\(\int \frac {1}{(a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{9/2}} \, dx\) [470]

Optimal result

Integrand size = 26, antiderivative size = 137 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=-\frac {3 a^2}{8 b^3 \left (a+b \sqrt [3]{x}\right )^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {6 a}{7 b^3 \left (a+b \sqrt [3]{x}\right )^6 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac {1}{2 b^3 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

[Out]

-3/8*a^2/b^3/(a+b*x^(1/3))^7/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)+6/7*a/b^3/(a+b*x^(1/3))^6/(a^2+2*a*b*x^(1/3
)+b^2*x^(2/3))^(1/2)-1/2/b^3/(a+b*x^(1/3))^5/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1355, 660, 45} \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=-\frac {3 a^2}{8 b^3 \left (a+b \sqrt [3]{x}\right )^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {6 a}{7 b^3 \left (a+b \sqrt [3]{x}\right )^6 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac {1}{2 b^3 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

[In]

Int[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(-9/2),x]

[Out]

(-3*a^2)/(8*b^3*(a + b*x^(1/3))^7*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) + (6*a)/(7*b^3*(a + b*x^(1/3))^6*Sq
rt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) - 1/(2*b^3*(a + b*x^(1/3))^5*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*} \text {integral}& = 3 \text {Subst}\left (\int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{9/2}} \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {\left (3 b^9 \left (a+b \sqrt [3]{x}\right )\right ) \text {Subst}\left (\int \frac {x^2}{\left (a b+b^2 x\right )^9} \, dx,x,\sqrt [3]{x}\right )}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \\ & = \frac {\left (3 b^9 \left (a+b \sqrt [3]{x}\right )\right ) \text {Subst}\left (\int \left (\frac {a^2}{b^{11} (a+b x)^9}-\frac {2 a}{b^{11} (a+b x)^8}+\frac {1}{b^{11} (a+b x)^7}\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}{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}}}+\frac {6 a}{7 b^3 \left (a+b \sqrt [3]{x}\right )^6 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac {1}{2 b^3 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=\frac {\left (a+b \sqrt [3]{x}\right ) \left (-a^2-8 a b \sqrt [3]{x}-28 b^2 x^{2/3}\right )}{56 b^3 \left (\left (a+b \sqrt [3]{x}\right )^2\right )^{9/2}} \]

[In]

Integrate[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(-9/2),x]

[Out]

((a + b*x^(1/3))*(-a^2 - 8*a*b*x^(1/3) - 28*b^2*x^(2/3)))/(56*b^3*((a + b*x^(1/3))^2)^(9/2))

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.31

method	result	size
derivativedivides	\(-\frac {\left (28 b^{2} x^{\frac {2}{3}}+8 a b \,x^{\frac {1}{3}}+a^{2}\right ) \left (a +b \,x^{\frac {1}{3}}\right )}{56 b^{3} {\left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )}^{\frac {9}{2}}}\)	\(43\)
default	\(\frac {\left (-a^{42}+202496 a^{27} b^{15} x^{5}+31696 a^{30} b^{12} x^{4}-11704 a^{33} b^{9} x^{3}-3844 a^{36} b^{6} x^{2}+40 a^{39} b^{3} x -46480 a^{9} b^{33} x^{11}-190568 a^{12} b^{30} x^{10}-276592 a^{15} b^{27} x^{9}-77037 a^{18} b^{24} x^{8}+270024 a^{21} b^{21} x^{7}+376320 a^{24} b^{18} x^{6}-28 x^{14} b^{42}-394632 x^{\frac {19}{3}} a^{23} b^{19}+2632 x^{13} a^{3} b^{39}+3878 x^{12} a^{6} b^{36}+216 x^{\frac {41}{3}} a \,b^{41}-4860 x^{\frac {38}{3}} a^{4} b^{38}-945 x^{\frac {40}{3}} a^{2} b^{40}-23976 x^{\frac {35}{3}} a^{7} b^{35}+4536 x^{\frac {37}{3}} a^{5} b^{37}-5859 x^{\frac {32}{3}} a^{10} b^{32}+46656 x^{\frac {34}{3}} a^{8} b^{34}+163296 x^{\frac {29}{3}} a^{13} b^{29}+105408 x^{\frac {31}{3}} a^{11} b^{31}+415044 x^{\frac {26}{3}} a^{16} b^{26}+19440 x^{\frac {28}{3}} a^{14} b^{28}+430920 x^{\frac {23}{3}} a^{19} b^{23}-285768 x^{\frac {25}{3}} a^{17} b^{25}+126846 x^{\frac {20}{3}} a^{22} b^{20}-519372 x^{\frac {22}{3}} a^{20} b^{22}-147744 x^{\frac {17}{3}} a^{25} b^{17}-96957 x^{\frac {16}{3}} a^{26} b^{16}-158544 x^{\frac {14}{3}} a^{28} b^{14}+49680 x^{\frac {13}{3}} a^{29} b^{13}-54432 x^{\frac {11}{3}} a^{31} b^{11}+36612 x^{\frac {10}{3}} a^{32} b^{10}-2835 x^{\frac {8}{3}} a^{34} b^{8}+5832 x^{\frac {7}{3}} a^{35} b^{7}+1512 x^{\frac {5}{3}} a^{37} b^{5}-378 x^{\frac {4}{3}} a^{38} b^{4}\right ) \left (a +b \,x^{\frac {1}{3}}\right )}{56 b^{3} \left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )^{8} \left (b^{3} x +a^{3}\right )^{8} \left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{\frac {9}{2}}}\)	\(503\)

[In]

int(1/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/56*(28*b^2*x^(2/3)+8*a*b*x^(1/3)+a^2)*(a+b*x^(1/3))/b^3/((a+b*x^(1/3))^2)^(9/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (107) = 214\).

Time = 0.41 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.01 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=-\frac {28 \, b^{18} x^{6} - 2856 \, a^{3} b^{15} x^{5} + 18186 \, a^{6} b^{12} x^{4} - 20608 \, a^{9} b^{9} x^{3} + 4200 \, a^{12} b^{6} x^{2} - 48 \, a^{15} b^{3} x + a^{18} - 27 \, {\left (8 \, a b^{17} x^{5} - 244 \, a^{4} b^{14} x^{4} + 840 \, a^{7} b^{11} x^{3} - 553 \, a^{10} b^{8} x^{2} + 56 \, a^{13} b^{5} x\right )} x^{\frac {2}{3}} + 27 \, {\left (35 \, a^{2} b^{16} x^{5} - 448 \, a^{5} b^{13} x^{4} + 876 \, a^{8} b^{10} x^{3} - 328 \, a^{11} b^{7} x^{2} + 14 \, a^{14} b^{4} x\right )} x^{\frac {1}{3}}}{56 \, {\left (b^{27} x^{8} + 8 \, a^{3} b^{24} x^{7} + 28 \, a^{6} b^{21} x^{6} + 56 \, a^{9} b^{18} x^{5} + 70 \, a^{12} b^{15} x^{4} + 56 \, a^{15} b^{12} x^{3} + 28 \, a^{18} b^{9} x^{2} + 8 \, a^{21} b^{6} x + a^{24} b^{3}\right )}} \]

[In]

integrate(1/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(9/2),x, algorithm="fricas")

[Out]

-1/56*(28*b^18*x^6 - 2856*a^3*b^15*x^5 + 18186*a^6*b^12*x^4 - 20608*a^9*b^9*x^3 + 4200*a^12*b^6*x^2 - 48*a^15*
b^3*x + a^18 - 27*(8*a*b^17*x^5 - 244*a^4*b^14*x^4 + 840*a^7*b^11*x^3 - 553*a^10*b^8*x^2 + 56*a^13*b^5*x)*x^(2
/3) + 27*(35*a^2*b^16*x^5 - 448*a^5*b^13*x^4 + 876*a^8*b^10*x^3 - 328*a^11*b^7*x^2 + 14*a^14*b^4*x)*x^(1/3))/(
b^27*x^8 + 8*a^3*b^24*x^7 + 28*a^6*b^21*x^6 + 56*a^9*b^18*x^5 + 70*a^12*b^15*x^4 + 56*a^15*b^12*x^3 + 28*a^18*
b^9*x^2 + 8*a^21*b^6*x + a^24*b^3)

Sympy [F]

\[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{\frac {9}{2}}}\, dx \]

[In]

integrate(1/(a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**(9/2),x)

[Out]

Integral((a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))**(-9/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=-\frac {1}{2 \, b^{9} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{6}} + \frac {6 \, a}{7 \, b^{10} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{7}} - \frac {3 \, a^{2}}{8 \, b^{11} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{8}} \]

[In]

integrate(1/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(9/2),x, algorithm="maxima")

[Out]

-1/2/(b^9*(x^(1/3) + a/b)^6) + 6/7*a/(b^10*(x^(1/3) + a/b)^7) - 3/8*a^2/(b^11*(x^(1/3) + a/b)^8)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.31 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=-\frac {28 \, b^{2} x^{\frac {2}{3}} + 8 \, a b x^{\frac {1}{3}} + a^{2}}{56 \, {\left (b x^{\frac {1}{3}} + a\right )}^{8} b^{3} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right )} \]

[In]

integrate(1/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(9/2),x, algorithm="giac")

[Out]

-1/56*(28*b^2*x^(2/3) + 8*a*b*x^(1/3) + a^2)/((b*x^(1/3) + a)^8*b^3*sgn(b*x^(1/3) + a))

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{9/2}} \, dx=-\frac {\sqrt {a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}}\,\left (a^2+28\,b^2\,x^{2/3}+8\,a\,b\,x^{1/3}\right )}{56\,b^3\,{\left (a+b\,x^{1/3}\right )}^9} \]

[In]

int(1/(a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(9/2),x)

[Out]

-((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(1/2)*(a^2 + 28*b^2*x^(2/3) + 8*a*b*x^(1/3)))/(56*b^3*(a + b*x^(1/3))^9)