Integrand size = 19, antiderivative size = 84 \[ \int \left (c+d x^{-1+n}\right ) \left (a+b x^n\right )^3 \, dx=a^3 c x+\frac {3 a^2 b c x^{1+n}}{1+n}+\frac {3 a b^2 c x^{1+2 n}}{1+2 n}+\frac {b^3 c x^{1+3 n}}{1+3 n}+\frac {d \left (a+b x^n\right )^4}{4 b n} \]
a^3*c*x+3*a^2*b*c*x^(1+n)/(1+n)+3*a*b^2*c*x^(1+2*n)/(1+2*n)+b^3*c*x^(1+3*n )/(1+3*n)+1/4*d*(a+b*x^n)^4/b/n
Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.29 \[ \int \left (c+d x^{-1+n}\right ) \left (a+b x^n\right )^3 \, dx=\frac {x \left (c+d x^{-1+n}\right ) \left (4 a^3 c x+\frac {12 a^2 b c x^{1+n}}{1+n}+\frac {12 a b^2 c x^{1+2 n}}{1+2 n}+\frac {4 b^3 c x^{1+3 n}}{1+3 n}+\frac {d \left (a+b x^n\right )^4}{b n}\right )}{4 \left (c x+d x^n\right )} \]
(x*(c + d*x^(-1 + n))*(4*a^3*c*x + (12*a^2*b*c*x^(1 + n))/(1 + n) + (12*a* b^2*c*x^(1 + 2*n))/(1 + 2*n) + (4*b^3*c*x^(1 + 3*n))/(1 + 3*n) + (d*(a + b *x^n)^4)/(b*n)))/(4*(c*x + d*x^n))
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2430, 775, 793, 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.
\(\displaystyle \int \left (a+b x^n\right )^3 \left (c+d x^{n-1}\right ) \, dx\) |
\(\Big \downarrow \) 2430 |
\(\displaystyle c \int \left (b x^n+a\right )^3dx+d \int x^{n-1} \left (b x^n+a\right )^3dx\) |
\(\Big \downarrow \) 775 |
\(\displaystyle c \int \left (3 a^2 b x^n+3 a b^2 x^{2 n}+b^3 x^{3 n}+a^3\right )dx+d \int x^{n-1} \left (b x^n+a\right )^3dx\) |
\(\Big \downarrow \) 793 |
\(\displaystyle c \int \left (3 a^2 b x^n+3 a b^2 x^{2 n}+b^3 x^{3 n}+a^3\right )dx+\frac {d \left (a+b x^n\right )^4}{4 b n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c \left (a^3 x+\frac {3 a^2 b x^{n+1}}{n+1}+\frac {3 a b^2 x^{2 n+1}}{2 n+1}+\frac {b^3 x^{3 n+1}}{3 n+1}\right )+\frac {d \left (a+b x^n\right )^4}{4 b n}\) |
(d*(a + b*x^n)^4)/(4*b*n) + c*(a^3*x + (3*a^2*b*x^(1 + n))/(1 + n) + (3*a* b^2*x^(1 + 2*n))/(1 + 2*n) + (b^3*x^(1 + 3*n))/(1 + 3*n))
3.6.77.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b* x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && IGtQ[p, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n) ^(p + 1)/(b*n*(p + 1)), x] /; FreeQ[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]
Int[((A_) + (B_.)*(x_)^(m_.))*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Simp[A Int[(a + b*x^n)^p, x], x] + Simp[B Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, A, B, m, n, p}, x] && EqQ[m - n + 1, 0]
Time = 1.76 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.40
method | result | size |
risch | \(a^{3} c x +\frac {b^{3} d \,x^{4 n}}{4 n}+\frac {b^{2} \left (n b c x +3 a d n +a d \right ) x^{3 n}}{n \left (1+3 n \right )}+\frac {3 a b \left (2 n b c x +2 a d n +a d \right ) x^{2 n}}{2 n \left (1+2 n \right )}+\frac {a^{2} \left (3 n b c x +a d n +a d \right ) x^{n}}{n \left (1+n \right )}\) | \(118\) |
norman | \(a^{3} c x +\frac {a^{3} d \,{\mathrm e}^{n \ln \left (x \right )}}{n}+\frac {a \,b^{2} d \,{\mathrm e}^{3 n \ln \left (x \right )}}{n}+\frac {b^{3} c x \,{\mathrm e}^{3 n \ln \left (x \right )}}{1+3 n}+\frac {b^{3} d \,{\mathrm e}^{4 n \ln \left (x \right )}}{4 n}+\frac {3 d \,a^{2} b \,{\mathrm e}^{2 n \ln \left (x \right )}}{2 n}+\frac {3 a \,b^{2} c x \,{\mathrm e}^{2 n \ln \left (x \right )}}{1+2 n}+\frac {3 a^{2} b c x \,{\mathrm e}^{n \ln \left (x \right )}}{1+n}\) | \(130\) |
parallelrisch | \(\frac {6 x \,x^{n} x^{-1+n} a^{2} b d +4 a^{3} c x n +66 x \,x^{n} x^{-1+n} a^{2} b d \,n^{2}+36 x \,x^{n} x^{-1+n} a^{2} b d n +36 x^{2 n} a \,b^{2} c \,n^{3} x +6 x \,x^{3 n} x^{-1+n} b^{3} d \,n^{3}+11 x \,x^{3 n} x^{-1+n} b^{3} d \,n^{2}+6 x \,x^{3 n} x^{-1+n} b^{3} d n +4 x \,x^{2 n} x^{-1+n} a \,b^{2} d +48 x^{2 n} a \,b^{2} c \,n^{2} x +12 x^{2 n} a \,b^{2} c n x +24 x \,x^{2 n} x^{-1+n} a \,b^{2} d n +24 x \,x^{2 n} x^{-1+n} a \,b^{2} d \,n^{3}+44 x \,x^{2 n} x^{-1+n} a \,b^{2} d \,n^{2}+36 x \,x^{n} x^{-1+n} a^{2} b d \,n^{3}+72 x^{n} a^{2} b c \,n^{3} x +60 x^{n} a^{2} b c \,n^{2} x +12 x^{n} a^{2} b c n x +4 x \,x^{-1+n} a^{3} d +24 a^{3} c \,n^{4} x +44 a^{3} c \,n^{3} x +24 a^{3} c \,n^{2} x +24 x \,x^{-1+n} a^{3} d \,n^{3}+44 x \,x^{-1+n} a^{3} d \,n^{2}+24 x \,x^{-1+n} a^{3} d n +12 x^{3 n} b^{3} c \,n^{2} x +4 x^{3 n} b^{3} c n x +8 x^{3 n} b^{3} c \,n^{3} x +x \,x^{3 n} x^{-1+n} b^{3} d}{4 n \left (1+3 n \right ) \left (1+2 n \right ) \left (1+n \right )}\) | \(472\) |
a^3*c*x+1/4*b^3*d/n*(x^n)^4+b^2*(b*c*n*x+3*a*d*n+a*d)/n/(1+3*n)*(x^n)^3+3/ 2*a*b*(2*b*c*n*x+2*a*d*n+a*d)/n/(1+2*n)*(x^n)^2+a^2*(3*b*c*n*x+a*d*n+a*d)/ n/(1+n)*x^n
Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.63 \[ \int \left (c+d x^{-1+n}\right ) \left (a+b x^n\right )^3 \, dx=\frac {4 \, {\left (6 \, a^{3} c n^{4} + 11 \, a^{3} c n^{3} + 6 \, a^{3} c n^{2} + a^{3} c n\right )} x + {\left (6 \, b^{3} d n^{3} + 11 \, b^{3} d n^{2} + 6 \, b^{3} d n + b^{3} d\right )} x^{4 \, n} + 4 \, {\left (6 \, a b^{2} d n^{3} + 11 \, a b^{2} d n^{2} + 6 \, a b^{2} d n + a b^{2} d + {\left (2 \, b^{3} c n^{3} + 3 \, b^{3} c n^{2} + b^{3} c n\right )} x\right )} x^{3 \, n} + 6 \, {\left (6 \, a^{2} b d n^{3} + 11 \, a^{2} b d n^{2} + 6 \, a^{2} b d n + a^{2} b d + 2 \, {\left (3 \, a b^{2} c n^{3} + 4 \, a b^{2} c n^{2} + a b^{2} c n\right )} x\right )} x^{2 \, n} + 4 \, {\left (6 \, a^{3} d n^{3} + 11 \, a^{3} d n^{2} + 6 \, a^{3} d n + a^{3} d + 3 \, {\left (6 \, a^{2} b c n^{3} + 5 \, a^{2} b c n^{2} + a^{2} b c n\right )} x\right )} x^{n}}{4 \, {\left (6 \, n^{4} + 11 \, n^{3} + 6 \, n^{2} + n\right )}} \]
1/4*(4*(6*a^3*c*n^4 + 11*a^3*c*n^3 + 6*a^3*c*n^2 + a^3*c*n)*x + (6*b^3*d*n ^3 + 11*b^3*d*n^2 + 6*b^3*d*n + b^3*d)*x^(4*n) + 4*(6*a*b^2*d*n^3 + 11*a*b ^2*d*n^2 + 6*a*b^2*d*n + a*b^2*d + (2*b^3*c*n^3 + 3*b^3*c*n^2 + b^3*c*n)*x )*x^(3*n) + 6*(6*a^2*b*d*n^3 + 11*a^2*b*d*n^2 + 6*a^2*b*d*n + a^2*b*d + 2* (3*a*b^2*c*n^3 + 4*a*b^2*c*n^2 + a*b^2*c*n)*x)*x^(2*n) + 4*(6*a^3*d*n^3 + 11*a^3*d*n^2 + 6*a^3*d*n + a^3*d + 3*(6*a^2*b*c*n^3 + 5*a^2*b*c*n^2 + a^2* b*c*n)*x)*x^n)/(6*n^4 + 11*n^3 + 6*n^2 + n)
Leaf count of result is larger than twice the leaf count of optimal. 1340 vs. \(2 (75) = 150\).
Time = 0.78 (sec) , antiderivative size = 1340, normalized size of antiderivative = 15.95 \[ \int \left (c+d x^{-1+n}\right ) \left (a+b x^n\right )^3 \, dx=\text {Too large to display} \]
Piecewise((a**3*c*x - a**3*d/x + 3*a**2*b*c*log(x) - 3*a**2*b*d/(2*x**2) - 3*a*b**2*c/x - a*b**2*d/x**3 - b**3*c/(2*x**2) - b**3*d/(4*x**4), Eq(n, - 1)), (a**3*c*x - 2*a**3*d/sqrt(x) + 6*a**2*b*c*sqrt(x) - 3*a**2*b*d/x + 3* a*b**2*c*log(x) - 2*a*b**2*d/x**(3/2) - 2*b**3*c/sqrt(x) - b**3*d/(2*x**2) , Eq(n, -1/2)), (a**3*c*x - 3*a**3*d/x**(1/3) + 9*a**2*b*c*x**(2/3)/2 - 9* a**2*b*d/(2*x**(2/3)) + 9*a*b**2*c*x**(1/3) - 3*a*b**2*d/x + b**3*c*log(x) - 3*b**3*d/(4*x**(4/3)), Eq(n, -1/3)), ((a + b)**3*(c*x + d*log(x)), Eq(n , 0)), (24*a**3*c*n**4*x/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 44*a**3*c*n **3*x/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 24*a**3*c*n**2*x/(24*n**4 + 44 *n**3 + 24*n**2 + 4*n) + 4*a**3*c*n*x/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 24*a**3*d*n**3*x*x**(n - 1)/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 44*a** 3*d*n**2*x*x**(n - 1)/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 24*a**3*d*n*x* x**(n - 1)/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 4*a**3*d*x*x**(n - 1)/(24 *n**4 + 44*n**3 + 24*n**2 + 4*n) + 72*a**2*b*c*n**3*x*x**n/(24*n**4 + 44*n **3 + 24*n**2 + 4*n) + 60*a**2*b*c*n**2*x*x**n/(24*n**4 + 44*n**3 + 24*n** 2 + 4*n) + 12*a**2*b*c*n*x*x**n/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 36*a **2*b*d*n**3*x*x**n*x**(n - 1)/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 66*a* *2*b*d*n**2*x*x**n*x**(n - 1)/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 36*a** 2*b*d*n*x*x**n*x**(n - 1)/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 6*a**2*b*d *x*x**n*x**(n - 1)/(24*n**4 + 44*n**3 + 24*n**2 + 4*n) + 36*a*b**2*c*n*...
Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.40 \[ \int \left (c+d x^{-1+n}\right ) \left (a+b x^n\right )^3 \, dx=a^{3} c x + \frac {b^{3} d x^{4 \, n}}{4 \, n} + \frac {a b^{2} d x^{3 \, n}}{n} + \frac {3 \, a^{2} b d x^{2 \, n}}{2 \, n} + \frac {b^{3} c x^{3 \, n + 1}}{3 \, n + 1} + \frac {3 \, a b^{2} c x^{2 \, n + 1}}{2 \, n + 1} + \frac {3 \, a^{2} b c x^{n + 1}}{n + 1} + \frac {a^{3} d x^{n}}{n} \]
a^3*c*x + 1/4*b^3*d*x^(4*n)/n + a*b^2*d*x^(3*n)/n + 3/2*a^2*b*d*x^(2*n)/n + b^3*c*x^(3*n + 1)/(3*n + 1) + 3*a*b^2*c*x^(2*n + 1)/(2*n + 1) + 3*a^2*b* c*x^(n + 1)/(n + 1) + a^3*d*x^n/n
Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (82) = 164\).
Time = 0.29 (sec) , antiderivative size = 392, normalized size of antiderivative = 4.67 \[ \int \left (c+d x^{-1+n}\right ) \left (a+b x^n\right )^3 \, dx=\frac {24 \, a^{3} c n^{4} x + 8 \, b^{3} c n^{3} x x^{3 \, n} + 36 \, a b^{2} c n^{3} x x^{2 \, n} + 72 \, a^{2} b c n^{3} x x^{n} + 44 \, a^{3} c n^{3} x + 6 \, b^{3} d n^{3} x^{4 \, n} + 24 \, a b^{2} d n^{3} x^{3 \, n} + 12 \, b^{3} c n^{2} x x^{3 \, n} + 36 \, a^{2} b d n^{3} x^{2 \, n} + 48 \, a b^{2} c n^{2} x x^{2 \, n} + 24 \, a^{3} d n^{3} x^{n} + 60 \, a^{2} b c n^{2} x x^{n} + 24 \, a^{3} c n^{2} x + 11 \, b^{3} d n^{2} x^{4 \, n} + 44 \, a b^{2} d n^{2} x^{3 \, n} + 4 \, b^{3} c n x x^{3 \, n} + 66 \, a^{2} b d n^{2} x^{2 \, n} + 12 \, a b^{2} c n x x^{2 \, n} + 44 \, a^{3} d n^{2} x^{n} + 12 \, a^{2} b c n x x^{n} + 4 \, a^{3} c n x + 6 \, b^{3} d n x^{4 \, n} + 24 \, a b^{2} d n x^{3 \, n} + 36 \, a^{2} b d n x^{2 \, n} + 24 \, a^{3} d n x^{n} + b^{3} d x^{4 \, n} + 4 \, a b^{2} d x^{3 \, n} + 6 \, a^{2} b d x^{2 \, n} + 4 \, a^{3} d x^{n}}{4 \, {\left (6 \, n^{4} + 11 \, n^{3} + 6 \, n^{2} + n\right )}} \]
1/4*(24*a^3*c*n^4*x + 8*b^3*c*n^3*x*x^(3*n) + 36*a*b^2*c*n^3*x*x^(2*n) + 7 2*a^2*b*c*n^3*x*x^n + 44*a^3*c*n^3*x + 6*b^3*d*n^3*x^(4*n) + 24*a*b^2*d*n^ 3*x^(3*n) + 12*b^3*c*n^2*x*x^(3*n) + 36*a^2*b*d*n^3*x^(2*n) + 48*a*b^2*c*n ^2*x*x^(2*n) + 24*a^3*d*n^3*x^n + 60*a^2*b*c*n^2*x*x^n + 24*a^3*c*n^2*x + 11*b^3*d*n^2*x^(4*n) + 44*a*b^2*d*n^2*x^(3*n) + 4*b^3*c*n*x*x^(3*n) + 66*a ^2*b*d*n^2*x^(2*n) + 12*a*b^2*c*n*x*x^(2*n) + 44*a^3*d*n^2*x^n + 12*a^2*b* c*n*x*x^n + 4*a^3*c*n*x + 6*b^3*d*n*x^(4*n) + 24*a*b^2*d*n*x^(3*n) + 36*a^ 2*b*d*n*x^(2*n) + 24*a^3*d*n*x^n + b^3*d*x^(4*n) + 4*a*b^2*d*x^(3*n) + 6*a ^2*b*d*x^(2*n) + 4*a^3*d*x^n)/(6*n^4 + 11*n^3 + 6*n^2 + n)
Time = 9.42 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.37 \[ \int \left (c+d x^{-1+n}\right ) \left (a+b x^n\right )^3 \, dx=a^3\,c\,x+\frac {a^3\,d\,x^n}{n}+\frac {b^3\,d\,x^{4\,n}}{4\,n}+\frac {b^3\,c\,x\,x^{3\,n}}{3\,n+1}+\frac {3\,a^2\,b\,d\,x^{2\,n}}{2\,n}+\frac {a\,b^2\,d\,x^{3\,n}}{n}+\frac {3\,a\,b^2\,c\,x\,x^{2\,n}}{2\,n+1}+\frac {3\,a^2\,b\,c\,x\,x^n}{n+1} \]