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

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

[Out]

(-60*a^2)/(b^6*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)])

________________________________________________________________________________________

Rubi [A]  time = 0.15092, antiderivative size = 268, normalized size of antiderivative = 1., 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 = {1341, 646, 43} \[ \frac{3 a^5}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt{a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac{10 a^4}{b^6 \left (a+b \sqrt [6]{x}\right )^2 \sqrt{a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac{30 a^3}{b^6 \left (a+b \sqrt [6]{x}\right ) \sqrt{a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac{60 a^2}{b^6 \sqrt{a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}+\frac{6 \sqrt [6]{x} \left (a+b \sqrt [6]{x}\right )}{b^5 \sqrt{a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}}-\frac{30 a \left (a+b \sqrt [6]{x}\right ) \log \left (a+b \sqrt [6]{x}\right )}{b^6 \sqrt{a^2+2 a b \sqrt [6]{x}+b^2 \sqrt [3]{x}}} \]

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 1341

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 646

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 43

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])

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 \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.0997186, size = 121, normalized size = 0.45 \[ \frac{-252 a^3 b^2 \sqrt [3]{x}-48 a^2 b^3 \sqrt{x}-248 a^4 b \sqrt [6]{x}-77 a^5+48 a b^4 x^{2/3}-60 a \left (a+b \sqrt [6]{x}\right )^4 \log \left (a+b \sqrt [6]{x}\right )+12 b^5 x^{5/6}}{2 b^6 \left (a+b \sqrt [6]{x}\right )^3 \sqrt{\left (a+b \sqrt [6]{x}\right )^2}} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.014, size = 174, normalized size = 0.7 \begin{align*}{\frac{1}{2\,{b}^{6}}\sqrt{{a}^{2}+2\,ab\sqrt [6]{x}+{b}^{2}\sqrt [3]{x}} \left ( 12\,{x}^{5/6}{b}^{5}-60\,{x}^{2/3}\ln \left ( a+b\sqrt [6]{x} \right ) a{b}^{4}+48\,{x}^{2/3}a{b}^{4}-240\,\sqrt{x}\ln \left ( a+b\sqrt [6]{x} \right ){a}^{2}{b}^{3}-48\,\sqrt{x}{a}^{2}{b}^{3}-360\,\sqrt [3]{x}\ln \left ( a+b\sqrt [6]{x} \right ){a}^{3}{b}^{2}-252\,\sqrt [3]{x}{a}^{3}{b}^{2}-240\,\sqrt [6]{x}\ln \left ( a+b\sqrt [6]{x} \right ){a}^{4}b-248\,\sqrt [6]{x}{a}^{4}b-60\,\ln \left ( a+b\sqrt [6]{x} \right ){a}^{5}-77\,{a}^{5} \right ) \left ( a+b\sqrt [6]{x} \right ) ^{-5}} \end{align*}

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)

[Out]

1/2*(a^2+2*a*b*x^(1/6)+b^2*x^(1/3))^(1/2)*(12*x^(5/6)*b^5-60*x^(2/3)*ln(a+b*x^(1/6))*a*b^4+48*x^(2/3)*a*b^4-24
0*x^(1/2)*ln(a+b*x^(1/6))*a^2*b^3-48*x^(1/2)*a^2*b^3-360*x^(1/3)*ln(a+b*x^(1/6))*a^3*b^2-252*x^(1/3)*a^3*b^2-2
40*x^(1/6)*ln(a+b*x^(1/6))*a^4*b-248*x^(1/6)*a^4*b-60*ln(a+b*x^(1/6))*a^5-77*a^5)/(a+b*x^(1/6))^5/b^6

________________________________________________________________________________________

Maxima [A]  time = 1.06323, size = 161, normalized size = 0.6 \begin{align*} \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{align*}

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

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 1.53807, size = 320, normalized size = 1.19 \begin{align*} \frac{3 \, a^{4}{\left | a \right |} \log \left ({\left | x^{\frac{1}{6}}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) +{\left | a \right |} \right |}\right )}{4 \,{\left (a^{3} b^{5}{\left | a \right |}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) - a^{4} b^{6}\right )}} + \frac{3 \,{\left (24 \, a^{5} b^{2}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) - 25 \, a^{4} b^{3}{\left | a \right |}\right )} \log \left ({\left | b x^{\frac{1}{6}} + a \right |}\right )}{4 \,{\left (a^{3} b^{8}{\left | a \right |}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) - a^{4} b^{9}\right )}} + \frac{6 \, x^{\frac{1}{6}}}{b^{4}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right )} + \frac{70 \, a^{5}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) - 70 \, a^{4} b{\left | a \right |} + 93 \,{\left (a^{3} b^{2}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) - a^{2} b^{3}{\left | a \right |}\right )} x^{\frac{1}{3}} + 159 \,{\left (a^{4} b{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) - a^{3} b^{2}{\left | a \right |}\right )} x^{\frac{1}{6}}}{4 \,{\left ({\left | a \right |}{\left | b \right |} \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right ) - a b\right )}{\left (b x^{\frac{1}{6}} + a\right )}^{3} b^{6}} \end{align*}

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]

3/4*a^4*abs(a)*log(abs(x^(1/6)*abs(b)*sgn(a)*sgn(b) + abs(a)))/(a^3*b^5*abs(a)*abs(b)*sgn(a)*sgn(b) - a^4*b^6)
+ 3/4*(24*a^5*b^2*abs(b)*sgn(a)*sgn(b) - 25*a^4*b^3*abs(a))*log(abs(b*x^(1/6) + a))/(a^3*b^8*abs(a)*abs(b)*sg
n(a)*sgn(b) - a^4*b^9) + 6*x^(1/6)/(b^4*abs(b)*sgn(a)*sgn(b)) + 1/4*(70*a^5*abs(b)*sgn(a)*sgn(b) - 70*a^4*b*ab
s(a) + 93*(a^3*b^2*abs(b)*sgn(a)*sgn(b) - a^2*b^3*abs(a))*x^(1/3) + 159*(a^4*b*abs(b)*sgn(a)*sgn(b) - a^3*b^2*
abs(a))*x^(1/6))/((abs(a)*abs(b)*sgn(a)*sgn(b) - a*b)*(b*x^(1/6) + a)^3*b^6)