Integrand size = 17, antiderivative size = 75 \[ \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx=\frac {a^2 \left (a+b x^n\right )^{1+p}}{b^3 n (1+p)}-\frac {2 a \left (a+b x^n\right )^{2+p}}{b^3 n (2+p)}+\frac {\left (a+b x^n\right )^{3+p}}{b^3 n (3+p)} \]
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx=\frac {\left (a+b x^n\right )^{1+p} \left (2 a^2-2 a b (1+p) x^n+b^2 \left (2+3 p+p^2\right ) x^{2 n}\right )}{b^3 n (1+p) (2+p) (3+p)} \]
((a + b*x^n)^(1 + p)*(2*a^2 - 2*a*b*(1 + p)*x^n + b^2*(2 + 3*p + p^2)*x^(2 *n)))/(b^3*n*(1 + p)*(2 + p)*(3 + p))
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {798, 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.
\(\displaystyle \int x^{3 n-1} \left (a+b x^n\right )^p \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int x^{2 n} \left (b x^n+a\right )^pdx^n}{n}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\int \left (\frac {a^2 \left (b x^n+a\right )^p}{b^2}-\frac {2 a \left (b x^n+a\right )^{p+1}}{b^2}+\frac {\left (b x^n+a\right )^{p+2}}{b^2}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^2 \left (a+b x^n\right )^{p+1}}{b^3 (p+1)}-\frac {2 a \left (a+b x^n\right )^{p+2}}{b^3 (p+2)}+\frac {\left (a+b x^n\right )^{p+3}}{b^3 (p+3)}}{n}\) |
((a^2*(a + b*x^n)^(1 + p))/(b^3*(1 + p)) - (2*a*(a + b*x^n)^(2 + p))/(b^3* (2 + p)) + (a + b*x^n)^(3 + p)/(b^3*(3 + p)))/n
3.28.22.3.1 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_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.40
method | result | size |
risch | \(\frac {\left (b^{3} p^{2} x^{3 n}+a \,b^{2} p^{2} x^{2 n}+3 b^{3} p \,x^{3 n}+a p \,x^{2 n} b^{2}+2 b^{3} x^{3 n}-2 a^{2} p \,x^{n} b +2 a^{3}\right ) \left (a +b \,x^{n}\right )^{p}}{\left (2+p \right ) \left (3+p \right ) \left (1+p \right ) n \,b^{3}}\) | \(105\) |
(b^3*p^2*(x^n)^3+a*b^2*p^2*(x^n)^2+3*b^3*p*(x^n)^3+a*p*(x^n)^2*b^2+2*b^3*( x^n)^3-2*a^2*p*x^n*b+2*a^3)/(2+p)/(3+p)/(1+p)/n/b^3*(a+b*x^n)^p
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.44 \[ \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx=-\frac {{\left (2 \, a^{2} b p x^{n} - 2 \, a^{3} - {\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{3 \, n} - {\left (a b^{2} p^{2} + a b^{2} p\right )} x^{2 \, n}\right )} {\left (b x^{n} + a\right )}^{p}}{b^{3} n p^{3} + 6 \, b^{3} n p^{2} + 11 \, b^{3} n p + 6 \, b^{3} n} \]
-(2*a^2*b*p*x^n - 2*a^3 - (b^3*p^2 + 3*b^3*p + 2*b^3)*x^(3*n) - (a*b^2*p^2 + a*b^2*p)*x^(2*n))*(b*x^n + a)^p/(b^3*n*p^3 + 6*b^3*n*p^2 + 11*b^3*n*p + 6*b^3*n)
Leaf count of result is larger than twice the leaf count of optimal. 770 vs. \(2 (61) = 122\).
Time = 91.65 (sec) , antiderivative size = 770, normalized size of antiderivative = 10.27 \[ \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx=\begin {cases} \frac {a^{p} x x^{3 n - 1}}{3 n} & \text {for}\: b = 0 \\\left (a + b\right )^{p} \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {2 a^{2} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a^{2} b^{3} n + 4 a b^{4} n x^{n} + 2 b^{5} n x^{2 n}} + \frac {3 a^{2}}{2 a^{2} b^{3} n + 4 a b^{4} n x^{n} + 2 b^{5} n x^{2 n}} + \frac {4 a b x^{n} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a^{2} b^{3} n + 4 a b^{4} n x^{n} + 2 b^{5} n x^{2 n}} + \frac {4 a b x^{n}}{2 a^{2} b^{3} n + 4 a b^{4} n x^{n} + 2 b^{5} n x^{2 n}} + \frac {2 b^{2} x^{2 n} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a^{2} b^{3} n + 4 a b^{4} n x^{n} + 2 b^{5} n x^{2 n}} & \text {for}\: p = -3 \\- \frac {2 a^{2} \log {\left (\frac {a}{b} + x^{n} \right )}}{a b^{3} n + b^{4} n x^{n}} - \frac {2 a^{2}}{a b^{3} n + b^{4} n x^{n}} - \frac {2 a b x^{n} \log {\left (\frac {a}{b} + x^{n} \right )}}{a b^{3} n + b^{4} n x^{n}} + \frac {b^{2} x^{2 n}}{a b^{3} n + b^{4} n x^{n}} & \text {for}\: p = -2 \\\frac {a^{2} \log {\left (\frac {a}{b} + x^{n} \right )}}{b^{3} n} - \frac {a x^{n}}{b^{2} n} + \frac {x^{2 n}}{2 b n} & \text {for}\: p = -1 \\\frac {2 a^{3} \left (a + b x^{n}\right )^{p}}{b^{3} n p^{3} + 6 b^{3} n p^{2} + 11 b^{3} n p + 6 b^{3} n} - \frac {2 a^{2} b p x^{n} \left (a + b x^{n}\right )^{p}}{b^{3} n p^{3} + 6 b^{3} n p^{2} + 11 b^{3} n p + 6 b^{3} n} + \frac {a b^{2} p^{2} x^{2 n} \left (a + b x^{n}\right )^{p}}{b^{3} n p^{3} + 6 b^{3} n p^{2} + 11 b^{3} n p + 6 b^{3} n} + \frac {a b^{2} p x^{2 n} \left (a + b x^{n}\right )^{p}}{b^{3} n p^{3} + 6 b^{3} n p^{2} + 11 b^{3} n p + 6 b^{3} n} + \frac {b^{3} p^{2} x^{3 n} \left (a + b x^{n}\right )^{p}}{b^{3} n p^{3} + 6 b^{3} n p^{2} + 11 b^{3} n p + 6 b^{3} n} + \frac {3 b^{3} p x^{3 n} \left (a + b x^{n}\right )^{p}}{b^{3} n p^{3} + 6 b^{3} n p^{2} + 11 b^{3} n p + 6 b^{3} n} + \frac {2 b^{3} x^{3 n} \left (a + b x^{n}\right )^{p}}{b^{3} n p^{3} + 6 b^{3} n p^{2} + 11 b^{3} n p + 6 b^{3} n} & \text {otherwise} \end {cases} \]
Piecewise((a**p*x*x**(3*n - 1)/(3*n), Eq(b, 0)), ((a + b)**p*log(x), Eq(n, 0)), (2*a**2*log(a/b + x**n)/(2*a**2*b**3*n + 4*a*b**4*n*x**n + 2*b**5*n* x**(2*n)) + 3*a**2/(2*a**2*b**3*n + 4*a*b**4*n*x**n + 2*b**5*n*x**(2*n)) + 4*a*b*x**n*log(a/b + x**n)/(2*a**2*b**3*n + 4*a*b**4*n*x**n + 2*b**5*n*x* *(2*n)) + 4*a*b*x**n/(2*a**2*b**3*n + 4*a*b**4*n*x**n + 2*b**5*n*x**(2*n)) + 2*b**2*x**(2*n)*log(a/b + x**n)/(2*a**2*b**3*n + 4*a*b**4*n*x**n + 2*b* *5*n*x**(2*n)), Eq(p, -3)), (-2*a**2*log(a/b + x**n)/(a*b**3*n + b**4*n*x* *n) - 2*a**2/(a*b**3*n + b**4*n*x**n) - 2*a*b*x**n*log(a/b + x**n)/(a*b**3 *n + b**4*n*x**n) + b**2*x**(2*n)/(a*b**3*n + b**4*n*x**n), Eq(p, -2)), (a **2*log(a/b + x**n)/(b**3*n) - a*x**n/(b**2*n) + x**(2*n)/(2*b*n), Eq(p, - 1)), (2*a**3*(a + b*x**n)**p/(b**3*n*p**3 + 6*b**3*n*p**2 + 11*b**3*n*p + 6*b**3*n) - 2*a**2*b*p*x**n*(a + b*x**n)**p/(b**3*n*p**3 + 6*b**3*n*p**2 + 11*b**3*n*p + 6*b**3*n) + a*b**2*p**2*x**(2*n)*(a + b*x**n)**p/(b**3*n*p* *3 + 6*b**3*n*p**2 + 11*b**3*n*p + 6*b**3*n) + a*b**2*p*x**(2*n)*(a + b*x* *n)**p/(b**3*n*p**3 + 6*b**3*n*p**2 + 11*b**3*n*p + 6*b**3*n) + b**3*p**2* x**(3*n)*(a + b*x**n)**p/(b**3*n*p**3 + 6*b**3*n*p**2 + 11*b**3*n*p + 6*b* *3*n) + 3*b**3*p*x**(3*n)*(a + b*x**n)**p/(b**3*n*p**3 + 6*b**3*n*p**2 + 1 1*b**3*n*p + 6*b**3*n) + 2*b**3*x**(3*n)*(a + b*x**n)**p/(b**3*n*p**3 + 6* b**3*n*p**2 + 11*b**3*n*p + 6*b**3*n), True))
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05 \[ \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx=\frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{3 \, n} + {\left (p^{2} + p\right )} a b^{2} x^{2 \, n} - 2 \, a^{2} b p x^{n} + 2 \, a^{3}\right )} {\left (b x^{n} + a\right )}^{p}}{{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3} n} \]
((p^2 + 3*p + 2)*b^3*x^(3*n) + (p^2 + p)*a*b^2*x^(2*n) - 2*a^2*b*p*x^n + 2 *a^3)*(b*x^n + a)^p/((p^3 + 6*p^2 + 11*p + 6)*b^3*n)
\[ \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx=\int { {\left (b x^{n} + a\right )}^{p} x^{3 \, n - 1} \,d x } \]
Timed out. \[ \int x^{-1+3 n} \left (a+b x^n\right )^p \, dx=\int x^{3\,n-1}\,{\left (a+b\,x^n\right )}^p \,d x \]