Integrand size = 15, antiderivative size = 200 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx=-\frac {a^{15}}{2 x^2}-\frac {9 a^{14} b}{x^{5/3}}-\frac {315 a^{13} b^2}{4 x^{4/3}}-\frac {455 a^{12} b^3}{x}-\frac {4095 a^{11} b^4}{2 x^{2/3}}-\frac {9009 a^{10} b^5}{\sqrt [3]{x}}+19305 a^8 b^7 \sqrt [3]{x}+\frac {19305}{2} a^7 b^8 x^{2/3}+5005 a^6 b^9 x+\frac {9009}{4} a^5 b^{10} x^{4/3}+819 a^4 b^{11} x^{5/3}+\frac {455}{2} a^3 b^{12} x^2+45 a^2 b^{13} x^{7/3}+\frac {45}{8} a b^{14} x^{8/3}+\frac {b^{15} x^3}{3}+5005 a^9 b^6 \log (x) \]
-1/2*a^15/x^2-9*a^14*b/x^(5/3)-315/4*a^13*b^2/x^(4/3)-455*a^12*b^3/x-4095/ 2*a^11*b^4/x^(2/3)-9009*a^10*b^5/x^(1/3)+19305*a^8*b^7*x^(1/3)+19305/2*a^7 *b^8*x^(2/3)+5005*a^6*b^9*x+9009/4*a^5*b^10*x^(4/3)+819*a^4*b^11*x^(5/3)+4 55/2*a^3*b^12*x^2+45*a^2*b^13*x^(7/3)+45/8*a*b^14*x^(8/3)+1/3*b^15*x^3+500 5*a^9*b^6*ln(x)
Time = 0.12 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx=\frac {-12 a^{15}-216 a^{14} b \sqrt [3]{x}-1890 a^{13} b^2 x^{2/3}-10920 a^{12} b^3 x-49140 a^{11} b^4 x^{4/3}-216216 a^{10} b^5 x^{5/3}+463320 a^8 b^7 x^{7/3}+231660 a^7 b^8 x^{8/3}+120120 a^6 b^9 x^3+54054 a^5 b^{10} x^{10/3}+19656 a^4 b^{11} x^{11/3}+5460 a^3 b^{12} x^4+1080 a^2 b^{13} x^{13/3}+135 a b^{14} x^{14/3}+8 b^{15} x^5}{24 x^2}+15015 a^9 b^6 \log \left (\sqrt [3]{x}\right ) \]
(-12*a^15 - 216*a^14*b*x^(1/3) - 1890*a^13*b^2*x^(2/3) - 10920*a^12*b^3*x - 49140*a^11*b^4*x^(4/3) - 216216*a^10*b^5*x^(5/3) + 463320*a^8*b^7*x^(7/3 ) + 231660*a^7*b^8*x^(8/3) + 120120*a^6*b^9*x^3 + 54054*a^5*b^10*x^(10/3) + 19656*a^4*b^11*x^(11/3) + 5460*a^3*b^12*x^4 + 1080*a^2*b^13*x^(13/3) + 1 35*a*b^14*x^(14/3) + 8*b^15*x^5)/(24*x^2) + 15015*a^9*b^6*Log[x^(1/3)]
Time = 0.34 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {798, 49, 2009}
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.
| \(\displaystyle \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle 3 \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^{7/3}}d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle 3 \int \left (\frac {a^{15}}{x^{7/3}}+\frac {15 b a^{14}}{x^2}+\frac {105 b^2 a^{13}}{x^{5/3}}+\frac {455 b^3 a^{12}}{x^{4/3}}+\frac {1365 b^4 a^{11}}{x}+\frac {3003 b^5 a^{10}}{x^{2/3}}+\frac {5005 b^6 a^9}{\sqrt [3]{x}}+6435 b^7 a^8+6435 b^8 \sqrt [3]{x} a^7+5005 b^9 x^{2/3} a^6+3003 b^{10} x a^5+1365 b^{11} x^{4/3} a^4+455 b^{12} x^{5/3} a^3+105 b^{13} x^2 a^2+15 b^{14} x^{7/3} a+b^{15} x^{8/3}\right )d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \left (-\frac {a^{15}}{6 x^2}-\frac {3 a^{14} b}{x^{5/3}}-\frac {105 a^{13} b^2}{4 x^{4/3}}-\frac {455 a^{12} b^3}{3 x}-\frac {1365 a^{11} b^4}{2 x^{2/3}}-\frac {3003 a^{10} b^5}{\sqrt [3]{x}}+5005 a^9 b^6 \log \left (\sqrt [3]{x}\right )+6435 a^8 b^7 \sqrt [3]{x}+\frac {6435}{2} a^7 b^8 x^{2/3}+\frac {5005}{3} a^6 b^9 x+\frac {3003}{4} a^5 b^{10} x^{4/3}+273 a^4 b^{11} x^{5/3}+\frac {455}{6} a^3 b^{12} x^2+15 a^2 b^{13} x^{7/3}+\frac {15}{8} a b^{14} x^{8/3}+\frac {b^{15} x^3}{9}\right )\) |
3*(-1/6*a^15/x^2 - (3*a^14*b)/x^(5/3) - (105*a^13*b^2)/(4*x^(4/3)) - (455* a^12*b^3)/(3*x) - (1365*a^11*b^4)/(2*x^(2/3)) - (3003*a^10*b^5)/x^(1/3) + 6435*a^8*b^7*x^(1/3) + (6435*a^7*b^8*x^(2/3))/2 + (5005*a^6*b^9*x)/3 + (30 03*a^5*b^10*x^(4/3))/4 + 273*a^4*b^11*x^(5/3) + (455*a^3*b^12*x^2)/6 + 15* a^2*b^13*x^(7/3) + (15*a*b^14*x^(8/3))/8 + (b^15*x^3)/9 + 5005*a^9*b^6*Log [x^(1/3)])
3.24.47.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 3.59 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(-\frac {a^{15}}{2 x^{2}}-\frac {9 a^{14} b}{x^{\frac {5}{3}}}-\frac {315 a^{13} b^{2}}{4 x^{\frac {4}{3}}}-\frac {455 a^{12} b^{3}}{x}-\frac {4095 a^{11} b^{4}}{2 x^{\frac {2}{3}}}-\frac {9009 a^{10} b^{5}}{x^{\frac {1}{3}}}+19305 a^{8} b^{7} x^{\frac {1}{3}}+\frac {19305 a^{7} b^{8} x^{\frac {2}{3}}}{2}+5005 a^{6} b^{9} x +\frac {9009 a^{5} b^{10} x^{\frac {4}{3}}}{4}+819 a^{4} b^{11} x^{\frac {5}{3}}+\frac {455 a^{3} b^{12} x^{2}}{2}+45 a^{2} b^{13} x^{\frac {7}{3}}+\frac {45 a \,b^{14} x^{\frac {8}{3}}}{8}+\frac {b^{15} x^{3}}{3}+5005 a^{9} b^{6} \ln \left (x \right )\) | \(165\) |
| default | \(-\frac {a^{15}}{2 x^{2}}-\frac {9 a^{14} b}{x^{\frac {5}{3}}}-\frac {315 a^{13} b^{2}}{4 x^{\frac {4}{3}}}-\frac {455 a^{12} b^{3}}{x}-\frac {4095 a^{11} b^{4}}{2 x^{\frac {2}{3}}}-\frac {9009 a^{10} b^{5}}{x^{\frac {1}{3}}}+19305 a^{8} b^{7} x^{\frac {1}{3}}+\frac {19305 a^{7} b^{8} x^{\frac {2}{3}}}{2}+5005 a^{6} b^{9} x +\frac {9009 a^{5} b^{10} x^{\frac {4}{3}}}{4}+819 a^{4} b^{11} x^{\frac {5}{3}}+\frac {455 a^{3} b^{12} x^{2}}{2}+45 a^{2} b^{13} x^{\frac {7}{3}}+\frac {45 a \,b^{14} x^{\frac {8}{3}}}{8}+\frac {b^{15} x^{3}}{3}+5005 a^{9} b^{6} \ln \left (x \right )\) | \(165\) |
| trager | \(\frac {\left (-1+x \right ) \left (2 b^{15} x^{4}+1365 a^{3} b^{12} x^{3}+2 b^{15} x^{3}+30030 a^{6} b^{9} x^{2}+1365 a^{3} b^{12} x^{2}+2 b^{15} x^{2}+3 x \,a^{15}+2730 a^{12} b^{3} x +3 a^{15}\right )}{6 x^{2}}-\frac {9 \left (-20 b^{12} x^{4}-1001 a^{3} b^{9} x^{3}-8580 a^{6} b^{6} x^{2}+910 a^{9} b^{3} x +4 a^{12}\right ) a^{2} b}{4 x^{\frac {5}{3}}}-\frac {9 \left (-5 b^{12} x^{4}-728 a^{3} b^{9} x^{3}-8580 a^{6} b^{6} x^{2}+8008 a^{9} b^{3} x +70 a^{12}\right ) a \,b^{2}}{8 x^{\frac {4}{3}}}-5005 a^{9} b^{6} \ln \left (\frac {1}{x}\right )\) | \(208\) |
-1/2*a^15/x^2-9*a^14*b/x^(5/3)-315/4*a^13*b^2/x^(4/3)-455*a^12*b^3/x-4095/ 2*a^11*b^4/x^(2/3)-9009*a^10*b^5/x^(1/3)+19305*a^8*b^7*x^(1/3)+19305/2*a^7 *b^8*x^(2/3)+5005*a^6*b^9*x+9009/4*a^5*b^10*x^(4/3)+819*a^4*b^11*x^(5/3)+4 55/2*a^3*b^12*x^2+45*a^2*b^13*x^(7/3)+45/8*a*b^14*x^(8/3)+1/3*b^15*x^3+500 5*a^9*b^6*ln(x)
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx=\frac {8 \, b^{15} x^{5} + 5460 \, a^{3} b^{12} x^{4} + 120120 \, a^{6} b^{9} x^{3} + 360360 \, a^{9} b^{6} x^{2} \log \left (x^{\frac {1}{3}}\right ) - 10920 \, a^{12} b^{3} x - 12 \, a^{15} + 27 \, {\left (5 \, a b^{14} x^{4} + 728 \, a^{4} b^{11} x^{3} + 8580 \, a^{7} b^{8} x^{2} - 8008 \, a^{10} b^{5} x - 70 \, a^{13} b^{2}\right )} x^{\frac {2}{3}} + 54 \, {\left (20 \, a^{2} b^{13} x^{4} + 1001 \, a^{5} b^{10} x^{3} + 8580 \, a^{8} b^{7} x^{2} - 910 \, a^{11} b^{4} x - 4 \, a^{14} b\right )} x^{\frac {1}{3}}}{24 \, x^{2}} \]
1/24*(8*b^15*x^5 + 5460*a^3*b^12*x^4 + 120120*a^6*b^9*x^3 + 360360*a^9*b^6 *x^2*log(x^(1/3)) - 10920*a^12*b^3*x - 12*a^15 + 27*(5*a*b^14*x^4 + 728*a^ 4*b^11*x^3 + 8580*a^7*b^8*x^2 - 8008*a^10*b^5*x - 70*a^13*b^2)*x^(2/3) + 5 4*(20*a^2*b^13*x^4 + 1001*a^5*b^10*x^3 + 8580*a^8*b^7*x^2 - 910*a^11*b^4*x - 4*a^14*b)*x^(1/3))/x^2
Time = 0.51 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx=- \frac {a^{15}}{2 x^{2}} - \frac {9 a^{14} b}{x^{\frac {5}{3}}} - \frac {315 a^{13} b^{2}}{4 x^{\frac {4}{3}}} - \frac {455 a^{12} b^{3}}{x} - \frac {4095 a^{11} b^{4}}{2 x^{\frac {2}{3}}} - \frac {9009 a^{10} b^{5}}{\sqrt [3]{x}} + 5005 a^{9} b^{6} \log {\left (x \right )} + 19305 a^{8} b^{7} \sqrt [3]{x} + \frac {19305 a^{7} b^{8} x^{\frac {2}{3}}}{2} + 5005 a^{6} b^{9} x + \frac {9009 a^{5} b^{10} x^{\frac {4}{3}}}{4} + 819 a^{4} b^{11} x^{\frac {5}{3}} + \frac {455 a^{3} b^{12} x^{2}}{2} + 45 a^{2} b^{13} x^{\frac {7}{3}} + \frac {45 a b^{14} x^{\frac {8}{3}}}{8} + \frac {b^{15} x^{3}}{3} \]
-a**15/(2*x**2) - 9*a**14*b/x**(5/3) - 315*a**13*b**2/(4*x**(4/3)) - 455*a **12*b**3/x - 4095*a**11*b**4/(2*x**(2/3)) - 9009*a**10*b**5/x**(1/3) + 50 05*a**9*b**6*log(x) + 19305*a**8*b**7*x**(1/3) + 19305*a**7*b**8*x**(2/3)/ 2 + 5005*a**6*b**9*x + 9009*a**5*b**10*x**(4/3)/4 + 819*a**4*b**11*x**(5/3 ) + 455*a**3*b**12*x**2/2 + 45*a**2*b**13*x**(7/3) + 45*a*b**14*x**(8/3)/8 + b**15*x**3/3
Time = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx=\frac {1}{3} \, b^{15} x^{3} + \frac {45}{8} \, a b^{14} x^{\frac {8}{3}} + 45 \, a^{2} b^{13} x^{\frac {7}{3}} + \frac {455}{2} \, a^{3} b^{12} x^{2} + 819 \, a^{4} b^{11} x^{\frac {5}{3}} + \frac {9009}{4} \, a^{5} b^{10} x^{\frac {4}{3}} + 5005 \, a^{6} b^{9} x + 5005 \, a^{9} b^{6} \log \left (x\right ) + \frac {19305}{2} \, a^{7} b^{8} x^{\frac {2}{3}} + 19305 \, a^{8} b^{7} x^{\frac {1}{3}} - \frac {36036 \, a^{10} b^{5} x^{\frac {5}{3}} + 8190 \, a^{11} b^{4} x^{\frac {4}{3}} + 1820 \, a^{12} b^{3} x + 315 \, a^{13} b^{2} x^{\frac {2}{3}} + 36 \, a^{14} b x^{\frac {1}{3}} + 2 \, a^{15}}{4 \, x^{2}} \]
1/3*b^15*x^3 + 45/8*a*b^14*x^(8/3) + 45*a^2*b^13*x^(7/3) + 455/2*a^3*b^12* x^2 + 819*a^4*b^11*x^(5/3) + 9009/4*a^5*b^10*x^(4/3) + 5005*a^6*b^9*x + 50 05*a^9*b^6*log(x) + 19305/2*a^7*b^8*x^(2/3) + 19305*a^8*b^7*x^(1/3) - 1/4* (36036*a^10*b^5*x^(5/3) + 8190*a^11*b^4*x^(4/3) + 1820*a^12*b^3*x + 315*a^ 13*b^2*x^(2/3) + 36*a^14*b*x^(1/3) + 2*a^15)/x^2
Time = 0.30 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx=\frac {1}{3} \, b^{15} x^{3} + \frac {45}{8} \, a b^{14} x^{\frac {8}{3}} + 45 \, a^{2} b^{13} x^{\frac {7}{3}} + \frac {455}{2} \, a^{3} b^{12} x^{2} + 819 \, a^{4} b^{11} x^{\frac {5}{3}} + \frac {9009}{4} \, a^{5} b^{10} x^{\frac {4}{3}} + 5005 \, a^{6} b^{9} x + 5005 \, a^{9} b^{6} \log \left ({\left | x \right |}\right ) + \frac {19305}{2} \, a^{7} b^{8} x^{\frac {2}{3}} + 19305 \, a^{8} b^{7} x^{\frac {1}{3}} - \frac {36036 \, a^{10} b^{5} x^{\frac {5}{3}} + 8190 \, a^{11} b^{4} x^{\frac {4}{3}} + 1820 \, a^{12} b^{3} x + 315 \, a^{13} b^{2} x^{\frac {2}{3}} + 36 \, a^{14} b x^{\frac {1}{3}} + 2 \, a^{15}}{4 \, x^{2}} \]
1/3*b^15*x^3 + 45/8*a*b^14*x^(8/3) + 45*a^2*b^13*x^(7/3) + 455/2*a^3*b^12* x^2 + 819*a^4*b^11*x^(5/3) + 9009/4*a^5*b^10*x^(4/3) + 5005*a^6*b^9*x + 50 05*a^9*b^6*log(abs(x)) + 19305/2*a^7*b^8*x^(2/3) + 19305*a^8*b^7*x^(1/3) - 1/4*(36036*a^10*b^5*x^(5/3) + 8190*a^11*b^4*x^(4/3) + 1820*a^12*b^3*x + 3 15*a^13*b^2*x^(2/3) + 36*a^14*b*x^(1/3) + 2*a^15)/x^2
Time = 5.97 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^3} \, dx=\frac {b^{15}\,x^3}{3}-\frac {\frac {a^{15}}{2}+455\,a^{12}\,b^3\,x+9\,a^{14}\,b\,x^{1/3}+\frac {315\,a^{13}\,b^2\,x^{2/3}}{4}+\frac {4095\,a^{11}\,b^4\,x^{4/3}}{2}+9009\,a^{10}\,b^5\,x^{5/3}}{x^2}+15015\,a^9\,b^6\,\ln \left (x^{1/3}\right )+5005\,a^6\,b^9\,x+\frac {45\,a\,b^{14}\,x^{8/3}}{8}+\frac {455\,a^3\,b^{12}\,x^2}{2}+19305\,a^8\,b^7\,x^{1/3}+\frac {19305\,a^7\,b^8\,x^{2/3}}{2}+\frac {9009\,a^5\,b^{10}\,x^{4/3}}{4}+819\,a^4\,b^{11}\,x^{5/3}+45\,a^2\,b^{13}\,x^{7/3} \]
(b^15*x^3)/3 - (a^15/2 + 455*a^12*b^3*x + 9*a^14*b*x^(1/3) + (315*a^13*b^2 *x^(2/3))/4 + (4095*a^11*b^4*x^(4/3))/2 + 9009*a^10*b^5*x^(5/3))/x^2 + 150 15*a^9*b^6*log(x^(1/3)) + 5005*a^6*b^9*x + (45*a*b^14*x^(8/3))/8 + (455*a^ 3*b^12*x^2)/2 + 19305*a^8*b^7*x^(1/3) + (19305*a^7*b^8*x^(2/3))/2 + (9009* a^5*b^10*x^(4/3))/4 + 819*a^4*b^11*x^(5/3) + 45*a^2*b^13*x^(7/3)