3.5.87 \(\int \frac {1}{(a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}})^{3/2}} \, dx\) [487]

Optimal. Leaf size=300 \[ \frac {3 b^5 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{2 a^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )^2}-\frac {15 b^4 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{a^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )}+\frac {18 b^2 \left (a+\frac {b}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}{a^5 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}-\frac {9 b \left (a+\frac {b}{\sqrt [3]{x}}\right ) x^{2/3}}{2 a^4 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}+\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right ) x}{a^3 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}-\frac {30 b^3 \left (a+\frac {b}{\sqrt [3]{x}}\right ) \log \left (b+a \sqrt [3]{x}\right )}{a^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}} \]

[Out]

3/2*b^5*(a+b/x^(1/3))/a^6/(b+a*x^(1/3))^2/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)-15*b^4*(a+b/x^(1/3))/a^6/(b+a*
x^(1/3))/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)+18*b^2*(a+b/x^(1/3))*x^(1/3)/a^5/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3)
)^(1/2)-9/2*b*(a+b/x^(1/3))*x^(2/3)/a^4/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)+(a+b/x^(1/3))*x/a^3/(a^2+b^2/x^(
2/3)+2*a*b/x^(1/3))^(1/2)-30*b^3*(a+b/x^(1/3))*ln(b+a*x^(1/3))/a^6/(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 300, normalized size of antiderivative = 1.00, 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 = {1355, 1369, 269, 45} \begin {gather*} \frac {3 b^5 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{2 a^6 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )^2}-\frac {15 b^4 \left (a+\frac {b}{\sqrt [3]{x}}\right )}{a^6 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )}-\frac {30 b^3 \left (a+\frac {b}{\sqrt [3]{x}}\right ) \log \left (a \sqrt [3]{x}+b\right )}{a^6 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}+\frac {18 b^2 \sqrt [3]{x} \left (a+\frac {b}{\sqrt [3]{x}}\right )}{a^5 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}-\frac {9 b x^{2/3} \left (a+\frac {b}{\sqrt [3]{x}}\right )}{2 a^4 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}}+\frac {x \left (a+\frac {b}{\sqrt [3]{x}}\right )}{a^3 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^5*(a + b/x^(1/3)))/(2*a^6*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))^2) - (15*b^4*(a + b/x
^(1/3)))/(a^6*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))) + (18*b^2*(a + b/x^(1/3))*x^(1/3))/(a
^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]) - (9*b*(a + b/x^(1/3))*x^(2/3))/(2*a^4*Sqrt[a^2 + b^2/x^(2/3) +
(2*a*b)/x^(1/3)]) + ((a + b/x^(1/3))*x)/(a^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]) - (30*b^3*(a + b/x^(1/
3))*Log[b + a*x^(1/3)])/(a^6*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])

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 269

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 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]

Rule 1369

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{3/2}} \, dx &=3 \text {Subst}\left (\int \frac {x^2}{\left (a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}\right )^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {\left (3 b^2 \left (a b+\frac {b^2}{\sqrt [3]{x}}\right )\right ) \text {Subst}\left (\int \frac {x^2}{\left (a b+\frac {b^2}{x}\right )^3} \, dx,x,\sqrt [3]{x}\right )}{\sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}\\ &=\frac {\left (3 b^2 \left (a b+\frac {b^2}{\sqrt [3]{x}}\right )\right ) \text {Subst}\left (\int \frac {x^5}{\left (b^2+a b x\right )^3} \, dx,x,\sqrt [3]{x}\right )}{\sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}\\ &=\frac {\left (3 b^2 \left (a b+\frac {b^2}{\sqrt [3]{x}}\right )\right ) \text {Subst}\left (\int \left (\frac {6}{a^5 b}-\frac {3 x}{a^4 b^2}+\frac {x^2}{a^3 b^3}-\frac {b^2}{a^5 (b+a x)^3}+\frac {5 b}{a^5 (b+a x)^2}-\frac {10}{a^5 (b+a x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}\\ &=\frac {3 \left (a b^5+\frac {b^6}{\sqrt [3]{x}}\right )}{2 a^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )^2}-\frac {15 \left (a b^4+\frac {b^5}{\sqrt [3]{x}}\right )}{a^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )}+\frac {18 \left (a b^2+\frac {b^3}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}{a^5 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}-\frac {9 \left (a b+\frac {b^2}{\sqrt [3]{x}}\right ) x^{2/3}}{2 a^4 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}+\frac {\left (a+\frac {b}{\sqrt [3]{x}}\right ) x}{a^3 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}-\frac {30 \left (a b^3+\frac {b^4}{\sqrt [3]{x}}\right ) \log \left (b+a \sqrt [3]{x}\right )}{a^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 126, normalized size = 0.42 \begin {gather*} \frac {\left (b+a \sqrt [3]{x}\right ) \left (-27 b^5+6 a b^4 \sqrt [3]{x}+63 a^2 b^3 x^{2/3}+20 a^3 b^2 x-5 a^4 b x^{4/3}+2 a^5 x^{5/3}-60 b^3 \left (b+a \sqrt [3]{x}\right )^2 \log \left (b+a \sqrt [3]{x}\right )\right )}{2 a^6 \left (\frac {\left (b+a \sqrt [3]{x}\right )^2}{x^{2/3}}\right )^{3/2} x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*x^(1/3))*(-27*b^5 + 6*a*b^4*x^(1/3) + 63*a^2*b^3*x^(2/3) + 20*a^3*b^2*x - 5*a^4*b*x^(4/3) + 2*a^5*x^(5
/3) - 60*b^3*(b + a*x^(1/3))^2*Log[b + a*x^(1/3)]))/(2*a^6*((b + a*x^(1/3))^2/x^(2/3))^(3/2)*x)

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Maple [A]
time = 0.04, size = 141, normalized size = 0.47

method	result	size
derivativedivides	\(-\frac {\left (-2 a^{5} x^{\frac {5}{3}}+5 b \,a^{4} x^{\frac {4}{3}}+60 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{2} b^{3} x^{\frac {2}{3}}-20 b^{2} a^{3} x +120 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a \,b^{4} x^{\frac {1}{3}}-63 b^{3} x^{\frac {2}{3}} a^{2}+60 \ln \left (b +a \,x^{\frac {1}{3}}\right ) b^{5}-6 b^{4} a \,x^{\frac {1}{3}}+27 b^{5}\right ) \left (b +a \,x^{\frac {1}{3}}\right )}{2 a^{6} x \left (\frac {a^{2} x^{\frac {2}{3}}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {3}{2}}}\)	\(141\)
default	\(\frac {\left (2 a^{5} x^{\frac {5}{3}}-5 b \,a^{4} x^{\frac {4}{3}}-60 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a^{2} b^{3} x^{\frac {2}{3}}+63 b^{3} x^{\frac {2}{3}} a^{2}-120 \ln \left (b +a \,x^{\frac {1}{3}}\right ) a \,b^{4} x^{\frac {1}{3}}+6 b^{4} a \,x^{\frac {1}{3}}-60 \ln \left (b +a \,x^{\frac {1}{3}}\right ) b^{5}+20 b^{2} a^{3} x -27 b^{5}\right ) \left (b +a \,x^{\frac {1}{3}}\right )}{2 \left (\frac {a^{2} x^{\frac {2}{3}}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {3}{2}} x \,a^{6}}\)	\(141\)

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,method=_RETURNVERBOSE)

[Out]

1/2/((a^2*x^(2/3)+2*a*b*x^(1/3)+b^2)/x^(2/3))^(3/2)/x*(2*a^5*x^(5/3)-5*b*a^4*x^(4/3)-60*ln(b+a*x^(1/3))*a^2*b^
3*x^(2/3)+63*b^3*x^(2/3)*a^2-120*ln(b+a*x^(1/3))*a*b^4*x^(1/3)+6*b^4*a*x^(1/3)-60*ln(b+a*x^(1/3))*b^5+20*b^2*a
^3*x-27*b^5)*(b+a*x^(1/3))/a^6

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Maxima [A]
time = 0.30, size = 97, normalized size = 0.32 \begin {gather*} \frac {2 \, a^{5} x^{\frac {5}{3}} - 5 \, a^{4} b x^{\frac {4}{3}} + 20 \, a^{3} b^{2} x + 63 \, a^{2} b^{3} x^{\frac {2}{3}} + 6 \, a b^{4} x^{\frac {1}{3}} - 27 \, b^{5}}{2 \, {\left (a^{8} x^{\frac {2}{3}} + 2 \, a^{7} b x^{\frac {1}{3}} + a^{6} b^{2}\right )}} - \frac {30 \, b^{3} \log \left (a x^{\frac {1}{3}} + b\right )}{a^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^5*x^(5/3) - 5*a^4*b*x^(4/3) + 20*a^3*b^2*x + 63*a^2*b^3*x^(2/3) + 6*a*b^4*x^(1/3) - 27*b^5)/(a^8*x^(2
/3) + 2*a^7*b*x^(1/3) + a^6*b^2) - 30*b^3*log(a*x^(1/3) + b)/a^6

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a^{2} + \frac {2 a b}{\sqrt [3]{x}} + \frac {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+b**2/x**(2/3)+2*a*b/x**(1/3))**(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 = 4.27, size = 127, normalized size = 0.42 \begin {gather*} -\frac {30 \, b^{3} \log \left ({\left | a x^{\frac {1}{3}} + b \right |}\right )}{a^{6} \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}} + 9 \, b^{5}\right )}}{2 \, {\left (a x^{\frac {1}{3}} + b\right )}^{2} a^{6} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )} + \frac {2 \, a^{6} x - 9 \, a^{5} b x^{\frac {2}{3}} + 36 \, a^{4} b^{2} x^{\frac {1}{3}}}{2 \, a^{9} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-30*b^3*log(abs(a*x^(1/3) + b))/(a^6*sgn(a*x + b*x^(2/3))*sgn(x)) - 3/2*(10*a*b^4*x^(1/3) + 9*b^5)/((a*x^(1/3)
+ b)^2*a^6*sgn(a*x + b*x^(2/3))*sgn(x)) + 1/2*(2*a^6*x - 9*a^5*b*x^(2/3) + 36*a^4*b^2*x^(1/3))/(a^9*sgn(a*x +
b*x^(2/3))*sgn(x))

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