Integrand size = 15, antiderivative size = 108 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {32 a^3 \left (a x^2+b x^3\right )^{5/2}}{1155 b^4 x^5}+\frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2} \] Output:
-32/1155*a^3*(b*x^3+a*x^2)^(5/2)/b^4/x^5+16/231*a^2*(b*x^3+a*x^2)^(5/2)/b^ 3/x^4-4/33*a*(b*x^3+a*x^2)^(5/2)/b^2/x^3+2/11*(b*x^3+a*x^2)^(5/2)/b/x^2
Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 x (a+b x)^3 \left (-16 a^3+40 a^2 b x-70 a b^2 x^2+105 b^3 x^3\right )}{1155 b^4 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*x*(a + b*x)^3*(-16*a^3 + 40*a^2*b*x - 70*a*b^2*x^2 + 105*b^3*x^3))/(115 5*b^4*Sqrt[x^2*(a + b*x)])
Time = 0.46 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {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.
\(\displaystyle \int \left (a x^2+b x^3\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1908 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x}dx}{11 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^2}dx}{9 b}\right )}{11 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^3}dx}{7 b}\right )}{9 b}\right )}{11 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}\right )}{9 b}\right )}{11 b}\) |
Input:
Int[(a*x^2 + b*x^3)^(3/2),x]
Output:
(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)
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 = 0.42 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.12
method | result | size |
pseudoelliptic | \(\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5 b}\) | \(13\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{1155 b^{4} x^{3}}\) | \(57\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{1155 b^{4} x^{3}}\) | \(57\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{1155 b^{4} x^{3}}\) | \(57\) |
risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-105 b^{5} x^{5}-140 a \,b^{4} x^{4}-5 a^{2} b^{3} x^{3}+6 a^{3} b^{2} x^{2}-8 a^{4} b x +16 a^{5}\right )}{1155 x \,b^{4}}\) | \(72\) |
trager | \(-\frac {2 \left (-105 b^{5} x^{5}-140 a \,b^{4} x^{4}-5 a^{2} b^{3} x^{3}+6 a^{3} b^{2} x^{2}-8 a^{4} b x +16 a^{5}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{1155 b^{4} x}\) | \(74\) |
Input:
int((b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5*(b*x+a)^(5/2)/b
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{1155 \, b^{4} x} \] Input:
integrate((b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^ 4*b*x - 16*a^5)*sqrt(b*x^3 + a*x^2)/(b^4*x)
\[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\int \left (a x^{2} + b x^{3}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((b*x**3+a*x**2)**(3/2),x)
Output:
Integral((a*x**2 + b*x**3)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.59 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x + a}}{1155 \, b^{4}} \] Input:
integrate((b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^ 4*b*x - 16*a^5)*sqrt(b*x + a)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (92) = 184\).
Time = 0.17 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.94 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {32 \, a^{\frac {11}{2}} \mathrm {sgn}\left (x\right )}{1155 \, b^{4}} + \frac {2 \, {\left (\frac {99 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2} \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {22 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {5 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} \mathrm {sgn}\left (x\right )}{b^{3}}\right )}}{3465 \, b} \] Input:
integrate((b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
32/1155*a^(11/2)*sgn(x)/b^4 + 2/3465*(99*(5*(b*x + a)^(7/2) - 21*(b*x + a) ^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*a^2*sgn(x)/b^3 + 22*(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*sgn(x)/b^3 + 5*(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)*sgn( x)/b^3)/b
Time = 8.78 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (16\,a^3-40\,a^2\,b\,x+70\,a\,b^2\,x^2-105\,b^3\,x^3\right )}{1155\,b^4\,x} \] Input:
int((a*x^2 + b*x^3)^(3/2),x)
Output:
-(2*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2*(16*a^3 - 105*b^3*x^3 + 70*a*b^2*x^2 - 40*a^2*b*x))/(1155*b^4*x)
Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.58 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \sqrt {b x +a}\, \left (105 b^{5} x^{5}+140 a \,b^{4} x^{4}+5 a^{2} b^{3} x^{3}-6 a^{3} b^{2} x^{2}+8 a^{4} b x -16 a^{5}\right )}{1155 b^{4}} \] Input:
int((b*x^3+a*x^2)^(3/2),x)
Output:
(2*sqrt(a + b*x)*( - 16*a**5 + 8*a**4*b*x - 6*a**3*b**2*x**2 + 5*a**2*b**3 *x**3 + 140*a*b**4*x**4 + 105*b**5*x**5))/(1155*b**4)