Integrand size = 19, antiderivative size = 101 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}} \] Output:
-1/7*x^(19/2)/b/(b*x^3+a*x)^(7/2)-6/35*x^(13/2)/b^2/(b*x^3+a*x)^(5/2)-8/35 *x^(7/2)/b^3/(b*x^3+a*x)^(3/2)-16/35*x^(1/2)/b^4/(b*x^3+a*x)^(1/2)
Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{9/2} \left (a+b x^2\right ) \left (-16 a^3-56 a^2 b x^2-70 a b^2 x^4-35 b^3 x^6\right )}{35 b^4 \left (x \left (a+b x^2\right )\right )^{9/2}} \] Input:
Integrate[x^(23/2)/(a*x + b*x^3)^(9/2),x]
Output:
(x^(9/2)*(a + b*x^2)*(-16*a^3 - 56*a^2*b*x^2 - 70*a*b^2*x^4 - 35*b^3*x^6)) /(35*b^4*(x*(a + b*x^2))^(9/2))
Time = 0.48 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1921, 1921, 1921, 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.
\(\displaystyle \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {6 \int \frac {x^{17/2}}{\left (b x^3+a x\right )^{7/2}}dx}{7 b}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {6 \left (\frac {4 \int \frac {x^{11/2}}{\left (b x^3+a x\right )^{5/2}}dx}{5 b}-\frac {x^{13/2}}{5 b \left (a x+b x^3\right )^{5/2}}\right )}{7 b}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {6 \left (\frac {4 \left (\frac {2 \int \frac {x^{5/2}}{\left (b x^3+a x\right )^{3/2}}dx}{3 b}-\frac {x^{7/2}}{3 b \left (a x+b x^3\right )^{3/2}}\right )}{5 b}-\frac {x^{13/2}}{5 b \left (a x+b x^3\right )^{5/2}}\right )}{7 b}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {6 \left (\frac {4 \left (-\frac {2 \sqrt {x}}{3 b^2 \sqrt {a x+b x^3}}-\frac {x^{7/2}}{3 b \left (a x+b x^3\right )^{3/2}}\right )}{5 b}-\frac {x^{13/2}}{5 b \left (a x+b x^3\right )^{5/2}}\right )}{7 b}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
Input:
Int[x^(23/2)/(a*x + b*x^3)^(9/2),x]
Output:
-1/7*x^(19/2)/(b*(a*x + b*x^3)^(7/2)) + (6*(-1/5*x^(13/2)/(b*(a*x + b*x^3) ^(5/2)) + (4*(-1/3*x^(7/2)/(b*(a*x + b*x^3)^(3/2)) - (2*Sqrt[x])/(3*b^2*Sq rt[a*x + b*x^3])))/(5*b)))/(7*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])
Time = 0.43 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right ) \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right ) x^{\frac {9}{2}}}{35 b^{4} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\) | \(59\) |
orering | \(-\frac {\left (b \,x^{2}+a \right ) \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right ) x^{\frac {9}{2}}}{35 b^{4} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\) | \(59\) |
default | \(-\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right )}{35 \sqrt {x}\, \left (b \,x^{2}+a \right )^{4} b^{4}}\) | \(61\) |
Input:
int(x^(23/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)
Output:
-1/35*(b*x^2+a)*(35*b^3*x^6+70*a*b^2*x^4+56*a^2*b*x^2+16*a^3)*x^(9/2)/b^4/ (b*x^3+a*x)^(9/2)
Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {{\left (35 \, b^{3} x^{6} + 70 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} + 16 \, a^{3}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (b^{8} x^{9} + 4 \, a b^{7} x^{7} + 6 \, a^{2} b^{6} x^{5} + 4 \, a^{3} b^{5} x^{3} + a^{4} b^{4} x\right )}} \] Input:
integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")
Output:
-1/35*(35*b^3*x^6 + 70*a*b^2*x^4 + 56*a^2*b*x^2 + 16*a^3)*sqrt(b*x^3 + a*x )*sqrt(x)/(b^8*x^9 + 4*a*b^7*x^7 + 6*a^2*b^6*x^5 + 4*a^3*b^5*x^3 + a^4*b^4 *x)
Timed out. \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \] Input:
integrate(x**(23/2)/(b*x**3+a*x)**(9/2),x)
Output:
Timed out
\[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {23}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")
Output:
integrate(x^(23/2)/(b*x^3 + a*x)^(9/2), x)
Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.54 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {35 \, {\left (b x^{2} + a\right )}^{3} - 35 \, {\left (b x^{2} + a\right )}^{2} a + 21 \, {\left (b x^{2} + a\right )} a^{2} - 5 \, a^{3}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}} \] Input:
integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")
Output:
-1/35*(35*(b*x^2 + a)^3 - 35*(b*x^2 + a)^2*a + 21*(b*x^2 + a)*a^2 - 5*a^3) /((b*x^2 + a)^(7/2)*b^4)
Timed out. \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{23/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \] Input:
int(x^(23/2)/(a*x + b*x^3)^(9/2),x)
Output:
int(x^(23/2)/(a*x + b*x^3)^(9/2), x)
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.86 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {\sqrt {b \,x^{2}+a}\, \left (-35 b^{3} x^{6}-70 a \,b^{2} x^{4}-56 a^{2} b \,x^{2}-16 a^{3}\right )}{35 b^{4} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(x^(23/2)/(b*x^3+a*x)^(9/2),x)
Output:
(sqrt(a + b*x**2)*( - 16*a**3 - 56*a**2*b*x**2 - 70*a*b**2*x**4 - 35*b**3* x**6))/(35*b**4*(a**4 + 4*a**3*b*x**2 + 6*a**2*b**2*x**4 + 4*a*b**3*x**6 + b**4*x**8))