Integrand size = 20, antiderivative size = 165 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=-\frac {A}{4 a x^4}+\frac {A b-a B}{a^2 x}-\frac {\sqrt [3]{b} (A b-a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{7/3}}-\frac {\sqrt [3]{b} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3}}+\frac {\sqrt [3]{b} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{7/3}} \] Output:
-1/4*A/a/x^4+(A*b-B*a)/a^2/x-1/3*b^(1/3)*(A*b-B*a)*arctan(1/3*(a^(1/3)-2*b ^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(7/3)-1/3*b^(1/3)*(A*b-B*a)*ln(a^(1/3 )+b^(1/3)*x)/a^(7/3)+1/6*b^(1/3)*(A*b-B*a)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^ (2/3)*x^2)/a^(7/3)
Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=\frac {-\frac {3 a^{4/3} A}{x^4}+\frac {12 \sqrt [3]{a} (A b-a B)}{x}-4 \sqrt {3} \sqrt [3]{b} (A b-a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 \sqrt [3]{b} (-A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt [3]{b} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{12 a^{7/3}} \] Input:
Integrate[(A + B*x^3)/(x^5*(a + b*x^3)),x]
Output:
((-3*a^(4/3)*A)/x^4 + (12*a^(1/3)*(A*b - a*B))/x - 4*Sqrt[3]*b^(1/3)*(A*b - a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 4*b^(1/3)*(-(A*b) + a *B)*Log[a^(1/3) + b^(1/3)*x] + 2*b^(1/3)*(A*b - a*B)*Log[a^(2/3) - a^(1/3) *b^(1/3)*x + b^(2/3)*x^2])/(12*a^(7/3))
Time = 0.56 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {955, 847, 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.
| \(\displaystyle \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx\) |
| \(\Big \downarrow \) 955 |
| \(\displaystyle -\frac {(A b-a B) \int \frac {1}{x^2 \left (b x^3+a\right )}dx}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \int \frac {x}{b x^3+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{a}-\frac {1}{a x}\right )}{a}-\frac {A}{4 a x^4}\) |
Input:
Int[(A + B*x^3)/(x^5*(a + b*x^3)),x]
Output:
-1/4*A/(a*x^4) - ((A*b - a*B)*(-(1/(a*x)) - (b*(-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))/Sqr t[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))))/a))/a
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[(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*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[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]
Time = 0.81 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.79
| method | result | size |
| default | \(-\frac {A}{4 a \,x^{4}}-\frac {-A b +B a}{a^{2} x}+\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) \left (A b -B a \right ) b}{a^{2}}\) | \(130\) |
| risch | \(\frac {\frac {\left (A b -B a \right ) x^{3}}{a^{2}}-\frac {A}{4 a}}{x^{4}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{3}+A^{3} b^{4}-3 A^{2} B a \,b^{3}+3 A \,B^{2} a^{2} b^{2}-B^{3} a^{3} b \right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{7} \textit {\_R}^{3}-3 A^{3} b^{4}+9 A^{2} B a \,b^{3}-9 A \,B^{2} a^{2} b^{2}+3 B^{3} a^{3} b \right ) x +\left (A \,a^{5} b -B \,a^{6}\right ) \textit {\_R}^{2}\right )\right )}{3}\) | \(151\) |
Input:
int((B*x^3+A)/x^5/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
-1/4*A/a/x^4-1/a^2*(-A*b+B*a)/x+(-1/3/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))+1/6/ b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+1/3*3^(1/2)/b/(a/b)^(1/3)* arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))*(A*b-B*a)/a^2*b
Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=-\frac {4 \, \sqrt {3} {\left (B a - A b\right )} x^{4} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - 2 \, {\left (B a - A b\right )} x^{4} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (-\frac {b}{a}\right )^{\frac {2}{3}} - a \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right ) + 4 \, {\left (B a - A b\right )} x^{4} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 12 \, {\left (B a - A b\right )} x^{3} + 3 \, A a}{12 \, a^{2} x^{4}} \] Input:
integrate((B*x^3+A)/x^5/(b*x^3+a),x, algorithm="fricas")
Output:
-1/12*(4*sqrt(3)*(B*a - A*b)*x^4*(-b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(-b/a)^ (1/3) + 1/3*sqrt(3)) - 2*(B*a - A*b)*x^4*(-b/a)^(1/3)*log(b*x^2 - a*x*(-b/ a)^(2/3) - a*(-b/a)^(1/3)) + 4*(B*a - A*b)*x^4*(-b/a)^(1/3)*log(b*x + a*(- b/a)^(2/3)) + 12*(B*a - A*b)*x^3 + 3*A*a)/(a^2*x^4)
Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.68 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=\operatorname {RootSum} {\left (27 t^{3} a^{7} + A^{3} b^{4} - 3 A^{2} B a b^{3} + 3 A B^{2} a^{2} b^{2} - B^{3} a^{3} b, \left ( t \mapsto t \log {\left (\frac {9 t^{2} a^{5}}{A^{2} b^{3} - 2 A B a b^{2} + B^{2} a^{2} b} + x \right )} \right )\right )} + \frac {- A a + x^{3} \cdot \left (4 A b - 4 B a\right )}{4 a^{2} x^{4}} \] Input:
integrate((B*x**3+A)/x**5/(b*x**3+a),x)
Output:
RootSum(27*_t**3*a**7 + A**3*b**4 - 3*A**2*B*a*b**3 + 3*A*B**2*a**2*b**2 - B**3*a**3*b, Lambda(_t, _t*log(9*_t**2*a**5/(A**2*b**3 - 2*A*B*a*b**2 + B **2*a**2*b) + x))) + (-A*a + x**3*(4*A*b - 4*B*a))/(4*a**2*x**4)
Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=-\frac {\sqrt {3} {\left (B a - A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (B a - A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (B a - A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {4 \, {\left (B a - A b\right )} x^{3} + A a}{4 \, a^{2} x^{4}} \] Input:
integrate((B*x^3+A)/x^5/(b*x^3+a),x, algorithm="maxima")
Output:
-1/3*sqrt(3)*(B*a - A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3 ))/(a^2*(a/b)^(1/3)) - 1/6*(B*a - A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/ 3))/(a^2*(a/b)^(1/3)) + 1/3*(B*a - A*b)*log(x + (a/b)^(1/3))/(a^2*(a/b)^(1 /3)) - 1/4*(4*(B*a - A*b)*x^3 + A*a)/(a^2*x^4)
Time = 0.13 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=\frac {{\left (B a b \left (-\frac {a}{b}\right )^{\frac {1}{3}} - A b^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{3}} + \frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{3} b} - \frac {{\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{3} b} - \frac {4 \, B a x^{3} - 4 \, A b x^{3} + A a}{4 \, a^{2} x^{4}} \] Input:
integrate((B*x^3+A)/x^5/(b*x^3+a),x, algorithm="giac")
Output:
1/3*(B*a*b*(-a/b)^(1/3) - A*b^2*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a /b)^(1/3)))/a^3 + 1/3*sqrt(3)*((-a*b^2)^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*ar ctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^3*b) - 1/6*((-a*b^2 )^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3)) /(a^3*b) - 1/4*(4*B*a*x^3 - 4*A*b*x^3 + A*a)/(a^2*x^4)
Time = 1.00 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=\frac {{\left (-b\right )}^{1/3}\,\ln \left (a^{1/3}\,{\left (-b\right )}^{8/3}+b^3\,x\right )\,\left (A\,b-B\,a\right )}{3\,a^{7/3}}-\frac {\frac {A}{4\,a}-\frac {x^3\,\left (A\,b-B\,a\right )}{a^2}}{x^4}+\frac {{\left (-b\right )}^{1/3}\,\ln \left (a^{1/3}\,{\left (-b\right )}^{8/3}-2\,b^3\,x+\sqrt {3}\,a^{1/3}\,{\left (-b\right )}^{8/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{7/3}}-\frac {{\left (-b\right )}^{1/3}\,\ln \left (2\,b^3\,x-a^{1/3}\,{\left (-b\right )}^{8/3}+\sqrt {3}\,a^{1/3}\,{\left (-b\right )}^{8/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{7/3}} \] Input:
int((A + B*x^3)/(x^5*(a + b*x^3)),x)
Output:
((-b)^(1/3)*log(a^(1/3)*(-b)^(8/3) + b^3*x)*(A*b - B*a))/(3*a^(7/3)) - (A/ (4*a) - (x^3*(A*b - B*a))/a^2)/x^4 + ((-b)^(1/3)*log(a^(1/3)*(-b)^(8/3) - 2*b^3*x + 3^(1/2)*a^(1/3)*(-b)^(8/3)*1i)*((3^(1/2)*1i)/2 - 1/2)*(A*b - B*a ))/(3*a^(7/3)) - ((-b)^(1/3)*log(2*b^3*x - a^(1/3)*(-b)^(8/3) + 3^(1/2)*a^ (1/3)*(-b)^(8/3)*1i)*((3^(1/2)*1i)/2 + 1/2)*(A*b - B*a))/(3*a^(7/3))
Time = 0.20 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.03 \[ \int \frac {A+B x^3}{x^5 \left (a+b x^3\right )} \, dx=-\frac {1}{4 x^{4}} \] Input:
int((B*x^3+A)/x^5/(b*x^3+a),x)
Output:
( - 1)/(4*x**4)