3.5.80 \(\int \frac {1}{(a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x})^{5/2}} \, dx\) [480]

Optimal. Leaf size=268 \[ -\frac {60 a^2}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {3 a^5}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {10 a^4}{b^6 \left (a+b \sqrt [6]{x}\right )^2 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {30 a^3}{b^6 \left (a+b \sqrt [6]{x}\right ) \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {6 \left (a+b \sqrt [6]{x}\right ) \sqrt [6]{x}}{b^5 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {30 a \left (a+b \sqrt [6]{x}\right ) \log \left (a+b \sqrt [6]{x}\right )}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}} \]

[Out]

-60*a^2/b^6/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)+3/2*a^5/b^6/(a+b*x^(1/6))^3/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^
(1/2)-10*a^4/b^6/(a+b*x^(1/6))^2/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)+30*a^3/b^6/(a+b*x^(1/6))/(a^2+2*a*b*x^(
1/6)+b^2*x^(1/3))^(1/2)+6*(a+b*x^(1/6))*x^(1/6)/b^5/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)-30*a*(a+b*x^(1/6))*l
n(a+b*x^(1/6))/b^6/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 268, 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 {60 a^2}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {30 a \left (a+b \sqrt [6]{x}\right ) \log \left (a+b \sqrt [6]{x}\right )}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {6 \sqrt [6]{x} \left (a+b \sqrt [6]{x}\right )}{b^5 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {3 a^5}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {10 a^4}{b^6 \left (a+b \sqrt [6]{x}\right )^2 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {30 a^3}{b^6 \left (a+b \sqrt [6]{x}\right ) \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^(1/6) + b^2*x^(1/3))^(-5/2),x]

[Out]

(-60*a^2)/(b^6*Sqrt[a^2 + 2*a*b*x^(1/6) + b^2*x^(1/3)]) + (3*a^5)/(2*b^6*(a + b*x^(1/6))^3*Sqrt[a^2 + 2*a*b*x^
(1/6) + b^2*x^(1/3)]) - (10*a^4)/(b^6*(a + b*x^(1/6))^2*Sqrt[a^2 + 2*a*b*x^(1/6) + b^2*x^(1/3)]) + (30*a^3)/(b
^6*(a + b*x^(1/6))*Sqrt[a^2 + 2*a*b*x^(1/6) + b^2*x^(1/3)]) + (6*(a + b*x^(1/6))*x^(1/6))/(b^5*Sqrt[a^2 + 2*a*
b*x^(1/6) + b^2*x^(1/3)]) - (30*a*(a + b*x^(1/6))*Log[a + b*x^(1/6)])/(b^6*Sqrt[a^2 + 2*a*b*x^(1/6) + b^2*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 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 [6]{x}+b^2 \sqrt [3]{x}\right )^{5/2}} \, dx &=6 \text {Subst}\left (\int \frac {x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,\sqrt [6]{x}\right )\\ &=\frac {\left (6 b^5 \left (a+b \sqrt [6]{x}\right )\right ) \text {Subst}\left (\int \frac {x^5}{\left (a b+b^2 x\right )^5} \, dx,x,\sqrt [6]{x}\right )}{\sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\\ &=\frac {\left (6 b^5 \left (a+b \sqrt [6]{x}\right )\right ) \text {Subst}\left (\int \left (\frac {1}{b^{10}}-\frac {a^5}{b^{10} (a+b x)^5}+\frac {5 a^4}{b^{10} (a+b x)^4}-\frac {10 a^3}{b^{10} (a+b x)^3}+\frac {10 a^2}{b^{10} (a+b x)^2}-\frac {5 a}{b^{10} (a+b x)}\right ) \, dx,x,\sqrt [6]{x}\right )}{\sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\\ &=-\frac {60 a^2}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {3 a^5}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {10 a^4}{b^6 \left (a+b \sqrt [6]{x}\right )^2 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {30 a^3}{b^6 \left (a+b \sqrt [6]{x}\right ) \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac {6 \left (a+b \sqrt [6]{x}\right ) \sqrt [6]{x}}{b^5 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac {30 a \left (a+b \sqrt [6]{x}\right ) \log \left (a+b \sqrt [6]{x}\right )}{b^6 \sqrt {a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 121, normalized size = 0.45 \begin {gather*} \frac {-77 a^5-248 a^4 b \sqrt [6]{x}-252 a^3 b^2 \sqrt [3]{x}-48 a^2 b^3 \sqrt {x}+48 a b^4 x^{2/3}+12 b^5 x^{5/6}-60 a \left (a+b \sqrt [6]{x}\right )^4 \log \left (a+b \sqrt [6]{x}\right )}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt {\left (a+b \sqrt [6]{x}\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^(1/6) + b^2*x^(1/3))^(-5/2),x]

[Out]

(-77*a^5 - 248*a^4*b*x^(1/6) - 252*a^3*b^2*x^(1/3) - 48*a^2*b^3*Sqrt[x] + 48*a*b^4*x^(2/3) + 12*b^5*x^(5/6) -
60*a*(a + b*x^(1/6))^4*Log[a + b*x^(1/6)])/(2*b^6*(a + b*x^(1/6))^3*Sqrt[(a + b*x^(1/6))^2])

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Maple [A]
time = 0.10, size = 174, normalized size = 0.65

method	result	size
derivativedivides	\(-\frac {\left (60 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a \,b^{4} x^{\frac {2}{3}}-12 b^{5} x^{\frac {5}{6}}+240 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{2} b^{3} \sqrt {x}-48 a \,b^{4} x^{\frac {2}{3}}+360 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{3} b^{2} x^{\frac {1}{3}}+48 a^{2} b^{3} \sqrt {x}+240 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{4} b \,x^{\frac {1}{6}}+252 a^{3} b^{2} x^{\frac {1}{3}}+60 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{5}+248 a^{4} b \,x^{\frac {1}{6}}+77 a^{5}\right ) \left (a +b \,x^{\frac {1}{6}}\right )}{2 b^{6} \left (\left (a +b \,x^{\frac {1}{6}}\right )^{2}\right )^{\frac {5}{2}}}\)	\(163\)
default	\(-\frac {\sqrt {a^{2}+2 a b \,x^{\frac {1}{6}}+b^{2} x^{\frac {1}{3}}}\, \left (60 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a \,b^{4} x^{\frac {2}{3}}-12 b^{5} x^{\frac {5}{6}}+240 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{2} b^{3} \sqrt {x}-48 a \,b^{4} x^{\frac {2}{3}}+360 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{3} b^{2} x^{\frac {1}{3}}+48 a^{2} b^{3} \sqrt {x}+240 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{4} b \,x^{\frac {1}{6}}+252 a^{3} b^{2} x^{\frac {1}{3}}+60 \ln \left (a +b \,x^{\frac {1}{6}}\right ) a^{5}+248 a^{4} b \,x^{\frac {1}{6}}+77 a^{5}\right )}{2 \left (a +b \,x^{\frac {1}{6}}\right )^{5} b^{6}}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)*(60*ln(a+b*x^(1/6))*a*b^4*x^(2/3)-12*b^5*x^(5/6)+240*ln(a+b*x^(1/6)
)*a^2*b^3*x^(1/2)-48*a*b^4*x^(2/3)+360*ln(a+b*x^(1/6))*a^3*b^2*x^(1/3)+48*a^2*b^3*x^(1/2)+240*ln(a+b*x^(1/6))*
a^4*b*x^(1/6)+252*a^3*b^2*x^(1/3)+60*ln(a+b*x^(1/6))*a^5+248*a^4*b*x^(1/6)+77*a^5)/(a+b*x^(1/6))^5/b^6

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Maxima [A]
time = 0.31, size = 119, normalized size = 0.44 \begin {gather*} \frac {12 \, b^{5} x^{\frac {5}{6}} + 48 \, a b^{4} x^{\frac {2}{3}} - 48 \, a^{2} b^{3} \sqrt {x} - 252 \, a^{3} b^{2} x^{\frac {1}{3}} - 248 \, a^{4} b x^{\frac {1}{6}} - 77 \, a^{5}}{2 \, {\left (b^{10} x^{\frac {2}{3}} + 4 \, a b^{9} \sqrt {x} + 6 \, a^{2} b^{8} x^{\frac {1}{3}} + 4 \, a^{3} b^{7} x^{\frac {1}{6}} + a^{4} b^{6}\right )}} - \frac {30 \, a \log \left (b x^{\frac {1}{6}} + a\right )}{b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x, algorithm="maxima")

[Out]

1/2*(12*b^5*x^(5/6) + 48*a*b^4*x^(2/3) - 48*a^2*b^3*sqrt(x) - 252*a^3*b^2*x^(1/3) - 248*a^4*b*x^(1/6) - 77*a^5
)/(b^10*x^(2/3) + 4*a*b^9*sqrt(x) + 6*a^2*b^8*x^(1/3) + 4*a^3*b^7*x^(1/6) + a^4*b^6) - 30*a*log(b*x^(1/6) + a)
/b^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+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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+2*a*b*x**(1/6)+b**2*x**(1/3))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 5.40, size = 105, normalized size = 0.39 \begin {gather*} -\frac {30 \, a \log \left ({\left | b x^{\frac {1}{6}} + a \right |}\right )}{b^{6} \mathrm {sgn}\left (b x^{\frac {1}{6}} + a\right )} + \frac {6 \, x^{\frac {1}{6}}}{b^{5} \mathrm {sgn}\left (b x^{\frac {1}{6}} + a\right )} - \frac {120 \, a^{2} b^{3} \sqrt {x} + 300 \, a^{3} b^{2} x^{\frac {1}{3}} + 260 \, a^{4} b x^{\frac {1}{6}} + 77 \, a^{5}}{2 \, {\left (b x^{\frac {1}{6}} + a\right )}^{4} b^{6} \mathrm {sgn}\left (b x^{\frac {1}{6}} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(5/2),x, algorithm="giac")

[Out]

-30*a*log(abs(b*x^(1/6) + a))/(b^6*sgn(b*x^(1/6) + a)) + 6*x^(1/6)/(b^5*sgn(b*x^(1/6) + a)) - 1/2*(120*a^2*b^3
*sqrt(x) + 300*a^3*b^2*x^(1/3) + 260*a^4*b*x^(1/6) + 77*a^5)/((b*x^(1/6) + a)^4*b^6*sgn(b*x^(1/6) + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a^2+b^2\,x^{1/3}+2\,a\,b\,x^{1/6}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^(1/3) + 2*a*b*x^(1/6))^(5/2),x)

[Out]

int(1/(a^2 + b^2*x^(1/3) + 2*a*b*x^(1/6))^(5/2), x)

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