Integrand size = 26, antiderivative size = 131 \[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {8 b^2 (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{315 c^4 x^3}+\frac {4 b (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x}-\frac {(2 b B-3 A c) x \left (b x^2+c x^4\right )^{3/2}}{21 c^2}+\frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c} \] Output:
-8/315*b^2*(-3*A*c+2*B*b)*(c*x^4+b*x^2)^(3/2)/c^4/x^3+4/105*b*(-3*A*c+2*B* b)*(c*x^4+b*x^2)^(3/2)/c^3/x-1/21*(-3*A*c+2*B*b)*x*(c*x^4+b*x^2)^(3/2)/c^2 +1/9*B*x^3*(c*x^4+b*x^2)^(3/2)/c
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.63 \[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (-16 b^3 B+24 b^2 c \left (A+B x^2\right )-6 b c^2 x^2 \left (6 A+5 B x^2\right )+5 c^3 x^4 \left (9 A+7 B x^2\right )\right )}{315 c^4 x^3} \] Input:
Integrate[x^4*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
Output:
((x^2*(b + c*x^2))^(3/2)*(-16*b^3*B + 24*b^2*c*(A + B*x^2) - 6*b*c^2*x^2*( 6*A + 5*B*x^2) + 5*c^3*x^4*(9*A + 7*B*x^2)))/(315*c^4*x^3)
Time = 0.48 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1945, 1421, 1421, 1398}
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.
\(\displaystyle \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac {(2 b B-3 A c) \int x^4 \sqrt {c x^4+b x^2}dx}{3 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac {(2 b B-3 A c) \left (\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {4 b \int x^2 \sqrt {c x^4+b x^2}dx}{7 c}\right )}{3 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac {(2 b B-3 A c) \left (\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {2 b \int \sqrt {c x^4+b x^2}dx}{5 c}\right )}{7 c}\right )}{3 c}\) |
\(\Big \downarrow \) 1398 |
\(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac {\left (\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {2 b \left (b x^2+c x^4\right )^{3/2}}{15 c^2 x^3}\right )}{7 c}\right ) (2 b B-3 A c)}{3 c}\) |
Input:
Int[x^4*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
Output:
(B*x^3*(b*x^2 + c*x^4)^(3/2))/(9*c) - ((2*b*B - 3*A*c)*((x*(b*x^2 + c*x^4) ^(3/2))/(7*c) - (4*b*((-2*b*(b*x^2 + c*x^4)^(3/2))/(15*c^2*x^3) + (b*x^2 + c*x^4)^(3/2)/(5*c*x)))/(7*c)))/(3*c)
Int[Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> Simp[(b*x^2 + c*x^4)^(3 /2)/(3*c*x^3), x] /; FreeQ[{b, c}, x]
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 = 91, normalized size of antiderivative = 0.69
method | result | size |
gosper | \(\frac {\left (c \,x^{2}+b \right ) \left (35 B \,c^{3} x^{6}+45 A \,c^{3} x^{4}-30 B b \,c^{2} x^{4}-36 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +24 A \,b^{2} c -16 B \,b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 c^{4} x}\) | \(91\) |
default | \(\frac {\left (c \,x^{2}+b \right ) \left (35 B \,c^{3} x^{6}+45 A \,c^{3} x^{4}-30 B b \,c^{2} x^{4}-36 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +24 A \,b^{2} c -16 B \,b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 c^{4} x}\) | \(91\) |
orering | \(\frac {\left (c \,x^{2}+b \right ) \left (35 B \,c^{3} x^{6}+45 A \,c^{3} x^{4}-30 B b \,c^{2} x^{4}-36 A b \,c^{2} x^{2}+24 x^{2} B \,b^{2} c +24 A \,b^{2} c -16 B \,b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 c^{4} x}\) | \(91\) |
trager | \(\frac {\left (35 B \,x^{8} c^{4}+45 A \,x^{6} c^{4}+5 B \,x^{6} b \,c^{3}+9 A \,x^{4} b \,c^{3}-6 B \,x^{4} b^{2} c^{2}-12 A \,b^{2} c^{2} x^{2}+8 B \,b^{3} c \,x^{2}+24 A \,b^{3} c -16 B \,b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 c^{4} x}\) | \(108\) |
risch | \(\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (35 B \,x^{8} c^{4}+45 A \,x^{6} c^{4}+5 B \,x^{6} b \,c^{3}+9 A \,x^{4} b \,c^{3}-6 B \,x^{4} b^{2} c^{2}-12 A \,b^{2} c^{2} x^{2}+8 B \,b^{3} c \,x^{2}+24 A \,b^{3} c -16 B \,b^{4}\right )}{315 x \,c^{4}}\) | \(108\) |
Input:
int(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/315*(c*x^2+b)*(35*B*c^3*x^6+45*A*c^3*x^4-30*B*b*c^2*x^4-36*A*b*c^2*x^2+2 4*B*b^2*c*x^2+24*A*b^2*c-16*B*b^3)*(c*x^4+b*x^2)^(1/2)/c^4/x
Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.81 \[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {{\left (35 \, B c^{4} x^{8} + 5 \, {\left (B b c^{3} + 9 \, A c^{4}\right )} x^{6} - 16 \, B b^{4} + 24 \, A b^{3} c - 3 \, {\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{4} + 4 \, {\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, c^{4} x} \] Input:
integrate(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
1/315*(35*B*c^4*x^8 + 5*(B*b*c^3 + 9*A*c^4)*x^6 - 16*B*b^4 + 24*A*b^3*c - 3*(2*B*b^2*c^2 - 3*A*b*c^3)*x^4 + 4*(2*B*b^3*c - 3*A*b^2*c^2)*x^2)*sqrt(c* x^4 + b*x^2)/(c^4*x)
\[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\int x^{4} \sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )\, dx \] Input:
integrate(x**4*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**4*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)
Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.81 \[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {{\left (15 \, c^{3} x^{6} + 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{2} + b} A}{105 \, c^{3}} + \frac {{\left (35 \, c^{4} x^{8} + 5 \, b c^{3} x^{6} - 6 \, b^{2} c^{2} x^{4} + 8 \, b^{3} c x^{2} - 16 \, b^{4}\right )} \sqrt {c x^{2} + b} B}{315 \, c^{4}} \] Input:
integrate(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/105*(15*c^3*x^6 + 3*b*c^2*x^4 - 4*b^2*c*x^2 + 8*b^3)*sqrt(c*x^2 + b)*A/c ^3 + 1/315*(35*c^4*x^8 + 5*b*c^3*x^6 - 6*b^2*c^2*x^4 + 8*b^3*c*x^2 - 16*b^ 4)*sqrt(c*x^2 + b)*B/c^4
Time = 0.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.07 \[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {8 \, {\left (2 \, B b^{\frac {9}{2}} - 3 \, A b^{\frac {7}{2}} c\right )} \mathrm {sgn}\left (x\right )}{315 \, c^{4}} + \frac {35 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B \mathrm {sgn}\left (x\right ) - 135 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b \mathrm {sgn}\left (x\right ) + 189 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{2} \mathrm {sgn}\left (x\right ) - 105 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{3} \mathrm {sgn}\left (x\right ) + 45 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A c \mathrm {sgn}\left (x\right ) - 126 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b c \mathrm {sgn}\left (x\right ) + 105 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b^{2} c \mathrm {sgn}\left (x\right )}{315 \, c^{4}} \] Input:
integrate(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
8/315*(2*B*b^(9/2) - 3*A*b^(7/2)*c)*sgn(x)/c^4 + 1/315*(35*(c*x^2 + b)^(9/ 2)*B*sgn(x) - 135*(c*x^2 + b)^(7/2)*B*b*sgn(x) + 189*(c*x^2 + b)^(5/2)*B*b ^2*sgn(x) - 105*(c*x^2 + b)^(3/2)*B*b^3*sgn(x) + 45*(c*x^2 + b)^(7/2)*A*c* sgn(x) - 126*(c*x^2 + b)^(5/2)*A*b*c*sgn(x) + 105*(c*x^2 + b)^(3/2)*A*b^2* c*sgn(x))/c^4
Time = 9.76 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79 \[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {B\,x^8}{9}-\frac {16\,B\,b^4-24\,A\,b^3\,c}{315\,c^4}+\frac {x^6\,\left (45\,A\,c^4+5\,B\,b\,c^3\right )}{315\,c^4}-\frac {4\,b^2\,x^2\,\left (3\,A\,c-2\,B\,b\right )}{315\,c^3}+\frac {b\,x^4\,\left (3\,A\,c-2\,B\,b\right )}{105\,c^2}\right )}{x} \] 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)*((B*x^8)/9 - (16*B*b^4 - 24*A*b^3*c)/(315*c^4) + (x ^6*(45*A*c^4 + 5*B*b*c^3))/(315*c^4) - (4*b^2*x^2*(3*A*c - 2*B*b))/(315*c^ 3) + (b*x^4*(3*A*c - 2*B*b))/(105*c^2)))/x
Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.74 \[ \int x^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (35 b \,c^{4} x^{8}+45 a \,c^{4} x^{6}+5 b^{2} c^{3} x^{6}+9 a b \,c^{3} x^{4}-6 b^{3} c^{2} x^{4}-12 a \,b^{2} c^{2} x^{2}+8 b^{4} c \,x^{2}+24 a \,b^{3} c -16 b^{5}\right )}{315 c^{4}} \] Input:
int(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x)
Output:
(sqrt(b + c*x**2)*(24*a*b**3*c - 12*a*b**2*c**2*x**2 + 9*a*b*c**3*x**4 + 4 5*a*c**4*x**6 - 16*b**5 + 8*b**4*c*x**2 - 6*b**3*c**2*x**4 + 5*b**2*c**3*x **6 + 35*b*c**4*x**8))/(315*c**4)