)*c/x^6 + 3/65*sqrt(c*x^2 + b*x)*b/x^7 - 1/5*(c*x^2 + b*x)^(3/2)/ x^8
Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (106) = 212\).
Time = 0.28 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.00 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx=\frac {2 \, {\left (48048 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{8} c^{4} + 240240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} b c^{\frac {7}{2}} + 531960 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} b^{2} c^{3} + 675675 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} b^{3} c^{\frac {5}{2}} + 535535 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} b^{4} c^{2} + 270270 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{5} c^{\frac {3}{2}} + 84630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{6} c + 15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{7} \sqrt {c} + 1155 \, b^{8}\right )}}{15015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{13}} \]
2/15015*(48048*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*c^4 + 240240*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*b*c^(7/2) + 531960*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6 *b^2*c^3 + 675675*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b^3*c^(5/2) + 535535*( sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^4*c^2 + 270270*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^5*c^(3/2) + 84630*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^6*c + 15 015*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^7*sqrt(c) + 1155*b^8)/(sqrt(c)*x - s qrt(c*x^2 + b*x))^13
Time = 10.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.15 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^9} \, dx=\frac {16\,c^3\,\sqrt {c\,x^2+b\,x}}{3003\,b^2\,x^4}-\frac {28\,c\,\sqrt {c\,x^2+b\,x}}{143\,x^6}-\frac {2\,c^2\,\sqrt {c\,x^2+b\,x}}{429\,b\,x^5}-\frac {2\,b\,\sqrt {c\,x^2+b\,x}}{13\,x^7}-\frac {32\,c^4\,\sqrt {c\,x^2+b\,x}}{5005\,b^3\,x^3}+\frac {128\,c^5\,\sqrt {c\,x^2+b\,x}}{15015\,b^4\,x^2}-\frac {256\,c^6\,\sqrt {c\,x^2+b\,x}}{15015\,b^5\,x} \]
(16*c^3*(b*x + c*x^2)^(1/2))/(3003*b^2*x^4) - (28*c*(b*x + c*x^2)^(1/2))/( 143*x^6) - (2*c^2*(b*x + c*x^2)^(1/2))/(429*b*x^5) - (2*b*(b*x + c*x^2)^(1 /2))/(13*x^7) - (32*c^4*(b*x + c*x^2)^(1/2))/(5005*b^3*x^3) + (128*c^5*(b* x + c*x^2)^(1/2))/(15015*b^4*x^2) - (256*c^6*(b*x + c*x^2)^(1/2))/(15015*b ^5*x)