3.5.66 \(\int \frac {1}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \, dx\) [466]

Optimal. Leaf size=147 \[ -\frac {3 a \left (a+b \sqrt [3]{x}\right ) \sqrt [3]{x}}{b^2 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {3 \left (a+b \sqrt [3]{x}\right ) x^{2/3}}{2 b \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {3 a^2 \left (a+b \sqrt [3]{x}\right ) \log \left (a+b \sqrt [3]{x}\right )}{b^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

[Out]

-3*a*(a+b*x^(1/3))*x^(1/3)/b^2/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)+3/2*(a+b*x^(1/3))*x^(2/3)/b/(a^2+2*a*b*x^
(1/3)+b^2*x^(2/3))^(1/2)+3*a^2*(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 = 147, 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 \sqrt [3]{x} \left (a+b \sqrt [3]{x}\right )}{b^2 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {3 x^{2/3} \left (a+b \sqrt [3]{x}\right )}{2 b \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {3 a^2 \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[1/Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)],x]

[Out]

(-3*a*(a + b*x^(1/3))*x^(1/3))/(b^2*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) + (3*(a + b*x^(1/3))*x^(2/3))/(2*
b*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) + (3*a^2*(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}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \, dx &=3 \text {Subst}\left (\int \frac {x^2}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {\left (3 b \left (a+b \sqrt [3]{x}\right )\right ) \text {Subst}\left (\int \frac {x^2}{a b+b^2 x} \, dx,x,\sqrt [3]{x}\right )}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=\frac {\left (3 b \left (a+b \sqrt [3]{x}\right )\right ) \text {Subst}\left (\int \left (-\frac {a}{b^3}+\frac {x}{b^2}+\frac {a^2}{b^3 (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 {3 a \left (a+b \sqrt [3]{x}\right ) \sqrt [3]{x}}{b^2 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {3 \left (a+b \sqrt [3]{x}\right ) x^{2/3}}{2 b \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {3 a^2 \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 = 0.03, size = 65, normalized size = 0.44 \begin {gather*} \frac {3 \left (a+b \sqrt [3]{x}\right ) \left (b \left (-2 a+b \sqrt [3]{x}\right ) \sqrt [3]{x}+2 a^2 \log \left (a+b \sqrt [3]{x}\right )\right )}{2 b^3 \sqrt {\left (a+b \sqrt [3]{x}\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 103, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {3 \left (a +b \,x^{\frac {1}{3}}\right ) \left (b^{2} x^{\frac {2}{3}}+2 a^{2} \ln \left (a +b \,x^{\frac {1}{3}}\right )-2 a b \,x^{\frac {1}{3}}\right )}{2 \sqrt {\left (a +b \,x^{\frac {1}{3}}\right )^{2}}\, b^{3}}\)	\(52\)
default	\(\frac {\sqrt {a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}}\, \left (3 b^{2} x^{\frac {2}{3}}-6 a b \,x^{\frac {1}{3}}+2 a^{2} \ln \left (b^{3} x +a^{3}\right )+4 a^{2} \ln \left (a +b \,x^{\frac {1}{3}}\right )-2 a^{2} \ln \left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )\right )}{2 \left (a +b \,x^{\frac {1}{3}}\right ) b^{3}}\)	\(103\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 36, normalized size = 0.24 \begin {gather*} \frac {3 \, a^{2} \log \left (x^{\frac {1}{3}} + \frac {a}{b}\right )}{b^{3}} + \frac {3 \, x^{\frac {2}{3}}}{2 \, b} - \frac {3 \, a x^{\frac {1}{3}}}{b^{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 33, normalized size = 0.22 \begin {gather*} \frac {3 \, {\left (2 \, a^{2} \log \left (b x^{\frac {1}{3}} + a\right ) + b^{2} x^{\frac {2}{3}} - 2 \, a b x^{\frac {1}{3}}\right )}}{2 \, b^{3}} \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))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}}}\, 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))**(1/2),x)

[Out]

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

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Giac [A]
time = 3.89, size = 61, normalized size = 0.41 \begin {gather*} \frac {3 \, {\left (b x^{\frac {2}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) - 2 \, a x^{\frac {1}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right )\right )}}{2 \, b^{2}} + \frac {3 \, a^{2} \log \left ({\left | b x^{\frac {1}{3}} + a \right |}\right )}{b^{3} \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))^(1/2),x, algorithm="giac")

[Out]

3/2*(b*x^(2/3)*sgn(b*x^(1/3) + a) - 2*a*x^(1/3)*sgn(b*x^(1/3) + a))/b^2 + 3*a^2*log(abs(b*x^(1/3) + a))/(b^3*s
gn(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}{\sqrt {a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}}} \,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))^(1/2),x)

[Out]

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

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