3.471 \(\int \frac{1}{(a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{11/2}} \, dx\)

Optimal. Leaf size=137 \[ -\frac{3 a^2}{10 b^3 \left (a+b \sqrt [3]{x}\right )^9 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{2 a}{3 b^3 \left (a+b \sqrt [3]{x}\right )^8 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac{3}{8 b^3 \left (a+b \sqrt [3]{x}\right )^7 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

[Out]

(-3*a^2)/(10*b^3*(a + b*x^(1/3))^9*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) + (2*a)/(3*b^3*(a + b*x^(1/3))^8*S
qrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) - 3/(8*b^3*(a + b*x^(1/3))^7*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])

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Rubi [A]  time = 0.0780041, antiderivative size = 137, 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} \[ -\frac{3 a^2}{10 b^3 \left (a+b \sqrt [3]{x}\right )^9 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{2 a}{3 b^3 \left (a+b \sqrt [3]{x}\right )^8 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac{3}{8 b^3 \left (a+b \sqrt [3]{x}\right )^7 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(-11/2),x]

[Out]

(-3*a^2)/(10*b^3*(a + b*x^(1/3))^9*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) + (2*a)/(3*b^3*(a + b*x^(1/3))^8*S
qrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) - 3/(8*b^3*(a + b*x^(1/3))^7*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])

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 [3]{x}+b^2 x^{2/3}\right )^{11/2}} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{11/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (3 b^{11} \left (a+b \sqrt [3]{x}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a b+b^2 x\right )^{11}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=\frac{\left (3 b^{11} \left (a+b \sqrt [3]{x}\right )\right ) \operatorname{Subst}\left (\int \left (\frac{a^2}{b^{13} (a+b x)^{11}}-\frac{2 a}{b^{13} (a+b x)^{10}}+\frac{1}{b^{13} (a+b x)^9}\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}{10 b^3 \left (a+b \sqrt [3]{x}\right )^9 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac{2 a}{3 b^3 \left (a+b \sqrt [3]{x}\right )^8 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac{3}{8 b^3 \left (a+b \sqrt [3]{x}\right )^7 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ \end{align*}

Mathematica [A]  time = 0.038088, size = 58, normalized size = 0.42 \[ \frac{-a^2-10 a b \sqrt [3]{x}-45 b^2 x^{2/3}}{120 b^3 \left (a+b \sqrt [3]{x}\right )^9 \sqrt{\left (a+b \sqrt [3]{x}\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(-11/2),x]

[Out]

(-a^2 - 10*a*b*x^(1/3) - 45*b^2*x^(2/3))/(120*b^3*(a + b*x^(1/3))^9*Sqrt[(a + b*x^(1/3))^2])

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Maple [A]  time = 0.007, size = 54, normalized size = 0.4 \begin{align*} -{\frac{1}{120\,{b}^{3}}\sqrt{{a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}}} \left ( 45\,{b}^{2}{x}^{2/3}+10\,ab\sqrt [3]{x}+{a}^{2} \right ) \left ( a+b\sqrt [3]{x} \right ) ^{-11}} \end{align*}

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))^(11/2),x)

[Out]

-1/120*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(45*b^2*x^(2/3)+10*a*b*x^(1/3)+a^2)/(a+b*x^(1/3))^11/b^3

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Maxima [A]  time = 1.01626, size = 85, normalized size = 0.62 \begin{align*} -\frac{3 \, a^{2} b^{2}}{10 \,{\left (b^{2}\right )}^{\frac{15}{2}}{\left (x^{\frac{1}{3}} + \frac{a}{b}\right )}^{10}} + \frac{2 \, a b}{3 \,{\left (b^{2}\right )}^{\frac{13}{2}}{\left (x^{\frac{1}{3}} + \frac{a}{b}\right )}^{9}} - \frac{3}{8 \,{\left (b^{2}\right )}^{\frac{11}{2}}{\left (x^{\frac{1}{3}} + \frac{a}{b}\right )}^{8}} \end{align*}

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))^(11/2),x, algorithm="maxima")

[Out]

-3/10*a^2*b^2/((b^2)^(15/2)*(x^(1/3) + a/b)^10) + 2/3*a*b/((b^2)^(13/2)*(x^(1/3) + a/b)^9) - 3/8/((b^2)^(11/2)
*(x^(1/3) + a/b)^8)

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Fricas [B]  time = 2.04098, size = 842, normalized size = 6.15 \begin{align*} \frac{440 \, a b^{21} x^{7} - 25630 \, a^{4} b^{18} x^{6} + 186252 \, a^{7} b^{15} x^{5} - 326150 \, a^{10} b^{12} x^{4} + 154000 \, a^{13} b^{9} x^{3} - 16005 \, a^{16} b^{6} x^{2} + 110 \, a^{19} b^{3} x - a^{22} - 27 \,{\left (88 \, a^{2} b^{20} x^{6} - 2200 \, a^{5} b^{17} x^{5} + 9625 \, a^{8} b^{14} x^{4} - 10910 \, a^{11} b^{11} x^{3} + 3245 \, a^{14} b^{8} x^{2} - 176 \, a^{17} b^{5} x\right )} x^{\frac{2}{3}} - 9 \,{\left (5 \, b^{22} x^{7} - 990 \, a^{3} b^{19} x^{6} + 12705 \, a^{6} b^{16} x^{5} - 34760 \, a^{9} b^{13} x^{4} + 25542 \, a^{12} b^{10} x^{3} - 4620 \, a^{15} b^{7} x^{2} + 110 \, a^{18} b^{4} x\right )} x^{\frac{1}{3}}}{120 \,{\left (b^{33} x^{10} + 10 \, a^{3} b^{30} x^{9} + 45 \, a^{6} b^{27} x^{8} + 120 \, a^{9} b^{24} x^{7} + 210 \, a^{12} b^{21} x^{6} + 252 \, a^{15} b^{18} x^{5} + 210 \, a^{18} b^{15} x^{4} + 120 \, a^{21} b^{12} x^{3} + 45 \, a^{24} b^{9} x^{2} + 10 \, a^{27} b^{6} x + a^{30} b^{3}\right )}} \end{align*}

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))^(11/2),x, algorithm="fricas")

[Out]

1/120*(440*a*b^21*x^7 - 25630*a^4*b^18*x^6 + 186252*a^7*b^15*x^5 - 326150*a^10*b^12*x^4 + 154000*a^13*b^9*x^3
- 16005*a^16*b^6*x^2 + 110*a^19*b^3*x - a^22 - 27*(88*a^2*b^20*x^6 - 2200*a^5*b^17*x^5 + 9625*a^8*b^14*x^4 - 1
0910*a^11*b^11*x^3 + 3245*a^14*b^8*x^2 - 176*a^17*b^5*x)*x^(2/3) - 9*(5*b^22*x^7 - 990*a^3*b^19*x^6 + 12705*a^
6*b^16*x^5 - 34760*a^9*b^13*x^4 + 25542*a^12*b^10*x^3 - 4620*a^15*b^7*x^2 + 110*a^18*b^4*x)*x^(1/3))/(b^33*x^1
0 + 10*a^3*b^30*x^9 + 45*a^6*b^27*x^8 + 120*a^9*b^24*x^7 + 210*a^12*b^21*x^6 + 252*a^15*b^18*x^5 + 210*a^18*b^
15*x^4 + 120*a^21*b^12*x^3 + 45*a^24*b^9*x^2 + 10*a^27*b^6*x + a^30*b^3)

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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**(1/3)+b**2*x**(2/3))**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

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))^(11/2),x, algorithm="giac")

[Out]

undef