Integrand size = 24, antiderivative size = 175 \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=-\frac {2 b^3 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}}}{\left (a+\frac {b}{\sqrt {x}}\right ) \sqrt {x}}+\frac {6 a^2 b \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} \sqrt {x}}{a+\frac {b}{\sqrt {x}}}+\frac {a^3 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} x}{a+\frac {b}{\sqrt {x}}}+\frac {3 a b^2 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} \log (x)}{a+\frac {b}{\sqrt {x}}} \] Output:
-2*b^3*(a^2+b^2/x+2*a*b/x^(1/2))^(1/2)/(a+b/x^(1/2))/x^(1/2)+6*a^2*b*(a^2+ b^2/x+2*a*b/x^(1/2))^(1/2)*x^(1/2)/(a+b/x^(1/2))+a^3*(a^2+b^2/x+2*a*b/x^(1 /2))^(1/2)*x/(a+b/x^(1/2))+3*a*b^2*(a^2+b^2/x+2*a*b/x^(1/2))^(1/2)*ln(x)/( a+b/x^(1/2))
Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.38 \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\frac {\sqrt {\frac {\left (b+a \sqrt {x}\right )^2}{x}} \left (-2 b^3+6 a^2 b x+a^3 x^{3/2}+3 a b^2 \sqrt {x} \log (x)\right )}{b+a \sqrt {x}} \] Input:
Integrate[(a^2 + b^2/x + (2*a*b)/Sqrt[x])^(3/2),x]
Output:
(Sqrt[(b + a*Sqrt[x])^2/x]*(-2*b^3 + 6*a^2*b*x + a^3*x^(3/2) + 3*a*b^2*Sqr t[x]*Log[x]))/(b + a*Sqrt[x])
Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.50, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.
\(\displaystyle \int \left (a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}} \int \left (\frac {b^2}{\sqrt {x}}+a b\right )^3dx}{a b^3+\frac {b^4}{\sqrt {x}}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {2 \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}} \int b^3 \left (a+\frac {b}{\sqrt {x}}\right )^3 \sqrt {x}d\sqrt {x}}{a b^3+\frac {b^4}{\sqrt {x}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b^3 \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}} \int \left (a+\frac {b}{\sqrt {x}}\right )^3 \sqrt {x}d\sqrt {x}}{a b^3+\frac {b^4}{\sqrt {x}}}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {2 b^3 \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}} \int \frac {\left (\sqrt {x} a+b\right )^3}{x}d\sqrt {x}}{a b^3+\frac {b^4}{\sqrt {x}}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {2 b^3 \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}} \int \left (\sqrt {x} a^3+3 b a^2+\frac {3 b^2 a}{\sqrt {x}}+\frac {b^3}{x}\right )d\sqrt {x}}{a b^3+\frac {b^4}{\sqrt {x}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 b^3 \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}} \left (\frac {a^3 x}{2}+3 a^2 b \sqrt {x}+3 a b^2 \log \left (\sqrt {x}\right )-\frac {b^3}{\sqrt {x}}\right )}{a b^3+\frac {b^4}{\sqrt {x}}}\) |
Input:
Int[(a^2 + b^2/x + (2*a*b)/Sqrt[x])^(3/2),x]
Output:
(2*b^3*Sqrt[a^2 + b^2/x + (2*a*b)/Sqrt[x]]*(-(b^3/Sqrt[x]) + 3*a^2*b*Sqrt[ x] + (a^3*x)/2 + 3*a*b^2*Log[Sqrt[x]]))/(a*b^3 + b^4/Sqrt[x])
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 = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.37
method | result | size |
derivativedivides | \(\frac {\left (\frac {x \,a^{2}+b^{2}+2 a b \sqrt {x}}{x}\right )^{\frac {3}{2}} x \left (a^{3} x^{\frac {3}{2}}+3 b^{2} a \ln \left (x \right ) \sqrt {x}+6 b \,a^{2} x -2 b^{3}\right )}{\left (\sqrt {x}\, a +b \right )^{3}}\) | \(65\) |
default | \(\frac {\left (\frac {a^{2} x^{\frac {3}{2}}+b^{2} \sqrt {x}+2 a b x}{x^{\frac {3}{2}}}\right )^{\frac {3}{2}} \left (x^{\frac {5}{2}} a^{3}+3 \ln \left (x \right ) x^{\frac {3}{2}} a \,b^{2}+6 a^{2} b \,x^{2}-2 b^{3} x \right )}{\left (\sqrt {x}\, a +b \right )^{3}}\) | \(71\) |
Input:
int((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x,method=_RETURNVERBOSE)
Output:
((x*a^2+b^2+2*a*b*x^(1/2))/x)^(3/2)*x*(a^3*x^(3/2)+3*b^2*a*ln(x)*x^(1/2)+6 *b*a^2*x-2*b^3)/(x^(1/2)*a+b)^3
Timed out. \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\text {Timed out} \] Input:
integrate((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\int \left (a^{2} + \frac {2 a b}{\sqrt {x}} + \frac {b^{2}}{x}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a**2+b**2/x+2*a*b/x**(1/2))**(3/2),x)
Output:
Integral((a**2 + 2*a*b/sqrt(x) + b**2/x)**(3/2), x)
\[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\int { {\left (a^{2} + \frac {2 \, a b}{\sqrt {x}} + \frac {b^{2}}{x}\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x, algorithm="maxima")
Output:
a^3*x + 3*a*b^2*integrate(1/x, x) + 6*a^2*b*sqrt(x) - 2*b^3/sqrt(x)
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.46 \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=a^{3} x \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right ) + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right ) + 6 \, a^{2} b \sqrt {x} \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right ) - \frac {2 \, b^{3} \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {x}} \] Input:
integrate((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x, algorithm="giac")
Output:
a^3*x*sgn(a*x + b*sqrt(x))*sgn(x) + 3*a*b^2*log(abs(x))*sgn(a*x + b*sqrt(x ))*sgn(x) + 6*a^2*b*sqrt(x)*sgn(a*x + b*sqrt(x))*sgn(x) - 2*b^3*sgn(a*x + b*sqrt(x))*sgn(x)/sqrt(x)
Timed out. \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\int {\left (a^2+\frac {b^2}{x}+\frac {2\,a\,b}{\sqrt {x}}\right )}^{3/2} \,d x \] Input:
int((a^2 + b^2/x + (2*a*b)/x^(1/2))^(3/2),x)
Output:
int((a^2 + b^2/x + (2*a*b)/x^(1/2))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.21 \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\frac {6 \sqrt {x}\, \mathrm {log}\left (\sqrt {x}\right ) a \,b^{2}+\sqrt {x}\, a^{3} x +6 a^{2} b x -2 b^{3}}{\sqrt {x}} \] Input:
int((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x)
Output:
(6*sqrt(x)*log(sqrt(x))*a*b**2 + sqrt(x)*a**3*x + 6*a**2*b*x - 2*b**3)/sqr t(x)