3.5.68 \(\int \frac {1}{(a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{5/2}} \, dx\) [468]

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]

-3/4*a^2/b^3/(a+b*x^(1/3))^3/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)+2*a/b^3/(a+b*x^(1/3))^2/(a^2+2*a*b*x^(1/3)+
b^2*x^(2/3))^(1/2)-3/2/b^3/(a+b*x^(1/3))/(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 = 135, 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}{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 {gather*}

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 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 )^{5/2}} \, dx &=3 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.04, size = 56, normalized size = 0.41 \begin {gather*} \frac {\left (a+b \sqrt [3]{x}\right ) \left (-a^2-4 a b \sqrt [3]{x}-6 b^2 x^{2/3}\right )}{4 b^3 \left (\left (a+b \sqrt [3]{x}\right )^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 54, normalized size = 0.40

method	result	size
derivativedivides	\(-\frac {\left (6 b^{2} x^{\frac {2}{3}}+4 a b \,x^{\frac {1}{3}}+a^{2}\right ) \left (a +b \,x^{\frac {1}{3}}\right )}{4 b^{3} \left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )^{\frac {5}{2}}}\)	\(43\)
default	\(-\frac {\sqrt {a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}}\, \left (6 b^{2} x^{\frac {2}{3}}+4 a b \,x^{\frac {1}{3}}+a^{2}\right )}{4 \left (a +b \,x^{\frac {1}{3}}\right )^{5} b^{3}}\)	\(54\)

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,method=_RETURNVERBOSE)

[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 = 0.28, size = 53, normalized size = 0.39 \begin {gather*} -\frac {3}{2 \, b^{5} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{2}} + \frac {2 \, a}{b^{6} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{2}}{4 \, b^{7} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{4}} \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))^(5/2),x, algorithm="maxima")

[Out]

-3/2/(b^5*(x^(1/3) + a/b)^2) + 2*a/(b^6*(x^(1/3) + a/b)^3) - 3/4*a^2/(b^7*(x^(1/3) + a/b)^4)

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Fricas [A]
time = 0.38, size = 136, normalized size = 1.01 \begin {gather*} \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 {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))^(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 = 0.00, 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 {5}{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))**(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 [A]
time = 2.60, size = 43, normalized size = 0.32 \begin {gather*} -\frac {6 \, b^{2} x^{\frac {2}{3}} + 4 \, a b x^{\frac {1}{3}} + a^{2}}{4 \, {\left (b x^{\frac {1}{3}} + a\right )}^{4} 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))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.80, size = 53, normalized size = 0.39 \begin {gather*} -\frac {\sqrt {a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}}\,\left (a^2+6\,b^2\,x^{2/3}+4\,a\,b\,x^{1/3}\right )}{4\,b^3\,{\left (a+b\,x^{1/3}\right )}^5} \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))^(5/2),x)

[Out]

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

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