Integrand size = 26, antiderivative size = 138 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {A}{7 b x^6 \sqrt {b x^2+c x^4}}-\frac {7 b B-8 A c}{35 b^2 x^4 \sqrt {b x^2+c x^4}}+\frac {2 c (7 b B-8 A c)}{35 b^3 x^2 \sqrt {b x^2+c x^4}}-\frac {8 c^2 (7 b B-8 A c) \left (b+2 c x^2\right )}{35 b^5 \sqrt {b x^2+c x^4}} \]
-1/7*A/b/x^6/(c*x^4+b*x^2)^(1/2)+1/35*(8*A*c-7*B*b)/b^2/x^4/(c*x^4+b*x^2)^ (1/2)+2/35*c*(-8*A*c+7*B*b)/b^3/x^2/(c*x^4+b*x^2)^(1/2)-8/35*c^2*(-8*A*c+7 *B*b)*(2*c*x^2+b)/b^5/(c*x^4+b*x^2)^(1/2)
Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-7 b B x^2 \left (b^3-2 b^2 c x^2+8 b c^2 x^4+16 c^3 x^6\right )+A \left (-5 b^4+8 b^3 c x^2-16 b^2 c^2 x^4+64 b c^3 x^6+128 c^4 x^8\right )}{35 b^5 x^6 \sqrt {x^2 \left (b+c x^2\right )}} \]
(-7*b*B*x^2*(b^3 - 2*b^2*c*x^2 + 8*b*c^2*x^4 + 16*c^3*x^6) + A*(-5*b^4 + 8 *b^3*c*x^2 - 16*b^2*c^2*x^4 + 64*b*c^3*x^6 + 128*c^4*x^8))/(35*b^5*x^6*Sqr t[x^2*(b + c*x^2)])
Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1940, 1220, 1129, 1129, 1088}
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.
\(\displaystyle \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^6 \left (c x^4+b x^2\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (\frac {(7 b B-8 A c) \int \frac {1}{x^4 \left (c x^4+b x^2\right )^{3/2}}dx^2}{7 b}-\frac {2 A}{7 b x^6 \sqrt {b x^2+c x^4}}\right )\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {1}{2} \left (\frac {(7 b B-8 A c) \left (-\frac {6 c \int \frac {1}{x^2 \left (c x^4+b x^2\right )^{3/2}}dx^2}{5 b}-\frac {2}{5 b x^4 \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {2 A}{7 b x^6 \sqrt {b x^2+c x^4}}\right )\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {1}{2} \left (\frac {(7 b B-8 A c) \left (-\frac {6 c \left (-\frac {4 c \int \frac {1}{\left (c x^4+b x^2\right )^{3/2}}dx^2}{3 b}-\frac {2}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {2}{5 b x^4 \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {2 A}{7 b x^6 \sqrt {b x^2+c x^4}}\right )\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (-\frac {6 c \left (\frac {8 c \left (b+2 c x^2\right )}{3 b^3 \sqrt {b x^2+c x^4}}-\frac {2}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {2}{5 b x^4 \sqrt {b x^2+c x^4}}\right ) (7 b B-8 A c)}{7 b}-\frac {2 A}{7 b x^6 \sqrt {b x^2+c x^4}}\right )\) |
((-2*A)/(7*b*x^6*Sqrt[b*x^2 + c*x^4]) + ((7*b*B - 8*A*c)*(-2/(5*b*x^4*Sqrt [b*x^2 + c*x^4]) - (6*c*(-2/(3*b*x^2*Sqrt[b*x^2 + c*x^4]) + (8*c*(b + 2*c* x^2))/(3*b^3*Sqrt[b*x^2 + c*x^4])))/(5*b)))/(7*b))/2
3.2.52.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 = 1.80 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.70
method | result | size |
pseudoelliptic | \(-\frac {\left (\frac {7 x^{2} B}{5}+A \right ) b^{4}-\frac {8 x^{2} \left (\frac {7 x^{2} B}{4}+A \right ) c \,b^{3}}{5}+\frac {16 x^{4} \left (\frac {7 x^{2} B}{2}+A \right ) c^{2} b^{2}}{5}-\frac {64 \left (-\frac {7 x^{2} B}{4}+A \right ) x^{6} c^{3} b}{5}-\frac {128 A \,x^{8} c^{4}}{5}}{7 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, x^{6} b^{5}}\) | \(97\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-128 A \,x^{8} c^{4}+112 B \,x^{8} b \,c^{3}-64 A \,x^{6} b \,c^{3}+56 B \,x^{6} b^{2} c^{2}+16 A \,b^{2} c^{2} x^{4}-14 B \,b^{3} c \,x^{4}-8 A \,x^{2} b^{3} c +7 B \,x^{2} b^{4}+5 A \,b^{4}\right )}{35 x^{4} b^{5} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\) | \(118\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-128 A \,x^{8} c^{4}+112 B \,x^{8} b \,c^{3}-64 A \,x^{6} b \,c^{3}+56 B \,x^{6} b^{2} c^{2}+16 A \,b^{2} c^{2} x^{4}-14 B \,b^{3} c \,x^{4}-8 A \,x^{2} b^{3} c +7 B \,x^{2} b^{4}+5 A \,b^{4}\right )}{35 x^{4} b^{5} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\) | \(118\) |
trager | \(-\frac {\left (-128 A \,x^{8} c^{4}+112 B \,x^{8} b \,c^{3}-64 A \,x^{6} b \,c^{3}+56 B \,x^{6} b^{2} c^{2}+16 A \,b^{2} c^{2} x^{4}-14 B \,b^{3} c \,x^{4}-8 A \,x^{2} b^{3} c +7 B \,x^{2} b^{4}+5 A \,b^{4}\right ) \sqrt {x^{4} c +b \,x^{2}}}{35 \left (c \,x^{2}+b \right ) b^{5} x^{8}}\) | \(120\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (-93 A \,c^{3} x^{6}+77 x^{6} B b \,c^{2}+29 A b \,c^{2} x^{4}-21 x^{4} B \,b^{2} c -13 A \,b^{2} c \,x^{2}+7 b^{3} B \,x^{2}+5 b^{3} A \right )}{35 b^{5} x^{6} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {x^{2} c^{3} \left (A c -B b \right )}{b^{5} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(126\) |
-1/7/(x^2*(c*x^2+b))^(1/2)*((7/5*x^2*B+A)*b^4-8/5*x^2*(7/4*x^2*B+A)*c*b^3+ 16/5*x^4*(7/2*x^2*B+A)*c^2*b^2-64/5*(-7/4*x^2*B+A)*x^6*c^3*b-128/5*A*x^8*c ^4)/x^6/b^5
Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {{\left (16 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{8} + 8 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{6} + 5 \, A b^{4} - 2 \, {\left (7 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{4} + {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, {\left (b^{5} c x^{10} + b^{6} x^{8}\right )}} \]
-1/35*(16*(7*B*b*c^3 - 8*A*c^4)*x^8 + 8*(7*B*b^2*c^2 - 8*A*b*c^3)*x^6 + 5* A*b^4 - 2*(7*B*b^3*c - 8*A*b^2*c^2)*x^4 + (7*B*b^4 - 8*A*b^3*c)*x^2)*sqrt( c*x^4 + b*x^2)/(b^5*c*x^10 + b^6*x^8)
\[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {A + B x^{2}}{x^{5} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.51 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {1}{5} \, B {\left (\frac {16 \, c^{3} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{4}} + \frac {8 \, c^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} - \frac {2 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} + \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{4}}\right )} + \frac {1}{35} \, A {\left (\frac {128 \, c^{4} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{5}} + \frac {64 \, c^{3}}{\sqrt {c x^{4} + b x^{2}} b^{4}} - \frac {16 \, c^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3} x^{2}} + \frac {8 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2} x^{4}} - \frac {5}{\sqrt {c x^{4} + b x^{2}} b x^{6}}\right )} \]
-1/5*B*(16*c^3*x^2/(sqrt(c*x^4 + b*x^2)*b^4) + 8*c^2/(sqrt(c*x^4 + b*x^2)* b^3) - 2*c/(sqrt(c*x^4 + b*x^2)*b^2*x^2) + 1/(sqrt(c*x^4 + b*x^2)*b*x^4)) + 1/35*A*(128*c^4*x^2/(sqrt(c*x^4 + b*x^2)*b^5) + 64*c^3/(sqrt(c*x^4 + b*x ^2)*b^4) - 16*c^2/(sqrt(c*x^4 + b*x^2)*b^3*x^2) + 8*c/(sqrt(c*x^4 + b*x^2) *b^2*x^4) - 5/(sqrt(c*x^4 + b*x^2)*b*x^6))
Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (122) = 244\).
Time = 1.28 (sec) , antiderivative size = 415, normalized size of antiderivative = 3.01 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {{\left (B b c^{3} - A c^{4}\right )} x}{\sqrt {c x^{2} + b} b^{5} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B b c^{\frac {5}{2}} - 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} A c^{\frac {7}{2}} - 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b^{2} c^{\frac {5}{2}} + 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A b c^{\frac {7}{2}} + 1015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{3} c^{\frac {5}{2}} - 1015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b^{2} c^{\frac {7}{2}} - 1680 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{4} c^{\frac {5}{2}} + 2240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{3} c^{\frac {7}{2}} + 1337 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{5} c^{\frac {5}{2}} - 1673 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{4} c^{\frac {7}{2}} - 504 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{6} c^{\frac {5}{2}} + 616 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{5} c^{\frac {7}{2}} + 77 \, B b^{7} c^{\frac {5}{2}} - 93 \, A b^{6} c^{\frac {7}{2}}\right )}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7} b^{4} \mathrm {sgn}\left (x\right )} \]
-(B*b*c^3 - A*c^4)*x/(sqrt(c*x^2 + b)*b^5*sgn(x)) + 2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*b*c^(5/2) - 35*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*c^ (7/2) - 280*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b^2*c^(5/2) + 280*(sqrt(c)* x - sqrt(c*x^2 + b))^10*A*b*c^(7/2) + 1015*(sqrt(c)*x - sqrt(c*x^2 + b))^8 *B*b^3*c^(5/2) - 1015*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b^2*c^(7/2) - 1680 *(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^4*c^(5/2) + 2240*(sqrt(c)*x - sqrt(c* x^2 + b))^6*A*b^3*c^(7/2) + 1337*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^5*c^( 5/2) - 1673*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^4*c^(7/2) - 504*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^6*c^(5/2) + 616*(sqrt(c)*x - sqrt(c*x^2 + b))^2* A*b^5*c^(7/2) + 77*B*b^7*c^(5/2) - 93*A*b^6*c^(7/2))/(((sqrt(c)*x - sqrt(c *x^2 + b))^2 - b)^7*b^4*sgn(x))
Time = 9.41 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x^2}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\left (7\,B\,b^2-13\,A\,b\,c\right )\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^6}-\frac {\left (x^2\,\left (\frac {58\,A\,c^4-42\,B\,b\,c^3}{35\,b^5}-\frac {2\,c^3\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^5}\right )-\frac {c^2\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^4}\right )\,\sqrt {c\,x^4+b\,x^2}}{x^2\,\left (c\,x^2+b\right )}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{7\,b^2\,x^8}-\frac {c\,\left (29\,A\,c-21\,B\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^4} \]
- ((7*B*b^2 - 13*A*b*c)*(b*x^2 + c*x^4)^(1/2))/(35*b^4*x^6) - ((x^2*((58*A *c^4 - 42*B*b*c^3)/(35*b^5) - (2*c^3*(93*A*c - 77*B*b))/(35*b^5)) - (c^2*( 93*A*c - 77*B*b))/(35*b^4))*(b*x^2 + c*x^4)^(1/2))/(x^2*(b + c*x^2)) - (A* (b*x^2 + c*x^4)^(1/2))/(7*b^2*x^8) - (c*(29*A*c - 21*B*b)*(b*x^2 + c*x^4)^ (1/2))/(35*b^4*x^4)