\(\int \frac {x^3}{(a+\frac {b}{x^3})^2} \, dx\) [1980]

Optimal result

Integrand size = 13, antiderivative size = 155 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=-\frac {7 b x}{3 a^3}+\frac {7 x^4}{12 a^2}-\frac {x^7}{3 a \left (b+a x^3\right )}-\frac {7 b^{4/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{10/3}}+\frac {7 b^{4/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{10/3}}-\frac {7 b^{4/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{10/3}} \]

[Out]

-7/3*b*x/a^3+7/12*x^4/a^2-1/3*x^7/a/(a*x^3+b)+7/9*b^(4/3)*ln(b^(1/3)+a^(1/3)*x)/a^(10/3)-7/18*b^(4/3)*ln(b^(2/
3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/a^(10/3)-7/9*b^(4/3)*arctan(1/3*(b^(1/3)-2*a^(1/3)*x)/b^(1/3)*3^(1/2))/a^(10
/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {269, 294, 308, 206, 31, 648, 631, 210, 642} \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=-\frac {7 b^{4/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{10/3}}-\frac {7 b^{4/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{18 a^{10/3}}+\frac {7 b^{4/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{10/3}}-\frac {7 b x}{3 a^3}+\frac {7 x^4}{12 a^2}-\frac {x^7}{3 a \left (a x^3+b\right )} \]

[In]

Int[x^3/(a + b/x^3)^2,x]

[Out]

(-7*b*x)/(3*a^3) + (7*x^4)/(12*a^2) - x^7/(3*a*(b + a*x^3)) - (7*b^(4/3)*ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqrt[
3]*b^(1/3))])/(3*Sqrt[3]*a^(10/3)) + (7*b^(4/3)*Log[b^(1/3) + a^(1/3)*x])/(9*a^(10/3)) - (7*b^(4/3)*Log[b^(2/3
) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2])/(18*a^(10/3))

Rule 31

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

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 210

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

Rule 269

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 294

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 308

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 631

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 642

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 648

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*} \text {integral}& = \int \frac {x^9}{\left (b+a x^3\right )^2} \, dx \\ & = -\frac {x^7}{3 a \left (b+a x^3\right )}+\frac {7 \int \frac {x^6}{b+a x^3} \, dx}{3 a} \\ & = -\frac {x^7}{3 a \left (b+a x^3\right )}+\frac {7 \int \left (-\frac {b}{a^2}+\frac {x^3}{a}+\frac {b^2}{a^2 \left (b+a x^3\right )}\right ) \, dx}{3 a} \\ & = -\frac {7 b x}{3 a^3}+\frac {7 x^4}{12 a^2}-\frac {x^7}{3 a \left (b+a x^3\right )}+\frac {\left (7 b^2\right ) \int \frac {1}{b+a x^3} \, dx}{3 a^3} \\ & = -\frac {7 b x}{3 a^3}+\frac {7 x^4}{12 a^2}-\frac {x^7}{3 a \left (b+a x^3\right )}+\frac {\left (7 b^{4/3}\right ) \int \frac {1}{\sqrt [3]{b}+\sqrt [3]{a} x} \, dx}{9 a^3}+\frac {\left (7 b^{4/3}\right ) \int \frac {2 \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^3} \\ & = -\frac {7 b x}{3 a^3}+\frac {7 x^4}{12 a^2}-\frac {x^7}{3 a \left (b+a x^3\right )}+\frac {7 b^{4/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{10/3}}-\frac {\left (7 b^{4/3}\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}{18 a^{10/3}}+\frac {\left (7 b^{5/3}\right ) \int \frac {1}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{6 a^3} \\ & = -\frac {7 b x}{3 a^3}+\frac {7 x^4}{12 a^2}-\frac {x^7}{3 a \left (b+a x^3\right )}+\frac {7 b^{4/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{10/3}}-\frac {7 b^{4/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{10/3}}+\frac {\left (7 b^{4/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}\right )}{3 a^{10/3}} \\ & = -\frac {7 b x}{3 a^3}+\frac {7 x^4}{12 a^2}-\frac {x^7}{3 a \left (b+a x^3\right )}-\frac {7 b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{10/3}}+\frac {7 b^{4/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{10/3}}-\frac {7 b^{4/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{10/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {-72 \sqrt [3]{a} b x+9 a^{4/3} x^4-\frac {12 \sqrt [3]{a} b^2 x}{b+a x^3}-28 \sqrt {3} b^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )+28 b^{4/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )-14 b^{4/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{36 a^{10/3}} \]

[In]

Integrate[x^3/(a + b/x^3)^2,x]

[Out]

(-72*a^(1/3)*b*x + 9*a^(4/3)*x^4 - (12*a^(1/3)*b^2*x)/(b + a*x^3) - 28*Sqrt[3]*b^(4/3)*ArcTan[(1 - (2*a^(1/3)*
x)/b^(1/3))/Sqrt[3]] + 28*b^(4/3)*Log[b^(1/3) + a^(1/3)*x] - 14*b^(4/3)*Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2
/3)*x^2])/(36*a^(10/3))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.41

method	result	size
risch	\(\frac {x^{4}}{4 a^{2}}-\frac {2 b x}{a^{3}}-\frac {b^{2} x}{3 \left (a \,x^{3}+b \right ) a^{3}}+\frac {7 b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{3}+b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{9 a^{4}}\)	\(64\)
default	\(\frac {\frac {1}{4} a \,x^{4}-2 b x}{a^{3}}+\frac {b^{2} \left (-\frac {x}{3 \left (a \,x^{3}+b \right )}+\frac {7 \ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {7 \ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {7 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}\right )}{a^{3}}\)	\(126\)

[In]

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

[Out]

1/4*x^4/a^2-2/a^3*b*x-1/3*b^2*x/(a*x^3+b)/a^3+7/9/a^4*b^2*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3*a+b))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {9 \, a^{2} x^{7} - 63 \, a b x^{4} - 84 \, b^{2} x + 28 \, \sqrt {3} {\left (a b x^{3} + b^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b}{a}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 14 \, {\left (a b x^{3} + b^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 28 \, {\left (a b x^{3} + b^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{36 \, {\left (a^{4} x^{3} + a^{3} b\right )}} \]

[In]

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

[Out]

1/36*(9*a^2*x^7 - 63*a*b*x^4 - 84*b^2*x + 28*sqrt(3)*(a*b*x^3 + b^2)*(b/a)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(b/
a)^(2/3) - sqrt(3)*b)/b) - 14*(a*b*x^3 + b^2)*(b/a)^(1/3)*log(x^2 - x*(b/a)^(1/3) + (b/a)^(2/3)) + 28*(a*b*x^3
+ b^2)*(b/a)^(1/3)*log(x + (b/a)^(1/3)))/(a^4*x^3 + a^3*b)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.42 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=- \frac {b^{2} x}{3 a^{4} x^{3} + 3 a^{3} b} + \operatorname {RootSum} {\left (729 t^{3} a^{10} - 343 b^{4}, \left ( t \mapsto t \log {\left (\frac {9 t a^{3}}{7 b} + x \right )} \right )\right )} + \frac {x^{4}}{4 a^{2}} - \frac {2 b x}{a^{3}} \]

[In]

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

[Out]

-b**2*x/(3*a**4*x**3 + 3*a**3*b) + RootSum(729*_t**3*a**10 - 343*b**4, Lambda(_t, _t*log(9*_t*a**3/(7*b) + x))
) + x**4/(4*a**2) - 2*b*x/a**3

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=-\frac {b^{2} x}{3 \, {\left (a^{4} x^{3} + a^{3} b\right )}} + \frac {7 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4} \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {7 \, b^{2} \log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{4} \left (\frac {b}{a}\right )^{\frac {2}{3}}} + \frac {7 \, b^{2} \log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \, a^{4} \left (\frac {b}{a}\right )^{\frac {2}{3}}} + \frac {a x^{4} - 8 \, b x}{4 \, a^{3}} \]

[In]

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

[Out]

-1/3*b^2*x/(a^4*x^3 + a^3*b) + 7/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*x - (b/a)^(1/3))/(b/a)^(1/3))/(a^4*(b/a)^
(2/3)) - 7/18*b^2*log(x^2 - x*(b/a)^(1/3) + (b/a)^(2/3))/(a^4*(b/a)^(2/3)) + 7/9*b^2*log(x + (b/a)^(1/3))/(a^4
*(b/a)^(2/3)) + 1/4*(a*x^4 - 8*b*x)/a^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=-\frac {7 \, b \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3}} - \frac {b^{2} x}{3 \, {\left (a x^{3} + b\right )} a^{3}} + \frac {7 \, \sqrt {3} \left (-a^{2} b\right )^{\frac {1}{3}} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4}} + \frac {7 \, \left (-a^{2} b\right )^{\frac {1}{3}} b \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{4}} + \frac {a^{6} x^{4} - 8 \, a^{5} b x}{4 \, a^{8}} \]

[In]

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

[Out]

-7/9*b*(-b/a)^(1/3)*log(abs(x - (-b/a)^(1/3)))/a^3 - 1/3*b^2*x/((a*x^3 + b)*a^3) + 7/9*sqrt(3)*(-a^2*b)^(1/3)*
b*arctan(1/3*sqrt(3)*(2*x + (-b/a)^(1/3))/(-b/a)^(1/3))/a^4 + 7/18*(-a^2*b)^(1/3)*b*log(x^2 + x*(-b/a)^(1/3) +
(-b/a)^(2/3))/a^4 + 1/4*(a^6*x^4 - 8*a^5*b*x)/a^8

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.98 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {x^4}{4\,a^2}-\frac {b^2\,x}{3\,\left (a^4\,x^3+b\,a^3\right )}+\frac {7\,b^{4/3}\,\ln \left (\frac {7\,b^{7/3}}{a^{4/3}}+\frac {7\,b^2\,x}{a}\right )}{9\,a^{10/3}}-\frac {2\,b\,x}{a^3}+\frac {7\,b^{4/3}\,\ln \left (\frac {7\,b^2\,x}{a}+\frac {7\,b^{7/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{10/3}}-\frac {7\,b^{4/3}\,\ln \left (\frac {7\,b^2\,x}{a}-\frac {7\,b^{7/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{10/3}} \]

[In]

int(x^3/(a + b/x^3)^2,x)

[Out]

x^4/(4*a^2) - (b^2*x)/(3*(a^3*b + a^4*x^3)) + (7*b^(4/3)*log((7*b^(7/3))/a^(4/3) + (7*b^2*x)/a))/(9*a^(10/3))
- (2*b*x)/a^3 + (7*b^(4/3)*log((7*b^2*x)/a + (7*b^(7/3)*((3^(1/2)*1i)/2 - 1/2))/a^(4/3))*((3^(1/2)*1i)/2 - 1/2
))/(9*a^(10/3)) - (7*b^(4/3)*log((7*b^2*x)/a - (7*b^(7/3)*((3^(1/2)*1i)/2 + 1/2))/a^(4/3))*((3^(1/2)*1i)/2 + 1
/2))/(9*a^(10/3))