Integrand size = 26, antiderivative size = 410 \[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=\frac {3 b^7 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{4 a^8 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )^4}-\frac {7 b^6 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{a^8 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )^3}+\frac {63 b^5 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{2 a^8 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )^2}-\frac {105 b^4 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{a^8 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )}+\frac {45 b^2 \left (a+\frac {b}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}{a^7 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}-\frac {15 b \left (a+\frac {b}{\sqrt [3]{x}}\right ) x^{2/3}}{2 a^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}+\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right ) x}{a^5 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}-\frac {105 b^3 \left (a+\frac {b}{\sqrt [3]{x}}\right ) \log \left (b+a \sqrt [3]{x}\right )}{a^8 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}} \] Output:
3/4*b^7*(a+b/x^(1/3))/a^8/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(b+a*x^(1/ 3))^4-7*b^6*(a+b/x^(1/3))/a^8/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(b+a*x ^(1/3))^3+63/2*b^5*(a+b/x^(1/3))/a^8/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2) /(b+a*x^(1/3))^2-105*b^4*(a+b/x^(1/3))/a^8/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3)) ^(1/2)/(b+a*x^(1/3))+45*b^2*(a+b/x^(1/3))*x^(1/3)/a^7/(a^2+b^2/x^(2/3)+2*a *b/x^(1/3))^(1/2)-15/2*b*(a+b/x^(1/3))*x^(2/3)/a^6/(a^2+b^2/x^(2/3)+2*a*b/ x^(1/3))^(1/2)+(a+b/x^(1/3))*x/a^5/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)-1 05*b^3*(a+b/x^(1/3))*ln(b+a*x^(1/3))/a^8/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^( 1/2)
Time = 0.12 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.37 \[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=\frac {\left (b+a \sqrt [3]{x}\right ) \left (-319 b^7-856 a b^6 \sqrt [3]{x}-444 a^2 b^5 x^{2/3}+544 a^3 b^4 x+556 a^4 b^3 x^{4/3}+84 a^5 b^2 x^{5/3}-14 a^6 b x^2+4 a^7 x^{7/3}-420 b^3 \left (b+a \sqrt [3]{x}\right )^4 \log \left (b+a \sqrt [3]{x}\right )\right )}{4 a^8 \left (\frac {\left (b+a \sqrt [3]{x}\right )^2}{x^{2/3}}\right )^{5/2} x^{5/3}} \] Input:
Integrate[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(-5/2),x]
Output:
((b + a*x^(1/3))*(-319*b^7 - 856*a*b^6*x^(1/3) - 444*a^2*b^5*x^(2/3) + 544 *a^3*b^4*x + 556*a^4*b^3*x^(4/3) + 84*a^5*b^2*x^(5/3) - 14*a^6*b*x^2 + 4*a ^7*x^(7/3) - 420*b^3*(b + a*x^(1/3))^4*Log[b + a*x^(1/3)]))/(4*a^8*((b + a *x^(1/3))^2/x^(2/3))^(5/2)*x^(5/3))
Time = 0.34 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.44, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1384, 774, 27, 795, 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.
| \(\displaystyle \int \frac {1}{\left (a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {\left (a b^5+\frac {b^6}{\sqrt [3]{x}}\right ) \int \frac {1}{\left (\frac {b^2}{\sqrt [3]{x}}+a b\right )^5}dx}{\sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}\) |
| \(\Big \downarrow \) 774 |
| \(\displaystyle \frac {3 \left (a b^5+\frac {b^6}{\sqrt [3]{x}}\right ) \int \frac {x^{2/3}}{b^5 \left (a+\frac {b}{\sqrt [3]{x}}\right )^5}d\sqrt [3]{x}}{\sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \left (a b^5+\frac {b^6}{\sqrt [3]{x}}\right ) \int \frac {x^{2/3}}{\left (a+\frac {b}{\sqrt [3]{x}}\right )^5}d\sqrt [3]{x}}{b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}\) |
| \(\Big \downarrow \) 795 |
| \(\displaystyle \frac {3 \left (a b^5+\frac {b^6}{\sqrt [3]{x}}\right ) \int \frac {x^{7/3}}{\left (\sqrt [3]{x} a+b\right )^5}d\sqrt [3]{x}}{b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {3 \left (a b^5+\frac {b^6}{\sqrt [3]{x}}\right ) \int \left (-\frac {b^7}{a^7 \left (\sqrt [3]{x} a+b\right )^5}+\frac {7 b^6}{a^7 \left (\sqrt [3]{x} a+b\right )^4}-\frac {21 b^5}{a^7 \left (\sqrt [3]{x} a+b\right )^3}+\frac {35 b^4}{a^7 \left (\sqrt [3]{x} a+b\right )^2}-\frac {35 b^3}{a^7 \left (\sqrt [3]{x} a+b\right )}+\frac {15 b^2}{a^7}-\frac {5 \sqrt [3]{x} b}{a^6}+\frac {x^{2/3}}{a^5}\right )d\sqrt [3]{x}}{b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \left (a b^5+\frac {b^6}{\sqrt [3]{x}}\right ) \left (\frac {b^7}{4 a^8 \left (a \sqrt [3]{x}+b\right )^4}-\frac {7 b^6}{3 a^8 \left (a \sqrt [3]{x}+b\right )^3}+\frac {21 b^5}{2 a^8 \left (a \sqrt [3]{x}+b\right )^2}-\frac {35 b^4}{a^8 \left (a \sqrt [3]{x}+b\right )}-\frac {35 b^3 \log \left (a \sqrt [3]{x}+b\right )}{a^8}+\frac {15 b^2 \sqrt [3]{x}}{a^7}-\frac {5 b x^{2/3}}{2 a^6}+\frac {x}{3 a^5}\right )}{b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}\) |
Input:
Int[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(-5/2),x]
Output:
(3*(a*b^5 + b^6/x^(1/3))*(b^7/(4*a^8*(b + a*x^(1/3))^4) - (7*b^6)/(3*a^8*( b + a*x^(1/3))^3) + (21*b^5)/(2*a^8*(b + a*x^(1/3))^2) - (35*b^4)/(a^8*(b + a*x^(1/3))) + (15*b^2*x^(1/3))/a^7 - (5*b*x^(2/3))/(2*a^6) + x/(3*a^5) - (35*b^3*Log[b + a*x^(1/3)])/a^8))/(b^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x ^(1/3)])
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}, x] && IGtQ[m, 0] && IGtQ[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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[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 = 0.05 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.49
| method | result | size |
| derivativedivides | \(-\frac {\left (-4 a^{7} x^{\frac {7}{3}}+14 a^{6} b \,x^{2}+420 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{4} b^{3} x^{\frac {4}{3}}-84 a^{5} b^{2} x^{\frac {5}{3}}+1680 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{3} b^{4} x -556 a^{4} b^{3} x^{\frac {4}{3}}+2520 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{2} b^{5} x^{\frac {2}{3}}-544 a^{3} b^{4} x +1680 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a \,b^{6} x^{\frac {1}{3}}+444 a^{2} b^{5} x^{\frac {2}{3}}+420 \ln \left (b +a \,x^{\frac {1}{3}}\right ) b^{7}+856 a \,b^{6} x^{\frac {1}{3}}+319 b^{7}\right ) \left (b +a \,x^{\frac {1}{3}}\right )}{4 a^{8} x^{\frac {5}{3}} \left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}}}\) | \(199\) |
| default | \(\frac {\left (4 a^{7} x^{\frac {7}{3}}+84 a^{5} b^{2} x^{\frac {5}{3}}-420 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{4} b^{3} x^{\frac {4}{3}}+556 a^{4} b^{3} x^{\frac {4}{3}}-2520 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{2} b^{5} x^{\frac {2}{3}}-444 a^{2} b^{5} x^{\frac {2}{3}}-1680 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a \,b^{6} x^{\frac {1}{3}}-1680 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{3} b^{4} x -14 a^{6} b \,x^{2}-856 a \,b^{6} x^{\frac {1}{3}}-420 \ln \left (b +a \,x^{\frac {1}{3}}\right ) b^{7}+544 a^{3} b^{4} x -319 b^{7}\right ) \left (b +a \,x^{\frac {1}{3}}\right )}{4 a^{8} x^{\frac {5}{3}} \left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}}}\) | \(199\) |
Input:
int(1/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/4*(-4*a^7*x^(7/3)+14*a^6*b*x^2+420*ln(b+a*x^(1/3))*a^4*b^3*x^(4/3)-84*a ^5*b^2*x^(5/3)+1680*ln(b+a*x^(1/3))*a^3*b^4*x-556*a^4*b^3*x^(4/3)+2520*ln( b+a*x^(1/3))*a^2*b^5*x^(2/3)-544*a^3*b^4*x+1680*ln(b+a*x^(1/3))*a*b^6*x^(1 /3)+444*a^2*b^5*x^(2/3)+420*ln(b+a*x^(1/3))*b^7+856*a*b^6*x^(1/3)+319*b^7) *(b+a*x^(1/3))/a^8/x^(5/3)/((x^(2/3)*a^2+2*a*b*x^(1/3)+b^2)/x^(2/3))^(5/2)
Timed out. \[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=\int \frac {1}{\left (a^{2} + \frac {2 a b}{\sqrt [3]{x}} + \frac {b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a**2+b**2/x**(2/3)+2*a*b/x**(1/3))**(5/2),x)
Output:
Integral((a**2 + 2*a*b/x**(1/3) + b**2/x**(2/3))**(-5/2), x)
Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=\frac {4 \, a^{7} x^{\frac {7}{3}} - 14 \, a^{6} b x^{2} + 84 \, a^{5} b^{2} x^{\frac {5}{3}} + 556 \, a^{4} b^{3} x^{\frac {4}{3}} + 544 \, a^{3} b^{4} x - 444 \, a^{2} b^{5} x^{\frac {2}{3}} - 856 \, a b^{6} x^{\frac {1}{3}} - 319 \, b^{7}}{4 \, {\left (a^{12} x^{\frac {4}{3}} + 4 \, a^{11} b x + 6 \, a^{10} b^{2} x^{\frac {2}{3}} + 4 \, a^{9} b^{3} x^{\frac {1}{3}} + a^{8} b^{4}\right )}} - \frac {105 \, b^{3} \log \left (a x^{\frac {1}{3}} + b\right )}{a^{8}} \] Input:
integrate(1/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="maxima")
Output:
1/4*(4*a^7*x^(7/3) - 14*a^6*b*x^2 + 84*a^5*b^2*x^(5/3) + 556*a^4*b^3*x^(4/ 3) + 544*a^3*b^4*x - 444*a^2*b^5*x^(2/3) - 856*a*b^6*x^(1/3) - 319*b^7)/(a ^12*x^(4/3) + 4*a^11*b*x + 6*a^10*b^2*x^(2/3) + 4*a^9*b^3*x^(1/3) + a^8*b^ 4) - 105*b^3*log(a*x^(1/3) + b)/a^8
Time = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=-\frac {105 \, b^{3} \log \left ({\left | a x^{\frac {1}{3}} + b \right |}\right )}{a^{8} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )} - \frac {420 \, a^{3} b^{4} x + 1134 \, a^{2} b^{5} x^{\frac {2}{3}} + 1036 \, a b^{6} x^{\frac {1}{3}} + 319 \, b^{7}}{4 \, {\left (a x^{\frac {1}{3}} + b\right )}^{4} a^{8} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )} + \frac {2 \, a^{10} x - 15 \, a^{9} b x^{\frac {2}{3}} + 90 \, a^{8} b^{2} x^{\frac {1}{3}}}{2 \, a^{15} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="giac")
Output:
-105*b^3*log(abs(a*x^(1/3) + b))/(a^8*sgn(a*x + b*x^(2/3))*sgn(x)) - 1/4*( 420*a^3*b^4*x + 1134*a^2*b^5*x^(2/3) + 1036*a*b^6*x^(1/3) + 319*b^7)/((a*x ^(1/3) + b)^4*a^8*sgn(a*x + b*x^(2/3))*sgn(x)) + 1/2*(2*a^10*x - 15*a^9*b* x^(2/3) + 90*a^8*b^2*x^(1/3))/(a^15*sgn(a*x + b*x^(2/3))*sgn(x))
Timed out. \[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=\int \frac {1}{{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2\,a\,b}{x^{1/3}}\right )}^{5/2}} \,d x \] Input:
int(1/(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2),x)
Output:
int(1/(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx=\frac {-2520 x^{\frac {2}{3}} \mathrm {log}\left (x^{\frac {1}{3}} a +b \right ) a^{2} b^{5}+84 x^{\frac {5}{3}} a^{5} b^{2}-1260 x^{\frac {2}{3}} a^{2} b^{5}-420 x^{\frac {4}{3}} \mathrm {log}\left (x^{\frac {1}{3}} a +b \right ) a^{4} b^{3}-1680 x^{\frac {1}{3}} \mathrm {log}\left (x^{\frac {1}{3}} a +b \right ) a \,b^{6}+4 x^{\frac {7}{3}} a^{7}+420 x^{\frac {4}{3}} a^{4} b^{3}-1400 x^{\frac {1}{3}} a \,b^{6}-1680 \,\mathrm {log}\left (x^{\frac {1}{3}} a +b \right ) a^{3} b^{4} x -420 \,\mathrm {log}\left (x^{\frac {1}{3}} a +b \right ) b^{7}-14 a^{6} b \,x^{2}-455 b^{7}}{4 a^{8} \left (6 x^{\frac {2}{3}} a^{2} b^{2}+x^{\frac {4}{3}} a^{4}+4 x^{\frac {1}{3}} a \,b^{3}+4 a^{3} b x +b^{4}\right )} \] Input:
int(1/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x)
Output:
( - 2520*x**(2/3)*log(x**(1/3)*a + b)*a**2*b**5 + 84*x**(2/3)*a**5*b**2*x - 1260*x**(2/3)*a**2*b**5 - 420*x**(1/3)*log(x**(1/3)*a + b)*a**4*b**3*x - 1680*x**(1/3)*log(x**(1/3)*a + b)*a*b**6 + 4*x**(1/3)*a**7*x**2 + 420*x** (1/3)*a**4*b**3*x - 1400*x**(1/3)*a*b**6 - 1680*log(x**(1/3)*a + b)*a**3*b **4*x - 420*log(x**(1/3)*a + b)*b**7 - 14*a**6*b*x**2 - 455*b**7)/(4*a**8* (6*x**(2/3)*a**2*b**2 + x**(1/3)*a**4*x + 4*x**(1/3)*a*b**3 + 4*a**3*b*x + b**4))