Integrand size = 20, antiderivative size = 233 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=-\frac {a (7 A b-10 a B) x}{3 b^4}+\frac {(7 A b-10 a B) x^4}{12 b^3}-\frac {(7 A b-10 a B) x^7}{21 a b^2}+\frac {(A b-a B) x^{10}}{3 a b \left (a+b x^3\right )}-\frac {a^{4/3} (7 A b-10 a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{13/3}}+\frac {a^{4/3} (7 A b-10 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{13/3}}-\frac {a^{4/3} (7 A b-10 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{13/3}} \]
-1/3*a*(7*A*b-10*B*a)*x/b^4+1/12*(7*A*b-10*B*a)*x^4/b^3-1/21*(7*A*b-10*B*a )*x^7/a/b^2+1/3*(A*b-B*a)*x^10/a/b/(b*x^3+a)+1/9*a^(4/3)*(7*A*b-10*B*a)*ln (a^(1/3)+b^(1/3)*x)/b^(13/3)-1/18*a^(4/3)*(7*A*b-10*B*a)*ln(a^(2/3)-a^(1/3 )*b^(1/3)*x+b^(2/3)*x^2)/b^(13/3)-1/9*a^(4/3)*(7*A*b-10*B*a)*arctan(1/3*(a ^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/b^(13/3)*3^(1/2)
Time = 0.13 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.87 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {252 a \sqrt [3]{b} (-2 A b+3 a B) x+63 b^{4/3} (A b-2 a B) x^4+36 b^{7/3} B x^7+\frac {84 a^2 \sqrt [3]{b} (-A b+a B) x}{a+b x^3}+28 \sqrt {3} a^{4/3} (-7 A b+10 a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-28 a^{4/3} (-7 A b+10 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+14 a^{4/3} (-7 A b+10 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{252 b^{13/3}} \]
(252*a*b^(1/3)*(-2*A*b + 3*a*B)*x + 63*b^(4/3)*(A*b - 2*a*B)*x^4 + 36*b^(7 /3)*B*x^7 + (84*a^2*b^(1/3)*(-(A*b) + a*B)*x)/(a + b*x^3) + 28*Sqrt[3]*a^( 4/3)*(-7*A*b + 10*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 28*a^ (4/3)*(-7*A*b + 10*a*B)*Log[a^(1/3) + b^(1/3)*x] + 14*a^(4/3)*(-7*A*b + 10 *a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(252*b^(13/3))
Time = 0.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {957, 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.
| \(\displaystyle \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {x^{10} (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(7 A b-10 a B) \int \frac {x^9}{b x^3+a}dx}{3 a b}\) |
| \(\Big \downarrow \) 831 |
| \(\displaystyle \frac {x^{10} (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(7 A b-10 a B) \int \left (\frac {x^6}{b}-\frac {a x^3}{b^2}-\frac {a^3}{b^3 \left (b x^3+a\right )}+\frac {a^2}{b^3}\right )dx}{3 a b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^{10} (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(7 A b-10 a B) \left (\frac {a^{7/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{10/3}}+\frac {a^{7/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{10/3}}-\frac {a^{7/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{10/3}}+\frac {a^2 x}{b^3}-\frac {a x^4}{4 b^2}+\frac {x^7}{7 b}\right )}{3 a b}\) |
((A*b - a*B)*x^10)/(3*a*b*(a + b*x^3)) - ((7*A*b - 10*a*B)*((a^2*x)/b^3 - (a*x^4)/(4*b^2) + x^7/(7*b) + (a^(7/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqr t[3]*a^(1/3))])/(Sqrt[3]*b^(10/3)) - (a^(7/3)*Log[a^(1/3) + b^(1/3)*x])/(3 *b^(10/3)) + (a^(7/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b ^(10/3))))/(3*a*b)
3.1.71.3.1 Defintions of rubi rules used
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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.49
| method | result | size |
| risch | \(\frac {B \,x^{7}}{7 b^{2}}+\frac {A \,x^{4}}{4 b^{2}}-\frac {B a \,x^{4}}{2 b^{3}}-\frac {2 a A x}{b^{3}}+\frac {3 a^{2} B x}{b^{4}}+\frac {\left (-\frac {1}{3} a^{2} b A +\frac {1}{3} a^{3} B \right ) x}{b^{4} \left (b \,x^{3}+a \right )}+\frac {a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (7 A b -10 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{9 b^{5}}\) | \(114\) |
| default | \(-\frac {-\frac {1}{7} b^{2} B \,x^{7}-\frac {1}{4} A \,b^{2} x^{4}+\frac {1}{2} B a b \,x^{4}+2 a A b x -3 a^{2} B x}{b^{4}}+\frac {a^{2} \left (\frac {\left (-\frac {A b}{3}+\frac {B a}{3}\right ) x}{b \,x^{3}+a}+\frac {\left (7 A b -10 B a \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}\right )}{b^{4}}\) | \(176\) |
1/7*B*x^7/b^2+1/4/b^2*A*x^4-1/2/b^3*B*a*x^4-2/b^3*a*A*x+3/b^4*a^2*B*x+(-1/ 3*a^2*b*A+1/3*a^3*B)*x/b^4/(b*x^3+a)+1/9/b^5*a^2*sum((7*A*b-10*B*a)/_R^2*l n(x-_R),_R=RootOf(_Z^3*b+a))
Time = 0.33 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.16 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {36 \, B b^{3} x^{10} - 9 \, {\left (10 \, B a b^{2} - 7 \, A b^{3}\right )} x^{7} + 63 \, {\left (10 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{4} - 28 \, \sqrt {3} {\left (10 \, B a^{3} - 7 \, A a^{2} b + {\left (10 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) + 14 \, {\left (10 \, B a^{3} - 7 \, A a^{2} b + {\left (10 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right ) - 28 \, {\left (10 \, B a^{3} - 7 \, A a^{2} b + {\left (10 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right ) + 84 \, {\left (10 \, B a^{3} - 7 \, A a^{2} b\right )} x}{252 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \]
1/252*(36*B*b^3*x^10 - 9*(10*B*a*b^2 - 7*A*b^3)*x^7 + 63*(10*B*a^2*b - 7*A *a*b^2)*x^4 - 28*sqrt(3)*(10*B*a^3 - 7*A*a^2*b + (10*B*a^2*b - 7*A*a*b^2)* x^3)*(a/b)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(a/b)^(2/3) - sqrt(3)*a)/a) + 1 4*(10*B*a^3 - 7*A*a^2*b + (10*B*a^2*b - 7*A*a*b^2)*x^3)*(a/b)^(1/3)*log(x^ 2 - x*(a/b)^(1/3) + (a/b)^(2/3)) - 28*(10*B*a^3 - 7*A*a^2*b + (10*B*a^2*b - 7*A*a*b^2)*x^3)*(a/b)^(1/3)*log(x + (a/b)^(1/3)) + 84*(10*B*a^3 - 7*A*a^ 2*b)*x)/(b^5*x^3 + a*b^4)
Time = 0.58 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.67 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {B x^{7}}{7 b^{2}} + x^{4} \left (\frac {A}{4 b^{2}} - \frac {B a}{2 b^{3}}\right ) + x \left (- \frac {2 A a}{b^{3}} + \frac {3 B a^{2}}{b^{4}}\right ) + \frac {x \left (- A a^{2} b + B a^{3}\right )}{3 a b^{4} + 3 b^{5} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} b^{13} - 343 A^{3} a^{4} b^{3} + 1470 A^{2} B a^{5} b^{2} - 2100 A B^{2} a^{6} b + 1000 B^{3} a^{7}, \left ( t \mapsto t \log {\left (- \frac {9 t b^{4}}{- 7 A a b + 10 B a^{2}} + x \right )} \right )\right )} \]
B*x**7/(7*b**2) + x**4*(A/(4*b**2) - B*a/(2*b**3)) + x*(-2*A*a/b**3 + 3*B* a**2/b**4) + x*(-A*a**2*b + B*a**3)/(3*a*b**4 + 3*b**5*x**3) + RootSum(729 *_t**3*b**13 - 343*A**3*a**4*b**3 + 1470*A**2*B*a**5*b**2 - 2100*A*B**2*a* *6*b + 1000*B**3*a**7, Lambda(_t, _t*log(-9*_t*b**4/(-7*A*a*b + 10*B*a**2) + x)))
Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.94 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {{\left (B a^{3} - A a^{2} b\right )} x}{3 \, {\left (b^{5} x^{3} + a b^{4}\right )}} + \frac {4 \, B b^{2} x^{7} - 7 \, {\left (2 \, B a b - A b^{2}\right )} x^{4} + 28 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} x}{28 \, b^{4}} - \frac {\sqrt {3} {\left (10 \, B a^{3} - 7 \, A a^{2} b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{5} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (10 \, B a^{3} - 7 \, A a^{2} b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{5} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (10 \, B a^{3} - 7 \, A a^{2} b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{5} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]
1/3*(B*a^3 - A*a^2*b)*x/(b^5*x^3 + a*b^4) + 1/28*(4*B*b^2*x^7 - 7*(2*B*a*b - A*b^2)*x^4 + 28*(3*B*a^2 - 2*A*a*b)*x)/b^4 - 1/9*sqrt(3)*(10*B*a^3 - 7* A*a^2*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^5*(a/b)^(2 /3)) + 1/18*(10*B*a^3 - 7*A*a^2*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/ (b^5*(a/b)^(2/3)) - 1/9*(10*B*a^3 - 7*A*a^2*b)*log(x + (a/b)^(1/3))/(b^5*( a/b)^(2/3))
Time = 0.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.05 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=-\frac {\sqrt {3} {\left (10 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a^{2} - 7 \, \left (-a b^{2}\right )^{\frac {1}{3}} A a b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{5}} + \frac {{\left (10 \, B a^{3} - 7 \, A a^{2} b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{4}} - \frac {{\left (10 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a^{2} - 7 \, \left (-a b^{2}\right )^{\frac {1}{3}} A a b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{5}} + \frac {B a^{3} x - A a^{2} b x}{3 \, {\left (b x^{3} + a\right )} b^{4}} + \frac {4 \, B b^{12} x^{7} - 14 \, B a b^{11} x^{4} + 7 \, A b^{12} x^{4} + 84 \, B a^{2} b^{10} x - 56 \, A a b^{11} x}{28 \, b^{14}} \]
-1/9*sqrt(3)*(10*(-a*b^2)^(1/3)*B*a^2 - 7*(-a*b^2)^(1/3)*A*a*b)*arctan(1/3 *sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^5 + 1/9*(10*B*a^3 - 7*A*a^2* b)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^4) - 1/18*(10*(-a*b^2)^(1/ 3)*B*a^2 - 7*(-a*b^2)^(1/3)*A*a*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3) )/b^5 + 1/3*(B*a^3*x - A*a^2*b*x)/((b*x^3 + a)*b^4) + 1/28*(4*B*b^12*x^7 - 14*B*a*b^11*x^4 + 7*A*b^12*x^4 + 84*B*a^2*b^10*x - 56*A*a*b^11*x)/b^14
Time = 6.80 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.90 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=x^4\,\left (\frac {A}{4\,b^2}-\frac {B\,a}{2\,b^3}\right )-x\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )+\frac {B\,x^7}{7\,b^2}+\frac {x\,\left (\frac {B\,a^3}{3}-\frac {A\,a^2\,b}{3}\right )}{b^5\,x^3+a\,b^4}+\frac {a^{4/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (7\,A\,b-10\,B\,a\right )}{9\,b^{13/3}}-\frac {a^{4/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (7\,A\,b-10\,B\,a\right )}{9\,b^{13/3}}+\frac {a^{4/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (7\,A\,b-10\,B\,a\right )}{9\,b^{13/3}} \]
x^4*(A/(4*b^2) - (B*a)/(2*b^3)) - x*((2*a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^ 2)/b^4) + (B*x^7)/(7*b^2) + (x*((B*a^3)/3 - (A*a^2*b)/3))/(a*b^4 + b^5*x^3 ) + (a^(4/3)*log(b^(1/3)*x + a^(1/3))*(7*A*b - 10*B*a))/(9*b^(13/3)) - (a^ (4/3)*log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/ 2)*(7*A*b - 10*B*a))/(9*b^(13/3)) + (a^(4/3)*log(3^(1/2)*a^(1/3)*1i + 2*b^ (1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(7*A*b - 10*B*a))/(9*b^(13/3))