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

Optimal. Leaf size=130 \[ -\frac{3 a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right ) \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{6 a}{b^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{3 \left (a+b \sqrt [3]{x}\right ) \log \left (a+b \sqrt [3]{x}\right )}{b^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

[Out]

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

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Rubi [A]  time = 0.0720178, antiderivative size = 130, 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} \[ -\frac{3 a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right ) \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{6 a}{b^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{3 \left (a+b \sqrt [3]{x}\right ) \log \left (a+b \sqrt [3]{x}\right )}{b^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a)/(b^3*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) - (3*a^2)/(2*b^3*(a + b*x^(1/3))*Sqrt[a^2 + 2*a*b*x^(1/3)
+ b^2*x^(2/3)]) + (3*(a + b*x^(1/3))*Log[a + b*x^(1/3)])/(b^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 )^{3/2}} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (3 b^3 \left (a+b \sqrt [3]{x}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a b+b^2 x\right )^3} \, dx,x,\sqrt [3]{x}\right )}{\sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=\frac{\left (3 b^3 \left (a+b \sqrt [3]{x}\right )\right ) \operatorname{Subst}\left (\int \left (\frac{a^2}{b^5 (a+b x)^3}-\frac{2 a}{b^5 (a+b x)^2}+\frac{1}{b^5 (a+b x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=\frac{6 a}{b^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac{3 a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right ) \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{3 \left (a+b \sqrt [3]{x}\right ) \log \left (a+b \sqrt [3]{x}\right )}{b^3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ \end{align*}

Mathematica [A]  time = 0.0463178, size = 72, normalized size = 0.55 \[ \frac{3 a \left (3 a+4 b \sqrt [3]{x}\right )+6 \left (a+b \sqrt [3]{x}\right )^2 \log \left (a+b \sqrt [3]{x}\right )}{2 b^3 \left (a+b \sqrt [3]{x}\right ) \sqrt{\left (a+b \sqrt [3]{x}\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 92, normalized size = 0.7 \begin{align*}{\frac{3}{2\,{b}^{3}}\sqrt{{a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}}} \left ( 2\,\ln \left ( a+b\sqrt [3]{x} \right ){x}^{2/3}{b}^{2}+4\,\ln \left ( a+b\sqrt [3]{x} \right ) \sqrt [3]{x}ab+2\,{a}^{2}\ln \left ( a+b\sqrt [3]{x} \right ) +4\,ab\sqrt [3]{x}+3\,{a}^{2} \right ) \left ( a+b\sqrt [3]{x} \right ) ^{-3}} \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))^(3/2),x)

[Out]

3/2*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(2*ln(a+b*x^(1/3))*x^(2/3)*b^2+4*ln(a+b*x^(1/3))*x^(1/3)*a*b+2*a^2*l
n(a+b*x^(1/3))+4*a*b*x^(1/3)+3*a^2)/(a+b*x^(1/3))^3/b^3

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Maxima [A]  time = 1.07779, size = 88, normalized size = 0.68 \begin{align*} \frac{3 \, \log \left (x^{\frac{1}{3}} + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{9 \, a^{2} b^{2}}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x^{\frac{1}{3}} + \frac{a}{b}\right )}^{2}} + \frac{6 \, a b x^{\frac{1}{3}}}{{\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))^(3/2),x, algorithm="maxima")

[Out]

3*log(x^(1/3) + a/b)/(b^2)^(3/2) + 9/2*a^2*b^2/((b^2)^(7/2)*(x^(1/3) + a/b)^2) + 6*a*b*x^(1/3)/((b^2)^(5/2)*(x
^(1/3) + a/b)^2)

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Fricas [A]  time = 1.86619, size = 243, normalized size = 1.87 \begin{align*} \frac{3 \,{\left (6 \, a^{3} b^{3} x + 3 \, a^{6} + 2 \,{\left (b^{6} x^{2} + 2 \, a^{3} b^{3} x + a^{6}\right )} \log \left (b x^{\frac{1}{3}} + a\right ) +{\left (4 \, a b^{5} x + a^{4} b^{2}\right )} x^{\frac{2}{3}} -{\left (5 \, a^{2} b^{4} x + 2 \, a^{5} b\right )} x^{\frac{1}{3}}\right )}}{2 \,{\left (b^{9} x^{2} + 2 \, a^{3} b^{6} x + a^{6} 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))^(3/2),x, algorithm="fricas")

[Out]

3/2*(6*a^3*b^3*x + 3*a^6 + 2*(b^6*x^2 + 2*a^3*b^3*x + a^6)*log(b*x^(1/3) + a) + (4*a*b^5*x + a^4*b^2)*x^(2/3)
- (5*a^2*b^4*x + 2*a^5*b)*x^(1/3))/(b^9*x^2 + 2*a^3*b^6*x + a^6*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{3}{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))**(3/2),x)

[Out]

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

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Giac [A]  time = 1.17455, size = 86, normalized size = 0.66 \begin{align*} \frac{3 \, \log \left ({\left | b x^{\frac{1}{3}} + a \right |}\right )}{b^{3} \mathrm{sgn}\left (b x^{\frac{1}{3}} + a\right )} + \frac{3 \,{\left (4 \, a x^{\frac{1}{3}} + \frac{3 \, a^{2}}{b}\right )}}{2 \,{\left (b x^{\frac{1}{3}} + a\right )}^{2} b^{2} \mathrm{sgn}\left (b x^{\frac{1}{3}} + a\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))^(3/2),x, algorithm="giac")

[Out]

3*log(abs(b*x^(1/3) + a))/(b^3*sgn(b*x^(1/3) + a)) + 3/2*(4*a*x^(1/3) + 3*a^2/b)/((b*x^(1/3) + a)^2*b^2*sgn(b*
x^(1/3) + a))