Integrand size = 21, antiderivative size = 102 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 a^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )}{7 b^3}-\frac {4 a \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^2}{11 b^3}+\frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^3}{15 b^3} \] Output:
2/7*a^2*(c*(b*x^2+a)^(1/2))^(3/2)*(b*x^2+a)/b^3-4/11*a*(c*(b*x^2+a)^(1/2)) ^(3/2)*(b*x^2+a)^2/b^3+2/15*(c*(b*x^2+a)^(1/2))^(3/2)*(b*x^2+a)^3/b^3
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.51 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right ) \left (32 a^2-56 a b x^2+77 b^2 x^4\right )}{1155 b^3} \] Input:
Integrate[x^5*(c*Sqrt[a + b*x^2])^(3/2),x]
Output:
(2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)*(32*a^2 - 56*a*b*x^2 + 77*b^2*x^4 ))/(1155*b^3)
Time = 0.44 (sec) , antiderivative size = 108, 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.
\(\displaystyle \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int x^5 \left (\frac {b x^2}{a}+1\right )^{3/4}dx}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int x^4 \left (\frac {b x^2}{a}+1\right )^{3/4}dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int \left (\frac {a^2 \left (\frac {b x^2}{a}+1\right )^{11/4}}{b^2}-\frac {2 a^2 \left (\frac {b x^2}{a}+1\right )^{7/4}}{b^2}+\frac {a^2 \left (\frac {b x^2}{a}+1\right )^{3/4}}{b^2}\right )dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (\frac {4 a^3 \left (\frac {b x^2}{a}+1\right )^{15/4}}{15 b^3}-\frac {8 a^3 \left (\frac {b x^2}{a}+1\right )^{11/4}}{11 b^3}+\frac {4 a^3 \left (\frac {b x^2}{a}+1\right )^{7/4}}{7 b^3}\right ) \left (c \sqrt {a+b x^2}\right )^{3/2}}{2 \left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
Input:
Int[x^5*(c*Sqrt[a + b*x^2])^(3/2),x]
Output:
((c*Sqrt[a + b*x^2])^(3/2)*((4*a^3*(1 + (b*x^2)/a)^(7/4))/(7*b^3) - (8*a^3 *(1 + (b*x^2)/a)^(11/4))/(11*b^3) + (4*a^3*(1 + (b*x^2)/a)^(15/4))/(15*b^3 )))/(2*(1 + (b*x^2)/a)^(3/4))
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 = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.46
method | result | size |
gosper | \(\frac {2 \left (b \,x^{2}+a \right ) \left (77 b^{2} x^{4}-56 a b \,x^{2}+32 a^{2}\right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{1155 b^{3}}\) | \(47\) |
orering | \(\frac {2 \left (b \,x^{2}+a \right ) \left (77 b^{2} x^{4}-56 a b \,x^{2}+32 a^{2}\right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{1155 b^{3}}\) | \(47\) |
Input:
int(x^5*(c*(b*x^2+a)^(1/2))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/1155*(b*x^2+a)*(77*b^2*x^4-56*a*b*x^2+32*a^2)*(c*(b*x^2+a)^(1/2))^(3/2)/ b^3
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \, {\left (77 \, b^{3} c x^{6} + 21 \, a b^{2} c x^{4} - 24 \, a^{2} b c x^{2} + 32 \, a^{3} c\right )} \sqrt {b x^{2} + a} \sqrt {\sqrt {b x^{2} + a} c}}{1155 \, b^{3}} \] Input:
integrate(x^5*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="fricas")
Output:
2/1155*(77*b^3*c*x^6 + 21*a*b^2*c*x^4 - 24*a^2*b*c*x^2 + 32*a^3*c)*sqrt(b* x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)/b^3
Time = 6.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.14 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\begin {cases} \frac {64 a^{3} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{1155 b^{3}} - \frac {16 a^{2} x^{2} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{385 b^{2}} + \frac {2 a x^{4} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{55 b} + \frac {2 x^{6} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{15} & \text {for}\: b \neq 0 \\\frac {x^{6} \left (\sqrt {a} c\right )^{\frac {3}{2}}}{6} & \text {otherwise} \end {cases} \] 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)/(1155*b**3) - 16*a**2*x**2* (c*sqrt(a + b*x**2))**(3/2)/(385*b**2) + 2*a*x**4*(c*sqrt(a + b*x**2))**(3 /2)/(55*b) + 2*x**6*(c*sqrt(a + b*x**2))**(3/2)/15, Ne(b, 0)), (x**6*(sqrt (a)*c)**(3/2)/6, True))
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \, {\left (165 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {7}{2}} a^{2} c^{4} - 210 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {11}{2}} a c^{2} + 77 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {15}{2}}\right )}}{1155 \, b^{3} c^{6}} \] Input:
integrate(x^5*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="maxima")
Output:
2/1155*(165*(sqrt(b*x^2 + a)*c)^(7/2)*a^2*c^4 - 210*(sqrt(b*x^2 + a)*c)^(1 1/2)*a*c^2 + 77*(sqrt(b*x^2 + a)*c)^(15/2))/(b^3*c^6)
Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.07 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \, c^{\frac {3}{2}} {\left (\frac {5 \, {\left (21 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} - 66 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{2}\right )} a}{b^{2}} + \frac {77 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{3}}{b^{2}}\right )}}{1155 \, b} \] Input:
integrate(x^5*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="giac")
Output:
2/1155*c^(3/2)*(5*(21*(b*x^2 + a)^(11/4) - 66*(b*x^2 + a)^(7/4)*a + 77*(b* x^2 + a)^(3/4)*a^2)*a/b^2 + (77*(b*x^2 + a)^(15/4) - 315*(b*x^2 + a)^(11/4 )*a + 495*(b*x^2 + a)^(7/4)*a^2 - 385*(b*x^2 + a)^(3/4)*a^3)/b^2)/b
Time = 0.53 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.86 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\sqrt {c\,\sqrt {b\,x^2+a}}\,\left (\frac {2\,c\,x^6\,\sqrt {b\,x^2+a}}{15}+\frac {64\,a^3\,c\,\sqrt {b\,x^2+a}}{1155\,b^3}+\frac {2\,a\,c\,x^4\,\sqrt {b\,x^2+a}}{55\,b}-\frac {16\,a^2\,c\,x^2\,\sqrt {b\,x^2+a}}{385\,b^2}\right ) \] 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^6*(a + b*x^2)^(1/2))/15 + (64*a^3*c*(a + b*x^2)^(1/2))/(1155*b^3) + (2*a*c*x^4*(a + b*x^2)^(1/2))/(55*b) - (16*a ^2*c*x^2*(a + b*x^2)^(1/2))/(385*b^2))
Time = 0.19 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.57 \[ \int x^5 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \sqrt {c}\, \sqrt {\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +a +b \,x^{2}}\, \sqrt {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}\, c \left (-32 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{3} x +24 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{3}-21 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{5}-77 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, b^{3} x^{7}+32 a^{4}+8 a^{3} b \,x^{2}-3 a^{2} b^{2} x^{4}+98 a \,b^{3} x^{6}+77 b^{4} x^{8}\right )}{1155 a \,b^{3}} \] 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(b)*sqrt(a + b*x**2)*a**3*x + 24*sqrt(b)*sq rt(a + b*x**2)*a**2*b*x**3 - 21*sqrt(b)*sqrt(a + b*x**2)*a*b**2*x**5 - 77* sqrt(b)*sqrt(a + b*x**2)*b**3*x**7 + 32*a**4 + 8*a**3*b*x**2 - 3*a**2*b**2 *x**4 + 98*a*b**3*x**6 + 77*b**4*x**8))/(1155*a*b**3)