Integrand size = 26, antiderivative size = 289 \[ \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx=-\frac {4 b^5 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}}}{\left (a+\frac {b}{\sqrt [4]{x}}\right ) \sqrt [4]{x}}+\frac {40 a^2 b^3 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \sqrt [4]{x}}{a+\frac {b}{\sqrt [4]{x}}}+\frac {20 a^3 b^2 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \sqrt {x}}{a+\frac {b}{\sqrt [4]{x}}}+\frac {20 a^4 b \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} x^{3/4}}{3 \left (a+\frac {b}{\sqrt [4]{x}}\right )}+\frac {a^5 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} x}{a+\frac {b}{\sqrt [4]{x}}}+\frac {20 a b^4 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \log \left (\sqrt [4]{x}\right )}{a+\frac {b}{\sqrt [4]{x}}} \]
-4*b^5*(a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))/x^(1/4)+40*a^2* b^3*x^(1/4)*(a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))+20/3*a^4*b *x^(3/4)*(a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))+a^5*x*(a^2+2* a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))+5*a*b^4*ln(x)*(a^2+2*a*b/x^(1 /4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))+20*a^3*b^2*x^(1/2)*(a^2+2*a*b/x^(1/4) +b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))
Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.34 \[ \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx=\frac {\sqrt {\frac {\left (b+a \sqrt [4]{x}\right )^2}{\sqrt {x}}} \left (-12 b^5+120 a^2 b^3 \sqrt {x}+60 a^3 b^2 x^{3/4}+20 a^4 b x+3 a^5 x^{5/4}+15 a b^4 \sqrt [4]{x} \log (x)\right )}{3 \left (b+a \sqrt [4]{x}\right )} \]
(Sqrt[(b + a*x^(1/4))^2/Sqrt[x]]*(-12*b^5 + 120*a^2*b^3*Sqrt[x] + 60*a^3*b ^2*x^(3/4) + 20*a^4*b*x + 3*a^5*x^(5/4) + 15*a*b^4*x^(1/4)*Log[x]))/(3*(b + a*x^(1/4)))
Time = 0.26 (sec) , antiderivative size = 118, 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.
\(\displaystyle \int \left (a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}} \int \left (\frac {b^2}{\sqrt [4]{x}}+a b\right )^5dx}{a b^5+\frac {b^6}{\sqrt [4]{x}}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {4 \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}} \int b^5 \left (a+\frac {b}{\sqrt [4]{x}}\right )^5 x^{3/4}d\sqrt [4]{x}}{a b^5+\frac {b^6}{\sqrt [4]{x}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}} \int \left (a+\frac {b}{\sqrt [4]{x}}\right )^5 x^{3/4}d\sqrt [4]{x}}{a b^5+\frac {b^6}{\sqrt [4]{x}}}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {4 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}} \int \frac {\left (\sqrt [4]{x} a+b\right )^5}{\sqrt {x}}d\sqrt [4]{x}}{a b^5+\frac {b^6}{\sqrt [4]{x}}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {4 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}} \int \left (x^{3/4} a^5+5 b \sqrt {x} a^4+10 b^2 \sqrt [4]{x} a^3+10 b^3 a^2+\frac {5 b^4 a}{\sqrt [4]{x}}+\frac {b^5}{\sqrt {x}}\right )d\sqrt [4]{x}}{a b^5+\frac {b^6}{\sqrt [4]{x}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}} \left (\frac {a^5 x}{4}+\frac {5}{3} a^4 b x^{3/4}+5 a^3 b^2 \sqrt {x}+10 a^2 b^3 \sqrt [4]{x}+5 a b^4 \log \left (\sqrt [4]{x}\right )-\frac {b^5}{\sqrt [4]{x}}\right )}{a b^5+\frac {b^6}{\sqrt [4]{x}}}\) |
(4*b^5*Sqrt[a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/4)]*(-(b^5/x^(1/4)) + 10*a^2* b^3*x^(1/4) + 5*a^3*b^2*Sqrt[x] + (5*a^4*b*x^(3/4))/3 + (a^5*x)/4 + 5*a*b^ 4*Log[x^(1/4)]))/(a*b^5 + b^6/x^(1/4))
3.5.89.3.1 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 = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.31
method | result | size |
derivativedivides | \(\frac {\left (\frac {a^{2} \sqrt {x}+b^{2}+2 a b \,x^{\frac {1}{4}}}{\sqrt {x}}\right )^{\frac {5}{2}} x \left (3 a^{5} x^{\frac {5}{4}}+20 x \,a^{4} b +60 a^{3} b^{2} x^{\frac {3}{4}}+15 b^{4} a \ln \left (x \right ) x^{\frac {1}{4}}+120 a^{2} b^{3} \sqrt {x}-12 b^{5}\right )}{3 \left (a \,x^{\frac {1}{4}}+b \right )^{5}}\) | \(91\) |
default | \(\frac {\left (\frac {a^{2} x^{\frac {3}{4}}+b^{2} x^{\frac {1}{4}}+2 a b \sqrt {x}}{x^{\frac {3}{4}}}\right )^{\frac {5}{2}} x \left (3 a^{5} x^{\frac {5}{4}}+20 x \,a^{4} b +60 a^{3} b^{2} x^{\frac {3}{4}}+15 b^{4} a \ln \left (x \right ) x^{\frac {1}{4}}+120 a^{2} b^{3} \sqrt {x}-12 b^{5}\right )}{3 \left (a \,x^{\frac {1}{4}}+b \right )^{5}}\) | \(95\) |
1/3*((a^2*x^(1/2)+b^2+2*a*b*x^(1/4))/x^(1/2))^(5/2)*x*(3*a^5*x^(5/4)+20*x* a^4*b+60*a^3*b^2*x^(3/4)+15*b^4*a*ln(x)*x^(1/4)+120*a^2*b^3*x^(1/2)-12*b^5 )/(a*x^(1/4)+b)^5
Timed out. \[ \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx=\text {Timed out} \]
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.20 \[ \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx=5 \, a b^{4} \log \left (x\right ) + \frac {3 \, a^{5} x^{\frac {5}{4}} + 20 \, a^{4} b x + 60 \, a^{3} b^{2} x^{\frac {3}{4}} + 120 \, a^{2} b^{3} \sqrt {x} - 12 \, b^{5}}{3 \, x^{\frac {1}{4}}} \]
5*a*b^4*log(x) + 1/3*(3*a^5*x^(5/4) + 20*a^4*b*x + 60*a^3*b^2*x^(3/4) + 12 0*a^2*b^3*sqrt(x) - 12*b^5)/x^(1/4)
Time = 0.31 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.44 \[ \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx=a^{5} x \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + 5 \, a b^{4} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + \frac {20}{3} \, a^{4} b x^{\frac {3}{4}} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + 20 \, a^{3} b^{2} \sqrt {x} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + 40 \, a^{2} b^{3} x^{\frac {1}{4}} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) - \frac {4 \, b^{5} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right )}{x^{\frac {1}{4}}} \]
a^5*x*sgn(a*x + b*x^(3/4))*sgn(x) + 5*a*b^4*log(abs(x))*sgn(a*x + b*x^(3/4 ))*sgn(x) + 20/3*a^4*b*x^(3/4)*sgn(a*x + b*x^(3/4))*sgn(x) + 20*a^3*b^2*sq rt(x)*sgn(a*x + b*x^(3/4))*sgn(x) + 40*a^2*b^3*x^(1/4)*sgn(a*x + b*x^(3/4) )*sgn(x) - 4*b^5*sgn(a*x + b*x^(3/4))*sgn(x)/x^(1/4)
Timed out. \[ \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx=\int {\left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2\,a\,b}{x^{1/4}}\right )}^{5/2} \,d x \]