Integrand size = 26, antiderivative size = 94 \[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {2 b (4 b B-7 A c) \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac {(4 b B-7 A c) \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c} \] Output:
2/105*b*(-7*A*c+4*B*b)*(c*x^4+b*x^2)^(3/2)/c^3/x^3-1/35*(-7*A*c+4*B*b)*(c* x^4+b*x^2)^(3/2)/c^2/x+1/7*B*x*(c*x^4+b*x^2)^(3/2)/c
Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68 \[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (8 b^2 B+3 c^2 x^2 \left (7 A+5 B x^2\right )-2 b c \left (7 A+6 B x^2\right )\right )}{105 c^3 x^3} \] Input:
Integrate[x^2*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
Output:
((x^2*(b + c*x^2))^(3/2)*(8*b^2*B + 3*c^2*x^2*(7*A + 5*B*x^2) - 2*b*c*(7*A + 6*B*x^2)))/(105*c^3*x^3)
Time = 0.42 (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, 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.
\(\displaystyle \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {(4 b B-7 A c) \int x^2 \sqrt {c x^4+b x^2}dx}{7 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {(4 b B-7 A c) \left (\frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {2 b \int \sqrt {c x^4+b x^2}dx}{5 c}\right )}{7 c}\) |
\(\Big \downarrow \) 1398 |
\(\displaystyle \frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {\left (\frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {2 b \left (b x^2+c x^4\right )^{3/2}}{15 c^2 x^3}\right ) (4 b B-7 A c)}{7 c}\) |
Input:
Int[x^2*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]
Output:
(B*x*(b*x^2 + c*x^4)^(3/2))/(7*c) - ((4*b*B - 7*A*c)*((-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)
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 = 0.75 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-15 B \,c^{2} x^{4}-21 A \,c^{2} x^{2}+12 x^{2} B b c +14 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x}\) | \(67\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-15 B \,c^{2} x^{4}-21 A \,c^{2} x^{2}+12 x^{2} B b c +14 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x}\) | \(67\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-15 B \,c^{2} x^{4}-21 A \,c^{2} x^{2}+12 x^{2} B b c +14 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x}\) | \(67\) |
trager | \(-\frac {\left (-15 B \,c^{3} x^{6}-21 A \,c^{3} x^{4}-3 B b \,c^{2} x^{4}-7 A b \,c^{2} x^{2}+4 x^{2} B \,b^{2} c +14 A \,b^{2} c -8 B \,b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x}\) | \(84\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-15 B \,c^{3} x^{6}-21 A \,c^{3} x^{4}-3 B b \,c^{2} x^{4}-7 A b \,c^{2} x^{2}+4 x^{2} B \,b^{2} c +14 A \,b^{2} c -8 B \,b^{3}\right )}{105 x \,c^{3}}\) | \(84\) |
Input:
int(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/105*(c*x^2+b)*(-15*B*c^2*x^4-21*A*c^2*x^2+12*B*b*c*x^2+14*A*b*c-8*B*b^2 )*(c*x^4+b*x^2)^(1/2)/c^3/x
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {{\left (15 \, B c^{3} x^{6} + 3 \, {\left (B b c^{2} + 7 \, A c^{3}\right )} x^{4} + 8 \, B b^{3} - 14 \, A b^{2} c - {\left (4 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{105 \, c^{3} x} \] Input:
integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
1/105*(15*B*c^3*x^6 + 3*(B*b*c^2 + 7*A*c^3)*x^4 + 8*B*b^3 - 14*A*b^2*c - ( 4*B*b^2*c - 7*A*b*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^3*x)
\[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\int x^{2} \sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )\, dx \] Input:
integrate(x**2*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**2*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt {c x^{2} + b} A}{15 \, c^{2}} + \frac {{\left (15 \, c^{3} x^{6} + 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{2} + b} B}{105 \, c^{3}} \] Input:
integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
1/15*(3*c^2*x^4 + b*c*x^2 - 2*b^2)*sqrt(c*x^2 + b)*A/c^2 + 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)*B/c^3
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12 \[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {2 \, {\left (4 \, B b^{\frac {7}{2}} - 7 \, A b^{\frac {5}{2}} c\right )} \mathrm {sgn}\left (x\right )}{105 \, c^{3}} + \frac {15 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B \mathrm {sgn}\left (x\right ) - 42 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b \mathrm {sgn}\left (x\right ) + 35 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{2} \mathrm {sgn}\left (x\right ) + 21 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A c \mathrm {sgn}\left (x\right ) - 35 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b c \mathrm {sgn}\left (x\right )}{105 \, c^{3}} \] Input:
integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
-2/105*(4*B*b^(7/2) - 7*A*b^(5/2)*c)*sgn(x)/c^3 + 1/105*(15*(c*x^2 + b)^(7 /2)*B*sgn(x) - 42*(c*x^2 + b)^(5/2)*B*b*sgn(x) + 35*(c*x^2 + b)^(3/2)*B*b^ 2*sgn(x) + 21*(c*x^2 + b)^(5/2)*A*c*sgn(x) - 35*(c*x^2 + b)^(3/2)*A*b*c*sg n(x))/c^3
Time = 9.62 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {B\,x^6}{7}+\frac {8\,B\,b^3-14\,A\,b^2\,c}{105\,c^3}+\frac {x^4\,\left (21\,A\,c^3+3\,B\,b\,c^2\right )}{105\,c^3}+\frac {b\,x^2\,\left (7\,A\,c-4\,B\,b\right )}{105\,c^2}\right )}{x} \] Input:
int(x^2*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)
Output:
((b*x^2 + c*x^4)^(1/2)*((B*x^6)/7 + (8*B*b^3 - 14*A*b^2*c)/(105*c^3) + (x^ 4*(21*A*c^3 + 3*B*b*c^2))/(105*c^3) + (b*x^2*(7*A*c - 4*B*b))/(105*c^2)))/ x
Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.79 \[ \int x^2 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (15 b \,c^{3} x^{6}+21 a \,c^{3} x^{4}+3 b^{2} c^{2} x^{4}+7 a b \,c^{2} x^{2}-4 b^{3} c \,x^{2}-14 a \,b^{2} c +8 b^{4}\right )}{105 c^{3}} \] Input:
int(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x)
Output:
(sqrt(b + c*x**2)*( - 14*a*b**2*c + 7*a*b*c**2*x**2 + 21*a*c**3*x**4 + 8*b **4 - 4*b**3*c*x**2 + 3*b**2*c**2*x**4 + 15*b*c**3*x**6))/(105*c**3)