Integrand size = 15, antiderivative size = 92 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=-\frac {2 (a+b x)^{3/2}}{9 a x^{9/2}}+\frac {4 b (a+b x)^{3/2}}{21 a^2 x^{7/2}}-\frac {16 b^2 (a+b x)^{3/2}}{105 a^3 x^{5/2}}+\frac {32 b^3 (a+b x)^{3/2}}{315 a^4 x^{3/2}} \] Output:
-2/9*(b*x+a)^(3/2)/a/x^(9/2)+4/21*b*(b*x+a)^(3/2)/a^2/x^(7/2)-16/105*b^2*( b*x+a)^(3/2)/a^3/x^(5/2)+32/315*b^3*(b*x+a)^(3/2)/a^4/x^(3/2)
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=-\frac {2 \sqrt {a+b x} \left (35 a^4+5 a^3 b x-6 a^2 b^2 x^2+8 a b^3 x^3-16 b^4 x^4\right )}{315 a^4 x^{9/2}} \] Input:
Integrate[Sqrt[a + b*x]/x^(11/2),x]
Output:
(-2*Sqrt[a + b*x]*(35*a^4 + 5*a^3*b*x - 6*a^2*b^2*x^2 + 8*a*b^3*x^3 - 16*b ^4*x^4))/(315*a^4*x^(9/2))
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.
\(\displaystyle \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {2 b \int \frac {\sqrt {a+b x}}{x^{9/2}}dx}{3 a}-\frac {2 (a+b x)^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {2 b \left (-\frac {4 b \int \frac {\sqrt {a+b x}}{x^{7/2}}dx}{7 a}-\frac {2 (a+b x)^{3/2}}{7 a x^{7/2}}\right )}{3 a}-\frac {2 (a+b x)^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {2 b \left (-\frac {4 b \left (-\frac {2 b \int \frac {\sqrt {a+b x}}{x^{5/2}}dx}{5 a}-\frac {2 (a+b x)^{3/2}}{5 a x^{5/2}}\right )}{7 a}-\frac {2 (a+b x)^{3/2}}{7 a x^{7/2}}\right )}{3 a}-\frac {2 (a+b x)^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {2 b \left (-\frac {4 b \left (\frac {4 b (a+b x)^{3/2}}{15 a^2 x^{3/2}}-\frac {2 (a+b x)^{3/2}}{5 a x^{5/2}}\right )}{7 a}-\frac {2 (a+b x)^{3/2}}{7 a x^{7/2}}\right )}{3 a}-\frac {2 (a+b x)^{3/2}}{9 a x^{9/2}}\) |
Input:
Int[Sqrt[a + b*x]/x^(11/2),x]
Output:
(-2*(a + b*x)^(3/2))/(9*a*x^(9/2)) - (2*b*((-2*(a + b*x)^(3/2))/(7*a*x^(7/ 2)) - (4*b*((-2*(a + b*x)^(3/2))/(5*a*x^(5/2)) + (4*b*(a + b*x)^(3/2))/(15 *a^2*x^(3/2))))/(7*a)))/(3*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 = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-16 b^{3} x^{3}+24 a \,b^{2} x^{2}-30 a^{2} b x +35 a^{3}\right )}{315 x^{\frac {9}{2}} a^{4}}\) | \(46\) |
orering | \(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-16 b^{3} x^{3}+24 a \,b^{2} x^{2}-30 a^{2} b x +35 a^{3}\right )}{315 x^{\frac {9}{2}} a^{4}}\) | \(46\) |
risch | \(-\frac {2 \sqrt {b x +a}\, \left (-16 b^{4} x^{4}+8 a \,x^{3} b^{3}-6 a^{2} b^{2} x^{2}+5 a^{3} b x +35 a^{4}\right )}{315 x^{\frac {9}{2}} a^{4}}\) | \(57\) |
default | \(-\frac {\sqrt {b x +a}}{4 x^{\frac {9}{2}}}-\frac {a \left (-\frac {2 \sqrt {b x +a}}{9 a \,x^{\frac {9}{2}}}-\frac {8 b \left (-\frac {2 \sqrt {b x +a}}{7 a \,x^{\frac {7}{2}}}-\frac {6 b \left (-\frac {2 \sqrt {b x +a}}{5 a \,x^{\frac {5}{2}}}-\frac {4 b \left (-\frac {2 \sqrt {b x +a}}{3 a \,x^{\frac {3}{2}}}+\frac {4 b \sqrt {b x +a}}{3 a^{2} \sqrt {x}}\right )}{5 a}\right )}{7 a}\right )}{9 a}\right )}{8}\) | \(115\) |
Input:
int((b*x+a)^(1/2)/x^(11/2),x,method=_RETURNVERBOSE)
Output:
-2/315*(b*x+a)^(3/2)*(-16*b^3*x^3+24*a*b^2*x^2-30*a^2*b*x+35*a^3)/x^(9/2)/ a^4
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=\frac {2 \, {\left (16 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{4} x^{\frac {9}{2}}} \] Input:
integrate((b*x+a)^(1/2)/x^(11/2),x, algorithm="fricas")
Output:
2/315*(16*b^4*x^4 - 8*a*b^3*x^3 + 6*a^2*b^2*x^2 - 5*a^3*b*x - 35*a^4)*sqrt (b*x + a)/(a^4*x^(9/2))
Leaf count of result is larger than twice the leaf count of optimal. 559 vs. \(2 (87) = 174\).
Time = 15.82 (sec) , antiderivative size = 559, normalized size of antiderivative = 6.08 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=- \frac {70 a^{7} b^{\frac {19}{2}} \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} - \frac {220 a^{6} b^{\frac {21}{2}} x \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} - \frac {228 a^{5} b^{\frac {23}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} - \frac {80 a^{4} b^{\frac {25}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} + \frac {10 a^{3} b^{\frac {27}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} + \frac {60 a^{2} b^{\frac {29}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} + \frac {80 a b^{\frac {31}{2}} x^{6} \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} + \frac {32 b^{\frac {33}{2}} x^{7} \sqrt {\frac {a}{b x} + 1}}{315 a^{7} b^{9} x^{4} + 945 a^{6} b^{10} x^{5} + 945 a^{5} b^{11} x^{6} + 315 a^{4} b^{12} x^{7}} \] Input:
integrate((b*x+a)**(1/2)/x**(11/2),x)
Output:
-70*a**7*b**(19/2)*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10* x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7) - 220*a**6*b**(21/2)*x*s qrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**1 1*x**6 + 315*a**4*b**12*x**7) - 228*a**5*b**(23/2)*x**2*sqrt(a/(b*x) + 1)/ (315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4 *b**12*x**7) - 80*a**4*b**(25/2)*x**3*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x** 4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7) + 10* a**3*b**(27/2)*x**4*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10 *x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7) + 60*a**2*b**(29/2)*x** 5*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b **11*x**6 + 315*a**4*b**12*x**7) + 80*a*b**(31/2)*x**6*sqrt(a/(b*x) + 1)/( 315*a**7*b**9*x**4 + 945*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4* b**12*x**7) + 32*b**(33/2)*x**7*sqrt(a/(b*x) + 1)/(315*a**7*b**9*x**4 + 94 5*a**6*b**10*x**5 + 945*a**5*b**11*x**6 + 315*a**4*b**12*x**7)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=\frac {2 \, {\left (\frac {105 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3}}{x^{\frac {3}{2}}} - \frac {189 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{2}}{x^{\frac {5}{2}}} + \frac {135 \, {\left (b x + a\right )}^{\frac {7}{2}} b}{x^{\frac {7}{2}}} - \frac {35 \, {\left (b x + a\right )}^{\frac {9}{2}}}{x^{\frac {9}{2}}}\right )}}{315 \, a^{4}} \] Input:
integrate((b*x+a)^(1/2)/x^(11/2),x, algorithm="maxima")
Output:
2/315*(105*(b*x + a)^(3/2)*b^3/x^(3/2) - 189*(b*x + a)^(5/2)*b^2/x^(5/2) + 135*(b*x + a)^(7/2)*b/x^(7/2) - 35*(b*x + a)^(9/2)/x^(9/2))/a^4
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=-\frac {2 \, {\left (\frac {105 \, b^{9}}{a} - 2 \, {\left (\frac {63 \, b^{9}}{a^{2}} + 4 \, {\left (\frac {2 \, {\left (b x + a\right )} b^{9}}{a^{4}} - \frac {9 \, b^{9}}{a^{3}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {3}{2}} b}{315 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {9}{2}} {\left | b \right |}} \] Input:
integrate((b*x+a)^(1/2)/x^(11/2),x, algorithm="giac")
Output:
-2/315*(105*b^9/a - 2*(63*b^9/a^2 + 4*(2*(b*x + a)*b^9/a^4 - 9*b^9/a^3)*(b *x + a))*(b*x + a))*(b*x + a)^(3/2)*b/(((b*x + a)*b - a*b)^(9/2)*abs(b))
Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {16\,b^3\,x^3}{315\,a^3}-\frac {4\,b^2\,x^2}{105\,a^2}-\frac {32\,b^4\,x^4}{315\,a^4}+\frac {2\,b\,x}{63\,a}+\frac {2}{9}\right )}{x^{9/2}} \] Input:
int((a + b*x)^(1/2)/x^(11/2),x)
Output:
-((a + b*x)^(1/2)*((16*b^3*x^3)/(315*a^3) - (4*b^2*x^2)/(105*a^2) - (32*b^ 4*x^4)/(315*a^4) + (2*b*x)/(63*a) + 2/9))/x^(9/2)
Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a+b x}}{x^{11/2}} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{4}}{9}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b x}{63}+\frac {4 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} x^{2}}{105}-\frac {16 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} x^{3}}{315}+\frac {32 \sqrt {x}\, \sqrt {b x +a}\, b^{4} x^{4}}{315}-\frac {32 \sqrt {b}\, b^{4} x^{5}}{315}}{a^{4} x^{5}} \] Input:
int((b*x+a)^(1/2)/x^(11/2),x)
Output:
(2*( - 35*sqrt(x)*sqrt(a + b*x)*a**4 - 5*sqrt(x)*sqrt(a + b*x)*a**3*b*x + 6*sqrt(x)*sqrt(a + b*x)*a**2*b**2*x**2 - 8*sqrt(x)*sqrt(a + b*x)*a*b**3*x* *3 + 16*sqrt(x)*sqrt(a + b*x)*b**4*x**4 - 16*sqrt(b)*b**4*x**5))/(315*a**4 *x**5)