3.483 \(\int (a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}})^{5/2} \, dx\)

Optimal. Leaf size=291 \[ \frac{a^5 x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{15 a^4 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{30 a^3 b^2 \sqrt [3]{x} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}-\frac{15 a b^4 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{\sqrt [3]{x} \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{3 b^5 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 x^{2/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{30 a^2 b^3 \log \left (\sqrt [3]{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

[Out]

(-3*b^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/(2*(a + b/x^(1/3))*x^(2/3)) - (15*a*b^4*Sqrt[a^2 + b^2/x^(2
/3) + (2*a*b)/x^(1/3)])/((a + b/x^(1/3))*x^(1/3)) + (30*a^3*b^2*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(1
/3))/(a + b/x^(1/3)) + (15*a^4*b*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(2/3))/(2*(a + b/x^(1/3))) + (a^5
*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x)/(a + b/x^(1/3)) + (30*a^2*b^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x
^(1/3)]*Log[x^(1/3)])/(a + b/x^(1/3))

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Rubi [A]  time = 0.137337, antiderivative size = 291, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1341, 1355, 263, 43} \[ \frac{a^5 x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{15 a^4 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{30 a^3 b^2 \sqrt [3]{x} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}-\frac{15 a b^4 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{\sqrt [3]{x} \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{3 b^5 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 x^{2/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{30 a^2 b^3 \log \left (\sqrt [3]{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2),x]

[Out]

(-3*b^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/(2*(a + b/x^(1/3))*x^(2/3)) - (15*a*b^4*Sqrt[a^2 + b^2/x^(2
/3) + (2*a*b)/x^(1/3)])/((a + b/x^(1/3))*x^(1/3)) + (30*a^3*b^2*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(1
/3))/(a + b/x^(1/3)) + (15*a^4*b*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(2/3))/(2*(a + b/x^(1/3))) + (a^5
*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x)/(a + b/x^(1/3)) + (30*a^2*b^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x
^(1/3)]*Log[x^(1/3)])/(a + b/x^(1/3))

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^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}\right )^{5/2} \, dx &=3 \operatorname{Subst}\left (\int \left (a^2+\frac{b^2}{x^2}+\frac{2 a b}{x}\right )^{5/2} x^2 \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \left (a b+\frac{b^2}{x}\right )^5 x^2 \, dx,x,\sqrt [3]{x}\right )}{b^4 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \frac{\left (b^2+a b x\right )^5}{x^3} \, dx,x,\sqrt [3]{x}\right )}{b^4 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \left (10 a^3 b^7+\frac{b^{10}}{x^3}+\frac{5 a b^9}{x^2}+\frac{10 a^2 b^8}{x}+5 a^4 b^6 x+a^5 b^5 x^2\right ) \, dx,x,\sqrt [3]{x}\right )}{b^4 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}\\ &=-\frac{3 b^6 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}{2 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right ) x^{2/3}}-\frac{15 a b^5 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}{\left (a b+\frac{b^2}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}+\frac{30 a^3 b^3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \sqrt [3]{x}}{a b+\frac{b^2}{\sqrt [3]{x}}}+\frac{15 a^4 b^2 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} x^{2/3}}{2 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}+\frac{a^5 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} x}{a+\frac{b}{\sqrt [3]{x}}}+\frac{10 a^2 b^4 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \log (x)}{a b+\frac{b^2}{\sqrt [3]{x}}}\\ \end{align*}

Mathematica [A]  time = 0.0591039, size = 99, normalized size = 0.34 \[ \frac{\left (a \sqrt [3]{x}+b\right ) \left (20 a^2 b^3 x^{2/3} \log (x)+60 a^3 b^2 x+15 a^4 b x^{4/3}+2 a^5 x^{5/3}-30 a b^4 \sqrt [3]{x}-3 b^5\right )}{2 x \sqrt{\frac{\left (a \sqrt [3]{x}+b\right )^2}{x^{2/3}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(5/2),x]

[Out]

((b + a*x^(1/3))*(-3*b^5 - 30*a*b^4*x^(1/3) + 60*a^3*b^2*x + 15*a^4*b*x^(4/3) + 2*a^5*x^(5/3) + 20*a^2*b^3*x^(
2/3)*Log[x]))/(2*Sqrt[(b + a*x^(1/3))^2/x^(2/3)]*x)

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Maple [A]  time = 0.009, size = 91, normalized size = 0.3 \begin{align*}{\frac{x}{2} \left ({ \left ({a}^{2}{x}^{{\frac{2}{3}}}+2\,ab\sqrt [3]{x}+{b}^{2} \right ){x}^{-{\frac{2}{3}}}} \right ) ^{{\frac{5}{2}}} \left ( 15\,{a}^{4}b{x}^{4/3}+60\,{a}^{3}{b}^{2}x+20\,{a}^{2}{b}^{3}\ln \left ( x \right ){x}^{2/3}+2\,{a}^{5}{x}^{5/3}-30\,a{b}^{4}\sqrt [3]{x}-3\,{b}^{5} \right ) \left ( b+a\sqrt [3]{x} \right ) ^{-5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x)

[Out]

1/2*((a^2*x^(2/3)+2*a*b*x^(1/3)+b^2)/x^(2/3))^(5/2)*x*(15*a^4*b*x^(4/3)+60*a^3*b^2*x+20*a^2*b^3*ln(x)*x^(2/3)+
2*a^5*x^(5/3)-30*a*b^4*x^(1/3)-3*b^5)/(b+a*x^(1/3))^5

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Maxima [A]  time = 1.03974, size = 77, normalized size = 0.26 \begin{align*} 10 \, a^{2} b^{3} \log \left (x\right ) + \frac{2 \, a^{5} x^{\frac{5}{3}} + 15 \, a^{4} b x^{\frac{4}{3}} + 60 \, a^{3} b^{2} x - 30 \, a b^{4} x^{\frac{1}{3}} - 3 \, b^{5}}{2 \, x^{\frac{2}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="maxima")

[Out]

10*a^2*b^3*log(x) + 1/2*(2*a^5*x^(5/3) + 15*a^4*b*x^(4/3) + 60*a^3*b^2*x - 30*a*b^4*x^(1/3) - 3*b^5)/x^(2/3)

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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^(2/3)+2*a*b/x^(1/3))^(5/2),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 [3]{x}} + \frac{b^{2}}{x^{\frac{2}{3}}}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2+b**2/x**(2/3)+2*a*b/x**(1/3))**(5/2),x)

[Out]

Integral((a**2 + 2*a*b/x**(1/3) + b**2/x**(2/3))**(5/2), x)

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Giac [A]  time = 1.26051, size = 173, normalized size = 0.59 \begin{align*} a^{5} x \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + 10 \, a^{2} b^{3} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + \frac{15}{2} \, a^{4} b x^{\frac{2}{3}} \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + 30 \, a^{3} b^{2} x^{\frac{1}{3}} \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) - \frac{3 \,{\left (10 \, a b^{4} x^{\frac{1}{3}} \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right ) + b^{5} \mathrm{sgn}\left (a x + b x^{\frac{2}{3}}\right ) \mathrm{sgn}\left (x\right )\right )}}{2 \, x^{\frac{2}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(5/2),x, algorithm="giac")

[Out]

a^5*x*sgn(a*x + b*x^(2/3))*sgn(x) + 10*a^2*b^3*log(abs(x))*sgn(a*x + b*x^(2/3))*sgn(x) + 15/2*a^4*b*x^(2/3)*sg
n(a*x + b*x^(2/3))*sgn(x) + 30*a^3*b^2*x^(1/3)*sgn(a*x + b*x^(2/3))*sgn(x) - 3/2*(10*a*b^4*x^(1/3)*sgn(a*x + b
*x^(2/3))*sgn(x) + b^5*sgn(a*x + b*x^(2/3))*sgn(x))/x^(2/3)