Integrand size = 19, antiderivative size = 76 \[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {4 x^{11/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 x^{9/2}}{105 a^3 \left (a x+b x^3\right )^{3/2}} \] Output:
1/7*x^(13/2)/a/(b*x^3+a*x)^(7/2)+4/35*x^(11/2)/a^2/(b*x^3+a*x)^(5/2)+8/105 *x^(9/2)/a^3/(b*x^3+a*x)^(3/2)
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{9/2} \left (a+b x^2\right ) \left (35 a^2 x^3+28 a b x^5+8 b^2 x^7\right )}{105 a^3 \left (x \left (a+b x^2\right )\right )^{9/2}} \] Input:
Integrate[x^(13/2)/(a*x + b*x^3)^(9/2),x]
Output:
(x^(9/2)*(a + b*x^2)*(35*a^2*x^3 + 28*a*b*x^5 + 8*b^2*x^7))/(105*a^3*(x*(a + b*x^2))^(9/2))
Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.
\(\displaystyle \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {4 \int \frac {x^{11/2}}{\left (b x^3+a x\right )^{7/2}}dx}{7 a}+\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {4 \left (\frac {2 \int \frac {x^{9/2}}{\left (b x^3+a x\right )^{5/2}}dx}{5 a}+\frac {x^{11/2}}{5 a \left (a x+b x^3\right )^{5/2}}\right )}{7 a}+\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {4 \left (\frac {2 x^{9/2}}{15 a^2 \left (a x+b x^3\right )^{3/2}}+\frac {x^{11/2}}{5 a \left (a x+b x^3\right )^{5/2}}\right )}{7 a}+\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}\) |
Input:
Int[x^(13/2)/(a*x + b*x^3)^(9/2),x]
Output:
x^(13/2)/(7*a*(a*x + b*x^3)^(7/2)) + (4*(x^(11/2)/(5*a*(a*x + b*x^3)^(5/2) ) + (2*x^(9/2))/(15*a^2*(a*x + b*x^3)^(3/2))))/(7*a)
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 = 0.42 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {\left (b \,x^{2}+a \right ) x^{\frac {15}{2}} \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 a^{3} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\) | \(48\) |
orering | \(\frac {\left (b \,x^{2}+a \right ) x^{\frac {15}{2}} \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 a^{3} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\) | \(48\) |
default | \(\frac {x^{\frac {5}{2}} \sqrt {x \left (b \,x^{2}+a \right )}\, \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 a^{3} \left (b \,x^{2}+a \right )^{4}}\) | \(50\) |
Input:
int(x^(13/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/105*(b*x^2+a)*x^(15/2)*(8*b^2*x^4+28*a*b*x^2+35*a^2)/a^3/(b*x^3+a*x)^(9/ 2)
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.14 \[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {{\left (8 \, b^{2} x^{6} + 28 \, a b x^{4} + 35 \, a^{2} x^{2}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{105 \, {\left (a^{3} b^{4} x^{8} + 4 \, a^{4} b^{3} x^{6} + 6 \, a^{5} b^{2} x^{4} + 4 \, a^{6} b x^{2} + a^{7}\right )}} \] Input:
integrate(x^(13/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")
Output:
1/105*(8*b^2*x^6 + 28*a*b*x^4 + 35*a^2*x^2)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^3 *b^4*x^8 + 4*a^4*b^3*x^6 + 6*a^5*b^2*x^4 + 4*a^6*b*x^2 + a^7)
\[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{\frac {13}{2}}}{\left (x \left (a + b x^{2}\right )\right )^{\frac {9}{2}}}\, dx \] Input:
integrate(x**(13/2)/(b*x**3+a*x)**(9/2),x)
Output:
Integral(x**(13/2)/(x*(a + b*x**2))**(9/2), x)
\[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {13}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate(x^(13/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")
Output:
integrate(x^(13/2)/(b*x^3 + a*x)^(9/2), x)
Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.57 \[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {{\left (4 \, x^{2} {\left (\frac {2 \, b^{2} x^{2}}{a^{3}} + \frac {7 \, b}{a^{2}}\right )} + \frac {35}{a}\right )} x^{3}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \] Input:
integrate(x^(13/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")
Output:
1/105*(4*x^2*(2*b^2*x^2/a^3 + 7*b/a^2) + 35/a)*x^3/(b*x^2 + a)^(7/2)
Timed out. \[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{13/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \] Input:
int(x^(13/2)/(a*x + b*x^3)^(9/2),x)
Output:
int(x^(13/2)/(a*x + b*x^3)^(9/2), x)
Time = 0.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.04 \[ \int \frac {x^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {35 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{3}+28 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{5}+8 \sqrt {b \,x^{2}+a}\, b^{4} x^{7}-8 \sqrt {b}\, a^{4}-32 \sqrt {b}\, a^{3} b \,x^{2}-48 \sqrt {b}\, a^{2} b^{2} x^{4}-32 \sqrt {b}\, a \,b^{3} x^{6}-8 \sqrt {b}\, b^{4} x^{8}}{105 a^{3} b^{2} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(x^(13/2)/(b*x^3+a*x)^(9/2),x)
Output:
(35*sqrt(a + b*x**2)*a**2*b**2*x**3 + 28*sqrt(a + b*x**2)*a*b**3*x**5 + 8* sqrt(a + b*x**2)*b**4*x**7 - 8*sqrt(b)*a**4 - 32*sqrt(b)*a**3*b*x**2 - 48* sqrt(b)*a**2*b**2*x**4 - 32*sqrt(b)*a*b**3*x**6 - 8*sqrt(b)*b**4*x**8)/(10 5*a**3*b**2*(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))