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

Optimal. Leaf size=410 \[ \frac{3 b^7 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{4 a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )^4}-\frac{7 b^6 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )^3}+\frac{63 b^5 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{2 a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )^2}-\frac{105 b^4 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )}+\frac{45 b^2 \sqrt [3]{x} \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^7 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}-\frac{15 b x^{2/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}{2 a^6 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}+\frac{x \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^5 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}-\frac{105 b^3 \left (a+\frac{b}{\sqrt [3]{x}}\right ) \log \left (a \sqrt [3]{x}+b\right )}{a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}} \]

[Out]

(3*b^7*(a + b/x^(1/3)))/(4*a^8*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))^4) - (7*b^6*(a + b/x^
(1/3)))/(a^8*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))^3) + (63*b^5*(a + b/x^(1/3)))/(2*a^8*Sq
rt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))^2) - (105*b^4*(a + b/x^(1/3)))/(a^8*Sqrt[a^2 + b^2/x^(
2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))) + (45*b^2*(a + b/x^(1/3))*x^(1/3))/(a^7*Sqrt[a^2 + b^2/x^(2/3) + (2*a
*b)/x^(1/3)]) - (15*b*(a + b/x^(1/3))*x^(2/3))/(2*a^6*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]) + ((a + b/x^(
1/3))*x)/(a^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]) - (105*b^3*(a + b/x^(1/3))*Log[b + a*x^(1/3)])/(a^8*S
qrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])

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Rubi [A]  time = 0.267721, antiderivative size = 410, normalized size of antiderivative = 1., 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 = {1341, 1355, 263, 43} \[ \frac{3 b^7 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{4 a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )^4}-\frac{7 b^6 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )^3}+\frac{63 b^5 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{2 a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )^2}-\frac{105 b^4 \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}} \left (a \sqrt [3]{x}+b\right )}+\frac{45 b^2 \sqrt [3]{x} \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^7 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}-\frac{15 b x^{2/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}{2 a^6 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}+\frac{x \left (a+\frac{b}{\sqrt [3]{x}}\right )}{a^5 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}-\frac{105 b^3 \left (a+\frac{b}{\sqrt [3]{x}}\right ) \log \left (a \sqrt [3]{x}+b\right )}{a^8 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^7*(a + b/x^(1/3)))/(4*a^8*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))^4) - (7*b^6*(a + b/x^
(1/3)))/(a^8*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))^3) + (63*b^5*(a + b/x^(1/3)))/(2*a^8*Sq
rt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))^2) - (105*b^4*(a + b/x^(1/3)))/(a^8*Sqrt[a^2 + b^2/x^(
2/3) + (2*a*b)/x^(1/3)]*(b + a*x^(1/3))) + (45*b^2*(a + b/x^(1/3))*x^(1/3))/(a^7*Sqrt[a^2 + b^2/x^(2/3) + (2*a
*b)/x^(1/3)]) - (15*b*(a + b/x^(1/3))*x^(2/3))/(2*a^6*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]) + ((a + b/x^(
1/3))*x)/(a^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]) - (105*b^3*(a + b/x^(1/3))*Log[b + a*x^(1/3)])/(a^8*S
qrt[a^2 + b^2/x^(2/3) + (2*a*b)/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 1355

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]

Rule 263

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 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+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}\right )^{5/2}} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2+\frac{b^2}{x^2}+\frac{2 a b}{x}\right )^{5/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (3 b^4 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a b+\frac{b^2}{x}\right )^5} \, 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^4 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )\right ) \operatorname{Subst}\left (\int \frac{x^7}{\left (b^2+a b x\right )^5} \, 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^4 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )\right ) \operatorname{Subst}\left (\int \left (\frac{15}{a^7 b^3}-\frac{5 x}{a^6 b^4}+\frac{x^2}{a^5 b^5}-\frac{b^2}{a^7 (b+a x)^5}+\frac{7 b}{a^7 (b+a x)^4}-\frac{21}{a^7 (b+a x)^3}+\frac{35}{a^7 b (b+a x)^2}-\frac{35}{a^7 b^2 (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^7+\frac{b^8}{\sqrt [3]{x}}\right )}{4 a^8 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )^4}-\frac{7 \left (a b^6+\frac{b^7}{\sqrt [3]{x}}\right )}{a^8 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )^3}+\frac{63 \left (a b^5+\frac{b^6}{\sqrt [3]{x}}\right )}{2 a^8 \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{105 \left (a b^4+\frac{b^5}{\sqrt [3]{x}}\right )}{a^8 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \left (b+a \sqrt [3]{x}\right )}+\frac{45 \left (a b^2+\frac{b^3}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}{a^7 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}-\frac{15 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right ) x^{2/3}}{2 a^6 \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^5 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}-\frac{105 \left (a b^3+\frac{b^4}{\sqrt [3]{x}}\right ) \log \left (b+a \sqrt [3]{x}\right )}{a^8 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}\\ \end{align*}

Mathematica [A]  time = 0.138987, size = 152, normalized size = 0.37 \[ \frac{\left (a \sqrt [3]{x}+b\right ) \left (84 a^5 b^2 x^{5/3}+556 a^4 b^3 x^{4/3}-444 a^2 b^5 x^{2/3}+544 a^3 b^4 x-14 a^6 b x^2+4 a^7 x^{7/3}-856 a b^6 \sqrt [3]{x}-420 b^3 \left (a \sqrt [3]{x}+b\right )^4 \log \left (a \sqrt [3]{x}+b\right )-319 b^7\right )}{4 a^8 x^{5/3} \left (\frac{\left (a \sqrt [3]{x}+b\right )^2}{x^{2/3}}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*x^(1/3))*(-319*b^7 - 856*a*b^6*x^(1/3) - 444*a^2*b^5*x^(2/3) + 544*a^3*b^4*x + 556*a^4*b^3*x^(4/3) + 8
4*a^5*b^2*x^(5/3) - 14*a^6*b*x^2 + 4*a^7*x^(7/3) - 420*b^3*(b + a*x^(1/3))^4*Log[b + a*x^(1/3)]))/(4*a^8*((b +
a*x^(1/3))^2/x^(2/3))^(5/2)*x^(5/3))

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Maple [A]  time = 0.011, size = 199, normalized size = 0.5 \begin{align*} -{\frac{1}{4\,{a}^{8}} \left ( 14\,{a}^{6}b{x}^{2}+420\,\ln \left ( b+a\sqrt [3]{x} \right ){x}^{4/3}{a}^{4}{b}^{3}-84\,{a}^{5}{b}^{2}{x}^{5/3}-4\,{x}^{7/3}{a}^{7}+1680\,\ln \left ( b+a\sqrt [3]{x} \right ) x{a}^{3}{b}^{4}-556\,{x}^{4/3}{a}^{4}{b}^{3}+2520\,\ln \left ( b+a\sqrt [3]{x} \right ){x}^{2/3}{a}^{2}{b}^{5}-544\,x{a}^{3}{b}^{4}+1680\,\ln \left ( b+a\sqrt [3]{x} \right ) \sqrt [3]{x}a{b}^{6}+444\,{x}^{2/3}{a}^{2}{b}^{5}+420\,\ln \left ( b+a\sqrt [3]{x} \right ){b}^{7}+856\,\sqrt [3]{x}a{b}^{6}+319\,{b}^{7} \right ) \left ( b+a\sqrt [3]{x} \right ) \left ({ \left ({a}^{2}{x}^{{\frac{2}{3}}}+2\,ab\sqrt [3]{x}+{b}^{2} \right ){x}^{-{\frac{2}{3}}}} \right ) ^{-{\frac{5}{2}}}{x}^{-{\frac{5}{3}}}} \end{align*}

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]

-1/4/((a^2*x^(2/3)+2*a*b*x^(1/3)+b^2)/x^(2/3))^(5/2)/x^(5/3)*(14*a^6*b*x^2+420*ln(b+a*x^(1/3))*x^(4/3)*a^4*b^3
-84*a^5*b^2*x^(5/3)-4*x^(7/3)*a^7+1680*ln(b+a*x^(1/3))*x*a^3*b^4-556*x^(4/3)*a^4*b^3+2520*ln(b+a*x^(1/3))*x^(2
/3)*a^2*b^5-544*x*a^3*b^4+1680*ln(b+a*x^(1/3))*x^(1/3)*a*b^6+444*x^(2/3)*a^2*b^5+420*ln(b+a*x^(1/3))*b^7+856*x
^(1/3)*a*b^6+319*b^7)*(b+a*x^(1/3))/a^8

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Maxima [A]  time = 1.00687, size = 188, normalized size = 0.46 \begin{align*} \frac{4 \, a^{7} x^{\frac{7}{3}} - 14 \, a^{6} b x^{2} + 84 \, a^{5} b^{2} x^{\frac{5}{3}} + 556 \, a^{4} b^{3} x^{\frac{4}{3}} + 544 \, a^{3} b^{4} x - 444 \, a^{2} b^{5} x^{\frac{2}{3}} - 856 \, a b^{6} x^{\frac{1}{3}} - 319 \, b^{7}}{4 \,{\left (a^{12} x^{\frac{4}{3}} + 4 \, a^{11} b x + 6 \, a^{10} b^{2} x^{\frac{2}{3}} + 4 \, a^{9} b^{3} x^{\frac{1}{3}} + a^{8} b^{4}\right )}} - \frac{105 \, b^{3} \log \left (a x^{\frac{1}{3}} + b\right )}{a^{8}} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

1/4*(4*a^7*x^(7/3) - 14*a^6*b*x^2 + 84*a^5*b^2*x^(5/3) + 556*a^4*b^3*x^(4/3) + 544*a^3*b^4*x - 444*a^2*b^5*x^(
2/3) - 856*a*b^6*x^(1/3) - 319*b^7)/(a^12*x^(4/3) + 4*a^11*b*x + 6*a^10*b^2*x^(2/3) + 4*a^9*b^3*x^(1/3) + a^8*
b^4) - 105*b^3*log(a*x^(1/3) + b)/a^8

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a^{2} + \frac{2 a b}{\sqrt [3]{x}} + \frac{b^{2}}{x^{\frac{2}{3}}}\right )^{\frac{5}{2}}}\, dx \end{align*}

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))**(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 = 1.21239, size = 190, normalized size = 0.46 \begin{align*} -\frac{105 \, b^{3} \log \left ({\left | a x^{\frac{1}{3}} + b \right |}\right )}{a^{8} \mathrm{sgn}\left (a x^{\frac{2}{3}} + b x^{\frac{1}{3}}\right )} - \frac{420 \, a^{3} b^{4} x + 1134 \, a^{2} b^{5} x^{\frac{2}{3}} + 1036 \, a b^{6} x^{\frac{1}{3}} + 319 \, b^{7}}{4 \,{\left (a x^{\frac{1}{3}} + b\right )}^{4} a^{8} \mathrm{sgn}\left (a x^{\frac{2}{3}} + b x^{\frac{1}{3}}\right )} + \frac{2 \, a^{10} x - 15 \, a^{9} b x^{\frac{2}{3}} + 90 \, a^{8} b^{2} x^{\frac{1}{3}}}{2 \, a^{15} \mathrm{sgn}\left (a x^{\frac{2}{3}} + b x^{\frac{1}{3}}\right )} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

-105*b^3*log(abs(a*x^(1/3) + b))/(a^8*sgn(a*x^(2/3) + b*x^(1/3))) - 1/4*(420*a^3*b^4*x + 1134*a^2*b^5*x^(2/3)
+ 1036*a*b^6*x^(1/3) + 319*b^7)/((a*x^(1/3) + b)^4*a^8*sgn(a*x^(2/3) + b*x^(1/3))) + 1/2*(2*a^10*x - 15*a^9*b*
x^(2/3) + 90*a^8*b^2*x^(1/3))/(a^15*sgn(a*x^(2/3) + b*x^(1/3)))