Integrand size = 24, antiderivative size = 78 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {A}{5 b x^5}-\frac {b B-A c}{3 b^2 x^3}+\frac {c (b B-A c)}{b^3 x}+\frac {c^{3/2} (b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}} \] Output:
-1/5*A/b/x^5-1/3*(-A*c+B*b)/b^2/x^3+c*(-A*c+B*b)/b^3/x+c^(3/2)*(-A*c+B*b)* arctan(c^(1/2)*x/b^(1/2))/b^(7/2)
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {A}{5 b x^5}+\frac {-b B+A c}{3 b^2 x^3}+\frac {c (b B-A c)}{b^3 x}+\frac {c^{3/2} (b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}} \] Input:
Integrate[(A + B*x^2)/(x^4*(b*x^2 + c*x^4)),x]
Output:
-1/5*A/(b*x^5) + (-(b*B) + A*c)/(3*b^2*x^3) + (c*(b*B - A*c))/(b^3*x) + (c ^(3/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(7/2)
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {9, 359, 264, 264, 218}
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.
\(\displaystyle \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{x^6 \left (b+c x^2\right )}dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {(b B-A c) \int \frac {1}{x^4 \left (c x^2+b\right )}dx}{b}-\frac {A}{5 b x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {(b B-A c) \left (-\frac {c \int \frac {1}{x^2 \left (c x^2+b\right )}dx}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {A}{5 b x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {(b B-A c) \left (-\frac {c \left (-\frac {c \int \frac {1}{c x^2+b}dx}{b}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {A}{5 b x^5}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(b B-A c) \left (-\frac {c \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{b}-\frac {A}{5 b x^5}\) |
Input:
Int[(A + B*x^2)/(x^4*(b*x^2 + c*x^4)),x]
Output:
-1/5*A/(b*x^5) + ((b*B - A*c)*(-1/3*1/(b*x^3) - (c*(-(1/(b*x)) - (Sqrt[c]* ArcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(3/2)))/b))/b
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Time = 0.44 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {A}{5 b \,x^{5}}-\frac {-A c +B b}{3 b^{2} x^{3}}-\frac {\left (A c -B b \right ) c}{b^{3} x}-\frac {\left (A c -B b \right ) c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{b^{3} \sqrt {b c}}\) | \(74\) |
risch | \(\frac {-\frac {\left (A c -B b \right ) c \,x^{4}}{b^{3}}+\frac {\left (A c -B b \right ) x^{2}}{3 b^{2}}-\frac {A}{5 b}}{x^{5}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{7} \textit {\_Z}^{2}+A^{2} c^{5}-2 A B b \,c^{4}+B^{2} b^{2} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{7}+2 A^{2} c^{5}-4 A B b \,c^{4}+2 B^{2} b^{2} c^{3}\right ) x +\left (A \,b^{4} c^{2}-B \,b^{5} c \right ) \textit {\_R} \right )\right )}{2}\) | \(145\) |
Input:
int((B*x^2+A)/x^4/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)
Output:
-1/5*A/b/x^5-1/3*(-A*c+B*b)/b^2/x^3-(A*c-B*b)/b^3*c/x-(A*c-B*b)/b^3*c^2/(b *c)^(1/2)*arctan(c*x/(b*c)^(1/2))
Time = 0.10 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.36 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=\left [-\frac {15 \, {\left (B b c - A c^{2}\right )} x^{5} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) - 30 \, {\left (B b c - A c^{2}\right )} x^{4} + 6 \, A b^{2} + 10 \, {\left (B b^{2} - A b c\right )} x^{2}}{30 \, b^{3} x^{5}}, \frac {15 \, {\left (B b c - A c^{2}\right )} x^{5} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 15 \, {\left (B b c - A c^{2}\right )} x^{4} - 3 \, A b^{2} - 5 \, {\left (B b^{2} - A b c\right )} x^{2}}{15 \, b^{3} x^{5}}\right ] \] Input:
integrate((B*x^2+A)/x^4/(c*x^4+b*x^2),x, algorithm="fricas")
Output:
[-1/30*(15*(B*b*c - A*c^2)*x^5*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)) - 30*(B*b*c - A*c^2)*x^4 + 6*A*b^2 + 10*(B*b^2 - A*b*c)*x^ 2)/(b^3*x^5), 1/15*(15*(B*b*c - A*c^2)*x^5*sqrt(c/b)*arctan(x*sqrt(c/b)) + 15*(B*b*c - A*c^2)*x^4 - 3*A*b^2 - 5*(B*b^2 - A*b*c)*x^2)/(b^3*x^5)]
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (68) = 136\).
Time = 0.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.09 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=- \frac {\sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right ) \log {\left (- \frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right )}{- A c^{3} + B b c^{2}} + x \right )}}{2} + \frac {\sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right ) \log {\left (\frac {b^{4} \sqrt {- \frac {c^{3}}{b^{7}}} \left (- A c + B b\right )}{- A c^{3} + B b c^{2}} + x \right )}}{2} + \frac {- 3 A b^{2} + x^{4} \left (- 15 A c^{2} + 15 B b c\right ) + x^{2} \cdot \left (5 A b c - 5 B b^{2}\right )}{15 b^{3} x^{5}} \] Input:
integrate((B*x**2+A)/x**4/(c*x**4+b*x**2),x)
Output:
-sqrt(-c**3/b**7)*(-A*c + B*b)*log(-b**4*sqrt(-c**3/b**7)*(-A*c + B*b)/(-A *c**3 + B*b*c**2) + x)/2 + sqrt(-c**3/b**7)*(-A*c + B*b)*log(b**4*sqrt(-c* *3/b**7)*(-A*c + B*b)/(-A*c**3 + B*b*c**2) + x)/2 + (-3*A*b**2 + x**4*(-15 *A*c**2 + 15*B*b*c) + x**2*(5*A*b*c - 5*B*b**2))/(15*b**3*x**5)
Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=\frac {{\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} + \frac {15 \, {\left (B b c - A c^{2}\right )} x^{4} - 3 \, A b^{2} - 5 \, {\left (B b^{2} - A b c\right )} x^{2}}{15 \, b^{3} x^{5}} \] Input:
integrate((B*x^2+A)/x^4/(c*x^4+b*x^2),x, algorithm="maxima")
Output:
(B*b*c^2 - A*c^3)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) + 1/15*(15*(B*b*c - A*c^2)*x^4 - 3*A*b^2 - 5*(B*b^2 - A*b*c)*x^2)/(b^3*x^5)
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=\frac {{\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} + \frac {15 \, B b c x^{4} - 15 \, A c^{2} x^{4} - 5 \, B b^{2} x^{2} + 5 \, A b c x^{2} - 3 \, A b^{2}}{15 \, b^{3} x^{5}} \] Input:
integrate((B*x^2+A)/x^4/(c*x^4+b*x^2),x, algorithm="giac")
Output:
(B*b*c^2 - A*c^3)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) + 1/15*(15*B*b*c*x ^4 - 15*A*c^2*x^4 - 5*B*b^2*x^2 + 5*A*b*c*x^2 - 3*A*b^2)/(b^3*x^5)
Time = 9.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {\frac {A}{5\,b}-\frac {x^2\,\left (A\,c-B\,b\right )}{3\,b^2}+\frac {c\,x^4\,\left (A\,c-B\,b\right )}{b^3}}{x^5}-\frac {c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{7/2}} \] Input:
int((A + B*x^2)/(x^4*(b*x^2 + c*x^4)),x)
Output:
- (A/(5*b) - (x^2*(A*c - B*b))/(3*b^2) + (c*x^4*(A*c - B*b))/b^3)/x^5 - (c ^(3/2)*atan((c^(1/2)*x)/b^(1/2))*(A*c - B*b))/b^(7/2)
Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x^2}{x^4 \left (b x^2+c x^4\right )} \, dx=\frac {-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{2} x^{5}+15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c \,x^{5}-3 a \,b^{3}+5 a \,b^{2} c \,x^{2}-15 a b \,c^{2} x^{4}-5 b^{4} x^{2}+15 b^{3} c \,x^{4}}{15 b^{4} x^{5}} \] Input:
int((B*x^2+A)/x^4/(c*x^4+b*x^2),x)
Output:
( - 15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*c**2*x**5 + 15*sqrt (c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*c*x**5 - 3*a*b**3 + 5*a*b** 2*c*x**2 - 15*a*b*c**2*x**4 - 5*b**4*x**2 + 15*b**3*c*x**4)/(15*b**4*x**5)