Integrand size = 19, antiderivative size = 126 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {64 x^{5/2}}{35 b^4 \sqrt {a x+b x^3}}+\frac {128 \sqrt {a x+b x^3}}{35 b^5 \sqrt {x}} \] Output:
-1/7*x^(23/2)/b/(b*x^3+a*x)^(7/2)-8/35*x^(17/2)/b^2/(b*x^3+a*x)^(5/2)-16/3 5*x^(11/2)/b^3/(b*x^3+a*x)^(3/2)-64/35*x^(5/2)/b^4/(b*x^3+a*x)^(1/2)+128/3 5*(b*x^3+a*x)^(1/2)/b^5/x^(1/2)
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.54 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{7/2} \left (128 a^4+448 a^3 b x^2+560 a^2 b^2 x^4+280 a b^3 x^6+35 b^4 x^8\right )}{35 b^5 \left (x \left (a+b x^2\right )\right )^{7/2}} \] Input:
Integrate[x^(27/2)/(a*x + b*x^3)^(9/2),x]
Output:
(x^(7/2)*(128*a^4 + 448*a^3*b*x^2 + 560*a^2*b^2*x^4 + 280*a*b^3*x^6 + 35*b ^4*x^8))/(35*b^5*(x*(a + b*x^2))^(7/2))
Time = 0.58 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1921, 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.
\(\displaystyle \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {8 \int \frac {x^{21/2}}{\left (b x^3+a x\right )^{7/2}}dx}{7 b}-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {8 \left (\frac {6 \int \frac {x^{15/2}}{\left (b x^3+a x\right )^{5/2}}dx}{5 b}-\frac {x^{17/2}}{5 b \left (a x+b x^3\right )^{5/2}}\right )}{7 b}-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {8 \left (\frac {6 \left (\frac {4 \int \frac {x^{9/2}}{\left (b x^3+a x\right )^{3/2}}dx}{3 b}-\frac {x^{11/2}}{3 b \left (a x+b x^3\right )^{3/2}}\right )}{5 b}-\frac {x^{17/2}}{5 b \left (a x+b x^3\right )^{5/2}}\right )}{7 b}-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {8 \left (\frac {6 \left (\frac {4 \left (\frac {2 \int \frac {x^{3/2}}{\sqrt {b x^3+a x}}dx}{b}-\frac {x^{5/2}}{b \sqrt {a x+b x^3}}\right )}{3 b}-\frac {x^{11/2}}{3 b \left (a x+b x^3\right )^{3/2}}\right )}{5 b}-\frac {x^{17/2}}{5 b \left (a x+b x^3\right )^{5/2}}\right )}{7 b}-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {8 \left (\frac {6 \left (\frac {4 \left (\frac {2 \sqrt {a x+b x^3}}{b^2 \sqrt {x}}-\frac {x^{5/2}}{b \sqrt {a x+b x^3}}\right )}{3 b}-\frac {x^{11/2}}{3 b \left (a x+b x^3\right )^{3/2}}\right )}{5 b}-\frac {x^{17/2}}{5 b \left (a x+b x^3\right )^{5/2}}\right )}{7 b}-\frac {x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}\) |
Input:
Int[x^(27/2)/(a*x + b*x^3)^(9/2),x]
Output:
-1/7*x^(23/2)/(b*(a*x + b*x^3)^(7/2)) + (8*(-1/5*x^(17/2)/(b*(a*x + b*x^3) ^(5/2)) + (6*(-1/3*x^(11/2)/(b*(a*x + b*x^3)^(3/2)) + (4*(-(x^(5/2)/(b*Sqr t[a*x + b*x^3])) + (2*Sqrt[a*x + b*x^3])/(b^2*Sqrt[x])))/(3*b)))/(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 = 0.46 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56
method | result | size |
gosper | \(\frac {\left (b \,x^{2}+a \right ) \left (35 b^{4} x^{8}+280 a \,b^{3} x^{6}+560 a^{2} b^{2} x^{4}+448 a^{3} b \,x^{2}+128 a^{4}\right ) x^{\frac {9}{2}}}{35 b^{5} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\) | \(70\) |
orering | \(\frac {\left (b \,x^{2}+a \right ) \left (35 b^{4} x^{8}+280 a \,b^{3} x^{6}+560 a^{2} b^{2} x^{4}+448 a^{3} b \,x^{2}+128 a^{4}\right ) x^{\frac {9}{2}}}{35 b^{5} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\) | \(70\) |
default | \(\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (35 b^{4} x^{8}+280 a \,b^{3} x^{6}+560 a^{2} b^{2} x^{4}+448 a^{3} b \,x^{2}+128 a^{4}\right )}{35 \sqrt {x}\, \left (b \,x^{2}+a \right )^{4} b^{5}}\) | \(72\) |
risch | \(\frac {\left (b \,x^{2}+a \right ) \sqrt {x}}{b^{5} \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {\left (b \,x^{2}+a \right ) \left (140 b^{3} x^{6}+350 a \,b^{2} x^{4}+308 a^{2} b \,x^{2}+93 a^{3}\right ) a \sqrt {x}}{35 b^{5} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right ) \sqrt {x \left (b \,x^{2}+a \right )}}\) | \(128\) |
Input:
int(x^(27/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/35*(b*x^2+a)*(35*b^4*x^8+280*a*b^3*x^6+560*a^2*b^2*x^4+448*a^3*b*x^2+128 *a^4)*x^(9/2)/b^5/(b*x^3+a*x)^(9/2)
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {{\left (35 \, b^{4} x^{8} + 280 \, a b^{3} x^{6} + 560 \, a^{2} b^{2} x^{4} + 448 \, a^{3} b x^{2} + 128 \, a^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (b^{9} x^{9} + 4 \, a b^{8} x^{7} + 6 \, a^{2} b^{7} x^{5} + 4 \, a^{3} b^{6} x^{3} + a^{4} b^{5} x\right )}} \] Input:
integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")
Output:
1/35*(35*b^4*x^8 + 280*a*b^3*x^6 + 560*a^2*b^2*x^4 + 448*a^3*b*x^2 + 128*a ^4)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^9*x^9 + 4*a*b^8*x^7 + 6*a^2*b^7*x^5 + 4*a ^3*b^6*x^3 + a^4*b^5*x)
Timed out. \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \] Input:
integrate(x**(27/2)/(b*x**3+a*x)**(9/2),x)
Output:
Timed out
\[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {27}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")
Output:
integrate(x^(27/2)/(b*x^3 + a*x)^(9/2), x)
Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.61 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {\frac {35 \, \sqrt {b x^{2} + a}}{b} + \frac {140 \, {\left (b x^{2} + a\right )}^{3} a - 70 \, {\left (b x^{2} + a\right )}^{2} a^{2} + 28 \, {\left (b x^{2} + a\right )} a^{3} - 5 \, a^{4}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b}}{35 \, b^{4}} \] Input:
integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")
Output:
1/35*(35*sqrt(b*x^2 + a)/b + (140*(b*x^2 + a)^3*a - 70*(b*x^2 + a)^2*a^2 + 28*(b*x^2 + a)*a^3 - 5*a^4)/((b*x^2 + a)^(7/2)*b))/b^4
Timed out. \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{27/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \] Input:
int(x^(27/2)/(a*x + b*x^3)^(9/2),x)
Output:
int(x^(27/2)/(a*x + b*x^3)^(9/2), x)
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78 \[ \int \frac {x^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {\sqrt {b \,x^{2}+a}\, \left (35 b^{4} x^{8}+280 a \,b^{3} x^{6}+560 a^{2} b^{2} x^{4}+448 a^{3} b \,x^{2}+128 a^{4}\right )}{35 b^{5} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(x^(27/2)/(b*x^3+a*x)^(9/2),x)
Output:
(sqrt(a + b*x**2)*(128*a**4 + 448*a**3*b*x**2 + 560*a**2*b**2*x**4 + 280*a *b**3*x**6 + 35*b**4*x**8))/(35*b**5*(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))