Integrand size = 19, antiderivative size = 143 \[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\frac {a^3 c x^3 \sqrt {c \left (a+b x^2\right )^2}}{3 \left (a+b x^2\right )}+\frac {3 a^2 b c x^5 \sqrt {c \left (a+b x^2\right )^2}}{5 \left (a+b x^2\right )}+\frac {3 a b^2 c x^7 \sqrt {c \left (a+b x^2\right )^2}}{7 \left (a+b x^2\right )}+\frac {b^3 c x^9 \sqrt {c \left (a+b x^2\right )^2}}{9 \left (a+b x^2\right )} \] Output:
a^3*c*x^3*(c*(b*x^2+a)^2)^(1/2)/(3*b*x^2+3*a)+3*a^2*b*c*x^5*(c*(b*x^2+a)^2 )^(1/2)/(5*b*x^2+5*a)+3*a*b^2*c*x^7*(c*(b*x^2+a)^2)^(1/2)/(7*b*x^2+7*a)+b^ 3*c*x^9*(c*(b*x^2+a)^2)^(1/2)/(9*b*x^2+9*a)
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.44 \[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\frac {x^3 \left (c \left (a+b x^2\right )^2\right )^{3/2} \left (105 a^3+189 a^2 b x^2+135 a b^2 x^4+35 b^3 x^6\right )}{315 \left (a+b x^2\right )^3} \] Input:
Integrate[x^2*(c*(a + b*x^2)^2)^(3/2),x]
Output:
(x^3*(c*(a + b*x^2)^2)^(3/2)*(105*a^3 + 189*a^2*b*x^2 + 135*a*b^2*x^4 + 35 *b^3*x^6))/(315*(a + b*x^2)^3)
Time = 0.37 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2045, 27, 244, 2009}
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.
\(\displaystyle \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^2} \int \frac {x^2 \left (b x^2+a\right )^3}{a^3}dx}{a+b x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int x^2 \left (b x^2+a\right )^3dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \left (b^3 x^8+3 a b^2 x^6+3 a^2 b x^4+a^3 x^2\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \left (\frac {a^3 x^3}{3}+\frac {3}{5} a^2 b x^5+\frac {3}{7} a b^2 x^7+\frac {b^3 x^9}{9}\right ) \sqrt {c \left (a+b x^2\right )^2}}{a+b x^2}\) |
Input:
Int[x^2*(c*(a + b*x^2)^2)^(3/2),x]
Output:
(c*Sqrt[c*(a + b*x^2)^2]*((a^3*x^3)/3 + (3*a^2*b*x^5)/5 + (3*a*b^2*x^7)/7 + (b^3*x^9)/9))/(a + b*x^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
Time = 0.90 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.42
method | result | size |
gosper | \(\frac {x^{3} \left (35 b^{3} x^{6}+135 a \,b^{2} x^{4}+189 a^{2} b \,x^{2}+105 a^{3}\right ) {\left (c \left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{315 \left (b \,x^{2}+a \right )^{3}}\) | \(60\) |
default | \(\frac {x^{3} \left (35 b^{3} x^{6}+135 a \,b^{2} x^{4}+189 a^{2} b \,x^{2}+105 a^{3}\right ) {\left (c \left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{315 \left (b \,x^{2}+a \right )^{3}}\) | \(60\) |
orering | \(\frac {x^{3} \left (35 b^{3} x^{6}+135 a \,b^{2} x^{4}+189 a^{2} b \,x^{2}+105 a^{3}\right ) {\left (c \left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{315 \left (b \,x^{2}+a \right )^{3}}\) | \(60\) |
trager | \(\frac {c \,x^{3} \left (35 b^{3} x^{6}+135 a \,b^{2} x^{4}+189 a^{2} b \,x^{2}+105 a^{3}\right ) \sqrt {b^{2} c \,x^{4}+2 a b c \,x^{2}+a^{2} c}}{315 b \,x^{2}+315 a}\) | \(72\) |
risch | \(\frac {a^{3} c \,x^{3} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{3 b \,x^{2}+3 a}+\frac {3 c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, a^{2} b \,x^{5}}{5 \left (b \,x^{2}+a \right )}+\frac {3 c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, a \,b^{2} x^{7}}{7 \left (b \,x^{2}+a \right )}+\frac {b^{3} c \,x^{9} \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{9 b \,x^{2}+9 a}\) | \(128\) |
Input:
int(x^2*(c*(b*x^2+a)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/315*x^3*(35*b^3*x^6+135*a*b^2*x^4+189*a^2*b*x^2+105*a^3)*(c*(b*x^2+a)^2) ^(3/2)/(b*x^2+a)^3
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.52 \[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\frac {{\left (35 \, b^{3} c x^{9} + 135 \, a b^{2} c x^{7} + 189 \, a^{2} b c x^{5} + 105 \, a^{3} c x^{3}\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{315 \, {\left (b x^{2} + a\right )}} \] Input:
integrate(x^2*(c*(b*x^2+a)^2)^(3/2),x, algorithm="fricas")
Output:
1/315*(35*b^3*c*x^9 + 135*a*b^2*c*x^7 + 189*a^2*b*c*x^5 + 105*a^3*c*x^3)*s qrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*x^2 + a)
\[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\int x^{2} \left (c \left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \] Input:
integrate(x**2*(c*(b*x**2+a)**2)**(3/2),x)
Output:
Integral(x**2*(c*(a + b*x**2)**2)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.33 \[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\frac {1}{9} \, b^{3} c^{\frac {3}{2}} x^{9} + \frac {3}{7} \, a b^{2} c^{\frac {3}{2}} x^{7} + \frac {3}{5} \, a^{2} b c^{\frac {3}{2}} x^{5} + \frac {1}{3} \, a^{3} c^{\frac {3}{2}} x^{3} \] Input:
integrate(x^2*(c*(b*x^2+a)^2)^(3/2),x, algorithm="maxima")
Output:
1/9*b^3*c^(3/2)*x^9 + 3/7*a*b^2*c^(3/2)*x^7 + 3/5*a^2*b*c^(3/2)*x^5 + 1/3* a^3*c^(3/2)*x^3
Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.50 \[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\frac {1}{315} \, {\left (35 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + 135 \, a b^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 189 \, a^{2} b x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, a^{3} x^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )} c^{\frac {3}{2}} \] Input:
integrate(x^2*(c*(b*x^2+a)^2)^(3/2),x, algorithm="giac")
Output:
1/315*(35*b^3*x^9*sgn(b*x^2 + a) + 135*a*b^2*x^7*sgn(b*x^2 + a) + 189*a^2* b*x^5*sgn(b*x^2 + a) + 105*a^3*x^3*sgn(b*x^2 + a))*c^(3/2)
Timed out. \[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\int x^2\,{\left (c\,{\left (b\,x^2+a\right )}^2\right )}^{3/2} \,d x \] Input:
int(x^2*(c*(a + b*x^2)^2)^(3/2),x)
Output:
int(x^2*(c*(a + b*x^2)^2)^(3/2), x)
Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.28 \[ \int x^2 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx=\frac {\sqrt {c}\, c \,x^{3} \left (35 b^{3} x^{6}+135 a \,b^{2} x^{4}+189 a^{2} b \,x^{2}+105 a^{3}\right )}{315} \] Input:
int(x^2*(c*(b*x^2+a)^2)^(3/2),x)
Output:
(sqrt(c)*c*x**3*(105*a**3 + 189*a**2*b*x**2 + 135*a*b**2*x**4 + 35*b**3*x* *6))/315