Integrand size = 18, antiderivative size = 170 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\frac {B x}{b^2}+\frac {x \left (a B-A b x^2\right )}{3 b^2 \left (a+b x^3\right )}+\frac {4 \sqrt [3]{a} B \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{7/3}}-\frac {4 \sqrt [3]{a} B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{7/3}}+\frac {2 \sqrt [3]{a} B \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{9 b^{7/3}}+\frac {A \log \left (a+b x^3\right )}{3 b^2} \] Output:
B*x/b^2+1/3*x*(-A*b*x^2+B*a)/b^2/(b*x^3+a)+4/9*a^(1/3)*B*arctan(1/3*(a^(1/ 3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/b^(7/3)-4/9*a^(1/3)*B*ln(a^(1/3)+ b^(1/3)*x)/b^(7/3)+2/9*a^(1/3)*B*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2) /b^(7/3)+1/3*A*ln(b*x^3+a)/b^2
Time = 0.08 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\frac {9 \sqrt [3]{b} B x+\frac {3 a \sqrt [3]{b} (A+B x)}{a+b x^3}+4 \sqrt {3} \sqrt [3]{a} B \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-4 \sqrt [3]{a} B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt [3]{a} B \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+3 A \sqrt [3]{b} \log \left (a+b x^3\right )}{9 b^{7/3}} \] Input:
Integrate[(x^5*(A + B*x))/(a + b*x^3)^2,x]
Output:
(9*b^(1/3)*B*x + (3*a*b^(1/3)*(A + B*x))/(a + b*x^3) + 4*Sqrt[3]*a^(1/3)*B *ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 4*a^(1/3)*B*Log[a^(1/3) + b ^(1/3)*x] + 2*a^(1/3)*B*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] + 3 *A*b^(1/3)*Log[a + b*x^3])/(9*b^(7/3))
Time = 0.76 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2367, 2426, 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.
| \(\displaystyle \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 2367 |
| \(\displaystyle \frac {x \left (a B-A b x^2\right )}{3 b^2 \left (a+b x^3\right )}-\frac {\int \frac {-3 a b B x^3-3 a A b x^2+a^2 B}{b x^3+a}dx}{3 a b^2}\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle \frac {x \left (a B-A b x^2\right )}{3 b^2 \left (a+b x^3\right )}-\frac {\int \left (\frac {4 a^2 B-3 a A b x^2}{b x^3+a}-3 a B\right )dx}{3 a b^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x \left (a B-A b x^2\right )}{3 b^2 \left (a+b x^3\right )}-\frac {-\frac {4 a^{4/3} B \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {2 a^{4/3} B \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{3 \sqrt [3]{b}}+\frac {4 a^{4/3} B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{b}}-a A \log \left (a+b x^3\right )-3 a B x}{3 a b^2}\) |
Input:
Int[(x^5*(A + B*x))/(a + b*x^3)^2,x]
Output:
(x*(a*B - A*b*x^2))/(3*b^2*(a + b*x^3)) - (-3*a*B*x - (4*a^(4/3)*B*ArcTan[ (a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(1/3)) + (4*a^(4/3) *B*Log[a^(1/3) + b^(1/3)*x])/(3*b^(1/3)) - (2*a^(4/3)*B*Log[a^(2/3) - a^(1 /3)*b^(1/3)*x + b^(2/3)*x^2])/(3*b^(1/3)) - a*A*Log[a + b*x^3])/(3*a*b^2)
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1) *x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x]}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floo r[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) I nt[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0 ] && LtQ[p, -1] && IGtQ[m, 0]
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.41
| method | result | size |
| risch | \(\frac {B x}{b^{2}}+\frac {\frac {1}{3} B a x +\frac {1}{3} A a}{b^{2} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (3 A \,\textit {\_R}^{2} b -4 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 b^{3}}\) | \(69\) |
| default | \(\frac {B x}{b^{2}}+\frac {\frac {\frac {1}{3} B a x +\frac {1}{3} A a}{b \,x^{3}+a}-\frac {4 B a \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{3}+\frac {A \ln \left (b \,x^{3}+a \right )}{3}}{b^{2}}\) | \(138\) |
Input:
int(x^5*(B*x+A)/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
B*x/b^2+(1/3*B*a*x+1/3*A*a)/b^2/(b*x^3+a)+1/9/b^3*sum((3*A*_R^2*b-4*B*a)/_ R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))
Result contains complex when optimal does not.
Time = 0.80 (sec) , antiderivative size = 1081, normalized size of antiderivative = 6.36 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x^5*(B*x+A)/(b*x^3+a)^2,x, algorithm="fricas")
Output:
1/36*(36*B*b*x^4 + 48*B*a*x - 2*(b^3*x^3 + a*b^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b ^2)*log(1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64* B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b^2)*b^2 + 4*B*x + 3*A) + 12*A*a + (18* A*b*x^3 + (b^3*x^3 + a*b^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 2 7*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b^2) + 18*A*a - 3*sqrt( 1/3)*(b^3*x^3 + a*b^2)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b^2)^2*b^4 + 12*((1/2 )^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3*b) /b^7)^(1/3) - 6*A/b^2)*A*b^2 + 36*A^2)/b^4))*log(-1/2*((1/2)^(1/3)*(I*sqrt (3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6 *A/b^2)*b^2 + 3/2*sqrt(1/3)*b^2*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^ 3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b^2)^2*b^4 + 12*((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b^2)*A*b^2 + 36*A^2)/b^4) + 8*B*x - 3*A) + (18* A*b*x^3 + (b^3*x^3 + a*b^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 2 7*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b^2) + 18*A*a + 3*sqrt( 1/3)*(b^3*x^3 + a*b^2)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3*b)/b^7)^(1/3) - 6*A/b^2)^2*b^4 + 12*((1/2 )^(1/3)*(I*sqrt(3) + 1)*(64*B^3*a/b^7 - 27*A^3/b^6 - (64*B^3*a - 27*A^3...
Time = 0.49 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.53 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\frac {B x}{b^{2}} + \frac {A a + B a x}{3 a b^{2} + 3 b^{3} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} b^{7} - 729 t^{2} A b^{5} + 243 t A^{2} b^{3} - 27 A^{3} b + 64 B^{3} a, \left ( t \mapsto t \log {\left (x + \frac {- 9 t b^{2} + 3 A}{4 B} \right )} \right )\right )} \] Input:
integrate(x**5*(B*x+A)/(b*x**3+a)**2,x)
Output:
B*x/b**2 + (A*a + B*a*x)/(3*a*b**2 + 3*b**3*x**3) + RootSum(729*_t**3*b**7 - 729*_t**2*A*b**5 + 243*_t*A**2*b**3 - 27*A**3*b + 64*B**3*a, Lambda(_t, _t*log(x + (-9*_t*b**2 + 3*A)/(4*B))))
Time = 0.11 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\frac {B a x + A a}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} + \frac {B x}{b^{2}} - \frac {2 \, \sqrt {3} {\left ({\left (2 \, B \left (\frac {a}{b}\right )^{\frac {1}{3}} + A\right )} a - A a\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{2}} + \frac {{\left (3 \, A b \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, B a\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (3 \, A b \left (\frac {a}{b}\right )^{\frac {2}{3}} - 4 \, B a\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate(x^5*(B*x+A)/(b*x^3+a)^2,x, algorithm="maxima")
Output:
1/3*(B*a*x + A*a)/(b^3*x^3 + a*b^2) + B*x/b^2 - 2/9*sqrt(3)*((2*B*(a/b)^(1 /3) + A)*a - A*a)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b ^2) + 1/9*(3*A*b*(a/b)^(2/3) + 2*B*a)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3 ))/(b^3*(a/b)^(2/3)) + 1/9*(3*A*b*(a/b)^(2/3) - 4*B*a)*log(x + (a/b)^(1/3) )/(b^3*(a/b)^(2/3))
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.89 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\frac {4 \, B \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, b^{2}} + \frac {B x}{b^{2}} + \frac {A \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{2}} - \frac {4 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} B \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{3}} - \frac {2 \, \left (-a b^{2}\right )^{\frac {1}{3}} B \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, b^{3}} + \frac {B a x + A a}{3 \, {\left (b x^{3} + a\right )} b^{2}} \] Input:
integrate(x^5*(B*x+A)/(b*x^3+a)^2,x, algorithm="giac")
Output:
4/9*B*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/b^2 + B*x/b^2 + 1/3*A*log(ab s(b*x^3 + a))/b^2 - 4/9*sqrt(3)*(-a*b^2)^(1/3)*B*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^3 - 2/9*(-a*b^2)^(1/3)*B*log(x^2 + x*(-a/b) ^(1/3) + (-a/b)^(2/3))/b^3 + 1/3*(B*a*x + A*a)/((b*x^3 + a)*b^2)
Time = 6.35 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.12 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\left (\sum _{k=1}^3\ln \left (\frac {a\,\left (A-\mathrm {root}\left (729\,b^7\,z^3-729\,A\,b^5\,z^2+243\,A^2\,b^3\,z+64\,B^3\,a-27\,A^3\,b,z,k\right )\,b^2\,3\right )\,\left (3\,A+4\,B\,x-\mathrm {root}\left (729\,b^7\,z^3-729\,A\,b^5\,z^2+243\,A^2\,b^3\,z+64\,B^3\,a-27\,A^3\,b,z,k\right )\,b^2\,9\right )}{b^2\,3}\right )\,\mathrm {root}\left (729\,b^7\,z^3-729\,A\,b^5\,z^2+243\,A^2\,b^3\,z+64\,B^3\,a-27\,A^3\,b,z,k\right )\right )+\frac {\frac {A\,a}{3}+\frac {B\,a\,x}{3}}{b^3\,x^3+a\,b^2}+\frac {B\,x}{b^2} \] Input:
int((x^5*(A + B*x))/(a + b*x^3)^2,x)
Output:
symsum(log((a*(A - 3*root(729*b^7*z^3 - 729*A*b^5*z^2 + 243*A^2*b^3*z + 64 *B^3*a - 27*A^3*b, z, k)*b^2)*(3*A + 4*B*x - 9*root(729*b^7*z^3 - 729*A*b^ 5*z^2 + 243*A^2*b^3*z + 64*B^3*a - 27*A^3*b, z, k)*b^2))/(3*b^2))*root(729 *b^7*z^3 - 729*A*b^5*z^2 + 243*A^2*b^3*z + 64*B^3*a - 27*A^3*b, z, k), k, 1, 3) + ((A*a)/3 + (B*a*x)/3)/(a*b^2 + b^3*x^3) + (B*x)/b^2
Time = 0.15 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.71 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^3\right )^2} \, dx=\frac {4 a^{\frac {4}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b +4 a^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b^{2} x^{3}+2 a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b +2 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{2} x^{3}-4 a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b -4 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{2} x^{3}+3 b^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2}+3 b^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,x^{3}+3 b^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2}+3 b^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,x^{3}-3 b^{\frac {4}{3}} a \,x^{3}+12 b^{\frac {4}{3}} a x +9 b^{\frac {7}{3}} x^{4}}{9 b^{\frac {7}{3}} \left (b \,x^{3}+a \right )} \] Input:
int(x^5*(B*x+A)/(b*x^3+a)^2,x)
Output:
(4*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b + 4*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b **2*x**3 + 2*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)* a*b + 2*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**2* x**3 - 4*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*a*b - 4*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*b**2*x**3 + 3*b**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2 + 3*b**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b** (2/3)*x**2)*a*b*x**3 + 3*b**(1/3)*log(a**(1/3) + b**(1/3)*x)*a**2 + 3*b**( 1/3)*log(a**(1/3) + b**(1/3)*x)*a*b*x**3 - 3*b**(1/3)*a*b*x**3 + 12*b**(1/ 3)*a*b*x + 9*b**(1/3)*b**2*x**4)/(9*b**(1/3)*b**2*(a + b*x**3))