3.468 \(\int \frac{1}{(a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{5/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{3 a^2}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{2 a}{b^3 \left (a+b \sqrt [3]{x}\right )^2 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac{3}{2 b^3 \left (a+b \sqrt [3]{x}\right ) \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

[Out]

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

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Rubi [A]  time = 0.0746981, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1341, 646, 43} \[ -\frac{3 a^2}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{2 a}{b^3 \left (a+b \sqrt [3]{x}\right )^2 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac{3}{2 b^3 \left (a+b \sqrt [3]{x}\right ) \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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 \frac{1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2}} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (3 b^5 \left (a+b \sqrt [3]{x}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a b+b^2 x\right )^5} \, dx,x,\sqrt [3]{x}\right )}{\sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=\frac{\left (3 b^5 \left (a+b \sqrt [3]{x}\right )\right ) \operatorname{Subst}\left (\int \left (\frac{a^2}{b^7 (a+b x)^5}-\frac{2 a}{b^7 (a+b x)^4}+\frac{1}{b^7 (a+b x)^3}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=-\frac{3 a^2}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{2 a}{b^3 \left (a+b \sqrt [3]{x}\right )^2 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac{3}{2 b^3 \left (a+b \sqrt [3]{x}\right ) \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ \end{align*}

Mathematica [A]  time = 0.036853, size = 58, normalized size = 0.43 \[ \frac{-a^2-4 a b \sqrt [3]{x}-6 b^2 x^{2/3}}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt{\left (a+b \sqrt [3]{x}\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-a^2 - 4*a*b*x^(1/3) - 6*b^2*x^(2/3))/(4*b^3*(a + b*x^(1/3))^3*Sqrt[(a + b*x^(1/3))^2])

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Maple [A]  time = 0.006, size = 54, normalized size = 0.4 \begin{align*} -{\frac{1}{4\,{b}^{3}}\sqrt{{a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}}} \left ( 6\,{b}^{2}{x}^{2/3}+4\,ab\sqrt [3]{x}+{a}^{2} \right ) \left ( a+b\sqrt [3]{x} \right ) ^{-5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(6*b^2*x^(2/3)+4*a*b*x^(1/3)+a^2)/(a+b*x^(1/3))^5/b^3

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Maxima [A]  time = 1.09819, size = 85, normalized size = 0.63 \begin{align*} -\frac{3 \, a^{2} b^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x^{\frac{1}{3}} + \frac{a}{b}\right )}^{4}} + \frac{2 \, a b}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x^{\frac{1}{3}} + \frac{a}{b}\right )}^{3}} - \frac{3}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x^{\frac{1}{3}} + \frac{a}{b}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*a^2*b^2/((b^2)^(9/2)*(x^(1/3) + a/b)^4) + 2*a*b/((b^2)^(7/2)*(x^(1/3) + a/b)^3) - 3/2/((b^2)^(5/2)*(x^(1/
3) + a/b)^2)

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Fricas [A]  time = 1.87059, size = 285, normalized size = 2.11 \begin{align*} \frac{20 \, a b^{9} x^{3} - 60 \, a^{4} b^{6} x^{2} - a^{10} - 9 \,{\left (5 \, a^{2} b^{8} x^{2} - 4 \, a^{5} b^{5} x\right )} x^{\frac{2}{3}} - 3 \,{\left (2 \, b^{10} x^{3} - 20 \, a^{3} b^{7} x^{2} + 5 \, a^{6} b^{4} x\right )} x^{\frac{1}{3}}}{4 \,{\left (b^{15} x^{4} + 4 \, a^{3} b^{12} x^{3} + 6 \, a^{6} b^{9} x^{2} + 4 \, a^{9} b^{6} x + a^{12} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(20*a*b^9*x^3 - 60*a^4*b^6*x^2 - a^10 - 9*(5*a^2*b^8*x^2 - 4*a^5*b^5*x)*x^(2/3) - 3*(2*b^10*x^3 - 20*a^3*b
^7*x^2 + 5*a^6*b^4*x)*x^(1/3))/(b^15*x^4 + 4*a^3*b^12*x^3 + 6*a^6*b^9*x^2 + 4*a^9*b^6*x + a^12*b^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a^{2} + 2 a b \sqrt [3]{x} + 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(1/(a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef