Integrand size = 20, antiderivative size = 208 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {B x^2}{2 b^3}+\frac {a (A b-a B) x^2}{6 b^3 \left (a+b x^3\right )^2}-\frac {(4 A b-7 a B) x^2}{9 b^3 \left (a+b x^3\right )}-\frac {5 (A b-4 a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} \sqrt [3]{a} b^{11/3}}-\frac {5 (A b-4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 \sqrt [3]{a} b^{11/3}}+\frac {5 (A b-4 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 \sqrt [3]{a} b^{11/3}} \] Output:
1/2*B*x^2/b^3+1/6*a*(A*b-B*a)*x^2/b^3/(b*x^3+a)^2-1/9*(4*A*b-7*B*a)*x^2/b^ 3/(b*x^3+a)-5/27*(A*b-4*B*a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1 /3))*3^(1/2)/a^(1/3)/b^(11/3)-5/27*(A*b-4*B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(1/ 3)/b^(11/3)+5/54*(A*b-4*B*a)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^( 1/3)/b^(11/3)
Time = 0.18 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.93 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {27 b^{2/3} B x^2+\frac {9 a b^{2/3} (A b-a B) x^2}{\left (a+b x^3\right )^2}-\frac {6 b^{2/3} (4 A b-7 a B) x^2}{a+b x^3}+\frac {10 \sqrt {3} (-A b+4 a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {10 (-A b+4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}+\frac {5 (A b-4 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{a}}}{54 b^{11/3}} \] Input:
Integrate[(x^7*(A + B*x^3))/(a + b*x^3)^3,x]
Output:
(27*b^(2/3)*B*x^2 + (9*a*b^(2/3)*(A*b - a*B)*x^2)/(a + b*x^3)^2 - (6*b^(2/ 3)*(4*A*b - 7*a*B)*x^2)/(a + b*x^3) + (10*Sqrt[3]*(-(A*b) + 4*a*B)*ArcTan[ (1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a^(1/3) + (10*(-(A*b) + 4*a*B)*Log[a ^(1/3) + b^(1/3)*x])/a^(1/3) + (5*(A*b - 4*a*B)*Log[a^(2/3) - a^(1/3)*b^(1 /3)*x + b^(2/3)*x^2])/a^(1/3))/(54*b^(11/3))
Time = 0.68 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {957, 817, 843, 821, 16, 1142, 25, 27, 1082, 217, 1103}
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.
| \(\displaystyle \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \int \frac {x^7}{\left (b x^3+a\right )^2}dx}{3 a b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \int \frac {x^4}{b x^3+a}dx}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \int \frac {x}{b x^3+a}dx}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {x^8 (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(A b-4 a B) \left (\frac {5 \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 b}-\frac {x^5}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
Input:
Int[(x^7*(A + B*x^3))/(a + b*x^3)^3,x]
Output:
((A*b - a*B)*x^8)/(6*a*b*(a + b*x^3)^2) - ((A*b - 4*a*B)*(-1/3*x^5/(b*(a + b*x^3)) + (5*(x^2/(2*b) - (a*(-1/3*Log[a^(1/3) + b^(1/3)*x]/(a^(1/3)*b^(2 /3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) + Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(1/3)*b^ (1/3))))/b))/(3*b)))/(3*a*b)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 1) Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[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]
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*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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)]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.86 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.43
| method | result | size |
| risch | \(\frac {B \,x^{2}}{2 b^{3}}+\frac {\left (-\frac {4}{9} b^{2} A +\frac {7}{9} a b B \right ) x^{5}-\frac {a \left (5 A b -11 B a \right ) x^{2}}{18}}{b^{3} \left (b \,x^{3}+a \right )^{2}}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (A b -4 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{27 b^{4}}\) | \(90\) |
| default | \(\frac {B \,x^{2}}{2 b^{3}}+\frac {\frac {\left (-\frac {4}{9} b^{2} A +\frac {7}{9} a b B \right ) x^{5}-\frac {a \left (5 A b -11 B a \right ) x^{2}}{18}}{\left (b \,x^{3}+a \right )^{2}}+\left (\frac {5 A b}{9}-\frac {20 B a}{9}\right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{3}}\) | \(158\) |
Input:
int(x^7*(B*x^3+A)/(b*x^3+a)^3,x,method=_RETURNVERBOSE)
Output:
1/2*B*x^2/b^3+((-4/9*b^2*A+7/9*a*b*B)*x^5-1/18*a*(5*A*b-11*B*a)*x^2)/b^3/( b*x^3+a)^2+5/27/b^4*sum((A*b-4*B*a)/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))
Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (168) = 336\).
Time = 0.13 (sec) , antiderivative size = 792, normalized size of antiderivative = 3.81 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(x^7*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="fricas")
Output:
[1/54*(27*B*a*b^4*x^8 + 24*(4*B*a^2*b^3 - A*a*b^4)*x^5 + 15*(4*B*a^3*b^2 - A*a^2*b^3)*x^2 - 15*sqrt(1/3)*((4*B*a^2*b^3 - A*a*b^4)*x^6 + 4*B*a^4*b - A*a^3*b^2 + 2*(4*B*a^3*b^2 - A*a^2*b^3)*x^3)*sqrt((-a*b^2)^(1/3)/a)*log((2 *b^2*x^3 - a*b + 3*sqrt(1/3)*(a*b*x + 2*(-a*b^2)^(2/3)*x^2 + (-a*b^2)^(1/3 )*a)*sqrt((-a*b^2)^(1/3)/a) - 3*(-a*b^2)^(2/3)*x)/(b*x^3 + a)) - 5*((4*B*a *b^2 - A*b^3)*x^6 + 4*B*a^3 - A*a^2*b + 2*(4*B*a^2*b - A*a*b^2)*x^3)*(-a*b ^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) + 10*((4*B*a* b^2 - A*b^3)*x^6 + 4*B*a^3 - A*a^2*b + 2*(4*B*a^2*b - A*a*b^2)*x^3)*(-a*b^ 2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a*b^7*x^6 + 2*a^2*b^6*x^3 + a^3*b^5), 1/54*(27*B*a*b^4*x^8 + 24*(4*B*a^2*b^3 - A*a*b^4)*x^5 + 15*(4*B*a^3*b^2 - A*a^2*b^3)*x^2 - 30*sqrt(1/3)*((4*B*a^2*b^3 - A*a*b^4)*x^6 + 4*B*a^4*b - A*a^3*b^2 + 2*(4*B*a^3*b^2 - A*a^2*b^3)*x^3)*sqrt(-(-a*b^2)^(1/3)/a)*arcta n(sqrt(1/3)*(2*b*x + (-a*b^2)^(1/3))*sqrt(-(-a*b^2)^(1/3)/a)/b) - 5*((4*B* a*b^2 - A*b^3)*x^6 + 4*B*a^3 - A*a^2*b + 2*(4*B*a^2*b - A*a*b^2)*x^3)*(-a* b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) + 10*((4*B*a *b^2 - A*b^3)*x^6 + 4*B*a^3 - A*a^2*b + 2*(4*B*a^2*b - A*a*b^2)*x^3)*(-a*b ^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a*b^7*x^6 + 2*a^2*b^6*x^3 + a^3*b^5) ]
Time = 1.49 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.78 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {B x^{2}}{2 b^{3}} + \frac {x^{5} \left (- 8 A b^{2} + 14 B a b\right ) + x^{2} \left (- 5 A a b + 11 B a^{2}\right )}{18 a^{2} b^{3} + 36 a b^{4} x^{3} + 18 b^{5} x^{6}} + \operatorname {RootSum} {\left (19683 t^{3} a b^{11} + 125 A^{3} b^{3} - 1500 A^{2} B a b^{2} + 6000 A B^{2} a^{2} b - 8000 B^{3} a^{3}, \left ( t \mapsto t \log {\left (\frac {729 t^{2} a b^{7}}{25 A^{2} b^{2} - 200 A B a b + 400 B^{2} a^{2}} + x \right )} \right )\right )} \] Input:
integrate(x**7*(B*x**3+A)/(b*x**3+a)**3,x)
Output:
B*x**2/(2*b**3) + (x**5*(-8*A*b**2 + 14*B*a*b) + x**2*(-5*A*a*b + 11*B*a** 2))/(18*a**2*b**3 + 36*a*b**4*x**3 + 18*b**5*x**6) + RootSum(19683*_t**3*a *b**11 + 125*A**3*b**3 - 1500*A**2*B*a*b**2 + 6000*A*B**2*a**2*b - 8000*B* *3*a**3, Lambda(_t, _t*log(729*_t**2*a*b**7/(25*A**2*b**2 - 200*A*B*a*b + 400*B**2*a**2) + x)))
Time = 0.11 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.94 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {2 \, {\left (7 \, B a b - 4 \, A b^{2}\right )} x^{5} + {\left (11 \, B a^{2} - 5 \, A a b\right )} x^{2}}{18 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{3} + a^{2} b^{3}\right )}} + \frac {B x^{2}}{2 \, b^{3}} - \frac {5 \, \sqrt {3} {\left (4 \, B a - A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {5 \, {\left (4 \, B a - A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, {\left (4 \, B a - A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \] Input:
integrate(x^7*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")
Output:
1/18*(2*(7*B*a*b - 4*A*b^2)*x^5 + (11*B*a^2 - 5*A*a*b)*x^2)/(b^5*x^6 + 2*a *b^4*x^3 + a^2*b^3) + 1/2*B*x^2/b^3 - 5/27*sqrt(3)*(4*B*a - A*b)*arctan(1/ 3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^4*(a/b)^(1/3)) - 5/54*(4*B*a - A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^4*(a/b)^(1/3)) + 5/27*(4 *B*a - A*b)*log(x + (a/b)^(1/3))/(b^4*(a/b)^(1/3))
Time = 0.14 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.01 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {B x^{2}}{2 \, b^{3}} - \frac {5 \, \sqrt {3} {\left (4 \, B a - A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{3}} + \frac {5 \, {\left (4 \, B a - A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{3}} + \frac {5 \, {\left (4 \, B a \left (-\frac {a}{b}\right )^{\frac {1}{3}} - A b \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a b^{3}} + \frac {14 \, B a b x^{5} - 8 \, A b^{2} x^{5} + 11 \, B a^{2} x^{2} - 5 \, A a b x^{2}}{18 \, {\left (b x^{3} + a\right )}^{2} b^{3}} \] Input:
integrate(x^7*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="giac")
Output:
1/2*B*x^2/b^3 - 5/27*sqrt(3)*(4*B*a - A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b )^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(1/3)*b^3) + 5/54*(4*B*a - A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(1/3)*b^3) + 5/27*(4*B*a*(-a/b)^ (1/3) - A*b*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^3) + 1/18*(14*B*a*b*x^5 - 8*A*b^2*x^5 + 11*B*a^2*x^2 - 5*A*a*b*x^2)/((b*x^3 + a)^2*b^3)
Time = 0.98 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.90 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {x^2\,\left (\frac {11\,B\,a^2}{18}-\frac {5\,A\,a\,b}{18}\right )-x^5\,\left (\frac {4\,A\,b^2}{9}-\frac {7\,B\,a\,b}{9}\right )}{a^2\,b^3+2\,a\,b^4\,x^3+b^5\,x^6}+\frac {B\,x^2}{2\,b^3}-\frac {5\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-4\,B\,a\right )}{27\,a^{1/3}\,b^{11/3}}-\frac {5\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-4\,B\,a\right )}{27\,a^{1/3}\,b^{11/3}}+\frac {5\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-4\,B\,a\right )}{27\,a^{1/3}\,b^{11/3}} \] Input:
int((x^7*(A + B*x^3))/(a + b*x^3)^3,x)
Output:
(x^2*((11*B*a^2)/18 - (5*A*a*b)/18) - x^5*((4*A*b^2)/9 - (7*B*a*b)/9))/(a^ 2*b^3 + b^5*x^6 + 2*a*b^4*x^3) + (B*x^2)/(2*b^3) - (5*log(b^(1/3)*x + a^(1 /3))*(A*b - 4*B*a))/(27*a^(1/3)*b^(11/3)) - (5*log(3^(1/2)*a^(1/3)*1i - 2* b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(A*b - 4*B*a))/(27*a^(1/3)*b^( 11/3)) + (5*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/ 2 + 1/2)*(A*b - 4*B*a))/(27*a^(1/3)*b^(11/3))
Time = 0.22 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.87 \[ \int \frac {x^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {10 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a^{2}+10 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a b \,x^{3}+15 b^{\frac {2}{3}} a^{\frac {4}{3}} x^{2}+9 b^{\frac {5}{3}} a^{\frac {1}{3}} x^{5}-5 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2}-5 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a b \,x^{3}+10 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2}+10 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a b \,x^{3}}{18 b^{\frac {8}{3}} a^{\frac {1}{3}} \left (b \,x^{3}+a \right )} \] Input:
int(x^7*(B*x^3+A)/(b*x^3+a)^3,x)
Output:
(10*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2 + 10*s qrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*b*x**3 + 15*b* *(2/3)*a**(1/3)*a*x**2 + 9*b**(2/3)*a**(1/3)*b*x**5 - 5*log(a**(2/3) - b** (1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2 - 5*log(a**(2/3) - b**(1/3)*a**(1/3 )*x + b**(2/3)*x**2)*a*b*x**3 + 10*log(a**(1/3) + b**(1/3)*x)*a**2 + 10*lo g(a**(1/3) + b**(1/3)*x)*a*b*x**3)/(18*b**(2/3)*a**(1/3)*b**2*(a + b*x**3) )