Integrand size = 26, antiderivative size = 61 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}-\frac {(5 b B-2 A c) \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6} \] Output:
-1/5*A*(c*x^4+b*x^2)^(3/2)/b/x^8-1/15*(-2*A*c+5*B*b)*(c*x^4+b*x^2)^(3/2)/b ^2/x^6
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=-\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (3 A b+5 b B x^2-2 A c x^2\right )}{15 b^2 x^8} \] Input:
Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^7,x]
Output:
-1/15*((x^2*(b + c*x^2))^(3/2)*(3*A*b + 5*b*B*x^2 - 2*A*c*x^2))/(b^2*x^8)
Time = 0.40 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1940, 1220, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \sqrt {c x^4+b x^2}}{x^8}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (\frac {(5 b B-2 A c) \int \frac {\sqrt {c x^4+b x^2}}{x^6}dx^2}{5 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right )\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 \left (b x^2+c x^4\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^6}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right )\) |
Input:
Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^7,x]
Output:
((-2*A*(b*x^2 + c*x^4)^(3/2))/(5*b*x^8) - (2*(5*b*B - 2*A*c)*(b*x^2 + c*x^ 4)^(3/2))/(15*b^2*x^6))/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_)*((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 = 47, normalized size of antiderivative = 0.77
method | result | size |
pseudoelliptic | \(-\frac {\left (c \,x^{2}+b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (\left (\frac {5 B \,x^{2}}{3}+A \right ) b -\frac {2 A c \,x^{2}}{3}\right )}{5 b^{2} x^{6}}\) | \(47\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+5 B b \,x^{2}+3 A b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 b^{2} x^{6}}\) | \(48\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+5 B b \,x^{2}+3 A b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 b^{2} x^{6}}\) | \(48\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+5 B b \,x^{2}+3 A b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 b^{2} x^{6}}\) | \(48\) |
trager | \(-\frac {\left (-2 x^{4} A \,c^{2}+5 x^{4} B b c +A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 b^{2} x^{6}}\) | \(62\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-2 x^{4} A \,c^{2}+5 x^{4} B b c +A b c \,x^{2}+5 x^{2} B \,b^{2}+3 b^{2} A \right )}{15 x^{6} b^{2}}\) | \(62\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/5*(c*x^2+b)*(x^2*(c*x^2+b))^(1/2)*((5/3*B*x^2+A)*b-2/3*A*c*x^2)/b^2/x^6
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=-\frac {{\left ({\left (5 \, B b c - 2 \, A c^{2}\right )} x^{4} + 3 \, A b^{2} + {\left (5 \, B b^{2} + A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b^{2} x^{6}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="fricas")
Output:
-1/15*((5*B*b*c - 2*A*c^2)*x^4 + 3*A*b^2 + (5*B*b^2 + A*b*c)*x^2)*sqrt(c*x ^4 + b*x^2)/(b^2*x^6)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{7}}\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**7,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**7, x)
Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (53) = 106\).
Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.82 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=-\frac {1}{3} \, B {\left (\frac {\sqrt {c x^{4} + b x^{2}} c}{b x^{2}} + \frac {\sqrt {c x^{4} + b x^{2}}}{x^{4}}\right )} + \frac {1}{15} \, A {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{b x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}}}{x^{6}}\right )} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="maxima")
Output:
-1/3*B*(sqrt(c*x^4 + b*x^2)*c/(b*x^2) + sqrt(c*x^4 + b*x^2)/x^4) + 1/15*A* (2*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^2) - sqrt(c*x^4 + b*x^2)*c/(b*x^4) - 3*s qrt(c*x^4 + b*x^2)/x^6)
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (53) = 106\).
Time = 0.62 (sec) , antiderivative size = 250, normalized size of antiderivative = 4.10 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=\frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) - 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{2} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{3} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{2} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 5 \, B b^{4} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) - 2 \, A b^{3} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x, algorithm="giac")
Output:
2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*c^(3/2)*sgn(x) - 30*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b*c^(3/2)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + b)) ^6*A*c^(5/2)*sgn(x) + 20*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^2*c^(3/2)*sgn (x) + 10*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b*c^(5/2)*sgn(x) - 10*(sqrt(c)* x - sqrt(c*x^2 + b))^2*B*b^3*c^(3/2)*sgn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^2*c^(5/2)*sgn(x) + 5*B*b^4*c^(3/2)*sgn(x) - 2*A*b^3*c^(5/2)*sgn (x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5
Time = 9.74 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.85 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=\frac {\left (A\,c^2+B\,b\,c\right )\,\sqrt {c\,x^4+b\,x^2}}{5\,b^2\,x^2}-\frac {\left (5\,B\,b^2+A\,c\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{15\,b^2\,x^4}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{5\,x^6}-\frac {c\,\left (A\,c+8\,B\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{15\,b^2\,x^2} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^7,x)
Output:
((A*c^2 + B*b*c)*(b*x^2 + c*x^4)^(1/2))/(5*b^2*x^2) - ((5*B*b^2 + A*b*c)*( b*x^2 + c*x^4)^(1/2))/(15*b^2*x^4) - (A*(b*x^2 + c*x^4)^(1/2))/(5*x^6) - ( c*(A*c + 8*B*b)*(b*x^2 + c*x^4)^(1/2))/(15*b^2*x^2)
Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.82 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^7} \, dx=\frac {-3 \sqrt {c \,x^{2}+b}\, a \,b^{2}-\sqrt {c \,x^{2}+b}\, a b c \,x^{2}+2 \sqrt {c \,x^{2}+b}\, a \,c^{2} x^{4}-5 \sqrt {c \,x^{2}+b}\, b^{3} x^{2}-5 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{4}-2 \sqrt {c}\, a \,c^{2} x^{5}-\sqrt {c}\, b^{2} c \,x^{5}}{15 b^{2} x^{5}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^7,x)
Output:
( - 3*sqrt(b + c*x**2)*a*b**2 - sqrt(b + c*x**2)*a*b*c*x**2 + 2*sqrt(b + c *x**2)*a*c**2*x**4 - 5*sqrt(b + c*x**2)*b**3*x**2 - 5*sqrt(b + c*x**2)*b** 2*c*x**4 - 2*sqrt(c)*a*c**2*x**5 - sqrt(c)*b**2*c*x**5)/(15*b**2*x**5)