Integrand size = 19, antiderivative size = 105 \[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\frac {2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac {32 a^3 \left (a x^2+b x^3\right )^{3/2}}{315 b^4 x^3}+\frac {16 a^2 \left (a x^2+b x^3\right )^{3/2}}{105 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{3/2}}{21 b^2 x} \] Output:
2/9*(b*x^3+a*x^2)^(3/2)/b-32/315*a^3*(b*x^3+a*x^2)^(3/2)/b^4/x^3+16/105*a^ 2*(b*x^3+a*x^2)^(3/2)/b^3/x^2-4/21*a*(b*x^3+a*x^2)^(3/2)/b^2/x
Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.50 \[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\frac {2 \left (x^2 (a+b x)\right )^{3/2} \left (-16 a^3+24 a^2 b x-30 a b^2 x^2+35 b^3 x^3\right )}{315 b^4 x^3} \] Input:
Integrate[x^2*Sqrt[a*x^2 + b*x^3],x]
Output:
(2*(x^2*(a + b*x))^(3/2)*(-16*a^3 + 24*a^2*b*x - 30*a*b^2*x^2 + 35*b^3*x^3 ))/(315*b^4*x^3)
Time = 0.46 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1922, 1922, 1908, 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.
\(\displaystyle \int x^2 \sqrt {a x^2+b x^3} \, dx\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac {2 a \int x \sqrt {b x^3+a x^2}dx}{3 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac {2 a \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}-\frac {4 a \int \sqrt {b x^3+a x^2}dx}{7 b}\right )}{3 b}\) |
\(\Big \downarrow \) 1908 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac {2 a \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac {2 a \int \frac {\sqrt {b x^3+a x^2}}{x}dx}{5 b}\right )}{7 b}\right )}{3 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{3/2}}{9 b}-\frac {2 a \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{7 b x}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{3/2}}{5 b x^2}-\frac {4 a \left (a x^2+b x^3\right )^{3/2}}{15 b^2 x^3}\right )}{7 b}\right )}{3 b}\) |
Input:
Int[x^2*Sqrt[a*x^2 + b*x^3],x]
Output:
(2*(a*x^2 + b*x^3)^(3/2))/(9*b) - (2*a*((2*(a*x^2 + b*x^3)^(3/2))/(7*b*x) - (4*a*((-4*a*(a*x^2 + b*x^3)^(3/2))/(15*b^2*x^3) + (2*(a*x^2 + b*x^3)^(3/ 2))/(5*b*x^2)))/(7*b)))/(3*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 = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.30
method | result | size |
pseudoelliptic | \(\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (15 b^{2} x^{2}-12 a b x +8 a^{2}\right )}{105 b^{3}}\) | \(32\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} x^{3}+30 a \,b^{2} x^{2}-24 a^{2} b x +16 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(57\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} x^{3}+30 a \,b^{2} x^{2}-24 a^{2} b x +16 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(57\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-35 b^{3} x^{3}+30 a \,b^{2} x^{2}-24 a^{2} b x +16 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(57\) |
risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-35 b^{4} x^{4}-5 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}-8 a^{3} b x +16 a^{4}\right )}{315 x \,b^{4}}\) | \(61\) |
trager | \(-\frac {2 \left (-35 b^{4} x^{4}-5 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}-8 a^{3} b x +16 a^{4}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{4} x}\) | \(63\) |
Input:
int(x^2*(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/105*(b*x+a)^(3/2)*(15*b^2*x^2-12*a*b*x+8*a^2)/b^3
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.59 \[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{315 \, b^{4} x} \] Input:
integrate(x^2*(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt (b*x^3 + a*x^2)/(b^4*x)
\[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\int x^{2} \sqrt {x^{2} \left (a + b x\right )}\, dx \] Input:
integrate(x**2*(b*x**3+a*x**2)**(1/2),x)
Output:
Integral(x**2*sqrt(x**2*(a + b*x)), x)
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.50 \[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{4}} \] Input:
integrate(x^2*(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
2/315*(35*b^4*x^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^2 + 8*a^3*b*x - 16*a^4)*sqrt (b*x + a)/b^4
Time = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.25 \[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\frac {32 \, a^{\frac {9}{2}} \mathrm {sgn}\left (x\right )}{315 \, b^{4}} + \frac {2 \, {\left (\frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} \mathrm {sgn}\left (x\right )}{b^{3}}\right )}}{315 \, b} \] Input:
integrate(x^2*(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
32/315*a^(9/2)*sgn(x)/b^4 + 2/315*(9*(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*sgn(x)/b^3 + (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)*sgn(x)/b^3)/b
Time = 8.70 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.59 \[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (-16\,a^4+8\,a^3\,b\,x-6\,a^2\,b^2\,x^2+5\,a\,b^3\,x^3+35\,b^4\,x^4\right )}{315\,b^4\,x} \] Input:
int(x^2*(a*x^2 + b*x^3)^(1/2),x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(35*b^4*x^4 - 16*a^4 + 5*a*b^3*x^3 - 6*a^2*b^2*x^ 2 + 8*a^3*b*x))/(315*b^4*x)
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.50 \[ \int x^2 \sqrt {a x^2+b x^3} \, dx=\frac {2 \sqrt {b x +a}\, \left (35 b^{4} x^{4}+5 a \,b^{3} x^{3}-6 a^{2} b^{2} x^{2}+8 a^{3} b x -16 a^{4}\right )}{315 b^{4}} \] Input:
int(x^2*(b*x^3+a*x^2)^(1/2),x)
Output:
(2*sqrt(a + b*x)*( - 16*a**4 + 8*a**3*b*x - 6*a**2*b**2*x**2 + 5*a*b**3*x* *3 + 35*b**4*x**4))/(315*b**4)