Integrand size = 20, antiderivative size = 65 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx=-\frac {a c \sqrt {c x^2} (a+b x)^{1+n}}{b^2 (1+n) x}+\frac {c \sqrt {c x^2} (a+b x)^{2+n}}{b^2 (2+n) x} \]
Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx=\frac {c^2 x (a+b x)^{1+n} (-a+b (1+n) x)}{b^2 (1+n) (2+n) \sqrt {c x^2}} \]
Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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.
\(\displaystyle \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {c \sqrt {c x^2} \int x (a+b x)^ndx}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {c \sqrt {c x^2} \int \left (\frac {(a+b x)^{n+1}}{b}-\frac {a (a+b x)^n}{b}\right )dx}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \sqrt {c x^2} \left (\frac {(a+b x)^{n+2}}{b^2 (n+2)}-\frac {a (a+b x)^{n+1}}{b^2 (n+1)}\right )}{x}\) |
(c*Sqrt[c*x^2]*(-((a*(a + b*x)^(1 + n))/(b^2*(1 + n))) + (a + b*x)^(2 + n) /(b^2*(2 + n))))/x
3.10.33.3.1 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 = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{1+n} \left (-b n x -b x +a \right )}{b^{2} x^{3} \left (n^{2}+3 n +2\right )}\) | \(46\) |
risch | \(-\frac {c \sqrt {c \,x^{2}}\, \left (-b^{2} n \,x^{2}-a b n x -b^{2} x^{2}+a^{2}\right ) \left (b x +a \right )^{n}}{x \,b^{2} \left (2+n \right ) \left (1+n \right )}\) | \(61\) |
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx=\frac {{\left (a b c n x - a^{2} c + {\left (b^{2} c n + b^{2} c\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}\right )} x} \]
(a*b*c*n*x - a^2*c + (b^2*c*n + b^2*c)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^2* n^2 + 3*b^2*n + 2*b^2)*x)
\[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx=\begin {cases} \frac {a^{n} \left (c x^{2}\right )^{\frac {3}{2}}}{2 x} & \text {for}\: b = 0 \\\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{2} \left (a + b x\right )^{2}}\, dx & \text {for}\: n = -2 \\\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{2} \left (a + b x\right )}\, dx & \text {for}\: n = -1 \\- \frac {a^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} + \frac {a b n x \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} + \frac {b^{2} n x^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} + \frac {b^{2} x^{2} \left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )^{n}}{b^{2} n^{2} x^{3} + 3 b^{2} n x^{3} + 2 b^{2} x^{3}} & \text {otherwise} \end {cases} \]
Piecewise((a**n*(c*x**2)**(3/2)/(2*x), Eq(b, 0)), (Integral((c*x**2)**(3/2 )/(x**2*(a + b*x)**2), x), Eq(n, -2)), (Integral((c*x**2)**(3/2)/(x**2*(a + b*x)), x), Eq(n, -1)), (-a**2*(c*x**2)**(3/2)*(a + b*x)**n/(b**2*n**2*x* *3 + 3*b**2*n*x**3 + 2*b**2*x**3) + a*b*n*x*(c*x**2)**(3/2)*(a + b*x)**n/( b**2*n**2*x**3 + 3*b**2*n*x**3 + 2*b**2*x**3) + b**2*n*x**2*(c*x**2)**(3/2 )*(a + b*x)**n/(b**2*n**2*x**3 + 3*b**2*n*x**3 + 2*b**2*x**3) + b**2*x**2* (c*x**2)**(3/2)*(a + b*x)**n/(b**2*n**2*x**3 + 3*b**2*n*x**3 + 2*b**2*x**3 ), True))
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx=\frac {{\left (b^{2} c^{\frac {3}{2}} {\left (n + 1\right )} x^{2} + a b c^{\frac {3}{2}} n x - a^{2} c^{\frac {3}{2}}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{2} + 3 \, n + 2\right )} b^{2}} \]
Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.83 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx={\left (\frac {a^{2} a^{n} \mathrm {sgn}\left (x\right )}{b^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}} + \frac {{\left (b x + a\right )}^{n} b^{2} n x^{2} \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{n} a b n x \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{n} b^{2} x^{2} \mathrm {sgn}\left (x\right ) - {\left (b x + a\right )}^{n} a^{2} \mathrm {sgn}\left (x\right )}{b^{2} n^{2} + 3 \, b^{2} n + 2 \, b^{2}}\right )} c^{\frac {3}{2}} \]
(a^2*a^n*sgn(x)/(b^2*n^2 + 3*b^2*n + 2*b^2) + ((b*x + a)^n*b^2*n*x^2*sgn(x ) + (b*x + a)^n*a*b*n*x*sgn(x) + (b*x + a)^n*b^2*x^2*sgn(x) - (b*x + a)^n* a^2*sgn(x))/(b^2*n^2 + 3*b^2*n + 2*b^2))*c^(3/2)
Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.35 \[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx=\frac {{\left (a+b\,x\right )}^n\,\left (\frac {c\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{n^2+3\,n+2}-\frac {a^2\,c\,\sqrt {c\,x^2}}{b^2\,\left (n^2+3\,n+2\right )}+\frac {a\,c\,n\,x\,\sqrt {c\,x^2}}{b\,\left (n^2+3\,n+2\right )}\right )}{x} \]
((a + b*x)^n*((c*x^2*(c*x^2)^(1/2)*(n + 1))/(3*n + n^2 + 2) - (a^2*c*(c*x^ 2)^(1/2))/(b^2*(3*n + n^2 + 2)) + (a*c*n*x*(c*x^2)^(1/2))/(b*(3*n + n^2 + 2))))/x
\[ \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^2} \, dx=\frac {\sqrt {c}\, c \left (\left (b x +a \right )^{n} {| x |} a n +\left (b x +a \right )^{n} {| x |} b n x +\left (b x +a \right )^{n} {| x |} b x -\left (\int \frac {\left (b x +a \right )^{n} {| x |}}{b \,n^{2} x^{2}+a \,n^{2} x +3 b n \,x^{2}+3 a n x +2 b \,x^{2}+2 a x}d x \right ) a^{2} n^{3}-3 \left (\int \frac {\left (b x +a \right )^{n} {| x |}}{b \,n^{2} x^{2}+a \,n^{2} x +3 b n \,x^{2}+3 a n x +2 b \,x^{2}+2 a x}d x \right ) a^{2} n^{2}-2 \left (\int \frac {\left (b x +a \right )^{n} {| x |}}{b \,n^{2} x^{2}+a \,n^{2} x +3 b n \,x^{2}+3 a n x +2 b \,x^{2}+2 a x}d x \right ) a^{2} n \right )}{b \left (n^{2}+3 n +2\right )} \]
(sqrt(c)*c*((a + b*x)**n*abs(x)*a*n + (a + b*x)**n*abs(x)*b*n*x + (a + b*x )**n*abs(x)*b*x - int(((a + b*x)**n*abs(x))/(a*n**2*x + 3*a*n*x + 2*a*x + b*n**2*x**2 + 3*b*n*x**2 + 2*b*x**2),x)*a**2*n**3 - 3*int(((a + b*x)**n*ab s(x))/(a*n**2*x + 3*a*n*x + 2*a*x + b*n**2*x**2 + 3*b*n*x**2 + 2*b*x**2),x )*a**2*n**2 - 2*int(((a + b*x)**n*abs(x))/(a*n**2*x + 3*a*n*x + 2*a*x + b* n**2*x**2 + 3*b*n*x**2 + 2*b*x**2),x)*a**2*n))/(b*(n**2 + 3*n + 2))