Integrand size = 13, antiderivative size = 154 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=-\frac {2 b x}{a^3}+\frac {x^4}{4 a^2}-\frac {b^2 x}{3 a^3 \left (b+a x^3\right )}-\frac {7 b^{4/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{10/3}}+\frac {7 b^{4/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{10/3}}-\frac {7 b^{4/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{10/3}} \] Output:
-2*b*x/a^3+1/4*x^4/a^2-1/3*b^2*x/a^3/(a*x^3+b)-7/9*b^(4/3)*arctan(1/3*(b^( 1/3)-2*a^(1/3)*x)*3^(1/2)/b^(1/3))*3^(1/2)/a^(10/3)+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)
Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.91 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {-72 \sqrt [3]{a} b x+9 a^{4/3} x^4-\frac {12 \sqrt [3]{a} b^2 x}{b+a x^3}-28 \sqrt {3} b^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )+28 b^{4/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )-14 b^{4/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{36 a^{10/3}} \] Input:
Integrate[x^3/(a + b/x^3)^2,x]
Output:
(-72*a^(1/3)*b*x + 9*a^(4/3)*x^4 - (12*a^(1/3)*b^2*x)/(b + a*x^3) - 28*Sqr t[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))
Time = 0.46 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {795, 817, 831, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {x^9}{\left (a x^3+b\right )^2}dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {7 \int \frac {x^6}{a x^3+b}dx}{3 a}-\frac {x^7}{3 a \left (a x^3+b\right )}\) |
\(\Big \downarrow \) 831 |
\(\displaystyle \frac {7 \int \left (\frac {x^3}{a}-\frac {b}{a^2}+\frac {b^2}{a^2 \left (a x^3+b\right )}\right )dx}{3 a}-\frac {x^7}{3 a \left (a x^3+b\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {7 \left (-\frac {b^{4/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} a^{7/3}}-\frac {b^{4/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{6 a^{7/3}}+\frac {b^{4/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{3 a^{7/3}}-\frac {b x}{a^2}+\frac {x^4}{4 a}\right )}{3 a}-\frac {x^7}{3 a \left (a x^3+b\right )}\) |
Input:
Int[x^3/(a + b/x^3)^2,x]
Output:
-1/3*x^7/(a*(b + a*x^3)) + (7*(-((b*x)/a^2) + x^4/(4*a) - (b^(4/3)*ArcTan[ (b^(1/3) - 2*a^(1/3)*x)/(Sqrt[3]*b^(1/3))])/(Sqrt[3]*a^(7/3)) + (b^(4/3)*L og[b^(1/3) + a^(1/3)*x])/(3*a^(7/3)) - (b^(4/3)*Log[b^(2/3) - a^(1/3)*b^(1 /3)*x + a^(2/3)*x^2])/(6*a^(7/3))))/(3*a)
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]
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] - Simp[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]
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] && Gt Q[m, 2*n - 1]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.42
method | result | size |
risch | \(\frac {x^{4}}{4 a^{2}}-\frac {2 b x}{a^{3}}-\frac {b^{2} x}{3 a^{3} \left (a \,x^{3}+b \right )}+\frac {7 b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{3}+b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{9 a^{4}}\) | \(64\) |
default | \(\frac {\frac {1}{4} a \,x^{4}-2 b x}{a^{3}}+\frac {b^{2} \left (-\frac {x}{3 \left (a \,x^{3}+b \right )}+\frac {7 \ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {7 \ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {7 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}\right )}{a^{3}}\) | \(126\) |
Input:
int(x^3/(a+b/x^3)^2,x,method=_RETURNVERBOSE)
Output:
1/4*x^4/a^2-2/a^3*b*x-1/3*b^2*x/a^3/(a*x^3+b)+7/9/a^4*b^2*sum(1/_R^2*ln(x- _R),_R=RootOf(_Z^3*a+b))
Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {9 \, a^{2} x^{7} - 63 \, a b x^{4} - 84 \, b^{2} x + 28 \, \sqrt {3} {\left (a b x^{3} + b^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b}{a}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 14 \, {\left (a b x^{3} + b^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 28 \, {\left (a b x^{3} + b^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{36 \, {\left (a^{4} x^{3} + a^{3} b\right )}} \] Input:
integrate(x^3/(a+b/x^3)^2,x, algorithm="fricas")
Output:
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)
Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.42 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=- \frac {b^{2} x}{3 a^{4} x^{3} + 3 a^{3} b} + \operatorname {RootSum} {\left (729 t^{3} a^{10} - 343 b^{4}, \left ( t \mapsto t \log {\left (\frac {9 t a^{3}}{7 b} + x \right )} \right )\right )} + \frac {x^{4}}{4 a^{2}} - \frac {2 b x}{a^{3}} \] Input:
integrate(x**3/(a+b/x**3)**2,x)
Output:
-b**2*x/(3*a**4*x**3 + 3*a**3*b) + RootSum(729*_t**3*a**10 - 343*b**4, Lam bda(_t, _t*log(9*_t*a**3/(7*b) + x))) + x**4/(4*a**2) - 2*b*x/a**3
Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=-\frac {b^{2} x}{3 \, {\left (a^{4} x^{3} + a^{3} b\right )}} + \frac {7 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4} \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {7 \, b^{2} \log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{4} \left (\frac {b}{a}\right )^{\frac {2}{3}}} + \frac {7 \, b^{2} \log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \, a^{4} \left (\frac {b}{a}\right )^{\frac {2}{3}}} + \frac {a x^{4} - 8 \, b x}{4 \, a^{3}} \] Input:
integrate(x^3/(a+b/x^3)^2,x, algorithm="maxima")
Output:
-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
Time = 0.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=-\frac {7 \, b \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3}} - \frac {b^{2} x}{3 \, {\left (a x^{3} + b\right )} a^{3}} + \frac {7 \, \sqrt {3} \left (-a^{2} b\right )^{\frac {1}{3}} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{4}} + \frac {7 \, \left (-a^{2} b\right )^{\frac {1}{3}} b \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{4}} + \frac {a^{6} x^{4} - 8 \, a^{5} b x}{4 \, a^{8}} \] Input:
integrate(x^3/(a+b/x^3)^2,x, algorithm="giac")
Output:
-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
Time = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {x^4}{4\,a^2}-\frac {b^2\,x}{3\,\left (a^4\,x^3+b\,a^3\right )}+\frac {7\,b^{4/3}\,\ln \left (\frac {7\,b^{7/3}}{a^{4/3}}+\frac {7\,b^2\,x}{a}\right )}{9\,a^{10/3}}-\frac {2\,b\,x}{a^3}+\frac {7\,b^{4/3}\,\ln \left (\frac {7\,b^2\,x}{a}+\frac {7\,b^{7/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{10/3}}-\frac {7\,b^{4/3}\,\ln \left (\frac {7\,b^2\,x}{a}-\frac {7\,b^{7/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{10/3}} \] Input:
int(x^3/(a + b/x^3)^2,x)
Output:
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))
Time = 0.21 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.24 \[ \int \frac {x^3}{\left (a+\frac {b}{x^3}\right )^2} \, dx=\frac {28 b^{\frac {4}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {2 a^{\frac {1}{3}} x -b^{\frac {1}{3}}}{b^{\frac {1}{3}} \sqrt {3}}\right ) a \,x^{3}+28 b^{\frac {7}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {2 a^{\frac {1}{3}} x -b^{\frac {1}{3}}}{b^{\frac {1}{3}} \sqrt {3}}\right )+9 a^{\frac {7}{3}} x^{7}-63 a^{\frac {4}{3}} b \,x^{4}-84 a^{\frac {1}{3}} b^{2} x -14 b^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {2}{3}} x^{2}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}}\right ) a \,x^{3}-14 b^{\frac {7}{3}} \mathrm {log}\left (a^{\frac {2}{3}} x^{2}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}}\right )+28 b^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}} x +b^{\frac {1}{3}}\right ) a \,x^{3}+28 b^{\frac {7}{3}} \mathrm {log}\left (a^{\frac {1}{3}} x +b^{\frac {1}{3}}\right )}{36 a^{\frac {10}{3}} \left (a \,x^{3}+b \right )} \] Input:
int(x^3/(a+b/x^3)^2,x)
Output:
(28*b**(1/3)*sqrt(3)*atan((2*a**(1/3)*x - b**(1/3))/(b**(1/3)*sqrt(3)))*a* b*x**3 + 28*b**(1/3)*sqrt(3)*atan((2*a**(1/3)*x - b**(1/3))/(b**(1/3)*sqrt (3)))*b**2 + 9*a**(1/3)*a**2*x**7 - 63*a**(1/3)*a*b*x**4 - 84*a**(1/3)*b** 2*x - 14*b**(1/3)*log(a**(2/3)*x**2 - b**(1/3)*a**(1/3)*x + b**(2/3))*a*b* x**3 - 14*b**(1/3)*log(a**(2/3)*x**2 - b**(1/3)*a**(1/3)*x + b**(2/3))*b** 2 + 28*b**(1/3)*log(a**(1/3)*x + b**(1/3))*a*b*x**3 + 28*b**(1/3)*log(a**( 1/3)*x + b**(1/3))*b**2)/(36*a**(1/3)*a**3*(a*x**3 + b))