Integrand size = 17, antiderivative size = 136 \[ \int x \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {256 a^4 \left (a x^2+b x^3\right )^{5/2}}{15015 b^5 x^5}-\frac {128 a^3 \left (a x^2+b x^3\right )^{5/2}}{3003 b^4 x^4}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^3}-\frac {16 a \left (a x^2+b x^3\right )^{5/2}}{143 b^2 x^2}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x} \]
256/15015*a^4*(b*x^3+a*x^2)^(5/2)/b^5/x^5-128/3003*a^3*(b*x^3+a*x^2)^(5/2) /b^4/x^4+32/429*a^2*(b*x^3+a*x^2)^(5/2)/b^3/x^3-16/143*a*(b*x^3+a*x^2)^(5/ 2)/b^2/x^2+2/13*(b*x^3+a*x^2)^(5/2)/b/x
Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.51 \[ \int x \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 x (a+b x)^3 \left (128 a^4-320 a^3 b x+560 a^2 b^2 x^2-840 a b^3 x^3+1155 b^4 x^4\right )}{15015 b^5 \sqrt {x^2 (a+b x)}} \]
(2*x*(a + b*x)^3*(128*a^4 - 320*a^3*b*x + 560*a^2*b^2*x^2 - 840*a*b^3*x^3 + 1155*b^4*x^4))/(15015*b^5*Sqrt[x^2*(a + b*x)])
Time = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1922, 1908, 1922, 1922, 1920}
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.
\(\displaystyle \int x \left (a x^2+b x^3\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \int \left (b x^3+a x^2\right )^{3/2}dx}{13 b}\) |
\(\Big \downarrow \) 1908 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x}dx}{11 b}\right )}{13 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^2}dx}{9 b}\right )}{11 b}\right )}{13 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^3}dx}{7 b}\right )}{9 b}\right )}{11 b}\right )}{13 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}\right )}{9 b}\right )}{11 b}\right )}{13 b}\) |
(2*(a*x^2 + b*x^3)^(5/2))/(13*b*x) - (8*a*((2*(a*x^2 + b*x^3)^(5/2))/(11*b *x^2) - (6*a*((2*(a*x^2 + b*x^3)^(5/2))/(9*b*x^3) - (4*a*((-4*a*(a*x^2 + b *x^3)^(5/2))/(35*b^2*x^5) + (2*(a*x^2 + b*x^3)^(5/2))/(7*b*x^4)))/(9*b)))/ (11*b)))/(13*b)
3.3.41.3.1 Defintions of rubi rules used
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j - 1)), x] - Simp[b*((n*p + n - j + 1)/(a*( j*p + 1))) Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Time = 1.86 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.15
method | result | size |
pseudoelliptic | \(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-5 b x +2 a \right )}{35 b^{2}}\) | \(21\) |
gosper | \(\frac {2 \left (b x +a \right ) \left (1155 b^{4} x^{4}-840 a \,b^{3} x^{3}+560 a^{2} b^{2} x^{2}-320 a^{3} b x +128 a^{4}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{15015 x^{3} b^{5}}\) | \(68\) |
default | \(\frac {2 \left (b x +a \right ) \left (1155 b^{4} x^{4}-840 a \,b^{3} x^{3}+560 a^{2} b^{2} x^{2}-320 a^{3} b x +128 a^{4}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{15015 x^{3} b^{5}}\) | \(68\) |
risch | \(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (1155 b^{6} x^{6}+1470 a \,x^{5} b^{5}+35 a^{2} x^{4} b^{4}-40 a^{3} x^{3} b^{3}+48 a^{4} x^{2} b^{2}-64 a^{5} x b +128 a^{6}\right )}{15015 x \,b^{5}}\) | \(83\) |
trager | \(\frac {2 \left (1155 b^{6} x^{6}+1470 a \,x^{5} b^{5}+35 a^{2} x^{4} b^{4}-40 a^{3} x^{3} b^{3}+48 a^{4} x^{2} b^{2}-64 a^{5} x b +128 a^{6}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{15015 b^{5} x}\) | \(85\) |
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.62 \[ \int x \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt {b x^{3} + a x^{2}}}{15015 \, b^{5} x} \]
2/15015*(1155*b^6*x^6 + 1470*a*b^5*x^5 + 35*a^2*b^4*x^4 - 40*a^3*b^3*x^3 + 48*a^4*b^2*x^2 - 64*a^5*b*x + 128*a^6)*sqrt(b*x^3 + a*x^2)/(b^5*x)
\[ \int x \left (a x^2+b x^3\right )^{3/2} \, dx=\int x \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.55 \[ \int x \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt {b x + a}}{15015 \, b^{5}} \]
2/15015*(1155*b^6*x^6 + 1470*a*b^5*x^5 + 35*a^2*b^4*x^4 - 40*a^3*b^3*x^3 + 48*a^4*b^2*x^2 - 64*a^5*b*x + 128*a^6)*sqrt(b*x + a)/b^5
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (116) = 232\).
Time = 0.28 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.81 \[ \int x \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {256 \, a^{\frac {13}{2}} \mathrm {sgn}\left (x\right )}{15015 \, b^{5}} + \frac {2 \, {\left (\frac {143 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{2} \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {130 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {15 \, {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )} \mathrm {sgn}\left (x\right )}{b^{4}}\right )}}{45045 \, b} \]
-256/15015*a^(13/2)*sgn(x)/b^5 + 2/45045*(143*(35*(b*x + a)^(9/2) - 180*(b *x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315* sqrt(b*x + a)*a^4)*a^2*sgn(x)/b^4 + 130*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b* x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a*sgn(x)/b^4 + 15*(231*(b*x + a) ^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003 *sqrt(b*x + a)*a^6)*sgn(x)/b^4)/b
Time = 9.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.51 \[ \int x \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (128\,a^4-320\,a^3\,b\,x+560\,a^2\,b^2\,x^2-840\,a\,b^3\,x^3+1155\,b^4\,x^4\right )}{15015\,b^5\,x} \]