Integrand size = 19, antiderivative size = 66 \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=-\frac {a c \left (a+b x^2\right )^4 \sqrt {c \left (a+b x^2\right )^3}}{11 b^2}+\frac {c \left (a+b x^2\right )^5 \sqrt {c \left (a+b x^2\right )^3}}{13 b^2} \] Output:
-1/11*a*c*(b*x^2+a)^4*(c*(b*x^2+a)^3)^(1/2)/b^2+1/13*c*(b*x^2+a)^5*(c*(b*x ^2+a)^3)^(1/2)/b^2
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.59 \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^3\right )^{3/2} \left (-2 a+11 b x^2\right )}{143 b^2} \] Input:
Integrate[x^3*(c*(a + b*x^2)^3)^(3/2),x]
Output:
((a + b*x^2)*(c*(a + b*x^2)^3)^(3/2)*(-2*a + 11*b*x^2))/(143*b^2)
Time = 0.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.30, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2045, 243, 53, 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.
\(\displaystyle \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int x^3 \left (\frac {b x^2}{a}+1\right )^{9/2}dx}{\left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int x^2 \left (\frac {b x^2}{a}+1\right )^{9/2}dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^3} \int \left (\frac {a \left (\frac {b x^2}{a}+1\right )^{11/2}}{b}-\frac {a \left (\frac {b x^2}{a}+1\right )^{9/2}}{b}\right )dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 c \left (\frac {2 a^2 \left (\frac {b x^2}{a}+1\right )^{13/2}}{13 b^2}-\frac {2 a^2 \left (\frac {b x^2}{a}+1\right )^{11/2}}{11 b^2}\right ) \sqrt {c \left (a+b x^2\right )^3}}{2 \left (\frac {b x^2}{a}+1\right )^{3/2}}\) |
Input:
Int[x^3*(c*(a + b*x^2)^3)^(3/2),x]
Output:
(a^3*c*Sqrt[c*(a + b*x^2)^3]*((-2*a^2*(1 + (b*x^2)/a)^(11/2))/(11*b^2) + ( 2*a^2*(1 + (b*x^2)/a)^(13/2))/(13*b^2)))/(2*(1 + (b*x^2)/a)^(3/2))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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 = 2.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.55
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right ) \left (-11 b \,x^{2}+2 a \right ) {\left (c \left (b \,x^{2}+a \right )^{3}\right )}^{\frac {3}{2}}}{143 b^{2}}\) | \(36\) |
orering | \(-\frac {\left (b \,x^{2}+a \right ) \left (-11 b \,x^{2}+2 a \right ) {\left (c \left (b \,x^{2}+a \right )^{3}\right )}^{\frac {3}{2}}}{143 b^{2}}\) | \(36\) |
default | \(-\frac {\left (-11 b \,x^{2}+2 a \right ) \left (x^{2} b c +a c \right )^{\frac {5}{2}} {\left (c \left (b \,x^{2}+a \right )^{3}\right )}^{\frac {3}{2}}}{143 b^{2} c {\left (\left (b \,x^{2}+a \right ) c \right )}^{\frac {3}{2}}}\) | \(55\) |
risch | \(-\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{3}}\, \left (-11 b^{6} x^{12}-53 a \,b^{5} x^{10}-100 a^{2} b^{4} x^{8}-90 a^{3} b^{3} x^{6}-35 a^{4} b^{2} x^{4}-a^{5} b \,x^{2}+2 a^{6}\right )}{143 \left (b \,x^{2}+a \right ) b^{2}}\) | \(94\) |
trager | \(-\frac {c \left (-11 b^{5} x^{10}-42 a \,b^{4} x^{8}-58 a^{2} b^{3} x^{6}-32 a^{3} b^{2} x^{4}-3 a^{4} b \,x^{2}+2 a^{5}\right ) \sqrt {b^{3} c \,x^{6}+3 a \,b^{2} c \,x^{4}+3 a^{2} b c \,x^{2}+a^{3} c}}{143 b^{2}}\) | \(97\) |
Input:
int(x^3*(c*(b*x^2+a)^3)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/143*(b*x^2+a)*(-11*b*x^2+2*a)*(c*(b*x^2+a)^3)^(3/2)/b^2
Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.53 \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=\frac {{\left (11 \, b^{5} c x^{10} + 42 \, a b^{4} c x^{8} + 58 \, a^{2} b^{3} c x^{6} + 32 \, a^{3} b^{2} c x^{4} + 3 \, a^{4} b c x^{2} - 2 \, a^{5} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{143 \, b^{2}} \] Input:
integrate(x^3*(c*(b*x^2+a)^3)^(3/2),x, algorithm="fricas")
Output:
1/143*(11*b^5*c*x^10 + 42*a*b^4*c*x^8 + 58*a^2*b^3*c*x^6 + 32*a^3*b^2*c*x^ 4 + 3*a^4*b*c*x^2 - 2*a^5*c)*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^ 2 + a^3*c)/b^2
Timed out. \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=\text {Timed out} \] Input:
integrate(x**3*(c*(b*x**2+a)**3)**(3/2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.30 \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=\frac {{\left (11 \, b^{5} c^{\frac {3}{2}} x^{10} + 42 \, a b^{4} c^{\frac {3}{2}} x^{8} + 58 \, a^{2} b^{3} c^{\frac {3}{2}} x^{6} + 32 \, a^{3} b^{2} c^{\frac {3}{2}} x^{4} + 3 \, a^{4} b c^{\frac {3}{2}} x^{2} - 2 \, a^{5} c^{\frac {3}{2}}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}}}{143 \, b^{2}} \] Input:
integrate(x^3*(c*(b*x^2+a)^3)^(3/2),x, algorithm="maxima")
Output:
1/143*(11*b^5*c^(3/2)*x^10 + 42*a*b^4*c^(3/2)*x^8 + 58*a^2*b^3*c^(3/2)*x^6 + 32*a^3*b^2*c^(3/2)*x^4 + 3*a^4*b*c^(3/2)*x^2 - 2*a^5*c^(3/2))*(b*x^2 + a)^(3/2)/b^2
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83 \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=-\frac {13 \, {\left (b c x^{2} + a c\right )}^{\frac {11}{2}} a c \mathrm {sgn}\left (b x^{2} + a\right ) - 11 \, {\left (b c x^{2} + a c\right )}^{\frac {13}{2}} \mathrm {sgn}\left (b x^{2} + a\right )}{143 \, b^{2} c^{5}} \] Input:
integrate(x^3*(c*(b*x^2+a)^3)^(3/2),x, algorithm="giac")
Output:
-1/143*(13*(b*c*x^2 + a*c)^(11/2)*a*c*sgn(b*x^2 + a) - 11*(b*c*x^2 + a*c)^ (13/2)*sgn(b*x^2 + a))/(b^2*c^5)
Time = 0.61 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.12 \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=\sqrt {c\,{\left (b\,x^2+a\right )}^3}\,\left (\frac {32\,a^3\,c\,x^4}{143}-\frac {2\,a^5\,c}{143\,b^2}+\frac {b^3\,c\,x^{10}}{13}+\frac {3\,a^4\,c\,x^2}{143\,b}+\frac {58\,a^2\,b\,c\,x^6}{143}+\frac {42\,a\,b^2\,c\,x^8}{143}\right ) \] Input:
int(x^3*(c*(a + b*x^2)^3)^(3/2),x)
Output:
(c*(a + b*x^2)^3)^(1/2)*((32*a^3*c*x^4)/143 - (2*a^5*c)/(143*b^2) + (b^3*c *x^10)/13 + (3*a^4*c*x^2)/(143*b) + (58*a^2*b*c*x^6)/143 + (42*a*b^2*c*x^8 )/143)
Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.21 \[ \int x^3 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, dx=\frac {\sqrt {c}\, \sqrt {b \,x^{2}+a}\, c \left (11 b^{6} x^{12}+53 a \,b^{5} x^{10}+100 a^{2} b^{4} x^{8}+90 a^{3} b^{3} x^{6}+35 a^{4} b^{2} x^{4}+a^{5} b \,x^{2}-2 a^{6}\right )}{143 b^{2}} \] Input:
int(x^3*(c*(b*x^2+a)^3)^(3/2),x)
Output:
(sqrt(c)*sqrt(a + b*x**2)*c*( - 2*a**6 + a**5*b*x**2 + 35*a**4*b**2*x**4 + 90*a**3*b**3*x**6 + 100*a**2*b**4*x**8 + 53*a*b**5*x**10 + 11*b**6*x**12) )/(143*b**2)