Integrand size = 26, antiderivative size = 96 \[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=-\frac {A \sqrt {b x^2+c x^4}}{5 b x^6}-\frac {(5 b B-4 A c) \sqrt {b x^2+c x^4}}{15 b^2 x^4}+\frac {2 c (5 b B-4 A c) \sqrt {b x^2+c x^4}}{15 b^3 x^2} \] Output:
-1/5*A*(c*x^4+b*x^2)^(1/2)/b/x^6-1/15*(-4*A*c+5*B*b)*(c*x^4+b*x^2)^(1/2)/b ^2/x^4+2/15*c*(-4*A*c+5*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^2
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-5 b B x^2 \left (b-2 c x^2\right )+A \left (-3 b^2+4 b c x^2-8 c^2 x^4\right )\right )}{15 b^3 x^6} \] Input:
Integrate[(A + B*x^2)/(x^5*Sqrt[b*x^2 + c*x^4]),x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-5*b*B*x^2*(b - 2*c*x^2) + A*(-3*b^2 + 4*b*c*x^2 - 8*c^2*x^4)))/(15*b^3*x^6)
Time = 0.47 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1940, 1220, 1129, 1123}
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.
\(\displaystyle \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^6 \sqrt {c x^4+b x^2}}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (\frac {(5 b B-4 A c) \int \frac {1}{x^4 \sqrt {c x^4+b x^2}}dx^2}{5 b}-\frac {2 A \sqrt {b x^2+c x^4}}{5 b x^6}\right )\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {1}{2} \left (\frac {(5 b B-4 A c) \left (-\frac {2 c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx^2}{3 b}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^4}\right )}{5 b}-\frac {2 A \sqrt {b x^2+c x^4}}{5 b x^6}\right )\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (\frac {4 c \sqrt {b x^2+c x^4}}{3 b^2 x^2}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^4}\right ) (5 b B-4 A c)}{5 b}-\frac {2 A \sqrt {b x^2+c x^4}}{5 b x^6}\right )\) |
Input:
Int[(A + B*x^2)/(x^5*Sqrt[b*x^2 + c*x^4]),x]
Output:
((-2*A*Sqrt[b*x^2 + c*x^4])/(5*b*x^6) + ((5*b*B - 4*A*c)*((-2*Sqrt[b*x^2 + c*x^4])/(3*b*x^4) + (4*c*Sqrt[b*x^2 + c*x^4])/(3*b^2*x^2)))/(5*b))/2
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b *e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) )) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d , e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[Simplify[m + 2*p + 2], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 0.42 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66
method | result | size |
trager | \(-\frac {\left (8 x^{4} A \,c^{2}-10 x^{4} B b c -4 A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 b^{3} x^{6}}\) | \(63\) |
pseudoelliptic | \(-\frac {\left (\left (\frac {5 B \,x^{2}}{3}+A \right ) b^{2}-\frac {4 \left (\frac {5 B \,x^{2}}{2}+A \right ) c \,x^{2} b}{3}+\frac {8 x^{4} A \,c^{2}}{3}\right ) \left (c \,x^{2}+b \right )}{5 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, x^{4} b^{3}}\) | \(66\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 x^{4} A \,c^{2}-10 x^{4} B b c -4 A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right )}{15 x^{4} b^{3} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(70\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 x^{4} A \,c^{2}-10 x^{4} B b c -4 A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right )}{15 x^{4} b^{3} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(70\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 x^{4} A \,c^{2}-10 x^{4} B b c -4 A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right )}{15 x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b^{3}}\) | \(70\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 x^{4} A \,c^{2}-10 x^{4} B b c -4 A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right )}{15 x^{4} b^{3} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(70\) |
Input:
int((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(8*A*c^2*x^4-10*B*b*c*x^4-4*A*b*c*x^2+5*B*b^2*x^2+3*A*b^2)/b^3/x^6*( c*x^4+b*x^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=\frac {{\left (2 \, {\left (5 \, B b c - 4 \, A c^{2}\right )} x^{4} - 3 \, A b^{2} - {\left (5 \, B b^{2} - 4 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b^{3} x^{6}} \] Input:
integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
1/15*(2*(5*B*b*c - 4*A*c^2)*x^4 - 3*A*b^2 - (5*B*b^2 - 4*A*b*c)*x^2)*sqrt( c*x^4 + b*x^2)/(b^3*x^6)
\[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{x^{5} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate((B*x**2+A)/x**5/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral((A + B*x**2)/(x**5*sqrt(x**2*(b + c*x**2))), x)
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.24 \[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=\frac {1}{3} \, B {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}}}{b x^{4}}\right )} - \frac {1}{15} \, A {\left (\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{3} x^{2}} - \frac {4 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}}}{b x^{6}}\right )} \] Input:
integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/3*B*(2*sqrt(c*x^4 + b*x^2)*c/(b^2*x^2) - sqrt(c*x^4 + b*x^2)/(b*x^4)) - 1/15*A*(8*sqrt(c*x^4 + b*x^2)*c^2/(b^3*x^2) - 4*sqrt(c*x^4 + b*x^2)*c/(b^2 *x^4) + 3*sqrt(c*x^4 + b*x^2)/(b*x^6))
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (84) = 168\).
Time = 0.49 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.88 \[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=\frac {4 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B c^{\frac {3}{2}} - 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b c^{\frac {3}{2}} + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A c^{\frac {5}{2}} + 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{2} c^{\frac {3}{2}} - 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b c^{\frac {5}{2}} - 5 \, B b^{3} c^{\frac {3}{2}} + 4 \, A b^{2} c^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5} \mathrm {sgn}\left (x\right )} \] Input:
integrate((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
4/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*c^(3/2) - 35*(sqrt(c)*x - sqrt( c*x^2 + b))^4*B*b*c^(3/2) + 40*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*c^(5/2) + 25*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^2*c^(3/2) - 20*(sqrt(c)*x - sqrt(c *x^2 + b))^2*A*b*c^(5/2) - 5*B*b^3*c^(3/2) + 4*A*b^2*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5*sgn(x))
Time = 8.93 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}\,\left (5\,B\,b^2\,x^2+3\,A\,b^2-10\,B\,b\,c\,x^4-4\,A\,b\,c\,x^2+8\,A\,c^2\,x^4\right )}{15\,b^3\,x^6} \] Input:
int((A + B*x^2)/(x^5*(b*x^2 + c*x^4)^(1/2)),x)
Output:
-((b*x^2 + c*x^4)^(1/2)*(3*A*b^2 + 5*B*b^2*x^2 + 8*A*c^2*x^4 - 4*A*b*c*x^2 - 10*B*b*c*x^4))/(15*b^3*x^6)
Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x^2}{x^5 \sqrt {b x^2+c x^4}} \, dx=\frac {-3 \sqrt {c \,x^{2}+b}\, a \,b^{2}+4 \sqrt {c \,x^{2}+b}\, a b c \,x^{2}-8 \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{4}-5 \sqrt {c \,x^{2}+b}\, b^{3} x^{2}+10 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{4}+8 \sqrt {c}\, a \,c^{2} x^{5}-10 \sqrt {c}\, b^{2} c \,x^{5}}{15 b^{3} x^{5}} \] Input:
int((B*x^2+A)/x^5/(c*x^4+b*x^2)^(1/2),x)
Output:
( - 3*sqrt(b + c*x**2)*a*b**2 + 4*sqrt(b + c*x**2)*a*b*c*x**2 - 8*sqrt(b + c*x**2)*a*c**2*x**4 - 5*sqrt(b + c*x**2)*b**3*x**2 + 10*sqrt(b + c*x**2)* b**2*c*x**4 + 8*sqrt(c)*a*c**2*x**5 - 10*sqrt(c)*b**2*c*x**5)/(15*b**3*x** 5)