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

Optimal. Leaf size=179 \[ \frac{a^3 x \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{a+\frac{b}{\sqrt{x}}}+\frac{6 a^2 b \sqrt{x} \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{a+\frac{b}{\sqrt{x}}}-\frac{2 b^3 \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{\sqrt{x} \left (a+\frac{b}{\sqrt{x}}\right )}+\frac{6 a b^2 \log \left (\sqrt{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{a+\frac{b}{\sqrt{x}}} \]

[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])

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Rubi [A]  time = 0.0922638, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1341, 1355, 263, 43} \[ \frac{a^3 x \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{a+\frac{b}{\sqrt{x}}}+\frac{6 a^2 b \sqrt{x} \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{a+\frac{b}{\sqrt{x}}}-\frac{2 b^3 \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{\sqrt{x} \left (a+\frac{b}{\sqrt{x}}\right )}+\frac{6 a b^2 \log \left (\sqrt{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt{x}}+\frac{b^2}{x}}}{a+\frac{b}{\sqrt{x}}} \]

Antiderivative was successfully verified.

[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 1341

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 1355

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]

Rule 263

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 43

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])

Rubi steps

\begin{align*} \int \left (a^2+\frac{b^2}{x}+\frac{2 a b}{\sqrt{x}}\right )^{3/2} \, dx &=2 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.0335123, size = 66, normalized size = 0.37 \[ \frac{\sqrt{\frac{\left (a \sqrt{x}+b\right )^2}{x}} \left (6 a^2 b x+a^3 x^{3/2}+3 a b^2 \sqrt{x} \log (x)-2 b^3\right )}{a \sqrt{x}+b} \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.026, size = 68, normalized size = 0.4 \begin{align*}{\sqrt{{ \left ({a}^{2}{x}^{{\frac{3}{2}}}+{b}^{2}\sqrt{x}+2\,abx \right ){x}^{-{\frac{3}{2}}}}} \left ({x}^{{\frac{3}{2}}}{a}^{3}+6\,x{a}^{2}b+3\,\sqrt{x}\ln \left ( x \right ) a{b}^{2}-2\,{b}^{3} \right ) \left ( a\sqrt{x}+b \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*x^(3/2)+b^2*x^(1/2)+2*a*b*x)/x^(3/2))^(1/2)*(x^(3/2)*a^3+6*x*a^2*b+3*x^(1/2)*ln(x)*a*b^2-2*b^3)/(a*x^(1/
2)+b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} x + 3 \, a b^{2} \int \frac{1}{x}\,{d x} + 6 \, a^{2} b \sqrt{x} - \frac{2 \, b^{3}}{\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a^{2} + \frac{2 a b}{\sqrt{x}} + \frac{b^{2}}{x}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.18833, size = 108, normalized size = 0.6 \begin{align*} 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)