Integrand size = 18, antiderivative size = 96 \[ \int x \sqrt {c x^2} (a+b x)^n \, dx=\frac {a^2 \sqrt {c x^2} (a+b x)^{1+n}}{b^3 (1+n) x}-\frac {2 a \sqrt {c x^2} (a+b x)^{2+n}}{b^3 (2+n) x}+\frac {\sqrt {c x^2} (a+b x)^{3+n}}{b^3 (3+n) x} \]
a^2*(b*x+a)^(1+n)*(c*x^2)^(1/2)/b^3/(1+n)/x-2*a*(b*x+a)^(2+n)*(c*x^2)^(1/2 )/b^3/(2+n)/x+(b*x+a)^(3+n)*(c*x^2)^(1/2)/b^3/(3+n)/x
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.71 \[ \int x \sqrt {c x^2} (a+b x)^n \, dx=\frac {c x (a+b x)^{1+n} \left (2 a^2-2 a b (1+n) x+b^2 \left (2+3 n+n^2\right ) x^2\right )}{b^3 (1+n) (2+n) (3+n) \sqrt {c x^2}} \]
(c*x*(a + b*x)^(1 + n)*(2*a^2 - 2*a*b*(1 + n)*x + b^2*(2 + 3*n + n^2)*x^2) )/(b^3*(1 + n)*(2 + n)*(3 + n)*Sqrt[c*x^2])
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76, 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.
\(\displaystyle \int x \sqrt {c x^2} (a+b x)^n \, dx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {\sqrt {c x^2} \int x^2 (a+b x)^ndx}{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {c x^2} \int \left (\frac {a^2 (a+b x)^n}{b^2}-\frac {2 a (a+b x)^{n+1}}{b^2}+\frac {(a+b x)^{n+2}}{b^2}\right )dx}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c x^2} \left (\frac {a^2 (a+b x)^{n+1}}{b^3 (n+1)}-\frac {2 a (a+b x)^{n+2}}{b^3 (n+2)}+\frac {(a+b x)^{n+3}}{b^3 (n+3)}\right )}{x}\) |
(Sqrt[c*x^2]*((a^2*(a + b*x)^(1 + n))/(b^3*(1 + n)) - (2*a*(a + b*x)^(2 + n))/(b^3*(2 + n)) + (a + b*x)^(3 + n)/(b^3*(3 + n))))/x
3.10.24.3.1 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 = 83, normalized size of antiderivative = 0.86
method | result | size |
gosper | \(\frac {\sqrt {c \,x^{2}}\, \left (b x +a \right )^{1+n} \left (b^{2} n^{2} x^{2}+3 b^{2} n \,x^{2}-2 a b n x +2 b^{2} x^{2}-2 a b x +2 a^{2}\right )}{b^{3} x \left (n^{3}+6 n^{2}+11 n +6\right )}\) | \(83\) |
risch | \(\frac {\sqrt {c \,x^{2}}\, \left (b^{3} n^{2} x^{3}+a \,b^{2} n^{2} x^{2}+3 b^{3} n \,x^{3}+a \,b^{2} n \,x^{2}+2 b^{3} x^{3}-2 a^{2} b n x +2 a^{3}\right ) \left (b x +a \right )^{n}}{x \left (2+n \right ) \left (3+n \right ) \left (1+n \right ) b^{3}}\) | \(98\) |
1/b^3/x*(c*x^2)^(1/2)*(b*x+a)^(1+n)/(n^3+6*n^2+11*n+6)*(b^2*n^2*x^2+3*b^2* n*x^2-2*a*b*n*x+2*b^2*x^2-2*a*b*x+2*a^2)
Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10 \[ \int x \sqrt {c x^2} (a+b x)^n \, dx=-\frac {{\left (2 \, a^{2} b n x - {\left (b^{3} n^{2} + 3 \, b^{3} n + 2 \, b^{3}\right )} x^{3} - 2 \, a^{3} - {\left (a b^{2} n^{2} + a b^{2} n\right )} x^{2}\right )} \sqrt {c x^{2}} {\left (b x + a\right )}^{n}}{{\left (b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}\right )} x} \]
-(2*a^2*b*n*x - (b^3*n^2 + 3*b^3*n + 2*b^3)*x^3 - 2*a^3 - (a*b^2*n^2 + a*b ^2*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^ 3)*x)
\[ \int x \sqrt {c x^2} (a+b x)^n \, dx=\begin {cases} \frac {a^{n} x^{2} \sqrt {c x^{2}}}{3} & \text {for}\: b = 0 \\\int \frac {x \sqrt {c x^{2}}}{\left (a + b x\right )^{3}}\, dx & \text {for}\: n = -3 \\\int \frac {x \sqrt {c x^{2}}}{\left (a + b x\right )^{2}}\, dx & \text {for}\: n = -2 \\\int \frac {x \sqrt {c x^{2}}}{a + b x}\, dx & \text {for}\: n = -1 \\\frac {2 a^{3} \sqrt {c x^{2}} \left (a + b x\right )^{n}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} - \frac {2 a^{2} b n x \sqrt {c x^{2}} \left (a + b x\right )^{n}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {a b^{2} n^{2} x^{2} \sqrt {c x^{2}} \left (a + b x\right )^{n}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {a b^{2} n x^{2} \sqrt {c x^{2}} \left (a + b x\right )^{n}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {b^{3} n^{2} x^{3} \sqrt {c x^{2}} \left (a + b x\right )^{n}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {3 b^{3} n x^{3} \sqrt {c x^{2}} \left (a + b x\right )^{n}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} + \frac {2 b^{3} x^{3} \sqrt {c x^{2}} \left (a + b x\right )^{n}}{b^{3} n^{3} x + 6 b^{3} n^{2} x + 11 b^{3} n x + 6 b^{3} x} & \text {otherwise} \end {cases} \]
Piecewise((a**n*x**2*sqrt(c*x**2)/3, Eq(b, 0)), (Integral(x*sqrt(c*x**2)/( a + b*x)**3, x), Eq(n, -3)), (Integral(x*sqrt(c*x**2)/(a + b*x)**2, x), Eq (n, -2)), (Integral(x*sqrt(c*x**2)/(a + b*x), x), Eq(n, -1)), (2*a**3*sqrt (c*x**2)*(a + b*x)**n/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3* x) - 2*a**2*b*n*x*sqrt(c*x**2)*(a + b*x)**n/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + a*b**2*n**2*x**2*sqrt(c*x**2)*(a + b*x)**n/(b** 3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + a*b**2*n*x**2*sqrt(c* x**2)*(a + b*x)**n/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + b**3*n**2*x**3*sqrt(c*x**2)*(a + b*x)**n/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + 3*b**3*n*x**3*sqrt(c*x**2)*(a + b*x)**n/(b**3*n* *3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x) + 2*b**3*x**3*sqrt(c*x**2)* (a + b*x)**n/(b**3*n**3*x + 6*b**3*n**2*x + 11*b**3*n*x + 6*b**3*x), True) )
Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.83 \[ \int x \sqrt {c x^2} (a+b x)^n \, dx=\frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} \sqrt {c} x^{3} + {\left (n^{2} + n\right )} a b^{2} \sqrt {c} x^{2} - 2 \, a^{2} b \sqrt {c} n x + 2 \, a^{3} \sqrt {c}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}} \]
((n^2 + 3*n + 2)*b^3*sqrt(c)*x^3 + (n^2 + n)*a*b^2*sqrt(c)*x^2 - 2*a^2*b*s qrt(c)*n*x + 2*a^3*sqrt(c))*(b*x + a)^n/((n^3 + 6*n^2 + 11*n + 6)*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (90) = 180\).
Time = 0.30 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.08 \[ \int x \sqrt {c x^2} (a+b x)^n \, dx=-{\left (\frac {2 \, a^{3} a^{n} \mathrm {sgn}\left (x\right )}{b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}} - \frac {{\left (b x + a\right )}^{n} b^{3} n^{2} x^{3} \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{n} a b^{2} n^{2} x^{2} \mathrm {sgn}\left (x\right ) + 3 \, {\left (b x + a\right )}^{n} b^{3} n x^{3} \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{n} a b^{2} n x^{2} \mathrm {sgn}\left (x\right ) + 2 \, {\left (b x + a\right )}^{n} b^{3} x^{3} \mathrm {sgn}\left (x\right ) - 2 \, {\left (b x + a\right )}^{n} a^{2} b n x \mathrm {sgn}\left (x\right ) + 2 \, {\left (b x + a\right )}^{n} a^{3} \mathrm {sgn}\left (x\right )}{b^{3} n^{3} + 6 \, b^{3} n^{2} + 11 \, b^{3} n + 6 \, b^{3}}\right )} \sqrt {c} \]
-(2*a^3*a^n*sgn(x)/(b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^3) - ((b*x + a)^n *b^3*n^2*x^3*sgn(x) + (b*x + a)^n*a*b^2*n^2*x^2*sgn(x) + 3*(b*x + a)^n*b^3 *n*x^3*sgn(x) + (b*x + a)^n*a*b^2*n*x^2*sgn(x) + 2*(b*x + a)^n*b^3*x^3*sgn (x) - 2*(b*x + a)^n*a^2*b*n*x*sgn(x) + 2*(b*x + a)^n*a^3*sgn(x))/(b^3*n^3 + 6*b^3*n^2 + 11*b^3*n + 6*b^3))*sqrt(c)
Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.48 \[ \int x \sqrt {c x^2} (a+b x)^n \, dx=\frac {{\left (a+b\,x\right )}^n\,\left (\frac {2\,a^3\,\sqrt {c\,x^2}}{b^3\,\left (n^3+6\,n^2+11\,n+6\right )}+\frac {x^3\,\sqrt {c\,x^2}\,\left (n^2+3\,n+2\right )}{n^3+6\,n^2+11\,n+6}-\frac {2\,a^2\,n\,x\,\sqrt {c\,x^2}}{b^2\,\left (n^3+6\,n^2+11\,n+6\right )}+\frac {a\,n\,x^2\,\sqrt {c\,x^2}\,\left (n+1\right )}{b\,\left (n^3+6\,n^2+11\,n+6\right )}\right )}{x} \]