Integrand size = 19, antiderivative size = 98 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 x^4}{b \sqrt {a x^2+b x^3}}-\frac {16 a \sqrt {a x^2+b x^3}}{5 b^3}+\frac {32 a^2 \sqrt {a x^2+b x^3}}{5 b^4 x}+\frac {12 x \sqrt {a x^2+b x^3}}{5 b^2} \] Output:
-2*x^4/b/(b*x^3+a*x^2)^(1/2)-16/5*a*(b*x^3+a*x^2)^(1/2)/b^3+32/5*a^2*(b*x^ 3+a*x^2)^(1/2)/b^4/x+12/5*x*(b*x^3+a*x^2)^(1/2)/b^2
Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.51 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (16 a^3+8 a^2 b x-2 a b^2 x^2+b^3 x^3\right )}{5 b^4 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[x^6/(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*x*(16*a^3 + 8*a^2*b*x - 2*a*b^2*x^2 + b^3*x^3))/(5*b^4*Sqrt[x^2*(a + b* x)])
Time = 0.43 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1921, 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.
\(\displaystyle \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {6 \int \frac {x^3}{\sqrt {b x^3+a x^2}}dx}{b}-\frac {2 x^4}{b \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {6 \left (\frac {2 x \sqrt {a x^2+b x^3}}{5 b}-\frac {4 a \int \frac {x^2}{\sqrt {b x^3+a x^2}}dx}{5 b}\right )}{b}-\frac {2 x^4}{b \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {6 \left (\frac {2 x \sqrt {a x^2+b x^3}}{5 b}-\frac {4 a \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {2 a \int \frac {x}{\sqrt {b x^3+a x^2}}dx}{3 b}\right )}{5 b}\right )}{b}-\frac {2 x^4}{b \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {6 \left (\frac {2 x \sqrt {a x^2+b x^3}}{5 b}-\frac {4 a \left (\frac {2 \sqrt {a x^2+b x^3}}{3 b}-\frac {4 a \sqrt {a x^2+b x^3}}{3 b^2 x}\right )}{5 b}\right )}{b}-\frac {2 x^4}{b \sqrt {a x^2+b x^3}}\) |
Input:
Int[x^6/(a*x^2 + b*x^3)^(3/2),x]
Output:
(-2*x^4)/(b*Sqrt[a*x^2 + b*x^3]) + (6*((2*x*Sqrt[a*x^2 + b*x^3])/(5*b) - ( 4*a*((2*Sqrt[a*x^2 + b*x^3])/(3*b) - (4*a*Sqrt[a*x^2 + b*x^3])/(3*b^2*x))) /(5*b)))/b
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*(n - j )*(p + 1))), x] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1))) In t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n} , x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( n - j)], 0] && LtQ[p, -1] && (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 = 0.54 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(\frac {2 \left (b x +a \right ) \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) x^{3}}{5 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(56\) |
default | \(\frac {2 \left (b x +a \right ) \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) x^{3}}{5 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(56\) |
orering | \(\frac {2 \left (b x +a \right ) \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) x^{3}}{5 b^{4} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(56\) |
trager | \(\frac {2 \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{5 \left (b x +a \right ) b^{4} x}\) | \(58\) |
risch | \(\frac {2 \left (b^{2} x^{2}-3 a b x +11 a^{2}\right ) \left (b x +a \right ) x}{5 b^{4} \sqrt {x^{2} \left (b x +a \right )}}+\frac {2 a^{3} x}{b^{4} \sqrt {x^{2} \left (b x +a \right )}}\) | \(62\) |
pseudoelliptic | \(\frac {\frac {2}{11} b^{6} x^{6}-\frac {8}{33} a \,b^{5} x^{5}+\frac {80}{231} a^{2} b^{4} x^{4}-\frac {128}{231} a^{3} b^{3} x^{3}+\frac {256}{231} a^{4} b^{2} x^{2}-\frac {1024}{231} a^{5} b x -\frac {2048}{231} a^{6}}{b^{7} \sqrt {b x +a}}\) | \(76\) |
Input:
int(x^6/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5*(b*x+a)*(b^3*x^3-2*a*b^2*x^2+8*a^2*b*x+16*a^3)*x^3/b^4/(b*x^3+a*x^2)^( 3/2)
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.61 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{5 \, {\left (b^{5} x^{2} + a b^{4} x\right )}} \] Input:
integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
2/5*(b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x + 16*a^3)*sqrt(b*x^3 + a*x^2)/(b^5* x^2 + a*b^4*x)
\[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x^{6}}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**6/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(x**6/(x**2*(a + b*x))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.42 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )}}{5 \, \sqrt {b x + a} b^{4}} \] Input:
integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
2/5*(b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x + 16*a^3)/(sqrt(b*x + a)*b^4)
Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {32 \, a^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{5 \, b^{4}} + \frac {2 \, a^{3}}{\sqrt {b x + a} b^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} b^{16} - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{16} + 15 \, \sqrt {b x + a} a^{2} b^{16}\right )}}{5 \, b^{20} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
-32/5*a^(5/2)*sgn(x)/b^4 + 2*a^3/(sqrt(b*x + a)*b^4*sgn(x)) + 2/5*((b*x + a)^(5/2)*b^16 - 5*(b*x + a)^(3/2)*a*b^16 + 15*sqrt(b*x + a)*a^2*b^16)/(b^2 0*sgn(x))
Time = 9.46 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (16\,a^3+8\,a^2\,b\,x-2\,a\,b^2\,x^2+b^3\,x^3\right )}{5\,b^4\,x\,\left (a+b\,x\right )} \] Input:
int(x^6/(a*x^2 + b*x^3)^(3/2),x)
Output:
(2*(a*x^2 + b*x^3)^(1/2)*(16*a^3 + b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x))/(5* b^4*x*(a + b*x))
Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.43 \[ \int \frac {x^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\frac {2}{5} b^{3} x^{3}-\frac {4}{5} a \,b^{2} x^{2}+\frac {16}{5} a^{2} b x +\frac {32}{5} a^{3}}{\sqrt {b x +a}\, b^{4}} \] Input:
int(x^6/(b*x^3+a*x^2)^(3/2),x)
Output:
(2*(16*a**3 + 8*a**2*b*x - 2*a*b**2*x**2 + b**3*x**3))/(5*sqrt(a + b*x)*b* *4)