Integrand size = 21, antiderivative size = 100 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\frac {2 a^2 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )}{b^3}-\frac {4 a \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )^2}{5 b^3}+\frac {2 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \left (a+b x^2\right )^3}{9 b^3} \] Output:
2*a^2*(c/(b*x^2+a)^(1/2))^(3/2)*(b*x^2+a)/b^3-4/5*a*(c/(b*x^2+a)^(1/2))^(3 /2)*(b*x^2+a)^2/b^3+2/9*(c/(b*x^2+a)^(1/2))^(3/2)*(b*x^2+a)^3/b^3
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.48 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\frac {2 c^2 \left (32 a^2-8 a b x^2+5 b^2 x^4\right )}{45 b^3 \sqrt {\frac {c}{\sqrt {a+b x^2}}}} \] Input:
Integrate[x^5*(c/Sqrt[a + b*x^2])^(3/2),x]
Output:
(2*c^2*(32*a^2 - 8*a*b*x^2 + 5*b^2*x^4))/(45*b^3*Sqrt[c/Sqrt[a + b*x^2]])
Time = 0.43 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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.
\(\displaystyle \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \int \frac {x^5}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \int \frac {x^4}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \int \left (\frac {\left (\frac {b x^2}{a}+1\right )^{5/4} a^2}{b^2}-\frac {2 \sqrt [4]{\frac {b x^2}{a}+1} a^2}{b^2}+\frac {a^2}{b^2 \left (\frac {b x^2}{a}+1\right )^{3/4}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {b x^2}{a}+1\right )^{3/4} \left (\frac {4 a^3 \left (\frac {b x^2}{a}+1\right )^{9/4}}{9 b^3}-\frac {8 a^3 \left (\frac {b x^2}{a}+1\right )^{5/4}}{5 b^3}+\frac {4 a^3 \sqrt [4]{\frac {b x^2}{a}+1}}{b^3}\right ) \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2}\) |
Input:
Int[x^5*(c/Sqrt[a + b*x^2])^(3/2),x]
Output:
((c/Sqrt[a + b*x^2])^(3/2)*(1 + (b*x^2)/a)^(3/4)*((4*a^3*(1 + (b*x^2)/a)^( 1/4))/b^3 - (8*a^3*(1 + (b*x^2)/a)^(5/4))/(5*b^3) + (4*a^3*(1 + (b*x^2)/a) ^(9/4))/(9*b^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 = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.47
method | result | size |
gosper | \(\frac {2 \left (b \,x^{2}+a \right ) \left (5 b^{2} x^{4}-8 a b \,x^{2}+32 a^{2}\right ) {\left (\frac {c}{\sqrt {b \,x^{2}+a}}\right )}^{\frac {3}{2}}}{45 b^{3}}\) | \(47\) |
orering | \(\frac {2 \left (b \,x^{2}+a \right ) \left (5 b^{2} x^{4}-8 a b \,x^{2}+32 a^{2}\right ) {\left (\frac {c}{\sqrt {b \,x^{2}+a}}\right )}^{\frac {3}{2}}}{45 b^{3}}\) | \(47\) |
Input:
int(x^5*(c/(b*x^2+a)^(1/2))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/45*(b*x^2+a)*(5*b^2*x^4-8*a*b*x^2+32*a^2)*(c/(b*x^2+a)^(1/2))^(3/2)/b^3
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.51 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\frac {2 \, {\left (5 \, b^{2} c x^{4} - 8 \, a b c x^{2} + 32 \, a^{2} c\right )} \sqrt {b x^{2} + a} \sqrt {\frac {c}{\sqrt {b x^{2} + a}}}}{45 \, b^{3}} \] Input:
integrate(x^5*(c/(b*x^2+a)^(1/2))^(3/2),x, algorithm="fricas")
Output:
2/45*(5*b^2*c*x^4 - 8*a*b*c*x^2 + 32*a^2*c)*sqrt(b*x^2 + a)*sqrt(c/sqrt(b* x^2 + a))/b^3
Time = 5.77 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\begin {cases} \frac {64 a^{3} \left (\frac {c}{\sqrt {a + b x^{2}}}\right )^{\frac {3}{2}}}{45 b^{3}} + \frac {16 a^{2} x^{2} \left (\frac {c}{\sqrt {a + b x^{2}}}\right )^{\frac {3}{2}}}{15 b^{2}} - \frac {2 a x^{4} \left (\frac {c}{\sqrt {a + b x^{2}}}\right )^{\frac {3}{2}}}{15 b} + \frac {2 x^{6} \left (\frac {c}{\sqrt {a + b x^{2}}}\right )^{\frac {3}{2}}}{9} & \text {for}\: b \neq 0 \\\frac {x^{6} \left (\frac {c}{\sqrt {a}}\right )^{\frac {3}{2}}}{6} & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(c/(b*x**2+a)**(1/2))**(3/2),x)
Output:
Piecewise((64*a**3*(c/sqrt(a + b*x**2))**(3/2)/(45*b**3) + 16*a**2*x**2*(c /sqrt(a + b*x**2))**(3/2)/(15*b**2) - 2*a*x**4*(c/sqrt(a + b*x**2))**(3/2) /(15*b) + 2*x**6*(c/sqrt(a + b*x**2))**(3/2)/9, Ne(b, 0)), (x**6*(c/sqrt(a ))**(3/2)/6, True))
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.59 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=-\frac {2 \, {\left (\frac {18 \, a c^{4}}{b x^{2} + a} - \frac {45 \, a^{2} c^{4}}{{\left (b x^{2} + a\right )}^{2}} - 5 \, c^{4}\right )} c^{2}}{45 \, b^{3} \left (\frac {c}{\sqrt {b x^{2} + a}}\right )^{\frac {9}{2}}} \] Input:
integrate(x^5*(c/(b*x^2+a)^(1/2))^(3/2),x, algorithm="maxima")
Output:
-2/45*(18*a*c^4/(b*x^2 + a) - 45*a^2*c^4/(b*x^2 + a)^2 - 5*c^4)*c^2/(b^3*( c/sqrt(b*x^2 + a))^(9/2))
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.56 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\frac {2 \, {\left (5 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}} - 18 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} a + 45 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} a^{2}\right )} c^{\frac {3}{2}}}{45 \, b^{3} \sqrt {\mathrm {sgn}\left (b x^{2} + a\right )}} \] Input:
integrate(x^5*(c/(b*x^2+a)^(1/2))^(3/2),x, algorithm="giac")
Output:
2/45*(5*(b*x^2 + a)^(9/4) - 18*(b*x^2 + a)^(5/4)*a + 45*(b*x^2 + a)^(1/4)* a^2)*c^(3/2)/(b^3*sqrt(sgn(b*x^2 + a)))
Time = 0.55 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.70 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\sqrt {\frac {c}{\sqrt {b\,x^2+a}}}\,\left (\frac {2\,c\,x^4\,\sqrt {b\,x^2+a}}{9\,b}+\frac {64\,a^2\,c\,\sqrt {b\,x^2+a}}{45\,b^3}-\frac {16\,a\,c\,x^2\,\sqrt {b\,x^2+a}}{45\,b^2}\right ) \] Input:
int(x^5*(c/(a + b*x^2)^(1/2))^(3/2),x)
Output:
(c/(a + b*x^2)^(1/2))^(1/2)*((2*c*x^4*(a + b*x^2)^(1/2))/(9*b) + (64*a^2*c *(a + b*x^2)^(1/2))/(45*b^3) - (16*a*c*x^2*(a + b*x^2)^(1/2))/(45*b^2))
Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.17 \[ \int x^5 \left (\frac {c}{\sqrt {a+b x^2}}\right )^{3/2} \, dx=\frac {2 \sqrt {c}\, \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, c \left (32 \sqrt {b \,x^{2}+a}\, a^{2}-8 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+5 \sqrt {b \,x^{2}+a}\, b^{2} x^{4}-32 \sqrt {b}\, a^{2} x +8 \sqrt {b}\, a b \,x^{3}-5 \sqrt {b}\, b^{2} x^{5}\right )}{45 a \,b^{3}} \] Input:
int(x^5*(c/(b*x^2+a)^(1/2))^(3/2),x)
Output:
(2*sqrt(c)*sqrt(sqrt(b)*sqrt(a + b*x**2)*x + a + b*x**2)*sqrt(sqrt(a + b*x **2) + sqrt(b)*x)*c*(32*sqrt(a + b*x**2)*a**2 - 8*sqrt(a + b*x**2)*a*b*x** 2 + 5*sqrt(a + b*x**2)*b**2*x**4 - 32*sqrt(b)*a**2*x + 8*sqrt(b)*a*b*x**3 - 5*sqrt(b)*b**2*x**5))/(45*a*b**3)