Integrand size = 26, antiderivative size = 94 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 b (4 b B-5 A c) \sqrt {b x^2+c x^4}}{15 c^3 x}-\frac {(4 b B-5 A c) x \sqrt {b x^2+c x^4}}{15 c^2}+\frac {B x^3 \sqrt {b x^2+c x^4}}{5 c} \] Output:
2/15*b*(-5*A*c+4*B*b)*(c*x^4+b*x^2)^(1/2)/c^3/x-1/15*(-5*A*c+4*B*b)*x*(c*x ^4+b*x^2)^(1/2)/c^2+1/5*B*x^3*(c*x^4+b*x^2)^(1/2)/c
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.67 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (8 b^2 B-2 b c \left (5 A+2 B x^2\right )+c^2 x^2 \left (5 A+3 B x^2\right )\right )}{15 c^3 x} \] Input:
Integrate[(x^4*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(8*b^2*B - 2*b*c*(5*A + 2*B*x^2) + c^2*x^2*(5*A + 3 *B*x^2)))/(15*c^3*x)
Time = 0.41 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1945, 1421, 1420}
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.
\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {B x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {(4 b B-5 A c) \int \frac {x^4}{\sqrt {c x^4+b x^2}}dx}{5 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {B x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {(4 b B-5 A c) \left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \int \frac {x^2}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{5 c}\) |
\(\Big \downarrow \) 1420 |
\(\displaystyle \frac {B x^3 \sqrt {b x^2+c x^4}}{5 c}-\frac {\left (\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x}\right ) (4 b B-5 A c)}{5 c}\) |
Input:
Int[(x^4*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]
Output:
(B*x^3*Sqrt[b*x^2 + c*x^4])/(5*c) - ((4*b*B - 5*A*c)*((-2*b*Sqrt[b*x^2 + c *x^4])/(3*c^2*x) + (x*Sqrt[b*x^2 + c*x^4])/(3*c)))/(5*c)
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(2*c*(p + 1))), x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 2*p - 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && IGtQ[Simplify[(m + 2*p - 1)/2], 0] && NeQ[m + 4*p + 1, 0]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.86 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.64
method | result | size |
trager | \(-\frac {\left (-3 B \,c^{2} x^{4}-5 A \,c^{2} x^{2}+4 x^{2} B b c +10 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 c^{3} x}\) | \(60\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-3 B \,c^{2} x^{4}-5 A \,c^{2} x^{2}+4 x^{2} B b c +10 A b c -8 B \,b^{2}\right ) x}{15 c^{3} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(65\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-3 B \,c^{2} x^{4}-5 A \,c^{2} x^{2}+4 x^{2} B b c +10 A b c -8 B \,b^{2}\right ) x}{15 c^{3} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(65\) |
risch | \(-\frac {x \left (c \,x^{2}+b \right ) \left (-3 B \,c^{2} x^{4}-5 A \,c^{2} x^{2}+4 x^{2} B b c +10 A b c -8 B \,b^{2}\right )}{15 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{3}}\) | \(65\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-3 B \,c^{2} x^{4}-5 A \,c^{2} x^{2}+4 x^{2} B b c +10 A b c -8 B \,b^{2}\right ) x}{15 c^{3} \sqrt {c \,x^{4}+b \,x^{2}}}\) | \(65\) |
Input:
int(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(-3*B*c^2*x^4-5*A*c^2*x^2+4*B*b*c*x^2+10*A*b*c-8*B*b^2)/c^3/x*(c*x^4 +b*x^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.63 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {{\left (3 \, B c^{2} x^{4} + 8 \, B b^{2} - 10 \, A b c - {\left (4 \, B b c - 5 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, c^{3} x} \] Input:
integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
1/15*(3*B*c^2*x^4 + 8*B*b^2 - 10*A*b*c - (4*B*b*c - 5*A*c^2)*x^2)*sqrt(c*x ^4 + b*x^2)/(c^3*x)
\[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{4} \left (A + B x^{2}\right )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(x**4*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**4*(A + B*x**2)/sqrt(x**2*(b + c*x**2)), x)
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {{\left (c^{2} x^{4} - b c x^{2} - 2 \, b^{2}\right )} A}{3 \, \sqrt {c x^{2} + b} c^{2}} + \frac {{\left (3 \, c^{3} x^{6} - b c^{2} x^{4} + 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} B}{15 \, \sqrt {c x^{2} + b} c^{3}} \] Input:
integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/3*(c^2*x^4 - b*c*x^2 - 2*b^2)*A/(sqrt(c*x^2 + b)*c^2) + 1/15*(3*c^3*x^6 - b*c^2*x^4 + 4*b^2*c*x^2 + 8*b^3)*B/(sqrt(c*x^2 + b)*c^3)
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\left (4 \, B b^{\frac {5}{2}} - 5 \, A b^{\frac {3}{2}} c\right )} \mathrm {sgn}\left (x\right )}{15 \, c^{3}} + \frac {{\left (B b^{2} - A b c\right )} \sqrt {c x^{2} + b}}{c^{3} \mathrm {sgn}\left (x\right )} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B - 10 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b + 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c}{15 \, c^{3} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
-2/15*(4*B*b^(5/2) - 5*A*b^(3/2)*c)*sgn(x)/c^3 + (B*b^2 - A*b*c)*sqrt(c*x^ 2 + b)/(c^3*sgn(x)) + 1/15*(3*(c*x^2 + b)^(5/2)*B - 10*(c*x^2 + b)^(3/2)*B *b + 5*(c*x^2 + b)^(3/2)*A*c)/(c^3*sgn(x))
Time = 9.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {8\,B\,b^2-10\,A\,b\,c}{15\,c^3}+\frac {x^2\,\left (5\,A\,c^2-4\,B\,b\,c\right )}{15\,c^3}+\frac {B\,x^4}{5\,c}\right )}{x} \] Input:
int((x^4*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2),x)
Output:
((b*x^2 + c*x^4)^(1/2)*((8*B*b^2 - 10*A*b*c)/(15*c^3) + (x^2*(5*A*c^2 - 4* B*b*c))/(15*c^3) + (B*x^4)/(5*c)))/x
Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.54 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (3 b \,c^{2} x^{4}+5 a \,c^{2} x^{2}-4 b^{2} c \,x^{2}-10 a b c +8 b^{3}\right )}{15 c^{3}} \] Input:
int(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)
Output:
(sqrt(b + c*x**2)*( - 10*a*b*c + 5*a*c**2*x**2 + 8*b**3 - 4*b**2*c*x**2 + 3*b*c**2*x**4))/(15*c**3)