\(\int x^2 (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\) [185]

Optimal result

Integrand size = 26, antiderivative size = 131 \[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {8 b^2 (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{3465 c^4 x^5}+\frac {4 b (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{693 c^3 x^3}-\frac {(6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{99 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c} \] Output:

-8/3465*b^2*(-11*A*c+6*B*b)*(c*x^4+b*x^2)^(5/2)/c^4/x^5+4/693*b*(-11*A*c+6 
*B*b)*(c*x^4+b*x^2)^(5/2)/c^3/x^3-1/99*(-11*A*c+6*B*b)*(c*x^4+b*x^2)^(5/2) 
/c^2/x+1/11*B*x*(c*x^4+b*x^2)^(5/2)/c
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (b+c x^2\right ) \left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (-48 b^3 B+88 A b^2 c+120 b^2 B c x^2-220 A b c^2 x^2-210 b B c^2 x^4+385 A c^3 x^4+315 B c^3 x^6\right )}{3465 c^4 x^3} \] Input:

Integrate[x^2*(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)*(-48*b^3*B + 88*A*b^2*c + 120*b^2*B*c 
*x^2 - 220*A*b*c^2*x^2 - 210*b*B*c^2*x^4 + 385*A*c^3*x^4 + 315*B*c^3*x^6)) 
/(3465*c^4*x^3)
 

Rubi [A] (verified)

Time = 0.50 (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, 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.

\(\displaystyle \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx\)

\(\Big \downarrow \) 1945

\(\displaystyle \frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {(6 b B-11 A c) \int x^2 \left (c x^4+b x^2\right )^{3/2}dx}{11 c}\)

\(\Big \downarrow \) 1421

\(\displaystyle \frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {(6 b B-11 A c) \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \int \left (c x^4+b x^2\right )^{3/2}dx}{9 c}\right )}{11 c}\)

\(\Big \downarrow \) 1399

\(\displaystyle \frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {(6 b B-11 A c) \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^2}dx}{7 c}\right )}{9 c}\right )}{11 c}\)

\(\Big \downarrow \) 1420

\(\displaystyle \frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {\left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5}\right )}{9 c}\right ) (6 b B-11 A c)}{11 c}\)

Input:

Int[x^2*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(B*x*(b*x^2 + c*x^4)^(5/2))/(11*c) - ((6*b*B - 11*A*c)*((b*x^2 + c*x^4)^(5 
/2)/(9*c*x) - (4*b*((-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)))/(11*c)
 

Defintions of rubi rules used

rule 1399

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]
 

rule 1420

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]
 

rule 1421

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]
 

rule 1945

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])
 

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {\left (c \,x^{2}+b \right ) \left (315 B \,c^{3} x^{6}+385 A \,c^{3} x^{4}-210 B b \,c^{2} x^{4}-220 A b \,c^{2} x^{2}+120 x^{2} B \,b^{2} c +88 A \,b^{2} c -48 B \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 c^{4} x^{3}}\)	\(91\)
default	\(\frac {\left (c \,x^{2}+b \right ) \left (315 B \,c^{3} x^{6}+385 A \,c^{3} x^{4}-210 B b \,c^{2} x^{4}-220 A b \,c^{2} x^{2}+120 x^{2} B \,b^{2} c +88 A \,b^{2} c -48 B \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 c^{4} x^{3}}\)	\(91\)
orering	\(\frac {\left (c \,x^{2}+b \right ) \left (315 B \,c^{3} x^{6}+385 A \,c^{3} x^{4}-210 B b \,c^{2} x^{4}-220 A b \,c^{2} x^{2}+120 x^{2} B \,b^{2} c +88 A \,b^{2} c -48 B \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 c^{4} x^{3}}\)	\(91\)
trager	\(\frac {\left (315 B \,c^{5} x^{10}+385 A \,c^{5} x^{8}+420 B b \,c^{4} x^{8}+550 A b \,c^{4} x^{6}+15 B \,b^{2} c^{3} x^{6}+33 A \,b^{2} c^{3} x^{4}-18 B \,b^{3} c^{2} x^{4}-44 A \,b^{3} c^{2} x^{2}+24 B \,b^{4} c \,x^{2}+88 A \,b^{4} c -48 b^{5} B \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 c^{4} x}\)	\(132\)
risch	\(\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (315 B \,c^{5} x^{10}+385 A \,c^{5} x^{8}+420 B b \,c^{4} x^{8}+550 A b \,c^{4} x^{6}+15 B \,b^{2} c^{3} x^{6}+33 A \,b^{2} c^{3} x^{4}-18 B \,b^{3} c^{2} x^{4}-44 A \,b^{3} c^{2} x^{2}+24 B \,b^{4} c \,x^{2}+88 A \,b^{4} c -48 b^{5} B \right )}{3465 x \,c^{4}}\)	\(132\)

Input:

int(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/3465*(c*x^2+b)*(315*B*c^3*x^6+385*A*c^3*x^4-210*B*b*c^2*x^4-220*A*b*c^2* 
x^2+120*B*b^2*c*x^2+88*A*b^2*c-48*B*b^3)*(c*x^4+b*x^2)^(3/2)/c^4/x^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00 \[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (315 \, B c^{5} x^{10} + 35 \, {\left (12 \, B b c^{4} + 11 \, A c^{5}\right )} x^{8} + 5 \, {\left (3 \, B b^{2} c^{3} + 110 \, A b c^{4}\right )} x^{6} - 48 \, B b^{5} + 88 \, A b^{4} c - 3 \, {\left (6 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4} + 4 \, {\left (6 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3465 \, c^{4} x} \] Input:

integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

1/3465*(315*B*c^5*x^10 + 35*(12*B*b*c^4 + 11*A*c^5)*x^8 + 5*(3*B*b^2*c^3 + 
 110*A*b*c^4)*x^6 - 48*B*b^5 + 88*A*b^4*c - 3*(6*B*b^3*c^2 - 11*A*b^2*c^3) 
*x^4 + 4*(6*B*b^4*c - 11*A*b^3*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^4*x)
 

Sympy [F]

\[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\int x^{2} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )\, dx \] Input:

integrate(x**2*(B*x**2+A)*(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral(x**2*(x**2*(b + c*x**2))**(3/2)*(A + B*x**2), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (35 \, c^{4} x^{8} + 50 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} - 4 \, b^{3} c x^{2} + 8 \, b^{4}\right )} \sqrt {c x^{2} + b} A}{315 \, c^{3}} + \frac {{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt {c x^{2} + b} B}{1155 \, c^{4}} \] Input:

integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

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)*sq 
rt(c*x^2 + b)*A/c^3 + 1/1155*(105*c^5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 
 - 6*b^3*c^2*x^4 + 8*b^4*c*x^2 - 16*b^5)*sqrt(c*x^2 + b)*B/c^4
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.07 \[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {8 \, {\left (6 \, B b^{\frac {11}{2}} - 11 \, A b^{\frac {9}{2}} c\right )} \mathrm {sgn}\left (x\right )}{3465 \, c^{4}} + \frac {315 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} B \mathrm {sgn}\left (x\right ) - 1155 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B b \mathrm {sgn}\left (x\right ) + 1485 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b^{2} \mathrm {sgn}\left (x\right ) - 693 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{3} \mathrm {sgn}\left (x\right ) + 385 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} A c \mathrm {sgn}\left (x\right ) - 990 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A b c \mathrm {sgn}\left (x\right ) + 693 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b^{2} c \mathrm {sgn}\left (x\right )}{3465 \, c^{4}} \] Input:

integrate(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

8/3465*(6*B*b^(11/2) - 11*A*b^(9/2)*c)*sgn(x)/c^4 + 1/3465*(315*(c*x^2 + b 
)^(11/2)*B*sgn(x) - 1155*(c*x^2 + b)^(9/2)*B*b*sgn(x) + 1485*(c*x^2 + b)^( 
7/2)*B*b^2*sgn(x) - 693*(c*x^2 + b)^(5/2)*B*b^3*sgn(x) + 385*(c*x^2 + b)^( 
9/2)*A*c*sgn(x) - 990*(c*x^2 + b)^(7/2)*A*b*c*sgn(x) + 693*(c*x^2 + b)^(5/ 
2)*A*b^2*c*sgn(x))/c^4
 

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95 \[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {x^8\,\left (385\,A\,c^5+420\,B\,b\,c^4\right )}{3465\,c^4}-\frac {48\,B\,b^5-88\,A\,b^4\,c}{3465\,c^4}+\frac {B\,c\,x^{10}}{11}+\frac {b^2\,x^4\,\left (11\,A\,c-6\,B\,b\right )}{1155\,c^2}-\frac {4\,b^3\,x^2\,\left (11\,A\,c-6\,B\,b\right )}{3465\,c^3}+\frac {b\,x^6\,\left (110\,A\,c+3\,B\,b\right )}{693\,c}\right )}{x} \] Input:

int(x^2*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x)
 

Output:

((b*x^2 + c*x^4)^(1/2)*((x^8*(385*A*c^5 + 420*B*b*c^4))/(3465*c^4) - (48*B 
*b^5 - 88*A*b^4*c)/(3465*c^4) + (B*c*x^10)/11 + (b^2*x^4*(11*A*c - 6*B*b)) 
/(1155*c^2) - (4*b^3*x^2*(11*A*c - 6*B*b))/(3465*c^3) + (b*x^6*(110*A*c + 
3*B*b))/(693*c)))/x
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (315 b \,c^{5} x^{10}+385 a \,c^{5} x^{8}+420 b^{2} c^{4} x^{8}+550 a b \,c^{4} x^{6}+15 b^{3} c^{3} x^{6}+33 a \,b^{2} c^{3} x^{4}-18 b^{4} c^{2} x^{4}-44 a \,b^{3} c^{2} x^{2}+24 b^{5} c \,x^{2}+88 a \,b^{4} c -48 b^{6}\right )}{3465 c^{4}} \] Input:

int(x^2*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x)
 

Output:

(sqrt(b + c*x**2)*(88*a*b**4*c - 44*a*b**3*c**2*x**2 + 33*a*b**2*c**3*x**4 
 + 550*a*b*c**4*x**6 + 385*a*c**5*x**8 - 48*b**6 + 24*b**5*c*x**2 - 18*b** 
4*c**2*x**4 + 15*b**3*c**3*x**6 + 420*b**2*c**4*x**8 + 315*b*c**5*x**10))/ 
(3465*c**4)