Integrand size = 11, antiderivative size = 83 \[ \int x^3 (a+b x)^n \, dx=-\frac {a^3 (a+b x)^{1+n}}{b^4 (1+n)}+\frac {3 a^2 (a+b x)^{2+n}}{b^4 (2+n)}-\frac {3 a (a+b x)^{3+n}}{b^4 (3+n)}+\frac {(a+b x)^{4+n}}{b^4 (4+n)} \]
-a^3*(b*x+a)^(1+n)/b^4/(1+n)+3*a^2*(b*x+a)^(2+n)/b^4/(2+n)-3*a*(b*x+a)^(3+ n)/b^4/(3+n)+(b*x+a)^(4+n)/b^4/(4+n)
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int x^3 (a+b x)^n \, dx=\frac {(a+b x)^{1+n} \left (-\frac {a^3}{1+n}+\frac {3 a^2 (a+b x)}{2+n}-\frac {3 a (a+b x)^2}{3+n}+\frac {(a+b x)^3}{4+n}\right )}{b^4} \]
((a + b*x)^(1 + n)*(-(a^3/(1 + n)) + (3*a^2*(a + b*x))/(2 + n) - (3*a*(a + b*x)^2)/(3 + n) + (a + b*x)^3/(4 + n)))/b^4
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {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.
\(\displaystyle \int x^3 (a+b x)^n \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {a^3 (a+b x)^n}{b^3}+\frac {3 a^2 (a+b x)^{n+1}}{b^3}-\frac {3 a (a+b x)^{n+2}}{b^3}+\frac {(a+b x)^{n+3}}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 (a+b x)^{n+1}}{b^4 (n+1)}+\frac {3 a^2 (a+b x)^{n+2}}{b^4 (n+2)}-\frac {3 a (a+b x)^{n+3}}{b^4 (n+3)}+\frac {(a+b x)^{n+4}}{b^4 (n+4)}\) |
-((a^3*(a + b*x)^(1 + n))/(b^4*(1 + n))) + (3*a^2*(a + b*x)^(2 + n))/(b^4* (2 + n)) - (3*a*(a + b*x)^(3 + n))/(b^4*(3 + n)) + (a + b*x)^(4 + n)/(b^4* (4 + n))
3.8.32.3.1 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])
Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.52
method | result | size |
gosper | \(-\frac {\left (b x +a \right )^{1+n} \left (-b^{3} n^{3} x^{3}-6 b^{3} n^{2} x^{3}+3 a \,b^{2} n^{2} x^{2}-11 b^{3} n \,x^{3}+9 a \,b^{2} n \,x^{2}-6 b^{3} x^{3}-6 a^{2} b n x +6 a \,b^{2} x^{2}-6 a^{2} b x +6 a^{3}\right )}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\) | \(126\) |
risch | \(-\frac {\left (-b^{4} n^{3} x^{4}-a \,b^{3} n^{3} x^{3}-6 b^{4} n^{2} x^{4}-3 a \,b^{3} n^{2} x^{3}-11 b^{4} n \,x^{4}+3 a^{2} b^{2} n^{2} x^{2}-2 x^{3} a n \,b^{3}-6 b^{4} x^{4}+3 a^{2} n \,x^{2} b^{2}-6 x n \,a^{3} b +6 a^{4}\right ) \left (b x +a \right )^{n}}{\left (3+n \right ) \left (4+n \right ) \left (2+n \right ) \left (1+n \right ) b^{4}}\) | \(146\) |
norman | \(\frac {x^{4} {\mathrm e}^{n \ln \left (b x +a \right )}}{4+n}+\frac {a n \,x^{3} {\mathrm e}^{n \ln \left (b x +a \right )}}{b \left (n^{2}+7 n +12\right )}-\frac {6 a^{4} {\mathrm e}^{n \ln \left (b x +a \right )}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {3 a^{2} n \,x^{2} {\mathrm e}^{n \ln \left (b x +a \right )}}{b^{2} \left (n^{3}+9 n^{2}+26 n +24\right )}+\frac {6 n \,a^{3} x \,{\mathrm e}^{n \ln \left (b x +a \right )}}{b^{3} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\) | \(160\) |
parallelrisch | \(\frac {x^{4} \left (b x +a \right )^{n} b^{4} n^{3}+6 x^{4} \left (b x +a \right )^{n} b^{4} n^{2}+x^{3} \left (b x +a \right )^{n} a \,b^{3} n^{3}+11 x^{4} \left (b x +a \right )^{n} b^{4} n +3 x^{3} \left (b x +a \right )^{n} a \,b^{3} n^{2}+6 x^{4} \left (b x +a \right )^{n} b^{4}+2 x^{3} \left (b x +a \right )^{n} a \,b^{3} n -3 x^{2} \left (b x +a \right )^{n} a^{2} b^{2} n^{2}-3 x^{2} \left (b x +a \right )^{n} a^{2} b^{2} n +6 x \left (b x +a \right )^{n} a^{3} b n -6 \left (b x +a \right )^{n} a^{4}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\) | \(213\) |
-1/b^4*(b*x+a)^(1+n)/(n^4+10*n^3+35*n^2+50*n+24)*(-b^3*n^3*x^3-6*b^3*n^2*x ^3+3*a*b^2*n^2*x^2-11*b^3*n*x^3+9*a*b^2*n*x^2-6*b^3*x^3-6*a^2*b*n*x+6*a*b^ 2*x^2-6*a^2*b*x+6*a^3)
Time = 0.24 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.72 \[ \int x^3 (a+b x)^n \, dx=\frac {{\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} {\left (b x + a\right )}^{n}}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \]
(6*a^3*b*n*x + (b^4*n^3 + 6*b^4*n^2 + 11*b^4*n + 6*b^4)*x^4 - 6*a^4 + (a*b ^3*n^3 + 3*a*b^3*n^2 + 2*a*b^3*n)*x^3 - 3*(a^2*b^2*n^2 + a^2*b^2*n)*x^2)*( b*x + a)^n/(b^4*n^4 + 10*b^4*n^3 + 35*b^4*n^2 + 50*b^4*n + 24*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 1318 vs. \(2 (71) = 142\).
Time = 1.84 (sec) , antiderivative size = 1318, normalized size of antiderivative = 15.88 \[ \int x^3 (a+b x)^n \, dx=\text {Too large to display} \]
Piecewise((a**n*x**4/4, Eq(b, 0)), (6*a**3*log(a/b + x)/(6*a**3*b**4 + 18* a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 11*a**3/(6*a**3*b**4 + 18*a* *2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 18*a**2*b*x*log(a/b + x)/(6*a* *3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 27*a**2*b*x/(6* a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 18*a*b**2*x** 2*log(a/b + x)/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x** 3) + 18*a*b**2*x**2/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b** 7*x**3) + 6*b**3*x**3*log(a/b + x)/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b* *6*x**2 + 6*b**7*x**3), Eq(n, -4)), (-6*a**3*log(a/b + x)/(2*a**2*b**4 + 4 *a*b**5*x + 2*b**6*x**2) - 9*a**3/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) - 12*a**2*b*x*log(a/b + x)/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) - 12* a**2*b*x/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) - 6*a*b**2*x**2*log(a/b + x)/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) + 2*b**3*x**3/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2), Eq(n, -3)), (6*a**3*log(a/b + x)/(2*a*b**4 + 2 *b**5*x) + 6*a**3/(2*a*b**4 + 2*b**5*x) + 6*a**2*b*x*log(a/b + x)/(2*a*b** 4 + 2*b**5*x) - 3*a*b**2*x**2/(2*a*b**4 + 2*b**5*x) + b**3*x**3/(2*a*b**4 + 2*b**5*x), Eq(n, -2)), (-a**3*log(a/b + x)/b**4 + a**2*x/b**3 - a*x**2/( 2*b**2) + x**3/(3*b), Eq(n, -1)), (-6*a**4*(a + b*x)**n/(b**4*n**4 + 10*b* *4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 6*a**3*b*n*x*(a + b*x)**n/ (b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) - 3*a**...
Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.22 \[ \int x^3 (a+b x)^n \, dx=\frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} + {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} x^{3} - 3 \, {\left (n^{2} + n\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b n x - 6 \, a^{4}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4}} \]
((n^3 + 6*n^2 + 11*n + 6)*b^4*x^4 + (n^3 + 3*n^2 + 2*n)*a*b^3*x^3 - 3*(n^2 + n)*a^2*b^2*x^2 + 6*a^3*b*n*x - 6*a^4)*(b*x + a)^n/((n^4 + 10*n^3 + 35*n ^2 + 50*n + 24)*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (83) = 166\).
Time = 0.29 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.72 \[ \int x^3 (a+b x)^n \, dx=\frac {{\left (b x + a\right )}^{n} b^{4} n^{3} x^{4} + {\left (b x + a\right )}^{n} a b^{3} n^{3} x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} n^{2} x^{4} + 3 \, {\left (b x + a\right )}^{n} a b^{3} n^{2} x^{3} + 11 \, {\left (b x + a\right )}^{n} b^{4} n x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n^{2} x^{2} + 2 \, {\left (b x + a\right )}^{n} a b^{3} n x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n x^{2} + 6 \, {\left (b x + a\right )}^{n} a^{3} b n x - 6 \, {\left (b x + a\right )}^{n} a^{4}}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \]
((b*x + a)^n*b^4*n^3*x^4 + (b*x + a)^n*a*b^3*n^3*x^3 + 6*(b*x + a)^n*b^4*n ^2*x^4 + 3*(b*x + a)^n*a*b^3*n^2*x^3 + 11*(b*x + a)^n*b^4*n*x^4 - 3*(b*x + a)^n*a^2*b^2*n^2*x^2 + 2*(b*x + a)^n*a*b^3*n*x^3 + 6*(b*x + a)^n*b^4*x^4 - 3*(b*x + a)^n*a^2*b^2*n*x^2 + 6*(b*x + a)^n*a^3*b*n*x - 6*(b*x + a)^n*a^ 4)/(b^4*n^4 + 10*b^4*n^3 + 35*b^4*n^2 + 50*b^4*n + 24*b^4)
Time = 0.59 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.12 \[ \int x^3 (a+b x)^n \, dx={\left (a+b\,x\right )}^n\,\left (\frac {x^4\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}-\frac {6\,a^4}{b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {6\,a^3\,n\,x}{b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {a\,n\,x^3\,\left (n^2+3\,n+2\right )}{b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {3\,a^2\,n\,x^2\,\left (n+1\right )}{b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right ) \]
(a + b*x)^n*((x^4*(11*n + 6*n^2 + n^3 + 6))/(50*n + 35*n^2 + 10*n^3 + n^4 + 24) - (6*a^4)/(b^4*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (6*a^3*n*x)/(b ^3*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (a*n*x^3*(3*n + n^2 + 2))/(b*(50 *n + 35*n^2 + 10*n^3 + n^4 + 24)) - (3*a^2*n*x^2*(n + 1))/(b^2*(50*n + 35* n^2 + 10*n^3 + n^4 + 24)))