\(\int (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}})^{3/2} \, dx\) [481]

Optimal result

Integrand size = 24, antiderivative size = 179 \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=-\frac {2 b^3 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}}}{\left (a+\frac {b}{\sqrt {x}}\right ) \sqrt {x}}+\frac {6 a^2 b \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} \sqrt {x}}{a+\frac {b}{\sqrt {x}}}+\frac {a^3 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} x}{a+\frac {b}{\sqrt {x}}}+\frac {6 a b^2 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} \log \left (\sqrt {x}\right )}{a+\frac {b}{\sqrt {x}}} \]

[Out]

a^3*x*(a^2+b^2/x+2*a*b/x^(1/2))^(1/2)/(a+b/x^(1/2))+3*a*b^2*ln(x)*(a^2+b^2/x+2*a*b/x^(1/2))^(1/2)/(a+b/x^(1/2)
)-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*x^(1/2)*(a^2+b^2/x+2*a*b/x^(1/2))^(1/2)/
(a+b/x^(1/2))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1355, 1369, 269, 45} \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\frac {6 a^2 b \sqrt {x} \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}}}{a+\frac {b}{\sqrt {x}}}+\frac {6 a b^2 \log \left (\sqrt {x}\right ) \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}}}{a+\frac {b}{\sqrt {x}}}-\frac {2 b^3 \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}}}{\sqrt {x} \left (a+\frac {b}{\sqrt {x}}\right )}+\frac {a^3 x \sqrt {a^2+\frac {2 a b}{\sqrt {x}}+\frac {b^2}{x}}}{a+\frac {b}{\sqrt {x}}} \]

[In]

Int[(a^2 + b^2/x + (2*a*b)/Sqrt[x])^(3/2),x]

[Out]

(-2*b^3*Sqrt[a^2 + b^2/x + (2*a*b)/Sqrt[x]])/((a + b/Sqrt[x])*Sqrt[x]) + (6*a^2*b*Sqrt[a^2 + b^2/x + (2*a*b)/S
qrt[x]]*Sqrt[x])/(a + b/Sqrt[x]) + (a^3*Sqrt[a^2 + b^2/x + (2*a*b)/Sqrt[x]]*x)/(a + b/Sqrt[x]) + (6*a*b^2*Sqrt
[a^2 + b^2/x + (2*a*b)/Sqrt[x]]*Log[Sqrt[x]])/(a + b/Sqrt[x])

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 269

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]

Rule 1355

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \left (a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}\right )^{3/2} x \, dx,x,\sqrt {x}\right ) \\ & = \frac {\left (2 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}}\right ) \text {Subst}\left (\int \left (a b+\frac {b^2}{x}\right )^3 x \, dx,x,\sqrt {x}\right )}{b^2 \left (a b+\frac {b^2}{\sqrt {x}}\right )} \\ & = \frac {\left (2 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}}\right ) \text {Subst}\left (\int \frac {\left (b^2+a b x\right )^3}{x^2} \, dx,x,\sqrt {x}\right )}{b^2 \left (a b+\frac {b^2}{\sqrt {x}}\right )} \\ & = \frac {\left (2 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}}\right ) \text {Subst}\left (\int \left (3 a^2 b^4+\frac {b^6}{x^2}+\frac {3 a b^5}{x}+a^3 b^3 x\right ) \, dx,x,\sqrt {x}\right )}{b^2 \left (a b+\frac {b^2}{\sqrt {x}}\right )} \\ & = -\frac {2 b^4 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}}}{\left (a b+\frac {b^2}{\sqrt {x}}\right ) \sqrt {x}}+\frac {6 a^2 b^2 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} \sqrt {x}}{a b+\frac {b^2}{\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^3 \sqrt {a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}} \log (x)}{a b+\frac {b^2}{\sqrt {x}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.37 \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\frac {\sqrt {\frac {\left (b+a \sqrt {x}\right )^2}{x}} \left (-2 b^3+6 a^2 b x+a^3 x^{3/2}+3 a b^2 \sqrt {x} \log (x)\right )}{b+a \sqrt {x}} \]

[In]

Integrate[(a^2 + b^2/x + (2*a*b)/Sqrt[x])^(3/2),x]

[Out]

(Sqrt[(b + a*Sqrt[x])^2/x]*(-2*b^3 + 6*a^2*b*x + a^3*x^(3/2) + 3*a*b^2*Sqrt[x]*Log[x]))/(b + a*Sqrt[x])

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.36

method	result	size
derivativedivides	\(\frac {\left (\frac {a^{2} x +b^{2}+2 a b \sqrt {x}}{x}\right )^{\frac {3}{2}} x \left (a^{3} x^{\frac {3}{2}}+3 b^{2} a \ln \left (x \right ) \sqrt {x}+6 a^{2} b x -2 b^{3}\right )}{\left (a \sqrt {x}+b \right )^{3}}\)	\(65\)
default	\(\frac {\left (\frac {a^{2} x^{\frac {3}{2}}+b^{2} \sqrt {x}+2 a b x}{x^{\frac {3}{2}}}\right )^{\frac {3}{2}} \left (x^{\frac {5}{2}} a^{3}+3 x^{\frac {3}{2}} \ln \left (x \right ) a \,b^{2}+6 a^{2} b \,x^{2}-2 b^{3} x \right )}{\left (a \sqrt {x}+b \right )^{3}}\)	\(71\)

[In]

int((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x,method=_RETURNVERBOSE)

[Out]

((a^2*x+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*a^2*b*x-2*b^3)/(a*x^(1/2)+b)^3

Fricas [F(-1)]

Timed out. \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\text {Timed out} \]

[In]

integrate((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\int \left (a^{2} + \frac {2 a b}{\sqrt {x}} + \frac {b^{2}}{x}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate((a**2+b**2/x+2*a*b/x**(1/2))**(3/2),x)

[Out]

Integral((a**2 + 2*a*b/sqrt(x) + b**2/x)**(3/2), x)

Maxima [F]

\[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\int { {\left (a^{2} + \frac {2 \, a b}{\sqrt {x}} + \frac {b^{2}}{x}\right )}^{\frac {3}{2}} \,d x } \]

[In]

integrate((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x, algorithm="maxima")

[Out]

a^3*x + 3*a*b^2*integrate(1/x, x) + 6*a^2*b*sqrt(x) - 2*b^3/sqrt(x)

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.45 \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=a^{3} x \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right ) + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right ) + 6 \, a^{2} b \sqrt {x} \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right ) - \frac {2 \, b^{3} \mathrm {sgn}\left (a x + b \sqrt {x}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {x}} \]

[In]

integrate((a^2+b^2/x+2*a*b/x^(1/2))^(3/2),x, algorithm="giac")

[Out]

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)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+\frac {b^2}{x}+\frac {2 a b}{\sqrt {x}}\right )^{3/2} \, dx=\int {\left (a^2+\frac {b^2}{x}+\frac {2\,a\,b}{\sqrt {x}}\right )}^{3/2} \,d x \]

[In]

int((a^2 + b^2/x + (2*a*b)/x^(1/2))^(3/2),x)

[Out]

int((a^2 + b^2/x + (2*a*b)/x^(1/2))^(3/2), x)