Integrand size = 26, antiderivative size = 287 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=-\frac {3 b^5 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{2 \left (a+\frac {b}{\sqrt [3]{x}}\right ) x^{2/3}}-\frac {15 a b^4 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{\left (a+\frac {b}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}+\frac {30 a^3 b^2 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \sqrt [3]{x}}{a+\frac {b}{\sqrt [3]{x}}}+\frac {15 a^4 b \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} x^{2/3}}{2 \left (a+\frac {b}{\sqrt [3]{x}}\right )}+\frac {a^5 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} x}{a+\frac {b}{\sqrt [3]{x}}}+\frac {10 a^2 b^3 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \log (x)}{a+\frac {b}{\sqrt [3]{x}}} \] Output:
-3/2*b^5*(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))/x^(2/3)-15*a* b^4*(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))/x^(1/3)+30*a^3*b^2 *(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)*x^(1/3)/(a+b/x^(1/3))+15*a^4*b*(a^2 +b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)*x^(2/3)/(2*a+2*b/x^(1/3))+a^5*(a^2+b^2/x ^(2/3)+2*a*b/x^(1/3))^(1/2)*x/(a+b/x^(1/3))+10*a^2*b^3*(a^2+b^2/x^(2/3)+2* a*b/x^(1/3))^(1/2)*ln(x)/(a+b/x^(1/3))
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.34 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\frac {\left (b+a \sqrt [3]{x}\right ) \left (-3 b^5-30 a b^4 \sqrt [3]{x}+60 a^3 b^2 x+15 a^4 b x^{4/3}+2 a^5 x^{5/3}+20 a^2 b^3 x^{2/3} \log (x)\right )}{2 \sqrt {\frac {\left (b+a \sqrt [3]{x}\right )^2}{x^{2/3}}} x} \] Input:
Integrate[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2),x]
Output:
((b + a*x^(1/3))*(-3*b^5 - 30*a*b^4*x^(1/3) + 60*a^3*b^2*x + 15*a^4*b*x^(4 /3) + 2*a^5*x^(5/3) + 20*a^2*b^3*x^(2/3)*Log[x]))/(2*Sqrt[(b + a*x^(1/3))^ 2/x^(2/3)]*x)
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.42, 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.
\(\displaystyle \int \left (a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \left (\frac {b^2}{\sqrt [3]{x}}+a b\right )^5dx}{a b^5+\frac {b^6}{\sqrt [3]{x}}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {3 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int b^5 \left (a+\frac {b}{\sqrt [3]{x}}\right )^5 x^{2/3}d\sqrt [3]{x}}{a b^5+\frac {b^6}{\sqrt [3]{x}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^5 x^{2/3}d\sqrt [3]{x}}{a b^5+\frac {b^6}{\sqrt [3]{x}}}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {3 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \frac {\left (\sqrt [3]{x} a+b\right )^5}{x}d\sqrt [3]{x}}{a b^5+\frac {b^6}{\sqrt [3]{x}}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {3 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \left (x^{2/3} a^5+5 b \sqrt [3]{x} a^4+10 b^2 a^3+\frac {10 b^3 a^2}{\sqrt [3]{x}}+\frac {5 b^4 a}{x^{2/3}}+\frac {b^5}{x}\right )d\sqrt [3]{x}}{a b^5+\frac {b^6}{\sqrt [3]{x}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \left (\frac {a^5 x}{3}+\frac {5}{2} a^4 b x^{2/3}+10 a^3 b^2 \sqrt [3]{x}+10 a^2 b^3 \log \left (\sqrt [3]{x}\right )-\frac {5 a b^4}{\sqrt [3]{x}}-\frac {b^5}{2 x^{2/3}}\right )}{a b^5+\frac {b^6}{\sqrt [3]{x}}}\) |
Input:
Int[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2),x]
Output:
(3*b^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(-1/2*b^5/x^(2/3) - (5*a* b^4)/x^(1/3) + 10*a^3*b^2*x^(1/3) + (5*a^4*b*x^(2/3))/2 + (a^5*x)/3 + 10*a ^2*b^3*Log[x^(1/3)]))/(a*b^5 + b^6/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 = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.32
method | result | size |
derivativedivides | \(\frac {\left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}} x \left (2 a^{5} x^{\frac {5}{3}}+15 a^{4} b \,x^{\frac {4}{3}}+20 a^{2} b^{3} \ln \left (x \right ) x^{\frac {2}{3}}+60 a^{3} b^{2} x -30 a \,b^{4} x^{\frac {1}{3}}-3 b^{5}\right )}{2 \left (b +a \,x^{\frac {1}{3}}\right )^{5}}\) | \(91\) |
default | \(\frac {\left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}} x \left (2 a^{5} x^{\frac {5}{3}}+15 a^{4} b \,x^{\frac {4}{3}}+20 a^{2} b^{3} \ln \left (x \right ) x^{\frac {2}{3}}+60 a^{3} b^{2} x -30 a \,b^{4} x^{\frac {1}{3}}-3 b^{5}\right )}{2 \left (b +a \,x^{\frac {1}{3}}\right )^{5}}\) | \(91\) |
Input:
int((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/2*((x^(2/3)*a^2+2*a*b*x^(1/3)+b^2)/x^(2/3))^(5/2)*x*(2*a^5*x^(5/3)+15*a^ 4*b*x^(4/3)+20*a^2*b^3*ln(x)*x^(2/3)+60*a^3*b^2*x-30*a*b^4*x^(1/3)-3*b^5)/ (b+a*x^(1/3))^5
Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\int \left (a^{2} + \frac {2 a b}{\sqrt [3]{x}} + \frac {b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((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 = 57, normalized size of antiderivative = 0.20 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=10 \, a^{2} b^{3} \log \left (x\right ) + \frac {2 \, a^{5} x^{\frac {5}{3}} + 15 \, a^{4} b x^{\frac {4}{3}} + 60 \, a^{3} b^{2} x - 30 \, a b^{4} x^{\frac {1}{3}} - 3 \, b^{5}}{2 \, x^{\frac {2}{3}}} \] Input:
integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="maxima")
Output:
10*a^2*b^3*log(x) + 1/2*(2*a^5*x^(5/3) + 15*a^4*b*x^(4/3) + 60*a^3*b^2*x - 30*a*b^4*x^(1/3) - 3*b^5)/x^(2/3)
Time = 0.13 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.45 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=a^{5} x \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 10 \, a^{2} b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + \frac {15}{2} \, a^{4} b x^{\frac {2}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 30 \, a^{3} b^{2} x^{\frac {1}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) - \frac {3 \, {\left (10 \, a b^{4} x^{\frac {1}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + b^{5} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )\right )}}{2 \, x^{\frac {2}{3}}} \] Input:
integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="giac")
Output:
a^5*x*sgn(a*x + b*x^(2/3))*sgn(x) + 10*a^2*b^3*log(abs(x))*sgn(a*x + b*x^( 2/3))*sgn(x) + 15/2*a^4*b*x^(2/3)*sgn(a*x + b*x^(2/3))*sgn(x) + 30*a^3*b^2 *x^(1/3)*sgn(a*x + b*x^(2/3))*sgn(x) - 3/2*(10*a*b^4*x^(1/3)*sgn(a*x + b*x ^(2/3))*sgn(x) + b^5*sgn(a*x + b*x^(2/3))*sgn(x))/x^(2/3)
Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\int {\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2\,a\,b}{x^{1/3}}\right )}^{5/2} \,d x \] Input:
int((a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2),x)
Output:
int((a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.21 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx=\frac {20 x^{\frac {2}{3}} \mathrm {log}\left (x \right ) a^{2} b^{3}+2 x^{\frac {5}{3}} a^{5}+15 x^{\frac {4}{3}} a^{4} b -30 x^{\frac {1}{3}} a \,b^{4}+60 a^{3} b^{2} x -3 b^{5}}{2 x^{\frac {2}{3}}} \] Input:
int((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x)
Output:
(20*x**(2/3)*log(x)*a**2*b**3 + 2*x**(2/3)*a**5*x + 15*x**(1/3)*a**4*b*x - 30*x**(1/3)*a*b**4 + 60*a**3*b**2*x - 3*b**5)/(2*x**(2/3))