Integrand size = 15, antiderivative size = 92 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}+\frac {12 b \sqrt {a+b x}}{35 a^2 x^{5/2}}-\frac {16 b^2 \sqrt {a+b x}}{35 a^3 x^{3/2}}+\frac {32 b^3 \sqrt {a+b x}}{35 a^4 \sqrt {x}} \] Output:
-2/7*(b*x+a)^(1/2)/a/x^(7/2)+12/35*b*(b*x+a)^(1/2)/a^2/x^(5/2)-16/35*b^2*( b*x+a)^(1/2)/a^3/x^(3/2)+32/35*b^3*(b*x+a)^(1/2)/a^4/x^(1/2)
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=-\frac {2 \sqrt {a+b x} \left (5 a^3-6 a^2 b x+8 a b^2 x^2-16 b^3 x^3\right )}{35 a^4 x^{7/2}} \] Input:
Integrate[1/(x^(9/2)*Sqrt[a + b*x]),x]
Output:
(-2*Sqrt[a + b*x]*(5*a^3 - 6*a^2*b*x + 8*a*b^2*x^2 - 16*b^3*x^3))/(35*a^4* x^(7/2))
Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {55, 55, 55, 48}
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.
\(\displaystyle \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 b \int \frac {1}{x^{7/2} \sqrt {a+b x}}dx}{7 a}-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \int \frac {1}{x^{5/2} \sqrt {a+b x}}dx}{5 a}-\frac {2 \sqrt {a+b x}}{5 a x^{5/2}}\right )}{7 a}-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (-\frac {2 b \int \frac {1}{x^{3/2} \sqrt {a+b x}}dx}{3 a}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{5 a}-\frac {2 \sqrt {a+b x}}{5 a x^{5/2}}\right )}{7 a}-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (\frac {4 b \sqrt {a+b x}}{3 a^2 \sqrt {x}}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{5 a}-\frac {2 \sqrt {a+b x}}{5 a x^{5/2}}\right )}{7 a}-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}\) |
Input:
Int[1/(x^(9/2)*Sqrt[a + b*x]),x]
Output:
(-2*Sqrt[a + b*x])/(7*a*x^(7/2)) - (6*b*((-2*Sqrt[a + b*x])/(5*a*x^(5/2)) - (4*b*((-2*Sqrt[a + b*x])/(3*a*x^(3/2)) + (4*b*Sqrt[a + b*x])/(3*a^2*Sqrt [x])))/(5*a)))/(7*a)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(-\frac {2 \sqrt {b x +a}\, \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{\frac {7}{2}} a^{4}}\) | \(46\) |
risch | \(-\frac {2 \sqrt {b x +a}\, \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{\frac {7}{2}} a^{4}}\) | \(46\) |
orering | \(-\frac {2 \sqrt {b x +a}\, \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{\frac {7}{2}} a^{4}}\) | \(46\) |
default | \(-\frac {2 \sqrt {b x +a}}{7 a \,x^{\frac {7}{2}}}-\frac {6 b \left (-\frac {2 \sqrt {b x +a}}{5 a \,x^{\frac {5}{2}}}-\frac {4 b \left (-\frac {2 \sqrt {b x +a}}{3 a \,x^{\frac {3}{2}}}+\frac {4 b \sqrt {b x +a}}{3 a^{2} \sqrt {x}}\right )}{5 a}\right )}{7 a}\) | \(77\) |
Input:
int(1/x^(9/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/35*(b*x+a)^(1/2)*(-16*b^3*x^3+8*a*b^2*x^2-6*a^2*b*x+5*a^3)/x^(7/2)/a^4
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=\frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x + a}}{35 \, a^{4} x^{\frac {7}{2}}} \] Input:
integrate(1/x^(9/2)/(b*x+a)^(1/2),x, algorithm="fricas")
Output:
2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x + a)/(a^4*x^( 7/2))
Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (87) = 174\).
Time = 6.80 (sec) , antiderivative size = 488, normalized size of antiderivative = 5.30 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=- \frac {10 a^{6} b^{\frac {19}{2}} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} - \frac {18 a^{5} b^{\frac {21}{2}} x \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} - \frac {10 a^{4} b^{\frac {23}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {10 a^{3} b^{\frac {25}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {60 a^{2} b^{\frac {27}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {80 a b^{\frac {29}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {32 b^{\frac {31}{2}} x^{6} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} \] Input:
integrate(1/x**(9/2)/(b*x+a)**(1/2),x)
Output:
-10*a**6*b**(19/2)*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x **4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6) - 18*a**5*b**(21/2)*x*sqrt (a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x* *5 + 35*a**4*b**12*x**6) - 10*a**4*b**(23/2)*x**2*sqrt(a/(b*x) + 1)/(35*a* *7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x **6) + 10*a**3*b**(25/2)*x**3*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a **6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6) + 60*a**2*b**(2 7/2)*x**4*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105 *a**5*b**11*x**5 + 35*a**4*b**12*x**6) + 80*a*b**(29/2)*x**5*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a **4*b**12*x**6) + 32*b**(31/2)*x**6*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6)
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=\frac {2 \, {\left (\frac {35 \, \sqrt {b x + a} b^{3}}{\sqrt {x}} - \frac {35 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} + \frac {21 \, {\left (b x + a\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} - \frac {5 \, {\left (b x + a\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}\right )}}{35 \, a^{4}} \] Input:
integrate(1/x^(9/2)/(b*x+a)^(1/2),x, algorithm="maxima")
Output:
2/35*(35*sqrt(b*x + a)*b^3/sqrt(x) - 35*(b*x + a)^(3/2)*b^2/x^(3/2) + 21*( b*x + a)^(5/2)*b/x^(5/2) - 5*(b*x + a)^(7/2)/x^(7/2))/a^4
Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=\frac {2 \, {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b x + a\right )} b^{3}}{a^{4}} - \frac {7 \, b^{3}}{a^{3}}\right )} + \frac {35 \, b^{3}}{a^{2}}\right )} - \frac {35 \, b^{3}}{a}\right )} \sqrt {b x + a} b^{5}}{35 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}} {\left | b \right |}} \] Input:
integrate(1/x^(9/2)/(b*x+a)^(1/2),x, algorithm="giac")
Output:
2/35*(2*(b*x + a)*(4*(b*x + a)*(2*(b*x + a)*b^3/a^4 - 7*b^3/a^3) + 35*b^3/ a^2) - 35*b^3/a)*sqrt(b*x + a)*b^5/(((b*x + a)*b - a*b)^(7/2)*abs(b))
Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2}{7\,a}+\frac {16\,b^2\,x^2}{35\,a^3}-\frac {32\,b^3\,x^3}{35\,a^4}-\frac {12\,b\,x}{35\,a^2}\right )}{x^{7/2}} \] Input:
int(1/(x^(9/2)*(a + b*x)^(1/2)),x)
Output:
-((a + b*x)^(1/2)*(2/(7*a) + (16*b^2*x^2)/(35*a^3) - (32*b^3*x^3)/(35*a^4) - (12*b*x)/(35*a^2)))/x^(7/2)
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{3}}{7}+\frac {12 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b x}{35}-\frac {16 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} x^{2}}{35}+\frac {32 \sqrt {x}\, \sqrt {b x +a}\, b^{3} x^{3}}{35}-\frac {32 \sqrt {b}\, b^{3} x^{4}}{35}}{a^{4} x^{4}} \] Input:
int(1/x^(9/2)/(b*x+a)^(1/2),x)
Output:
(2*( - 5*sqrt(x)*sqrt(a + b*x)*a**3 + 6*sqrt(x)*sqrt(a + b*x)*a**2*b*x - 8 *sqrt(x)*sqrt(a + b*x)*a*b**2*x**2 + 16*sqrt(x)*sqrt(a + b*x)*b**3*x**3 - 16*sqrt(b)*b**3*x**4))/(35*a**4*x**4)