Integrand size = 23, antiderivative size = 96 \[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {2 b (4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{315 c^3 x^5}-\frac {(4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x} \] Output:
2/315*b*(-9*A*c+4*B*b)*(c*x^4+b*x^2)^(5/2)/c^3/x^5-1/63*(-9*A*c+4*B*b)*(c* x^4+b*x^2)^(5/2)/c^2/x^3+1/9*B*(c*x^4+b*x^2)^(5/2)/c/x
Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.73 \[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (b+c x^2\right ) \left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (8 b^2 B-18 A b c-20 b B c x^2+45 A c^2 x^2+35 B c^2 x^4\right )}{315 c^3 x^3} \] Input:
Integrate[(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]
Output:
((b + c*x^2)*(x^2*(b + c*x^2))^(3/2)*(8*b^2*B - 18*A*b*c - 20*b*B*c*x^2 + 45*A*c^2*x^2 + 35*B*c^2*x^4))/(315*c^3*x^3)
Time = 0.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1465, 1399, 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.
\(\displaystyle \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1465 |
\(\displaystyle \frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {(4 b B-9 A c) \int \left (c x^4+b x^2\right )^{3/2}dx}{9 c}\) |
\(\Big \downarrow \) 1399 |
\(\displaystyle \frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {(4 b B-9 A c) \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^2}dx}{7 c}\right )}{9 c}\) |
\(\Big \downarrow \) 1420 |
\(\displaystyle \frac {B \left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {\left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5}\right ) (4 b B-9 A c)}{9 c}\) |
Input:
Int[(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]
Output:
(B*(b*x^2 + c*x^4)^(5/2))/(9*c*x) - ((4*b*B - 9*A*c)*((-2*b*(b*x^2 + c*x^4 )^(5/2))/(35*c^2*x^5) + (b*x^2 + c*x^4)^(5/2)/(7*c*x^3)))/(9*c)
Int[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(b*x^2 + c*x^4)^( p + 1)/(c*(4*p + 1)*x^3), x] - Simp[b*((2*p - 1)/(c*(4*p + 1))) Int[(b*x^ 2 + c*x^4)^p/x^2, x], x] /; FreeQ[{b, c, p}, x] && IGtQ[p - 1/2, 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)/(2*c*(p + 1))), x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 2*p - 1, 0]
Int[((d_) + (e_.)*(x_)^2)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e*((b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 3)*x)), x] - Simp[(b*e*(2*p + 1) - c*d*(4*p + 3))/(c*(4*p + 3)) Int[(b*x^2 + c*x^4)^p, x], x] /; FreeQ[{b, c, d, e, p}, x] && !IntegerQ[p] && NeQ[4*p + 3, 0] && NeQ[b*e*(2*p + 1) - c*d*(4*p + 3), 0]
Time = 0.66 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-35 B \,c^{2} x^{4}-45 A \,c^{2} x^{2}+20 x^{2} B b c +18 A b c -8 B \,b^{2}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 c^{3} x^{3}}\) | \(67\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-35 B \,c^{2} x^{4}-45 A \,c^{2} x^{2}+20 x^{2} B b c +18 A b c -8 B \,b^{2}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 c^{3} x^{3}}\) | \(67\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-35 B \,c^{2} x^{4}-45 A \,c^{2} x^{2}+20 x^{2} B b c +18 A b c -8 B \,b^{2}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 c^{3} x^{3}}\) | \(67\) |
trager | \(-\frac {\left (-35 B \,x^{8} c^{4}-45 A \,x^{6} c^{4}-50 B \,x^{6} b \,c^{3}-72 A \,x^{4} b \,c^{3}-3 B \,x^{4} b^{2} c^{2}-9 A \,b^{2} c^{2} x^{2}+4 B \,b^{3} c \,x^{2}+18 A \,b^{3} c -8 B \,b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 c^{3} x}\) | \(108\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-35 B \,x^{8} c^{4}-45 A \,x^{6} c^{4}-50 B \,x^{6} b \,c^{3}-72 A \,x^{4} b \,c^{3}-3 B \,x^{4} b^{2} c^{2}-9 A \,b^{2} c^{2} x^{2}+4 B \,b^{3} c \,x^{2}+18 A \,b^{3} c -8 B \,b^{4}\right )}{315 x \,c^{3}}\) | \(108\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/315*(c*x^2+b)*(-35*B*c^2*x^4-45*A*c^2*x^2+20*B*b*c*x^2+18*A*b*c-8*B*b^2 )*(c*x^4+b*x^2)^(3/2)/c^3/x^3
Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10 \[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (35 \, B c^{4} x^{8} + 5 \, {\left (10 \, B b c^{3} + 9 \, A c^{4}\right )} x^{6} + 8 \, B b^{4} - 18 \, A b^{3} c + 3 \, {\left (B b^{2} c^{2} + 24 \, A b c^{3}\right )} x^{4} - {\left (4 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, c^{3} x} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
1/315*(35*B*c^4*x^8 + 5*(10*B*b*c^3 + 9*A*c^4)*x^6 + 8*B*b^4 - 18*A*b^3*c + 3*(B*b^2*c^2 + 24*A*b*c^3)*x^4 - (4*B*b^3*c - 9*A*b^2*c^2)*x^2)*sqrt(c*x ^4 + b*x^2)/(c^3*x)
\[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\int \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2),x)
Output:
Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2), x)
Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09 \[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (5 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} + b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{2} + b} A}{35 \, c^{2}} + \frac {{\left (35 \, c^{4} x^{8} + 50 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} - 4 \, b^{3} c x^{2} + 8 \, b^{4}\right )} \sqrt {c x^{2} + b} B}{315 \, c^{3}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
1/35*(5*c^3*x^6 + 8*b*c^2*x^4 + b^2*c*x^2 - 2*b^3)*sqrt(c*x^2 + b)*A/c^2 + 1/315*(35*c^4*x^8 + 50*b*c^3*x^6 + 3*b^2*c^2*x^4 - 4*b^3*c*x^2 + 8*b^4)*s qrt(c*x^2 + b)*B/c^3
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09 \[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {2 \, {\left (4 \, B b^{\frac {9}{2}} - 9 \, A b^{\frac {7}{2}} c\right )} \mathrm {sgn}\left (x\right )}{315 \, c^{3}} + \frac {35 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B \mathrm {sgn}\left (x\right ) - 90 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b \mathrm {sgn}\left (x\right ) + 63 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{2} \mathrm {sgn}\left (x\right ) + 45 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A c \mathrm {sgn}\left (x\right ) - 63 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b c \mathrm {sgn}\left (x\right )}{315 \, c^{3}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
-2/315*(4*B*b^(9/2) - 9*A*b^(7/2)*c)*sgn(x)/c^3 + 1/315*(35*(c*x^2 + b)^(9 /2)*B*sgn(x) - 90*(c*x^2 + b)^(7/2)*B*b*sgn(x) + 63*(c*x^2 + b)^(5/2)*B*b^ 2*sgn(x) + 45*(c*x^2 + b)^(7/2)*A*c*sgn(x) - 63*(c*x^2 + b)^(5/2)*A*b*c*sg n(x))/c^3
Time = 8.91 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.07 \[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {8\,B\,b^4-18\,A\,b^3\,c}{315\,c^3}+\frac {x^6\,\left (45\,A\,c^4+50\,B\,b\,c^3\right )}{315\,c^3}+\frac {B\,c\,x^8}{9}+\frac {b^2\,x^2\,\left (9\,A\,c-4\,B\,b\right )}{315\,c^2}+\frac {b\,x^4\,\left (24\,A\,c+B\,b\right )}{105\,c}\right )}{x} \] Input:
int((A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x)
Output:
((b*x^2 + c*x^4)^(1/2)*((8*B*b^4 - 18*A*b^3*c)/(315*c^3) + (x^6*(45*A*c^4 + 50*B*b*c^3))/(315*c^3) + (B*c*x^8)/9 + (b^2*x^2*(9*A*c - 4*B*b))/(315*c^ 2) + (b*x^4*(24*A*c + B*b))/(105*c)))/x
Time = 0.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01 \[ \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (35 b \,c^{4} x^{8}+45 a \,c^{4} x^{6}+50 b^{2} c^{3} x^{6}+72 a b \,c^{3} x^{4}+3 b^{3} c^{2} x^{4}+9 a \,b^{2} c^{2} x^{2}-4 b^{4} c \,x^{2}-18 a \,b^{3} c +8 b^{5}\right )}{315 c^{3}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(3/2),x)
Output:
(sqrt(b + c*x**2)*( - 18*a*b**3*c + 9*a*b**2*c**2*x**2 + 72*a*b*c**3*x**4 + 45*a*c**4*x**6 + 8*b**5 - 4*b**4*c*x**2 + 3*b**3*c**2*x**4 + 50*b**2*c** 3*x**6 + 35*b*c**4*x**8))/(315*c**3)