Integrand size = 26, antiderivative size = 268 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=-\frac {60 a^2}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {3 a^5}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {10 a^4}{b^6 \left (a+b \sqrt [6]{x}\right )^2 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {30 a^3}{b^6 \left (a+b \sqrt [6]{x}\right ) \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {6 \left (a+b \sqrt [6]{x}\right ) \sqrt [6]{x}}{b^5 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {30 a \left (a+b \sqrt [6]{x}\right ) \log \left (a+b \sqrt [6]{x}\right )}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}} \] Output:
-60*a^2/b^6/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)+3/2*a^5/b^6/(a+b*x^(1/6) )^3/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)-10*a^4/b^6/(a+b*x^(1/6))^2/(a^2+ 2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)+30*a^3/b^6/(a+b*x^(1/6))/(a^2+2*a*b*x^(1/ 6)+b^2*x^(1/3))^(1/2)+6*(a+b*x^(1/6))*x^(1/6)/b^5/(a^2+2*a*b*x^(1/6)+b^2*x ^(1/3))^(1/2)-30*a*(a+b*x^(1/6))*ln(a+b*x^(1/6))/b^6/(a^2+2*a*b*x^(1/6)+b^ 2*x^(1/3))^(1/2)
Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=\frac {-77 a^5-248 a^4 b \sqrt [6]{x}-252 a^3 b^2 \sqrt [3]{x}-48 a^2 b^3 \sqrt {x}+48 a b^4 x^{2/3}+12 b^5 x^{5/6}-60 a \left (a+b \sqrt [6]{x}\right )^4 \log \left (a+b \sqrt [6]{x}\right )}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt {\left (a+b \sqrt [6]{x}\right )^2}} \] Input:
Integrate[(a^2 + 2*a*b*x^(1/6) + b^2*x^(1/3))^(-5/2),x]
Output:
(-77*a^5 - 248*a^4*b*x^(1/6) - 252*a^3*b^2*x^(1/3) - 48*a^2*b^3*Sqrt[x] + 48*a*b^4*x^(2/3) + 12*b^5*x^(5/6) - 60*a*(a + b*x^(1/6))^4*Log[a + b*x^(1/ 6)])/(2*b^6*(a + b*x^(1/6))^3*Sqrt[(a + b*x^(1/6))^2])
Time = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.57, 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, 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.
\(\displaystyle \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\left (a b^5+b^6 \sqrt [6]{x}\right ) \int \frac {1}{\left (\sqrt [6]{x} b^2+a b\right )^5}dx}{\sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {6 \left (a b^5+b^6 \sqrt [6]{x}\right ) \int \frac {x^{5/6}}{b^5 \left (a+b \sqrt [6]{x}\right )^5}d\sqrt [6]{x}}{\sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6 \left (a b^5+b^6 \sqrt [6]{x}\right ) \int \frac {x^{5/6}}{\left (a+b \sqrt [6]{x}\right )^5}d\sqrt [6]{x}}{b^5 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {6 \left (a b^5+b^6 \sqrt [6]{x}\right ) \int \left (-\frac {a^5}{b^5 \left (a+b \sqrt [6]{x}\right )^5}+\frac {5 a^4}{b^5 \left (a+b \sqrt [6]{x}\right )^4}-\frac {10 a^3}{b^5 \left (a+b \sqrt [6]{x}\right )^3}+\frac {10 a^2}{b^5 \left (a+b \sqrt [6]{x}\right )^2}-\frac {5 a}{b^5 \left (a+b \sqrt [6]{x}\right )}+\frac {1}{b^5}\right )d\sqrt [6]{x}}{b^5 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 \left (a b^5+b^6 \sqrt [6]{x}\right ) \left (\frac {a^5}{4 b^6 \left (a+b \sqrt [6]{x}\right )^4}-\frac {5 a^4}{3 b^6 \left (a+b \sqrt [6]{x}\right )^3}+\frac {5 a^3}{b^6 \left (a+b \sqrt [6]{x}\right )^2}-\frac {10 a^2}{b^6 \left (a+b \sqrt [6]{x}\right )}-\frac {5 a \log \left (a+b \sqrt [6]{x}\right )}{b^6}+\frac {\sqrt [6]{x}}{b^5}\right )}{b^5 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\) |
Input:
Int[(a^2 + 2*a*b*x^(1/6) + b^2*x^(1/3))^(-5/2),x]
Output:
(6*(a*b^5 + b^6*x^(1/6))*(a^5/(4*b^6*(a + b*x^(1/6))^4) - (5*a^4)/(3*b^6*( a + b*x^(1/6))^3) + (5*a^3)/(b^6*(a + b*x^(1/6))^2) - (10*a^2)/(b^6*(a + b *x^(1/6))) + x^(1/6)/b^5 - (5*a*Log[a + b*x^(1/6)])/b^6))/(b^5*Sqrt[a^2 + 2*a*b*x^(1/6) + b^2*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[(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 = 3.74 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.61
method | result | size |
derivativedivides | \(-\frac {\left (60 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a \,b^{4} x^{\frac {2}{3}}-12 b^{5} x^{\frac {5}{6}}+240 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{2} b^{3} \sqrt {x}-48 a \,b^{4} x^{\frac {2}{3}}+360 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{3} b^{2} x^{\frac {1}{3}}+48 a^{2} b^{3} \sqrt {x}+240 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{4} b \,x^{\frac {1}{6}}+252 a^{3} b^{2} x^{\frac {1}{3}}+60 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{5}+248 a^{4} b \,x^{\frac {1}{6}}+77 a^{5}\right ) \left (a +b \,x^{\frac {1}{6}}\right )}{2 b^{6} {\left (\left (a +b \,x^{\frac {1}{6}}\right )^{2}\right )}^{\frac {5}{2}}}\) | \(163\) |
default | \(\text {Expression too large to display}\) | \(5002\) |
Input:
int(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/2*(60*ln(a+b*x^(1/6))*a*b^4*x^(2/3)-12*b^5*x^(5/6)+240*ln(a+b*x^(1/6))* a^2*b^3*x^(1/2)-48*a*b^4*x^(2/3)+360*ln(a+b*x^(1/6))*a^3*b^2*x^(1/3)+48*a^ 2*b^3*x^(1/2)+240*ln(a+b*x^(1/6))*a^4*b*x^(1/6)+252*a^3*b^2*x^(1/3)+60*ln( a+b*x^(1/6))*a^5+248*a^4*b*x^(1/6)+77*a^5)*(a+b*x^(1/6))/b^6/((a+b*x^(1/6) )^2)^(5/2)
Timed out. \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a**2+2*a*b*x**(1/6)+b**2*x**(1/3))**(5/2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=\frac {12 \, b^{5} x^{\frac {5}{6}} + 48 \, a b^{4} x^{\frac {2}{3}} - 48 \, a^{2} b^{3} \sqrt {x} - 252 \, a^{3} b^{2} x^{\frac {1}{3}} - 248 \, a^{4} b x^{\frac {1}{6}} - 77 \, a^{5}}{2 \, {\left (b^{10} x^{\frac {2}{3}} + 4 \, a b^{9} \sqrt {x} + 6 \, a^{2} b^{8} x^{\frac {1}{3}} + 4 \, a^{3} b^{7} x^{\frac {1}{6}} + a^{4} b^{6}\right )}} - \frac {30 \, a \log \left (b x^{\frac {1}{6}} + a\right )}{b^{6}} \] Input:
integrate(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x, algorithm="maxima")
Output:
1/2*(12*b^5*x^(5/6) + 48*a*b^4*x^(2/3) - 48*a^2*b^3*sqrt(x) - 252*a^3*b^2* x^(1/3) - 248*a^4*b*x^(1/6) - 77*a^5)/(b^10*x^(2/3) + 4*a*b^9*sqrt(x) + 6* a^2*b^8*x^(1/3) + 4*a^3*b^7*x^(1/6) + a^4*b^6) - 30*a*log(b*x^(1/6) + a)/b ^6
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=-\frac {30 \, a \log \left ({\left | b x^{\frac {1}{6}} + a \right |}\right )}{b^{6} \mathrm {sgn}\left (b x^{\frac {1}{6}} + a\right )} + \frac {6 \, x^{\frac {1}{6}}}{b^{5} \mathrm {sgn}\left (b x^{\frac {1}{6}} + a\right )} - \frac {120 \, a^{2} b^{3} \sqrt {x} + 300 \, a^{3} b^{2} x^{\frac {1}{3}} + 260 \, a^{4} b x^{\frac {1}{6}} + 77 \, a^{5}}{2 \, {\left (b x^{\frac {1}{6}} + a\right )}^{4} b^{6} \mathrm {sgn}\left (b x^{\frac {1}{6}} + a\right )} \] Input:
integrate(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x, algorithm="giac")
Output:
-30*a*log(abs(b*x^(1/6) + a))/(b^6*sgn(b*x^(1/6) + a)) + 6*x^(1/6)/(b^5*sg n(b*x^(1/6) + a)) - 1/2*(120*a^2*b^3*sqrt(x) + 300*a^3*b^2*x^(1/3) + 260*a ^4*b*x^(1/6) + 77*a^5)/((b*x^(1/6) + a)^4*b^6*sgn(b*x^(1/6) + a))
Timed out. \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=\int \frac {1}{{\left (a^2+b^2\,x^{1/3}+2\,a\,b\,x^{1/6}\right )}^{5/2}} \,d x \] Input:
int(1/(a^2 + b^2*x^(1/3) + 2*a*b*x^(1/6))^(5/2),x)
Output:
int(1/(a^2 + b^2*x^(1/3) + 2*a*b*x^(1/6))^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx=\frac {12 x^{\frac {5}{6}} b^{5}-240 x^{\frac {1}{6}} \mathrm {log}\left (x^{\frac {1}{6}} b +a \right ) a^{4} b -200 x^{\frac {1}{6}} a^{4} b -60 x^{\frac {2}{3}} \mathrm {log}\left (x^{\frac {1}{6}} b +a \right ) a \,b^{4}+60 x^{\frac {2}{3}} a \,b^{4}-360 x^{\frac {1}{3}} \mathrm {log}\left (x^{\frac {1}{6}} b +a \right ) a^{3} b^{2}-180 x^{\frac {1}{3}} a^{3} b^{2}-240 \sqrt {x}\, \mathrm {log}\left (x^{\frac {1}{6}} b +a \right ) a^{2} b^{3}-60 \,\mathrm {log}\left (x^{\frac {1}{6}} b +a \right ) a^{5}-65 a^{5}}{2 b^{6} \left (4 x^{\frac {1}{6}} a^{3} b +x^{\frac {2}{3}} b^{4}+6 x^{\frac {1}{3}} a^{2} b^{2}+4 \sqrt {x}\, a \,b^{3}+a^{4}\right )} \] Input:
int(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x)
Output:
(12*x**(5/6)*b**5 - 240*x**(1/6)*log(x**(1/6)*b + a)*a**4*b - 200*x**(1/6) *a**4*b - 60*x**(2/3)*log(x**(1/6)*b + a)*a*b**4 + 60*x**(2/3)*a*b**4 - 36 0*x**(1/3)*log(x**(1/6)*b + a)*a**3*b**2 - 180*x**(1/3)*a**3*b**2 - 240*sq rt(x)*log(x**(1/6)*b + a)*a**2*b**3 - 60*log(x**(1/6)*b + a)*a**5 - 65*a** 5)/(2*b**6*(4*x**(1/6)*a**3*b + x**(2/3)*b**4 + 6*x**(1/3)*a**2*b**2 + 4*s qrt(x)*a*b**3 + a**4))