Integrand size = 17, antiderivative size = 100 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=-\frac {32 b^3 \left (a+\frac {b}{x}\right )^{7/2} x^{7/2}}{3003 a^4}+\frac {16 b^2 \left (a+\frac {b}{x}\right )^{7/2} x^{9/2}}{429 a^3}-\frac {12 b \left (a+\frac {b}{x}\right )^{7/2} x^{11/2}}{143 a^2}+\frac {2 \left (a+\frac {b}{x}\right )^{7/2} x^{13/2}}{13 a} \] Output:
-32/3003*b^3*(a+b/x)^(7/2)*x^(7/2)/a^4+16/429*b^2*(a+b/x)^(7/2)*x^(9/2)/a^ 3-12/143*b*(a+b/x)^(7/2)*x^(11/2)/a^2+2/13*(a+b/x)^(7/2)*x^(13/2)/a
Time = 5.39 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.60 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=\frac {2 \sqrt {a+\frac {b}{x}} \sqrt {x} (b+a x)^3 \left (-16 b^3+56 a b^2 x-126 a^2 b x^2+231 a^3 x^3\right )}{3003 a^4} \] Input:
Integrate[(a + b/x)^(5/2)*x^(11/2),x]
Output:
(2*Sqrt[a + b/x]*Sqrt[x]*(b + a*x)^3*(-16*b^3 + 56*a*b^2*x - 126*a^2*b*x^2 + 231*a^3*x^3))/(3003*a^4)
Time = 0.35 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {803, 803, 803, 796}
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.
\(\displaystyle \int x^{11/2} \left (a+\frac {b}{x}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{13/2} \left (a+\frac {b}{x}\right )^{7/2}}{13 a}-\frac {6 b \int \left (a+\frac {b}{x}\right )^{5/2} x^{9/2}dx}{13 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{13/2} \left (a+\frac {b}{x}\right )^{7/2}}{13 a}-\frac {6 b \left (\frac {2 x^{11/2} \left (a+\frac {b}{x}\right )^{7/2}}{11 a}-\frac {4 b \int \left (a+\frac {b}{x}\right )^{5/2} x^{7/2}dx}{11 a}\right )}{13 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{13/2} \left (a+\frac {b}{x}\right )^{7/2}}{13 a}-\frac {6 b \left (\frac {2 x^{11/2} \left (a+\frac {b}{x}\right )^{7/2}}{11 a}-\frac {4 b \left (\frac {2 x^{9/2} \left (a+\frac {b}{x}\right )^{7/2}}{9 a}-\frac {2 b \int \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}dx}{9 a}\right )}{11 a}\right )}{13 a}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {2 x^{13/2} \left (a+\frac {b}{x}\right )^{7/2}}{13 a}-\frac {6 b \left (\frac {2 x^{11/2} \left (a+\frac {b}{x}\right )^{7/2}}{11 a}-\frac {4 b \left (\frac {2 x^{9/2} \left (a+\frac {b}{x}\right )^{7/2}}{9 a}-\frac {4 b x^{7/2} \left (a+\frac {b}{x}\right )^{7/2}}{63 a^2}\right )}{11 a}\right )}{13 a}\) |
Input:
Int[(a + b/x)^(5/2)*x^(11/2),x]
Output:
(2*(a + b/x)^(7/2)*x^(13/2))/(13*a) - (6*b*((2*(a + b/x)^(7/2)*x^(11/2))/( 11*a) - (4*b*((-4*b*(a + b/x)^(7/2)*x^(7/2))/(63*a^2) + (2*(a + b/x)^(7/2) *x^(9/2))/(9*a)))/(11*a)))/(13*a)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.53
method | result | size |
orering | \(\frac {2 \left (231 a^{3} x^{3}-126 a^{2} b \,x^{2}+56 a \,b^{2} x -16 b^{3}\right ) x^{\frac {5}{2}} \left (a x +b \right ) \left (a +\frac {b}{x}\right )^{\frac {5}{2}}}{3003 a^{4}}\) | \(53\) |
gosper | \(\frac {2 \left (a x +b \right ) \left (231 a^{3} x^{3}-126 a^{2} b \,x^{2}+56 a \,b^{2} x -16 b^{3}\right ) x^{\frac {5}{2}} \left (\frac {a x +b}{x}\right )^{\frac {5}{2}}}{3003 a^{4}}\) | \(55\) |
default | \(\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (a x +b \right )^{3} \left (231 a^{3} x^{3}-126 a^{2} b \,x^{2}+56 a \,b^{2} x -16 b^{3}\right )}{3003 a^{4}}\) | \(57\) |
risch | \(\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (231 a^{6} x^{6}+567 a^{5} b \,x^{5}+371 a^{4} b^{2} x^{4}+5 a^{3} x^{3} b^{3}-6 b^{4} x^{2} a^{2}+8 b^{5} x a -16 b^{6}\right )}{3003 a^{4}}\) | \(83\) |
Input:
int((a+b/x)^(5/2)*x^(11/2),x,method=_RETURNVERBOSE)
Output:
2/3003*(231*a^3*x^3-126*a^2*b*x^2+56*a*b^2*x-16*b^3)/a^4*x^(5/2)*(a*x+b)*( a+b/x)^(5/2)
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=\frac {2 \, {\left (231 \, a^{6} x^{6} + 567 \, a^{5} b x^{5} + 371 \, a^{4} b^{2} x^{4} + 5 \, a^{3} b^{3} x^{3} - 6 \, a^{2} b^{4} x^{2} + 8 \, a b^{5} x - 16 \, b^{6}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{3003 \, a^{4}} \] Input:
integrate((a+b/x)^(5/2)*x^(11/2),x, algorithm="fricas")
Output:
2/3003*(231*a^6*x^6 + 567*a^5*b*x^5 + 371*a^4*b^2*x^4 + 5*a^3*b^3*x^3 - 6* a^2*b^4*x^2 + 8*a*b^5*x - 16*b^6)*sqrt(x)*sqrt((a*x + b)/x)/a^4
Timed out. \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=\text {Timed out} \] Input:
integrate((a+b/x)**(5/2)*x**(11/2),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.69 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=\frac {2 \, {\left (231 \, {\left (a + \frac {b}{x}\right )}^{\frac {13}{2}} x^{\frac {13}{2}} - 819 \, {\left (a + \frac {b}{x}\right )}^{\frac {11}{2}} b x^{\frac {11}{2}} + 1001 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{2}} b^{2} x^{\frac {9}{2}} - 429 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{3} x^{\frac {7}{2}}\right )}}{3003 \, a^{4}} \] Input:
integrate((a+b/x)^(5/2)*x^(11/2),x, algorithm="maxima")
Output:
2/3003*(231*(a + b/x)^(13/2)*x^(13/2) - 819*(a + b/x)^(11/2)*b*x^(11/2) + 1001*(a + b/x)^(9/2)*b^2*x^(9/2) - 429*(a + b/x)^(7/2)*b^3*x^(7/2))/a^4
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (76) = 152\).
Time = 0.13 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.00 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=\frac {32 \, b^{\frac {13}{2}} \mathrm {sgn}\left (x\right )}{3003 \, a^{4}} + \frac {2 \, {\left (\frac {429 \, {\left (5 \, {\left (a x + b\right )}^{\frac {7}{2}} - 21 \, {\left (a x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{2} - 35 \, \sqrt {a x + b} b^{3}\right )} b^{3} \mathrm {sgn}\left (x\right )}{a^{3}} + \frac {143 \, {\left (35 \, {\left (a x + b\right )}^{\frac {9}{2}} - 180 \, {\left (a x + b\right )}^{\frac {7}{2}} b + 378 \, {\left (a x + b\right )}^{\frac {5}{2}} b^{2} - 420 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{3} + 315 \, \sqrt {a x + b} b^{4}\right )} b^{2} \mathrm {sgn}\left (x\right )}{a^{3}} + \frac {65 \, {\left (63 \, {\left (a x + b\right )}^{\frac {11}{2}} - 385 \, {\left (a x + b\right )}^{\frac {9}{2}} b + 990 \, {\left (a x + b\right )}^{\frac {7}{2}} b^{2} - 1386 \, {\left (a x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{4} - 693 \, \sqrt {a x + b} b^{5}\right )} b \mathrm {sgn}\left (x\right )}{a^{3}} + \frac {5 \, {\left (231 \, {\left (a x + b\right )}^{\frac {13}{2}} - 1638 \, {\left (a x + b\right )}^{\frac {11}{2}} b + 5005 \, {\left (a x + b\right )}^{\frac {9}{2}} b^{2} - 8580 \, {\left (a x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (a x + b\right )}^{\frac {5}{2}} b^{4} - 6006 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{5} + 3003 \, \sqrt {a x + b} b^{6}\right )} \mathrm {sgn}\left (x\right )}{a^{3}}\right )}}{15015 \, a} \] Input:
integrate((a+b/x)^(5/2)*x^(11/2),x, algorithm="giac")
Output:
32/3003*b^(13/2)*sgn(x)/a^4 + 2/15015*(429*(5*(a*x + b)^(7/2) - 21*(a*x + b)^(5/2)*b + 35*(a*x + b)^(3/2)*b^2 - 35*sqrt(a*x + b)*b^3)*b^3*sgn(x)/a^3 + 143*(35*(a*x + b)^(9/2) - 180*(a*x + b)^(7/2)*b + 378*(a*x + b)^(5/2)*b ^2 - 420*(a*x + b)^(3/2)*b^3 + 315*sqrt(a*x + b)*b^4)*b^2*sgn(x)/a^3 + 65* (63*(a*x + b)^(11/2) - 385*(a*x + b)^(9/2)*b + 990*(a*x + b)^(7/2)*b^2 - 1 386*(a*x + b)^(5/2)*b^3 + 1155*(a*x + b)^(3/2)*b^4 - 693*sqrt(a*x + b)*b^5 )*b*sgn(x)/a^3 + 5*(231*(a*x + b)^(13/2) - 1638*(a*x + b)^(11/2)*b + 5005* (a*x + b)^(9/2)*b^2 - 8580*(a*x + b)^(7/2)*b^3 + 9009*(a*x + b)^(5/2)*b^4 - 6006*(a*x + b)^(3/2)*b^5 + 3003*sqrt(a*x + b)*b^6)*sgn(x)/a^3)/a
Time = 0.77 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=\sqrt {a+\frac {b}{x}}\,\left (\frac {2\,a^2\,x^{13/2}}{13}+\frac {106\,b^2\,x^{9/2}}{429}+\frac {10\,b^3\,x^{7/2}}{3003\,a}-\frac {4\,b^4\,x^{5/2}}{1001\,a^2}+\frac {16\,b^5\,x^{3/2}}{3003\,a^3}-\frac {32\,b^6\,\sqrt {x}}{3003\,a^4}+\frac {54\,a\,b\,x^{11/2}}{143}\right ) \] Input:
int(x^(11/2)*(a + b/x)^(5/2),x)
Output:
(a + b/x)^(1/2)*((2*a^2*x^(13/2))/13 + (106*b^2*x^(9/2))/429 + (10*b^3*x^( 7/2))/(3003*a) - (4*b^4*x^(5/2))/(1001*a^2) + (16*b^5*x^(3/2))/(3003*a^3) - (32*b^6*x^(1/2))/(3003*a^4) + (54*a*b*x^(11/2))/143)
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^{11/2} \, dx=\frac {2 \sqrt {a x +b}\, \left (231 a^{6} x^{6}+567 a^{5} b \,x^{5}+371 a^{4} b^{2} x^{4}+5 a^{3} b^{3} x^{3}-6 a^{2} b^{4} x^{2}+8 a \,b^{5} x -16 b^{6}\right )}{3003 a^{4}} \] Input:
int((a+b/x)^(5/2)*x^(11/2),x)
Output:
(2*sqrt(a*x + b)*(231*a**6*x**6 + 567*a**5*b*x**5 + 371*a**4*b**2*x**4 + 5 *a**3*b**3*x**3 - 6*a**2*b**4*x**2 + 8*a*b**5*x - 16*b**6))/(3003*a**4)