Integrand size = 26, antiderivative size = 137 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx=-\frac {a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {6 a}{5 b^3 \left (a+b \sqrt [3]{x}\right )^4 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac {3}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]
-1/2*a^2/b^3/(a+b*x^(1/3))^5/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)+6/5*a/b ^3/(a+b*x^(1/3))^4/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)-3/4/b^3/(a+b*x^(1 /3))^3/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx=\frac {\left (a+b \sqrt [3]{x}\right ) \left (-a^2-6 a b \sqrt [3]{x}-15 b^2 x^{2/3}\right )}{20 b^3 \left (\left (a+b \sqrt [3]{x}\right )^2\right )^{7/2}} \]
((a + b*x^(1/3))*(-a^2 - 6*a*b*x^(1/3) - 15*b^2*x^(2/3)))/(20*b^3*((a + b* x^(1/3))^2)^(7/2))
Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 774, 27, 53, 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.
\(\displaystyle \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\left (a b^7+b^8 \sqrt [3]{x}\right ) \int \frac {1}{\left (\sqrt [3]{x} b^2+a b\right )^7}dx}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {3 \left (a b^7+b^8 \sqrt [3]{x}\right ) \int \frac {x^{2/3}}{b^7 \left (a+b \sqrt [3]{x}\right )^7}d\sqrt [3]{x}}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \left (a b^7+b^8 \sqrt [3]{x}\right ) \int \frac {x^{2/3}}{\left (a+b \sqrt [3]{x}\right )^7}d\sqrt [3]{x}}{b^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {3 \left (a b^7+b^8 \sqrt [3]{x}\right ) \int \left (\frac {a^2}{b^2 \left (a+b \sqrt [3]{x}\right )^7}-\frac {2 a}{b^2 \left (a+b \sqrt [3]{x}\right )^6}+\frac {1}{b^2 \left (a+b \sqrt [3]{x}\right )^5}\right )d\sqrt [3]{x}}{b^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \left (-\frac {a^2}{6 b^3 \left (a+b \sqrt [3]{x}\right )^6}+\frac {2 a}{5 b^3 \left (a+b \sqrt [3]{x}\right )^5}-\frac {1}{4 b^3 \left (a+b \sqrt [3]{x}\right )^4}\right ) \left (a b^7+b^8 \sqrt [3]{x}\right )}{b^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\) |
(3*(-1/6*a^2/(b^3*(a + b*x^(1/3))^6) + (2*a)/(5*b^3*(a + b*x^(1/3))^5) - 1 /(4*b^3*(a + b*x^(1/3))^4))*(a*b^7 + b^8*x^(1/3)))/(b^7*Sqrt[a^2 + 2*a*b*x ^(1/3) + b^2*x^(2/3)])
3.5.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 0.49 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.31
method | result | size |
derivativedivides | \(-\frac {\left (15 b^{2} x^{\frac {2}{3}}+6 a b \,x^{\frac {1}{3}}+a^{2}\right ) \left (a +b \,x^{\frac {1}{3}}\right )}{20 b^{3} {\left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )}^{\frac {7}{2}}}\) | \(43\) |
default | \(-\frac {\left (280 x^{9} a^{5} b^{27}-540 x^{\frac {29}{3}} a^{3} b^{29}-84 x^{\frac {31}{3}} a \,b^{31}-2106 x^{\frac {26}{3}} a^{6} b^{26}+567 x^{\frac {28}{3}} a^{4} b^{28}-792 x^{\frac {23}{3}} a^{9} b^{23}+3996 x^{\frac {25}{3}} a^{7} b^{25}+7344 x^{\frac {20}{3}} a^{12} b^{20}+7470 x^{\frac {22}{3}} a^{10} b^{22}+14580 x^{\frac {17}{3}} a^{15} b^{17}+3240 x^{\frac {19}{3}} a^{13} b^{19}-6264 x^{\frac {16}{3}} a^{16} b^{16}+280 x^{10} a^{2} b^{30}+a^{32}+2640 a^{20} b^{12} x^{4}+2820 a^{23} b^{9} x^{3}+666 a^{26} b^{6} x^{2}-8 a^{29} b^{3} x -3465 a^{8} b^{24} x^{8}-11004 a^{11} b^{21} x^{7}-12858 a^{14} b^{18} x^{6}-4824 a^{17} b^{15} x^{5}+15 x^{\frac {32}{3}} b^{32}+11250 x^{\frac {14}{3}} a^{18} b^{14}-9108 x^{\frac {13}{3}} a^{19} b^{13}+3024 x^{\frac {11}{3}} a^{21} b^{11}-4374 x^{\frac {10}{3}} a^{22} b^{10}-567 x^{\frac {8}{3}} a^{24} b^{8}-540 x^{\frac {7}{3}} a^{25} b^{7}-336 x^{\frac {5}{3}} a^{27} b^{5}+105 x^{\frac {4}{3}} a^{28} b^{4}\right ) \left (a +b \,x^{\frac {1}{3}}\right )}{20 b^{3} \left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )^{6} \left (b^{3} x +a^{3}\right )^{6} \left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{\frac {7}{2}}}\) | \(391\) |
Time = 0.34 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.53 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx=-\frac {280 \, a^{2} b^{12} x^{4} - 1400 \, a^{5} b^{9} x^{3} + 735 \, a^{8} b^{6} x^{2} - 14 \, a^{11} b^{3} x + a^{14} + 3 \, {\left (5 \, b^{14} x^{4} - 210 \, a^{3} b^{11} x^{3} + 483 \, a^{6} b^{8} x^{2} - 112 \, a^{9} b^{5} x\right )} x^{\frac {2}{3}} - 3 \, {\left (28 \, a b^{13} x^{4} - 357 \, a^{4} b^{10} x^{3} + 390 \, a^{7} b^{7} x^{2} - 35 \, a^{10} b^{4} x\right )} x^{\frac {1}{3}}}{20 \, {\left (b^{21} x^{6} + 6 \, a^{3} b^{18} x^{5} + 15 \, a^{6} b^{15} x^{4} + 20 \, a^{9} b^{12} x^{3} + 15 \, a^{12} b^{9} x^{2} + 6 \, a^{15} b^{6} x + a^{18} b^{3}\right )}} \]
-1/20*(280*a^2*b^12*x^4 - 1400*a^5*b^9*x^3 + 735*a^8*b^6*x^2 - 14*a^11*b^3 *x + a^14 + 3*(5*b^14*x^4 - 210*a^3*b^11*x^3 + 483*a^6*b^8*x^2 - 112*a^9*b ^5*x)*x^(2/3) - 3*(28*a*b^13*x^4 - 357*a^4*b^10*x^3 + 390*a^7*b^7*x^2 - 35 *a^10*b^4*x)*x^(1/3))/(b^21*x^6 + 6*a^3*b^18*x^5 + 15*a^6*b^15*x^4 + 20*a^ 9*b^12*x^3 + 15*a^12*b^9*x^2 + 6*a^15*b^6*x + a^18*b^3)
\[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{\frac {7}{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx=-\frac {3}{4 \, b^{7} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{4}} + \frac {6 \, a}{5 \, b^{8} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{5}} - \frac {a^{2}}{2 \, b^{9} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{6}} \]
-3/4/(b^7*(x^(1/3) + a/b)^4) + 6/5*a/(b^8*(x^(1/3) + a/b)^5) - 1/2*a^2/(b^ 9*(x^(1/3) + a/b)^6)
Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.31 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx=-\frac {15 \, b^{2} x^{\frac {2}{3}} + 6 \, a b x^{\frac {1}{3}} + a^{2}}{20 \, {\left (b x^{\frac {1}{3}} + a\right )}^{6} b^{3} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right )} \]
Time = 9.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx=-\frac {\sqrt {a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}}\,\left (a^2+15\,b^2\,x^{2/3}+6\,a\,b\,x^{1/3}\right )}{20\,b^3\,{\left (a+b\,x^{1/3}\right )}^7} \]