Integrand size = 18, antiderivative size = 169 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\frac {a^4 c \sqrt {c x^2} (a+b x)^{1+n}}{b^5 (1+n) x}-\frac {4 a^3 c \sqrt {c x^2} (a+b x)^{2+n}}{b^5 (2+n) x}+\frac {6 a^2 c \sqrt {c x^2} (a+b x)^{3+n}}{b^5 (3+n) x}-\frac {4 a c \sqrt {c x^2} (a+b x)^{4+n}}{b^5 (4+n) x}+\frac {c \sqrt {c x^2} (a+b x)^{5+n}}{b^5 (5+n) x} \]
a^4*c*(b*x+a)^(1+n)*(c*x^2)^(1/2)/b^5/(1+n)/x-4*a^3*c*(b*x+a)^(2+n)*(c*x^2 )^(1/2)/b^5/(2+n)/x+6*a^2*c*(b*x+a)^(3+n)*(c*x^2)^(1/2)/b^5/(3+n)/x-4*a*c* (b*x+a)^(4+n)*(c*x^2)^(1/2)/b^5/(4+n)/x+c*(b*x+a)^(5+n)*(c*x^2)^(1/2)/b^5/ (5+n)/x
Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\frac {\left (c x^2\right )^{3/2} (a+b x)^{1+n} \left (24 a^4-24 a^3 b (1+n) x+12 a^2 b^2 \left (2+3 n+n^2\right ) x^2-4 a b^3 \left (6+11 n+6 n^2+n^3\right ) x^3+b^4 \left (24+50 n+35 n^2+10 n^3+n^4\right ) x^4\right )}{b^5 (1+n) (2+n) (3+n) (4+n) (5+n) x^3} \]
((c*x^2)^(3/2)*(a + b*x)^(1 + n)*(24*a^4 - 24*a^3*b*(1 + n)*x + 12*a^2*b^2 *(2 + 3*n + n^2)*x^2 - 4*a*b^3*(6 + 11*n + 6*n^2 + n^3)*x^3 + b^4*(24 + 50 *n + 35*n^2 + 10*n^3 + n^4)*x^4))/(b^5*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*x^3)
Time = 0.23 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {30, 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.
| \(\displaystyle \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx\) |
| \(\Big \downarrow \) 30 |
| \(\displaystyle \frac {c \sqrt {c x^2} \int x^4 (a+b x)^ndx}{x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {c \sqrt {c x^2} \int \left (\frac {a^4 (a+b x)^n}{b^4}-\frac {4 a^3 (a+b x)^{n+1}}{b^4}+\frac {6 a^2 (a+b x)^{n+2}}{b^4}-\frac {4 a (a+b x)^{n+3}}{b^4}+\frac {(a+b x)^{n+4}}{b^4}\right )dx}{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c \sqrt {c x^2} \left (\frac {a^4 (a+b x)^{n+1}}{b^5 (n+1)}-\frac {4 a^3 (a+b x)^{n+2}}{b^5 (n+2)}+\frac {6 a^2 (a+b x)^{n+3}}{b^5 (n+3)}-\frac {4 a (a+b x)^{n+4}}{b^5 (n+4)}+\frac {(a+b x)^{n+5}}{b^5 (n+5)}\right )}{x}\) |
(c*Sqrt[c*x^2]*((a^4*(a + b*x)^(1 + n))/(b^5*(1 + n)) - (4*a^3*(a + b*x)^( 2 + n))/(b^5*(2 + n)) + (6*a^2*(a + b*x)^(3 + n))/(b^5*(3 + n)) - (4*a*(a + b*x)^(4 + n))/(b^5*(4 + n)) + (a + b*x)^(5 + n)/(b^5*(5 + n))))/x
3.10.30.3.1 Defintions of rubi rules used
Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & & !IntegerQ[p]
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])
Time = 0.14 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.18
| method | result | size |
| gosper | \(\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{1+n} \left (b^{4} n^{4} x^{4}+10 b^{4} n^{3} x^{4}-4 a \,b^{3} n^{3} x^{3}+35 b^{4} n^{2} x^{4}-24 a \,b^{3} n^{2} x^{3}+50 b^{4} n \,x^{4}+12 a^{2} b^{2} n^{2} x^{2}-44 x^{3} a n \,b^{3}+24 b^{4} x^{4}+36 a^{2} n \,x^{2} b^{2}-24 a \,b^{3} x^{3}-24 x \,a^{3} n b +24 a^{2} b^{2} x^{2}-24 a^{3} b x +24 a^{4}\right )}{b^{5} x^{3} \left (n^{5}+15 n^{4}+85 n^{3}+225 n^{2}+274 n +120\right )}\) | \(199\) |
| risch | \(\frac {c \sqrt {c \,x^{2}}\, \left (b^{5} n^{4} x^{5}+a \,b^{4} n^{4} x^{4}+10 b^{5} n^{3} x^{5}+6 a \,b^{4} n^{3} x^{4}+35 b^{5} n^{2} x^{5}-4 a^{2} b^{3} n^{3} x^{3}+11 a \,b^{4} n^{2} x^{4}+50 b^{5} n \,x^{5}-12 a^{2} b^{3} n^{2} x^{3}+6 x^{4} a n \,b^{4}+24 b^{5} x^{5}+12 a^{3} b^{2} n^{2} x^{2}-8 a^{2} n \,x^{3} b^{3}+12 x^{2} a^{3} n \,b^{2}-24 x \,a^{4} n b +24 a^{5}\right ) \left (b x +a \right )^{n}}{x \left (4+n \right ) \left (5+n \right ) \left (3+n \right ) \left (2+n \right ) \left (1+n \right ) b^{5}}\) | \(222\) |
1/b^5/x^3*(c*x^2)^(3/2)*(b*x+a)^(1+n)/(n^5+15*n^4+85*n^3+225*n^2+274*n+120 )*(b^4*n^4*x^4+10*b^4*n^3*x^4-4*a*b^3*n^3*x^3+35*b^4*n^2*x^4-24*a*b^3*n^2* x^3+50*b^4*n*x^4+12*a^2*b^2*n^2*x^2-44*a*b^3*n*x^3+24*b^4*x^4+36*a^2*b^2*n *x^2-24*a*b^3*x^3-24*a^3*b*n*x+24*a^2*b^2*x^2-24*a^3*b*x+24*a^4)
Time = 0.23 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.38 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx=-\frac {{\left (24 \, a^{4} b c n x - 24 \, a^{5} c - {\left (b^{5} c n^{4} + 10 \, b^{5} c n^{3} + 35 \, b^{5} c n^{2} + 50 \, b^{5} c n + 24 \, b^{5} c\right )} x^{5} - {\left (a b^{4} c n^{4} + 6 \, a b^{4} c n^{3} + 11 \, a b^{4} c n^{2} + 6 \, a b^{4} c n\right )} x^{4} + 4 \, {\left (a^{2} b^{3} c n^{3} + 3 \, a^{2} b^{3} c n^{2} + 2 \, a^{2} b^{3} c n\right )} x^{3} - 12 \, {\left (a^{3} b^{2} c n^{2} + a^{3} b^{2} c n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{5} n^{5} + 15 \, b^{5} n^{4} + 85 \, b^{5} n^{3} + 225 \, b^{5} n^{2} + 274 \, b^{5} n + 120 \, b^{5}\right )} x} \]
-(24*a^4*b*c*n*x - 24*a^5*c - (b^5*c*n^4 + 10*b^5*c*n^3 + 35*b^5*c*n^2 + 5 0*b^5*c*n + 24*b^5*c)*x^5 - (a*b^4*c*n^4 + 6*a*b^4*c*n^3 + 11*a*b^4*c*n^2 + 6*a*b^4*c*n)*x^4 + 4*(a^2*b^3*c*n^3 + 3*a^2*b^3*c*n^2 + 2*a^2*b^3*c*n)*x ^3 - 12*(a^3*b^2*c*n^2 + a^3*b^2*c*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^5*n ^5 + 15*b^5*n^4 + 85*b^5*n^3 + 225*b^5*n^2 + 274*b^5*n + 120*b^5)*x)
\[ \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\int x \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}\, dx \]
Time = 0.22 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.93 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\frac {{\left ({\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{5} c^{\frac {3}{2}} x^{5} + {\left (n^{4} + 6 \, n^{3} + 11 \, n^{2} + 6 \, n\right )} a b^{4} c^{\frac {3}{2}} x^{4} - 4 \, {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a^{2} b^{3} c^{\frac {3}{2}} x^{3} + 12 \, {\left (n^{2} + n\right )} a^{3} b^{2} c^{\frac {3}{2}} x^{2} - 24 \, a^{4} b c^{\frac {3}{2}} n x + 24 \, a^{5} c^{\frac {3}{2}}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{5} + 15 \, n^{4} + 85 \, n^{3} + 225 \, n^{2} + 274 \, n + 120\right )} b^{5}} \]
((n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*b^5*c^(3/2)*x^5 + (n^4 + 6*n^3 + 11*n ^2 + 6*n)*a*b^4*c^(3/2)*x^4 - 4*(n^3 + 3*n^2 + 2*n)*a^2*b^3*c^(3/2)*x^3 + 12*(n^2 + n)*a^3*b^2*c^(3/2)*x^2 - 24*a^4*b*c^(3/2)*n*x + 24*a^5*c^(3/2))* (b*x + a)^n/((n^5 + 15*n^4 + 85*n^3 + 225*n^2 + 274*n + 120)*b^5)
Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (159) = 318\).
Time = 0.29 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.52 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx=-{\left (\frac {24 \, a^{5} a^{n} \mathrm {sgn}\left (x\right )}{b^{5} n^{5} + 15 \, b^{5} n^{4} + 85 \, b^{5} n^{3} + 225 \, b^{5} n^{2} + 274 \, b^{5} n + 120 \, b^{5}} - \frac {{\left (b x + a\right )}^{n} b^{5} n^{4} x^{5} \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{n} a b^{4} n^{4} x^{4} \mathrm {sgn}\left (x\right ) + 10 \, {\left (b x + a\right )}^{n} b^{5} n^{3} x^{5} \mathrm {sgn}\left (x\right ) + 6 \, {\left (b x + a\right )}^{n} a b^{4} n^{3} x^{4} \mathrm {sgn}\left (x\right ) + 35 \, {\left (b x + a\right )}^{n} b^{5} n^{2} x^{5} \mathrm {sgn}\left (x\right ) - 4 \, {\left (b x + a\right )}^{n} a^{2} b^{3} n^{3} x^{3} \mathrm {sgn}\left (x\right ) + 11 \, {\left (b x + a\right )}^{n} a b^{4} n^{2} x^{4} \mathrm {sgn}\left (x\right ) + 50 \, {\left (b x + a\right )}^{n} b^{5} n x^{5} \mathrm {sgn}\left (x\right ) - 12 \, {\left (b x + a\right )}^{n} a^{2} b^{3} n^{2} x^{3} \mathrm {sgn}\left (x\right ) + 6 \, {\left (b x + a\right )}^{n} a b^{4} n x^{4} \mathrm {sgn}\left (x\right ) + 24 \, {\left (b x + a\right )}^{n} b^{5} x^{5} \mathrm {sgn}\left (x\right ) + 12 \, {\left (b x + a\right )}^{n} a^{3} b^{2} n^{2} x^{2} \mathrm {sgn}\left (x\right ) - 8 \, {\left (b x + a\right )}^{n} a^{2} b^{3} n x^{3} \mathrm {sgn}\left (x\right ) + 12 \, {\left (b x + a\right )}^{n} a^{3} b^{2} n x^{2} \mathrm {sgn}\left (x\right ) - 24 \, {\left (b x + a\right )}^{n} a^{4} b n x \mathrm {sgn}\left (x\right ) + 24 \, {\left (b x + a\right )}^{n} a^{5} \mathrm {sgn}\left (x\right )}{b^{5} n^{5} + 15 \, b^{5} n^{4} + 85 \, b^{5} n^{3} + 225 \, b^{5} n^{2} + 274 \, b^{5} n + 120 \, b^{5}}\right )} c^{\frac {3}{2}} \]
-(24*a^5*a^n*sgn(x)/(b^5*n^5 + 15*b^5*n^4 + 85*b^5*n^3 + 225*b^5*n^2 + 274 *b^5*n + 120*b^5) - ((b*x + a)^n*b^5*n^4*x^5*sgn(x) + (b*x + a)^n*a*b^4*n^ 4*x^4*sgn(x) + 10*(b*x + a)^n*b^5*n^3*x^5*sgn(x) + 6*(b*x + a)^n*a*b^4*n^3 *x^4*sgn(x) + 35*(b*x + a)^n*b^5*n^2*x^5*sgn(x) - 4*(b*x + a)^n*a^2*b^3*n^ 3*x^3*sgn(x) + 11*(b*x + a)^n*a*b^4*n^2*x^4*sgn(x) + 50*(b*x + a)^n*b^5*n* x^5*sgn(x) - 12*(b*x + a)^n*a^2*b^3*n^2*x^3*sgn(x) + 6*(b*x + a)^n*a*b^4*n *x^4*sgn(x) + 24*(b*x + a)^n*b^5*x^5*sgn(x) + 12*(b*x + a)^n*a^3*b^2*n^2*x ^2*sgn(x) - 8*(b*x + a)^n*a^2*b^3*n*x^3*sgn(x) + 12*(b*x + a)^n*a^3*b^2*n* x^2*sgn(x) - 24*(b*x + a)^n*a^4*b*n*x*sgn(x) + 24*(b*x + a)^n*a^5*sgn(x))/ (b^5*n^5 + 15*b^5*n^4 + 85*b^5*n^3 + 225*b^5*n^2 + 274*b^5*n + 120*b^5))*c ^(3/2)
Time = 0.60 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.82 \[ \int x \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\frac {{\left (a+b\,x\right )}^n\,\left (\frac {24\,a^5\,c\,\sqrt {c\,x^2}}{b^5\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {c\,x^5\,\sqrt {c\,x^2}\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}{n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120}-\frac {24\,a^4\,c\,n\,x\,\sqrt {c\,x^2}}{b^4\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {a\,c\,n\,x^4\,\sqrt {c\,x^2}\,\left (n^3+6\,n^2+11\,n+6\right )}{b\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}+\frac {12\,a^3\,c\,n\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{b^3\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}-\frac {4\,a^2\,c\,n\,x^3\,\sqrt {c\,x^2}\,\left (n^2+3\,n+2\right )}{b^2\,\left (n^5+15\,n^4+85\,n^3+225\,n^2+274\,n+120\right )}\right )}{x} \]
((a + b*x)^n*((24*a^5*c*(c*x^2)^(1/2))/(b^5*(274*n + 225*n^2 + 85*n^3 + 15 *n^4 + n^5 + 120)) + (c*x^5*(c*x^2)^(1/2)*(50*n + 35*n^2 + 10*n^3 + n^4 + 24))/(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120) - (24*a^4*c*n*x*(c*x^ 2)^(1/2))/(b^4*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120)) + (a*c*n*x ^4*(c*x^2)^(1/2)*(11*n + 6*n^2 + n^3 + 6))/(b*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120)) + (12*a^3*c*n*x^2*(c*x^2)^(1/2)*(n + 1))/(b^3*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120)) - (4*a^2*c*n*x^3*(c*x^2)^(1/2)*( 3*n + n^2 + 2))/(b^2*(274*n + 225*n^2 + 85*n^3 + 15*n^4 + n^5 + 120))))/x