3.5.67 \(\int \frac {1}{(a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{3/2}} \, dx\) [467]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 130, normalized size of antiderivative = 1.00, 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 = {1355, 660, 45} \begin {gather*} -\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}}} \end {gather*}

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 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 660

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 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]

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 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.05, size = 72, normalized size = 0.55 \begin {gather*} \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}} \end {gather*}

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 = 0.04, size = 92, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {3 \left (2 \ln \left (a +b \,x^{\frac {1}{3}}\right ) b^{2} x^{\frac {2}{3}}+4 \ln \left (a +b \,x^{\frac {1}{3}}\right ) a b \,x^{\frac {1}{3}}+2 a^{2} \ln \left (a +b \,x^{\frac {1}{3}}\right )+4 a b \,x^{\frac {1}{3}}+3 a^{2}\right ) \left (a +b \,x^{\frac {1}{3}}\right )}{2 b^{3} \left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )^{\frac {3}{2}}}\)	\(81\)
default	\(\frac {3 \sqrt {a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}}\, \left (2 \ln \left (a +b \,x^{\frac {1}{3}}\right ) b^{2} x^{\frac {2}{3}}+4 \ln \left (a +b \,x^{\frac {1}{3}}\right ) a b \,x^{\frac {1}{3}}+2 a^{2} \ln \left (a +b \,x^{\frac {1}{3}}\right )+4 a b \,x^{\frac {1}{3}}+3 a^{2}\right )}{2 \left (a +b \,x^{\frac {1}{3}}\right )^{3} b^{3}}\)	\(92\)

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,method=_RETURNVERBOSE)

[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))*b^2*x^(2/3)+4*ln(a+b*x^(1/3))*a*b*x^(1/3)+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 = 0.41, size = 55, normalized size = 0.42 \begin {gather*} \frac {3 \, \log \left (x^{\frac {1}{3}} + \frac {a}{b}\right )}{b^{3}} + \frac {6 \, a x^{\frac {1}{3}}}{b^{4} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{2}} + \frac {9 \, a^{2}}{2 \, b^{5} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{2}} \end {gather*}

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^3 + 6*a*x^(1/3)/(b^4*(x^(1/3) + a/b)^2) + 9/2*a^2/(b^5*(x^(1/3) + a/b)^2)

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Fricas [A]
time = 0.36, size = 113, normalized size = 0.87 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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 = 4.08, size = 64, normalized size = 0.49 \begin {gather*} \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 {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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