3.1979 \(\int \frac {x^4}{(a+\frac {b}{x^3})^2} \, dx\)

Optimal. Leaf size=157 \[ \frac {4 b^{5/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{9 a^{11/3}}-\frac {8 b^{5/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{11/3}}-\frac {8 b^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{11/3}}-\frac {4 b x^2}{3 a^3}+\frac {8 x^5}{15 a^2}-\frac {x^8}{3 a \left (a x^3+b\right )} \]

[Out]

-4/3*b*x^2/a^3+8/15*x^5/a^2-1/3*x^8/a/(a*x^3+b)-8/9*b^(5/3)*ln(b^(1/3)+a^(1/3)*x)/a^(11/3)+4/9*b^(5/3)*ln(b^(2
/3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/a^(11/3)-8/9*b^(5/3)*arctan(1/3*(b^(1/3)-2*a^(1/3)*x)/b^(1/3)*3^(1/2))/a^(1
1/3)*3^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {263, 288, 302, 292, 31, 634, 617, 204, 628} \[ \frac {4 b^{5/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{9 a^{11/3}}-\frac {8 b^{5/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{11/3}}-\frac {8 b^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{11/3}}-\frac {4 b x^2}{3 a^3}+\frac {8 x^5}{15 a^2}-\frac {x^8}{3 a \left (a x^3+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b/x^3)^2,x]

[Out]

(-4*b*x^2)/(3*a^3) + (8*x^5)/(15*a^2) - x^8/(3*a*(b + a*x^3)) - (8*b^(5/3)*ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqr
t[3]*b^(1/3))])/(3*Sqrt[3]*a^(11/3)) - (8*b^(5/3)*Log[b^(1/3) + a^(1/3)*x])/(9*a^(11/3)) + (4*b^(5/3)*Log[b^(2
/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2])/(9*a^(11/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {x^4}{\left (a+\frac {b}{x^3}\right )^2} \, dx &=\int \frac {x^{10}}{\left (b+a x^3\right )^2} \, dx\\ &=-\frac {x^8}{3 a \left (b+a x^3\right )}+\frac {8 \int \frac {x^7}{b+a x^3} \, dx}{3 a}\\ &=-\frac {x^8}{3 a \left (b+a x^3\right )}+\frac {8 \int \left (-\frac {b x}{a^2}+\frac {x^4}{a}+\frac {b^2 x}{a^2 \left (b+a x^3\right )}\right ) \, dx}{3 a}\\ &=-\frac {4 b x^2}{3 a^3}+\frac {8 x^5}{15 a^2}-\frac {x^8}{3 a \left (b+a x^3\right )}+\frac {\left (8 b^2\right ) \int \frac {x}{b+a x^3} \, dx}{3 a^3}\\ &=-\frac {4 b x^2}{3 a^3}+\frac {8 x^5}{15 a^2}-\frac {x^8}{3 a \left (b+a x^3\right )}-\frac {\left (8 b^{5/3}\right ) \int \frac {1}{\sqrt [3]{b}+\sqrt [3]{a} x} \, dx}{9 a^{10/3}}+\frac {\left (8 b^{5/3}\right ) \int \frac {\sqrt [3]{b}+\sqrt [3]{a} x}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{9 a^{10/3}}\\ &=-\frac {4 b x^2}{3 a^3}+\frac {8 x^5}{15 a^2}-\frac {x^8}{3 a \left (b+a x^3\right )}-\frac {8 b^{5/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{11/3}}+\frac {\left (4 b^{5/3}\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 a^{2/3} x}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{9 a^{11/3}}+\frac {\left (4 b^2\right ) \int \frac {1}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{3 a^{10/3}}\\ &=-\frac {4 b x^2}{3 a^3}+\frac {8 x^5}{15 a^2}-\frac {x^8}{3 a \left (b+a x^3\right )}-\frac {8 b^{5/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{11/3}}+\frac {4 b^{5/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{9 a^{11/3}}+\frac {\left (8 b^{5/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}\right )}{3 a^{11/3}}\\ &=-\frac {4 b x^2}{3 a^3}+\frac {8 x^5}{15 a^2}-\frac {x^8}{3 a \left (b+a x^3\right )}-\frac {8 b^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{11/3}}-\frac {8 b^{5/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{11/3}}+\frac {4 b^{5/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{9 a^{11/3}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 144, normalized size = 0.92 \[ \frac {20 b^{5/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )-\frac {15 a^{2/3} b^2 x^2}{a x^3+b}-45 a^{2/3} b x^2+9 a^{5/3} x^5-40 b^{5/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )-40 \sqrt {3} b^{5/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{45 a^{11/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b/x^3)^2,x]

[Out]

(-45*a^(2/3)*b*x^2 + 9*a^(5/3)*x^5 - (15*a^(2/3)*b^2*x^2)/(b + a*x^3) - 40*Sqrt[3]*b^(5/3)*ArcTan[(1 - (2*a^(1
/3)*x)/b^(1/3))/Sqrt[3]] - 40*b^(5/3)*Log[b^(1/3) + a^(1/3)*x] + 20*b^(5/3)*Log[b^(2/3) - a^(1/3)*b^(1/3)*x +
a^(2/3)*x^2])/(45*a^(11/3))

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fricas [A]  time = 0.63, size = 190, normalized size = 1.21 \[ \frac {9 \, a^{2} x^{8} - 36 \, a b x^{5} - 60 \, b^{2} x^{2} + 40 \, \sqrt {3} {\left (a b x^{3} + b^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + \sqrt {3} b}{3 \, b}\right ) - 20 \, {\left (a b x^{3} + b^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - b \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 40 \, {\left (a b x^{3} + b^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right )}{45 \, {\left (a^{4} x^{3} + a^{3} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^3)^2,x, algorithm="fricas")

[Out]

1/45*(9*a^2*x^8 - 36*a*b*x^5 - 60*b^2*x^2 + 40*sqrt(3)*(a*b*x^3 + b^2)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*
a*x*(-b^2/a^2)^(1/3) + sqrt(3)*b)/b) - 20*(a*b*x^3 + b^2)*(-b^2/a^2)^(1/3)*log(b*x^2 - a*x*(-b^2/a^2)^(2/3) -
b*(-b^2/a^2)^(1/3)) + 40*(a*b*x^3 + b^2)*(-b^2/a^2)^(1/3)*log(b*x + a*(-b^2/a^2)^(2/3)))/(a^4*x^3 + a^3*b)

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giac [A]  time = 0.20, size = 151, normalized size = 0.96 \[ -\frac {b^{2} x^{2}}{3 \, {\left (a x^{3} + b\right )} a^{3}} - \frac {8 \, b \left (-\frac {b}{a}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3}} - \frac {8 \, \sqrt {3} \left (-a^{2} b\right )^{\frac {2}{3}} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{5}} + \frac {4 \, \left (-a^{2} b\right )^{\frac {2}{3}} b \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{9 \, a^{5}} + \frac {a^{8} x^{5} - 5 \, a^{7} b x^{2}}{5 \, a^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^3)^2,x, algorithm="giac")

[Out]

-1/3*b^2*x^2/((a*x^3 + b)*a^3) - 8/9*b*(-b/a)^(2/3)*log(abs(x - (-b/a)^(1/3)))/a^3 - 8/9*sqrt(3)*(-a^2*b)^(2/3
)*b*arctan(1/3*sqrt(3)*(2*x + (-b/a)^(1/3))/(-b/a)^(1/3))/a^5 + 4/9*(-a^2*b)^(2/3)*b*log(x^2 + x*(-b/a)^(1/3)
+ (-b/a)^(2/3))/a^5 + 1/5*(a^8*x^5 - 5*a^7*b*x^2)/a^10

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maple [A]  time = 0.01, size = 137, normalized size = 0.87 \[ \frac {x^{5}}{5 a^{2}}-\frac {b^{2} x^{2}}{3 \left (a \,x^{3}+b \right ) a^{3}}-\frac {b \,x^{2}}{a^{3}}+\frac {8 \sqrt {3}\, b^{2} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {b}{a}\right )^{\frac {1}{3}} a^{4}}-\frac {8 b^{2} \ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {b}{a}\right )^{\frac {1}{3}} a^{4}}+\frac {4 b^{2} \ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{9 \left (\frac {b}{a}\right )^{\frac {1}{3}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b/x^3)^2,x)

[Out]

1/5/a^2*x^5-1/a^3*b*x^2-1/3/a^3*b^2*x^2/(a*x^3+b)-8/9/a^4*b^2/(1/a*b)^(1/3)*ln(x+(1/a*b)^(1/3))+4/9/a^4*b^2/(1
/a*b)^(1/3)*ln(x^2-(1/a*b)^(1/3)*x+(1/a*b)^(2/3))+8/9/a^4*b^2*3^(1/2)/(1/a*b)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/a
*b)^(1/3)*x-1))

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maxima [A]  time = 1.80, size = 147, normalized size = 0.94 \[ -\frac {b^{2} x^{2}}{3 \, {\left (a^{4} x^{3} + a^{3} b\right )}} + \frac {8 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}}} + \frac {4 \, b^{2} \log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{9 \, a^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}}} - \frac {8 \, b^{2} \log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \, a^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}}} + \frac {a x^{5} - 5 \, b x^{2}}{5 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^3)^2,x, algorithm="maxima")

[Out]

-1/3*b^2*x^2/(a^4*x^3 + a^3*b) + 8/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*x - (b/a)^(1/3))/(b/a)^(1/3))/(a^4*(b/a
)^(1/3)) + 4/9*b^2*log(x^2 - x*(b/a)^(1/3) + (b/a)^(2/3))/(a^4*(b/a)^(1/3)) - 8/9*b^2*log(x + (b/a)^(1/3))/(a^
4*(b/a)^(1/3)) + 1/5*(a*x^5 - 5*b*x^2)/a^3

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mupad [B]  time = 1.29, size = 172, normalized size = 1.10 \[ \frac {x^5}{5\,a^2}-\frac {b^2\,x^2}{3\,\left (a^4\,x^3+b\,a^3\right )}-\frac {b\,x^2}{a^3}+\frac {8\,{\left (-b\right )}^{5/3}\,\ln \left (\frac {64\,b^4\,x}{9\,a^5}-\frac {64\,{\left (-b\right )}^{13/3}}{9\,a^{16/3}}\right )}{9\,a^{11/3}}+\frac {8\,{\left (-b\right )}^{5/3}\,\ln \left (\frac {64\,b^4\,x}{9\,a^5}-\frac {64\,{\left (-b\right )}^{13/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{9\,a^{16/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{11/3}}-\frac {8\,{\left (-b\right )}^{5/3}\,\ln \left (\frac {64\,b^4\,x}{9\,a^5}-\frac {64\,{\left (-b\right )}^{13/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{9\,a^{16/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{11/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b/x^3)^2,x)

[Out]

x^5/(5*a^2) - (b^2*x^2)/(3*(a^3*b + a^4*x^3)) - (b*x^2)/a^3 + (8*(-b)^(5/3)*log((64*b^4*x)/(9*a^5) - (64*(-b)^
(13/3))/(9*a^(16/3))))/(9*a^(11/3)) + (8*(-b)^(5/3)*log((64*b^4*x)/(9*a^5) - (64*(-b)^(13/3)*((3^(1/2)*1i)/2 -
1/2)^2)/(9*a^(16/3)))*((3^(1/2)*1i)/2 - 1/2))/(9*a^(11/3)) - (8*(-b)^(5/3)*log((64*b^4*x)/(9*a^5) - (64*(-b)^
(13/3)*((3^(1/2)*1i)/2 + 1/2)^2)/(9*a^(16/3)))*((3^(1/2)*1i)/2 + 1/2))/(9*a^(11/3))

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sympy [A]  time = 0.41, size = 70, normalized size = 0.45 \[ - \frac {b^{2} x^{2}}{3 a^{4} x^{3} + 3 a^{3} b} + \operatorname {RootSum} {\left (729 t^{3} a^{11} + 512 b^{5}, \left (t \mapsto t \log {\left (\frac {81 t^{2} a^{7}}{64 b^{3}} + x \right )} \right )\right )} + \frac {x^{5}}{5 a^{2}} - \frac {b x^{2}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(a+b/x**3)**2,x)

[Out]

-b**2*x**2/(3*a**4*x**3 + 3*a**3*b) + RootSum(729*_t**3*a**11 + 512*b**5, Lambda(_t, _t*log(81*_t**2*a**7/(64*
b**3) + x))) + x**5/(5*a**2) - b*x**2/a**3

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