Integrand size = 26, antiderivative size = 291 \[ \int \left (a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}\right )^{5/2} \, dx=\frac {25 a b^4 \sqrt {a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}} \sqrt [5]{x}}{a+\frac {b}{\sqrt [5]{x}}}+\frac {25 a^2 b^3 \sqrt {a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}} x^{2/5}}{a+\frac {b}{\sqrt [5]{x}}}+\frac {50 a^3 b^2 \sqrt {a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}} x^{3/5}}{3 \left (a+\frac {b}{\sqrt [5]{x}}\right )}+\frac {25 a^4 b \sqrt {a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}} x^{4/5}}{4 \left (a+\frac {b}{\sqrt [5]{x}}\right )}+\frac {a^5 \sqrt {a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}} x}{a+\frac {b}{\sqrt [5]{x}}}+\frac {5 b^5 \sqrt {a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}} \log \left (\sqrt [5]{x}\right )}{a+\frac {b}{\sqrt [5]{x}}} \]
25*a*b^4*x^(1/5)*(a^2+b^2/x^(2/5)+2*a*b/x^(1/5))^(1/2)/(a+b/x^(1/5))+25*a^ 2*b^3*x^(2/5)*(a^2+b^2/x^(2/5)+2*a*b/x^(1/5))^(1/2)/(a+b/x^(1/5))+50/3*a^3 *b^2*x^(3/5)*(a^2+b^2/x^(2/5)+2*a*b/x^(1/5))^(1/2)/(a+b/x^(1/5))+25/4*a^4* b*x^(4/5)*(a^2+b^2/x^(2/5)+2*a*b/x^(1/5))^(1/2)/(a+b/x^(1/5))+a^5*x*(a^2+b ^2/x^(2/5)+2*a*b/x^(1/5))^(1/2)/(a+b/x^(1/5))+b^5*ln(x)*(a^2+b^2/x^(2/5)+2 *a*b/x^(1/5))^(1/2)/(a+b/x^(1/5))
Time = 0.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.35 \[ \int \left (a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}\right )^{5/2} \, dx=\frac {\left (b+a \sqrt [5]{x}\right ) \left (300 a b^4 \sqrt [5]{x}+300 a^2 b^3 x^{2/5}+200 a^3 b^2 x^{3/5}+75 a^4 b x^{4/5}+12 a^5 x+12 b^5 \log (x)\right )}{12 \sqrt {\frac {\left (b+a \sqrt [5]{x}\right )^2}{x^{2/5}}} \sqrt [5]{x}} \]
((b + a*x^(1/5))*(300*a*b^4*x^(1/5) + 300*a^2*b^3*x^(2/5) + 200*a^3*b^2*x^ (3/5) + 75*a^4*b*x^(4/5) + 12*a^5*x + 12*b^5*Log[x]))/(12*Sqrt[(b + a*x^(1 /5))^2/x^(2/5)]*x^(1/5))
Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.41, 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.
\(\displaystyle \int \left (a^2+\frac {2 a b}{\sqrt [5]{x}}+\frac {b^2}{x^{2/5}}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+\frac {2 a b}{\sqrt [5]{x}}+\frac {b^2}{x^{2/5}}} \int \left (\frac {b^2}{\sqrt [5]{x}}+a b\right )^5dx}{a b^5+\frac {b^6}{\sqrt [5]{x}}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {5 \sqrt {a^2+\frac {2 a b}{\sqrt [5]{x}}+\frac {b^2}{x^{2/5}}} \int b^5 \left (a+\frac {b}{\sqrt [5]{x}}\right )^5 x^{4/5}d\sqrt [5]{x}}{a b^5+\frac {b^6}{\sqrt [5]{x}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [5]{x}}+\frac {b^2}{x^{2/5}}} \int \left (a+\frac {b}{\sqrt [5]{x}}\right )^5 x^{4/5}d\sqrt [5]{x}}{a b^5+\frac {b^6}{\sqrt [5]{x}}}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {5 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [5]{x}}+\frac {b^2}{x^{2/5}}} \int \frac {\left (\sqrt [5]{x} a+b\right )^5}{\sqrt [5]{x}}d\sqrt [5]{x}}{a b^5+\frac {b^6}{\sqrt [5]{x}}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {5 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [5]{x}}+\frac {b^2}{x^{2/5}}} \int \left (x^{4/5} a^5+5 b x^{3/5} a^4+10 b^2 x^{2/5} a^3+10 b^3 \sqrt [5]{x} a^2+5 b^4 a+\frac {b^5}{\sqrt [5]{x}}\right )d\sqrt [5]{x}}{a b^5+\frac {b^6}{\sqrt [5]{x}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [5]{x}}+\frac {b^2}{x^{2/5}}} \left (\frac {a^5 x}{5}+\frac {5}{4} a^4 b x^{4/5}+\frac {10}{3} a^3 b^2 x^{3/5}+5 a^2 b^3 x^{2/5}+5 a b^4 \sqrt [5]{x}+b^5 \log \left (\sqrt [5]{x}\right )\right )}{a b^5+\frac {b^6}{\sqrt [5]{x}}}\) |
(5*b^5*Sqrt[a^2 + b^2/x^(2/5) + (2*a*b)/x^(1/5)]*(5*a*b^4*x^(1/5) + 5*a^2* b^3*x^(2/5) + (10*a^3*b^2*x^(3/5))/3 + (5*a^4*b*x^(4/5))/4 + (a^5*x)/5 + b ^5*Log[x^(1/5)]))/(a*b^5 + b^6/x^(1/5))
3.5.90.3.1 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}, 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 = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.31
method | result | size |
derivativedivides | \(\frac {\left (\frac {a^{2} x^{\frac {2}{5}}+2 a b \,x^{\frac {1}{5}}+b^{2}}{x^{\frac {2}{5}}}\right )^{\frac {5}{2}} x \left (12 a^{5} x +75 b \,a^{4} x^{\frac {4}{5}}+200 a^{3} b^{2} x^{\frac {3}{5}}+300 a^{2} b^{3} x^{\frac {2}{5}}+12 b^{5} \ln \left (x \right )+300 b^{4} a \,x^{\frac {1}{5}}\right )}{12 \left (a \,x^{\frac {1}{5}}+b \right )^{5}}\) | \(91\) |
default | \(\frac {\left (\frac {a^{2} x^{\frac {2}{5}}+2 a b \,x^{\frac {1}{5}}+b^{2}}{x^{\frac {2}{5}}}\right )^{\frac {5}{2}} x \left (12 a^{5} x +75 b \,a^{4} x^{\frac {4}{5}}+200 a^{3} b^{2} x^{\frac {3}{5}}+300 a^{2} b^{3} x^{\frac {2}{5}}+12 b^{5} \ln \left (x \right )+300 b^{4} a \,x^{\frac {1}{5}}\right )}{12 \left (a \,x^{\frac {1}{5}}+b \right )^{5}}\) | \(91\) |
1/12*((a^2*x^(2/5)+2*a*b*x^(1/5)+b^2)/x^(2/5))^(5/2)*x*(12*a^5*x+75*b*a^4* x^(4/5)+200*a^3*b^2*x^(3/5)+300*a^2*b^3*x^(2/5)+12*b^5*ln(x)+300*b^4*a*x^( 1/5))/(a*x^(1/5)+b)^5
Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}\right )^{5/2} \, dx=\text {Timed out} \]
\[ \int \left (a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}\right )^{5/2} \, dx=\int \left (a^{2} + \frac {2 a b}{\sqrt [5]{x}} + \frac {b^{2}}{x^{\frac {2}{5}}}\right )^{\frac {5}{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.18 \[ \int \left (a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}\right )^{5/2} \, dx=a^{5} x + b^{5} \log \left (x\right ) + \frac {25}{4} \, a^{4} b x^{\frac {4}{5}} + \frac {50}{3} \, a^{3} b^{2} x^{\frac {3}{5}} + 25 \, a^{2} b^{3} x^{\frac {2}{5}} + 25 \, a b^{4} x^{\frac {1}{5}} \]
a^5*x + b^5*log(x) + 25/4*a^4*b*x^(4/5) + 50/3*a^3*b^2*x^(3/5) + 25*a^2*b^ 3*x^(2/5) + 25*a*b^4*x^(1/5)
Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.43 \[ \int \left (a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}\right )^{5/2} \, dx=a^{5} x \mathrm {sgn}\left (a x + b x^{\frac {4}{5}}\right ) \mathrm {sgn}\left (x\right ) + b^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b x^{\frac {4}{5}}\right ) \mathrm {sgn}\left (x\right ) + \frac {25}{4} \, a^{4} b x^{\frac {4}{5}} \mathrm {sgn}\left (a x + b x^{\frac {4}{5}}\right ) \mathrm {sgn}\left (x\right ) + \frac {50}{3} \, a^{3} b^{2} x^{\frac {3}{5}} \mathrm {sgn}\left (a x + b x^{\frac {4}{5}}\right ) \mathrm {sgn}\left (x\right ) + 25 \, a^{2} b^{3} x^{\frac {2}{5}} \mathrm {sgn}\left (a x + b x^{\frac {4}{5}}\right ) \mathrm {sgn}\left (x\right ) + 25 \, a b^{4} x^{\frac {1}{5}} \mathrm {sgn}\left (a x + b x^{\frac {4}{5}}\right ) \mathrm {sgn}\left (x\right ) \]
a^5*x*sgn(a*x + b*x^(4/5))*sgn(x) + b^5*log(abs(x))*sgn(a*x + b*x^(4/5))*s gn(x) + 25/4*a^4*b*x^(4/5)*sgn(a*x + b*x^(4/5))*sgn(x) + 50/3*a^3*b^2*x^(3 /5)*sgn(a*x + b*x^(4/5))*sgn(x) + 25*a^2*b^3*x^(2/5)*sgn(a*x + b*x^(4/5))* sgn(x) + 25*a*b^4*x^(1/5)*sgn(a*x + b*x^(4/5))*sgn(x)
Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/5}}+\frac {2 a b}{\sqrt [5]{x}}\right )^{5/2} \, dx=\int {\left (a^2+\frac {b^2}{x^{2/5}}+\frac {2\,a\,b}{x^{1/5}}\right )}^{5/2} \,d x \]