Integrand size = 21, antiderivative size = 138 \[ \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=-\frac {2 a^3 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )}{7 b^4}+\frac {6 a^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^2}{11 b^4}-\frac {2 a \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^3}{5 b^4}+\frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^4}{19 b^4} \]
-2/7*a^3*(b*x^2+a)*(c*(b*x^2+a)^(1/2))^(3/2)/b^4+6/11*a^2*(b*x^2+a)^2*(c*( b*x^2+a)^(1/2))^(3/2)/b^4-2/5*a*(b*x^2+a)^3*(c*(b*x^2+a)^(1/2))^(3/2)/b^4+ 2/19*(b*x^2+a)^4*(c*(b*x^2+a)^(1/2))^(3/2)/b^4
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.46 \[ \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right ) \left (-128 a^3+224 a^2 b x^2-308 a b^2 x^4+385 b^3 x^6\right )}{7315 b^4} \]
(2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)*(-128*a^3 + 224*a^2*b*x^2 - 308*a *b^2*x^4 + 385*b^3*x^6))/(7315*b^4)
Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96, 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.
\(\displaystyle \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int x^7 \left (\frac {b x^2}{a}+1\right )^{3/4}dx}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int x^6 \left (\frac {b x^2}{a}+1\right )^{3/4}dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int \left (\frac {a^3 \left (\frac {b x^2}{a}+1\right )^{15/4}}{b^3}-\frac {3 a^3 \left (\frac {b x^2}{a}+1\right )^{11/4}}{b^3}+\frac {3 a^3 \left (\frac {b x^2}{a}+1\right )^{7/4}}{b^3}-\frac {a^3 \left (\frac {b x^2}{a}+1\right )^{3/4}}{b^3}\right )dx^2}{2 \left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (\frac {4 a^4 \left (\frac {b x^2}{a}+1\right )^{19/4}}{19 b^4}-\frac {4 a^4 \left (\frac {b x^2}{a}+1\right )^{15/4}}{5 b^4}+\frac {12 a^4 \left (\frac {b x^2}{a}+1\right )^{11/4}}{11 b^4}-\frac {4 a^4 \left (\frac {b x^2}{a}+1\right )^{7/4}}{7 b^4}\right ) \left (c \sqrt {a+b x^2}\right )^{3/2}}{2 \left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
((c*Sqrt[a + b*x^2])^(3/2)*((-4*a^4*(1 + (b*x^2)/a)^(7/4))/(7*b^4) + (12*a ^4*(1 + (b*x^2)/a)^(11/4))/(11*b^4) - (4*a^4*(1 + (b*x^2)/a)^(15/4))/(5*b^ 4) + (4*a^4*(1 + (b*x^2)/a)^(19/4))/(19*b^4)))/(2*(1 + (b*x^2)/a)^(3/4))
3.3.49.3.1 Defintions of rubi rules used
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 = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.42
method | result | size |
gosper | \(-\frac {2 \left (b \,x^{2}+a \right ) \left (-385 b^{3} x^{6}+308 b^{2} x^{4} a -224 a^{2} b \,x^{2}+128 a^{3}\right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{7315 b^{4}}\) | \(58\) |
-2/7315*(b*x^2+a)*(-385*b^3*x^6+308*a*b^2*x^4-224*a^2*b*x^2+128*a^3)*(c*(b *x^2+a)^(1/2))^(3/2)/b^4
Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54 \[ \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \, {\left (385 \, b^{4} c x^{8} + 77 \, a b^{3} c x^{6} - 84 \, a^{2} b^{2} c x^{4} + 96 \, a^{3} b c x^{2} - 128 \, a^{4} c\right )} \sqrt {b x^{2} + a} \sqrt {\sqrt {b x^{2} + a} c}}{7315 \, b^{4}} \]
2/7315*(385*b^4*c*x^8 + 77*a*b^3*c*x^6 - 84*a^2*b^2*c*x^4 + 96*a^3*b*c*x^2 - 128*a^4*c)*sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)/b^4
Time = 11.19 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04 \[ \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\begin {cases} - \frac {256 a^{4} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{7315 b^{4}} + \frac {192 a^{3} x^{2} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{7315 b^{3}} - \frac {24 a^{2} x^{4} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{1045 b^{2}} + \frac {2 a x^{6} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{95 b} + \frac {2 x^{8} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{19} & \text {for}\: b \neq 0 \\\frac {x^{8} \left (\sqrt {a} c\right )^{\frac {3}{2}}}{8} & \text {otherwise} \end {cases} \]
Piecewise((-256*a**4*(c*sqrt(a + b*x**2))**(3/2)/(7315*b**4) + 192*a**3*x* *2*(c*sqrt(a + b*x**2))**(3/2)/(7315*b**3) - 24*a**2*x**4*(c*sqrt(a + b*x* *2))**(3/2)/(1045*b**2) + 2*a*x**6*(c*sqrt(a + b*x**2))**(3/2)/(95*b) + 2* x**8*(c*sqrt(a + b*x**2))**(3/2)/19, Ne(b, 0)), (x**8*(sqrt(a)*c)**(3/2)/8 , True))
Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62 \[ \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=-\frac {2 \, {\left (1045 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {7}{2}} a^{3} c^{6} - 1995 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {11}{2}} a^{2} c^{4} + 1463 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {15}{2}} a c^{2} - 385 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {19}{2}}\right )}}{7315 \, b^{4} c^{8}} \]
-2/7315*(1045*(sqrt(b*x^2 + a)*c)^(7/2)*a^3*c^6 - 1995*(sqrt(b*x^2 + a)*c) ^(11/2)*a^2*c^4 + 1463*(sqrt(b*x^2 + a)*c)^(15/2)*a*c^2 - 385*(sqrt(b*x^2 + a)*c)^(19/2))/(b^4*c^8)
Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99 \[ \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2 \, c^{\frac {3}{2}} {\left (\frac {19 \, {\left (77 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{3}\right )} a}{b^{3}} + \frac {1155 \, {\left (b x^{2} + a\right )}^{\frac {19}{4}} - 5852 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}} a + 11970 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a^{2} - 12540 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{3} + 7315 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{4}}{b^{3}}\right )}}{21945 \, b} \]
2/21945*c^(3/2)*(19*(77*(b*x^2 + a)^(15/4) - 315*(b*x^2 + a)^(11/4)*a + 49 5*(b*x^2 + a)^(7/4)*a^2 - 385*(b*x^2 + a)^(3/4)*a^3)*a/b^3 + (1155*(b*x^2 + a)^(19/4) - 5852*(b*x^2 + a)^(15/4)*a + 11970*(b*x^2 + a)^(11/4)*a^2 - 1 2540*(b*x^2 + a)^(7/4)*a^3 + 7315*(b*x^2 + a)^(3/4)*a^4)/b^3)/b
Time = 18.88 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.79 \[ \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\sqrt {c\,\sqrt {b\,x^2+a}}\,\left (\frac {2\,c\,x^8\,\sqrt {b\,x^2+a}}{19}-\frac {256\,a^4\,c\,\sqrt {b\,x^2+a}}{7315\,b^4}+\frac {2\,a\,c\,x^6\,\sqrt {b\,x^2+a}}{95\,b}-\frac {24\,a^2\,c\,x^4\,\sqrt {b\,x^2+a}}{1045\,b^2}+\frac {192\,a^3\,c\,x^2\,\sqrt {b\,x^2+a}}{7315\,b^3}\right ) \]