3.2365 \(\int \frac {x^3}{(a+b \sqrt [3]{x})^2} \, dx\)

Optimal. Leaf size=171 \[ \frac {3 a^{11}}{b^{12} \left (a+b \sqrt [3]{x}\right )}+\frac {33 a^{10} \log \left (a+b \sqrt [3]{x}\right )}{b^{12}}-\frac {30 a^9 \sqrt [3]{x}}{b^{11}}+\frac {27 a^8 x^{2/3}}{2 b^{10}}-\frac {8 a^7 x}{b^9}+\frac {21 a^6 x^{4/3}}{4 b^8}-\frac {18 a^5 x^{5/3}}{5 b^7}+\frac {5 a^4 x^2}{2 b^6}-\frac {12 a^3 x^{7/3}}{7 b^5}+\frac {9 a^2 x^{8/3}}{8 b^4}-\frac {2 a x^3}{3 b^3}+\frac {3 x^{10/3}}{10 b^2} \]

[Out]

3*a^11/b^12/(a+b*x^(1/3))-30*a^9*x^(1/3)/b^11+27/2*a^8*x^(2/3)/b^10-8*a^7*x/b^9+21/4*a^6*x^(4/3)/b^8-18/5*a^5*
x^(5/3)/b^7+5/2*a^4*x^2/b^6-12/7*a^3*x^(7/3)/b^5+9/8*a^2*x^(8/3)/b^4-2/3*a*x^3/b^3+3/10*x^(10/3)/b^2+33*a^10*l
n(a+b*x^(1/3))/b^12

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Rubi [A]  time = 0.14, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac {27 a^8 x^{2/3}}{2 b^{10}}+\frac {21 a^6 x^{4/3}}{4 b^8}-\frac {18 a^5 x^{5/3}}{5 b^7}+\frac {5 a^4 x^2}{2 b^6}-\frac {12 a^3 x^{7/3}}{7 b^5}+\frac {9 a^2 x^{8/3}}{8 b^4}+\frac {3 a^{11}}{b^{12} \left (a+b \sqrt [3]{x}\right )}-\frac {30 a^9 \sqrt [3]{x}}{b^{11}}-\frac {8 a^7 x}{b^9}+\frac {33 a^{10} \log \left (a+b \sqrt [3]{x}\right )}{b^{12}}-\frac {2 a x^3}{3 b^3}+\frac {3 x^{10/3}}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^(1/3))^2,x]

[Out]

(3*a^11)/(b^12*(a + b*x^(1/3))) - (30*a^9*x^(1/3))/b^11 + (27*a^8*x^(2/3))/(2*b^10) - (8*a^7*x)/b^9 + (21*a^6*
x^(4/3))/(4*b^8) - (18*a^5*x^(5/3))/(5*b^7) + (5*a^4*x^2)/(2*b^6) - (12*a^3*x^(7/3))/(7*b^5) + (9*a^2*x^(8/3))
/(8*b^4) - (2*a*x^3)/(3*b^3) + (3*x^(10/3))/(10*b^2) + (33*a^10*Log[a + b*x^(1/3)])/b^12

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])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^3}{\left (a+b \sqrt [3]{x}\right )^2} \, dx &=3 \operatorname {Subst}\left (\int \frac {x^{11}}{(a+b x)^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (-\frac {10 a^9}{b^{11}}+\frac {9 a^8 x}{b^{10}}-\frac {8 a^7 x^2}{b^9}+\frac {7 a^6 x^3}{b^8}-\frac {6 a^5 x^4}{b^7}+\frac {5 a^4 x^5}{b^6}-\frac {4 a^3 x^6}{b^5}+\frac {3 a^2 x^7}{b^4}-\frac {2 a x^8}{b^3}+\frac {x^9}{b^2}-\frac {a^{11}}{b^{11} (a+b x)^2}+\frac {11 a^{10}}{b^{11} (a+b x)}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {3 a^{11}}{b^{12} \left (a+b \sqrt [3]{x}\right )}-\frac {30 a^9 \sqrt [3]{x}}{b^{11}}+\frac {27 a^8 x^{2/3}}{2 b^{10}}-\frac {8 a^7 x}{b^9}+\frac {21 a^6 x^{4/3}}{4 b^8}-\frac {18 a^5 x^{5/3}}{5 b^7}+\frac {5 a^4 x^2}{2 b^6}-\frac {12 a^3 x^{7/3}}{7 b^5}+\frac {9 a^2 x^{8/3}}{8 b^4}-\frac {2 a x^3}{3 b^3}+\frac {3 x^{10/3}}{10 b^2}+\frac {33 a^{10} \log \left (a+b \sqrt [3]{x}\right )}{b^{12}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 171, normalized size = 1.00 \[ \frac {3 a^{11}}{b^{12} \left (a+b \sqrt [3]{x}\right )}+\frac {33 a^{10} \log \left (a+b \sqrt [3]{x}\right )}{b^{12}}-\frac {30 a^9 \sqrt [3]{x}}{b^{11}}+\frac {27 a^8 x^{2/3}}{2 b^{10}}-\frac {8 a^7 x}{b^9}+\frac {21 a^6 x^{4/3}}{4 b^8}-\frac {18 a^5 x^{5/3}}{5 b^7}+\frac {5 a^4 x^2}{2 b^6}-\frac {12 a^3 x^{7/3}}{7 b^5}+\frac {9 a^2 x^{8/3}}{8 b^4}-\frac {2 a x^3}{3 b^3}+\frac {3 x^{10/3}}{10 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*x^(1/3))^2,x]

[Out]

(3*a^11)/(b^12*(a + b*x^(1/3))) - (30*a^9*x^(1/3))/b^11 + (27*a^8*x^(2/3))/(2*b^10) - (8*a^7*x)/b^9 + (21*a^6*
x^(4/3))/(4*b^8) - (18*a^5*x^(5/3))/(5*b^7) + (5*a^4*x^2)/(2*b^6) - (12*a^3*x^(7/3))/(7*b^5) + (9*a^2*x^(8/3))
/(8*b^4) - (2*a*x^3)/(3*b^3) + (3*x^(10/3))/(10*b^2) + (33*a^10*Log[a + b*x^(1/3)])/b^12

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fricas [A]  time = 0.60, size = 181, normalized size = 1.06 \[ -\frac {560 \, a b^{12} x^{4} - 1540 \, a^{4} b^{9} x^{3} + 4620 \, a^{7} b^{6} x^{2} + 6720 \, a^{10} b^{3} x - 2520 \, a^{13} - 27720 \, {\left (a^{10} b^{3} x + a^{13}\right )} \log \left (b x^{\frac {1}{3}} + a\right ) - 63 \, {\left (15 \, a^{2} b^{11} x^{3} - 33 \, a^{5} b^{8} x^{2} + 132 \, a^{8} b^{5} x + 220 \, a^{11} b^{2}\right )} x^{\frac {2}{3}} - 18 \, {\left (14 \, b^{13} x^{4} - 66 \, a^{3} b^{10} x^{3} + 165 \, a^{6} b^{7} x^{2} - 1155 \, a^{9} b^{4} x - 1540 \, a^{12} b\right )} x^{\frac {1}{3}}}{840 \, {\left (b^{15} x + a^{3} b^{12}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(560*a*b^12*x^4 - 1540*a^4*b^9*x^3 + 4620*a^7*b^6*x^2 + 6720*a^10*b^3*x - 2520*a^13 - 27720*(a^10*b^3*x
+ a^13)*log(b*x^(1/3) + a) - 63*(15*a^2*b^11*x^3 - 33*a^5*b^8*x^2 + 132*a^8*b^5*x + 220*a^11*b^2)*x^(2/3) - 1
8*(14*b^13*x^4 - 66*a^3*b^10*x^3 + 165*a^6*b^7*x^2 - 1155*a^9*b^4*x - 1540*a^12*b)*x^(1/3))/(b^15*x + a^3*b^12
)

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giac [A]  time = 0.17, size = 144, normalized size = 0.84 \[ \frac {33 \, a^{10} \log \left ({\left | b x^{\frac {1}{3}} + a \right |}\right )}{b^{12}} + \frac {3 \, a^{11}}{{\left (b x^{\frac {1}{3}} + a\right )} b^{12}} + \frac {252 \, b^{18} x^{\frac {10}{3}} - 560 \, a b^{17} x^{3} + 945 \, a^{2} b^{16} x^{\frac {8}{3}} - 1440 \, a^{3} b^{15} x^{\frac {7}{3}} + 2100 \, a^{4} b^{14} x^{2} - 3024 \, a^{5} b^{13} x^{\frac {5}{3}} + 4410 \, a^{6} b^{12} x^{\frac {4}{3}} - 6720 \, a^{7} b^{11} x + 11340 \, a^{8} b^{10} x^{\frac {2}{3}} - 25200 \, a^{9} b^{9} x^{\frac {1}{3}}}{840 \, b^{20}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

33*a^10*log(abs(b*x^(1/3) + a))/b^12 + 3*a^11/((b*x^(1/3) + a)*b^12) + 1/840*(252*b^18*x^(10/3) - 560*a*b^17*x
^3 + 945*a^2*b^16*x^(8/3) - 1440*a^3*b^15*x^(7/3) + 2100*a^4*b^14*x^2 - 3024*a^5*b^13*x^(5/3) + 4410*a^6*b^12*
x^(4/3) - 6720*a^7*b^11*x + 11340*a^8*b^10*x^(2/3) - 25200*a^9*b^9*x^(1/3))/b^20

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maple [A]  time = 0.01, size = 138, normalized size = 0.81 \[ \frac {3 x^{\frac {10}{3}}}{10 b^{2}}-\frac {2 a \,x^{3}}{3 b^{3}}+\frac {9 a^{2} x^{\frac {8}{3}}}{8 b^{4}}-\frac {12 a^{3} x^{\frac {7}{3}}}{7 b^{5}}+\frac {5 a^{4} x^{2}}{2 b^{6}}-\frac {18 a^{5} x^{\frac {5}{3}}}{5 b^{7}}+\frac {21 a^{6} x^{\frac {4}{3}}}{4 b^{8}}+\frac {3 a^{11}}{\left (b \,x^{\frac {1}{3}}+a \right ) b^{12}}+\frac {33 a^{10} \ln \left (b \,x^{\frac {1}{3}}+a \right )}{b^{12}}-\frac {8 a^{7} x}{b^{9}}+\frac {27 a^{8} x^{\frac {2}{3}}}{2 b^{10}}-\frac {30 a^{9} x^{\frac {1}{3}}}{b^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^11/b^12/(b*x^(1/3)+a)-30*a^9*x^(1/3)/b^11+27/2*a^8*x^(2/3)/b^10-8*a^7*x/b^9+21/4*a^6*x^(4/3)/b^8-18/5*a^5*
x^(5/3)/b^7+5/2*a^4*x^2/b^6-12/7*a^3*x^(7/3)/b^5+9/8*a^2*x^(8/3)/b^4-2/3*a*x^3/b^3+3/10*x^(10/3)/b^2+33*a^10*l
n(b*x^(1/3)+a)/b^12

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maxima [A]  time = 0.49, size = 197, normalized size = 1.15 \[ \frac {33 \, a^{10} \log \left (b x^{\frac {1}{3}} + a\right )}{b^{12}} + \frac {3 \, {\left (b x^{\frac {1}{3}} + a\right )}^{10}}{10 \, b^{12}} - \frac {11 \, {\left (b x^{\frac {1}{3}} + a\right )}^{9} a}{3 \, b^{12}} + \frac {165 \, {\left (b x^{\frac {1}{3}} + a\right )}^{8} a^{2}}{8 \, b^{12}} - \frac {495 \, {\left (b x^{\frac {1}{3}} + a\right )}^{7} a^{3}}{7 \, b^{12}} + \frac {165 \, {\left (b x^{\frac {1}{3}} + a\right )}^{6} a^{4}}{b^{12}} - \frac {1386 \, {\left (b x^{\frac {1}{3}} + a\right )}^{5} a^{5}}{5 \, b^{12}} + \frac {693 \, {\left (b x^{\frac {1}{3}} + a\right )}^{4} a^{6}}{2 \, b^{12}} - \frac {330 \, {\left (b x^{\frac {1}{3}} + a\right )}^{3} a^{7}}{b^{12}} + \frac {495 \, {\left (b x^{\frac {1}{3}} + a\right )}^{2} a^{8}}{2 \, b^{12}} - \frac {165 \, {\left (b x^{\frac {1}{3}} + a\right )} a^{9}}{b^{12}} + \frac {3 \, a^{11}}{{\left (b x^{\frac {1}{3}} + a\right )} b^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

33*a^10*log(b*x^(1/3) + a)/b^12 + 3/10*(b*x^(1/3) + a)^10/b^12 - 11/3*(b*x^(1/3) + a)^9*a/b^12 + 165/8*(b*x^(1
/3) + a)^8*a^2/b^12 - 495/7*(b*x^(1/3) + a)^7*a^3/b^12 + 165*(b*x^(1/3) + a)^6*a^4/b^12 - 1386/5*(b*x^(1/3) +
a)^5*a^5/b^12 + 693/2*(b*x^(1/3) + a)^4*a^6/b^12 - 330*(b*x^(1/3) + a)^3*a^7/b^12 + 495/2*(b*x^(1/3) + a)^2*a^
8/b^12 - 165*(b*x^(1/3) + a)*a^9/b^12 + 3*a^11/((b*x^(1/3) + a)*b^12)

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mupad [B]  time = 0.08, size = 143, normalized size = 0.84 \[ \frac {3\,x^{10/3}}{10\,b^2}+\frac {3\,a^{11}}{b\,\left (a\,b^{11}+b^{12}\,x^{1/3}\right )}-\frac {2\,a\,x^3}{3\,b^3}-\frac {8\,a^7\,x}{b^9}+\frac {33\,a^{10}\,\ln \left (a+b\,x^{1/3}\right )}{b^{12}}+\frac {5\,a^4\,x^2}{2\,b^6}+\frac {9\,a^2\,x^{8/3}}{8\,b^4}-\frac {12\,a^3\,x^{7/3}}{7\,b^5}-\frac {18\,a^5\,x^{5/3}}{5\,b^7}+\frac {21\,a^6\,x^{4/3}}{4\,b^8}+\frac {27\,a^8\,x^{2/3}}{2\,b^{10}}-\frac {30\,a^9\,x^{1/3}}{b^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b*x^(1/3))^2,x)

[Out]

(3*x^(10/3))/(10*b^2) + (3*a^11)/(b*(a*b^11 + b^12*x^(1/3))) - (2*a*x^3)/(3*b^3) - (8*a^7*x)/b^9 + (33*a^10*lo
g(a + b*x^(1/3)))/b^12 + (5*a^4*x^2)/(2*b^6) + (9*a^2*x^(8/3))/(8*b^4) - (12*a^3*x^(7/3))/(7*b^5) - (18*a^5*x^
(5/3))/(5*b^7) + (21*a^6*x^(4/3))/(4*b^8) + (27*a^8*x^(2/3))/(2*b^10) - (30*a^9*x^(1/3))/b^11

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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