Integrand size = 11, antiderivative size = 64 \[ \int x^3 (a+b x)^{10} \, dx=-\frac {a^3 (a+b x)^{11}}{11 b^4}+\frac {a^2 (a+b x)^{12}}{4 b^4}-\frac {3 a (a+b x)^{13}}{13 b^4}+\frac {(a+b x)^{14}}{14 b^4} \] Output:
-1/11*a^3*(b*x+a)^11/b^4+1/4*a^2*(b*x+a)^12/b^4-3/13*a*(b*x+a)^13/b^4+1/14 *(b*x+a)^14/b^4
Time = 0.00 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.00 \[ \int x^3 (a+b x)^{10} \, dx=\frac {a^{10} x^4}{4}+2 a^9 b x^5+\frac {15}{2} a^8 b^2 x^6+\frac {120}{7} a^7 b^3 x^7+\frac {105}{4} a^6 b^4 x^8+28 a^5 b^5 x^9+21 a^4 b^6 x^{10}+\frac {120}{11} a^3 b^7 x^{11}+\frac {15}{4} a^2 b^8 x^{12}+\frac {10}{13} a b^9 x^{13}+\frac {b^{10} x^{14}}{14} \] Input:
Integrate[x^3*(a + b*x)^10,x]
Output:
(a^10*x^4)/4 + 2*a^9*b*x^5 + (15*a^8*b^2*x^6)/2 + (120*a^7*b^3*x^7)/7 + (1 05*a^6*b^4*x^8)/4 + 28*a^5*b^5*x^9 + 21*a^4*b^6*x^10 + (120*a^3*b^7*x^11)/ 11 + (15*a^2*b^8*x^12)/4 + (10*a*b^9*x^13)/13 + (b^10*x^14)/14
Time = 0.18 (sec) , antiderivative size = 64, 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 = {49, 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.
\(\displaystyle \int x^3 (a+b x)^{10} \, dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (-\frac {a^3 (a+b x)^{10}}{b^3}+\frac {3 a^2 (a+b x)^{11}}{b^3}+\frac {(a+b x)^{13}}{b^3}-\frac {3 a (a+b x)^{12}}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 (a+b x)^{11}}{11 b^4}+\frac {a^2 (a+b x)^{12}}{4 b^4}+\frac {(a+b x)^{14}}{14 b^4}-\frac {3 a (a+b x)^{13}}{13 b^4}\) |
Input:
Int[x^3*(a + b*x)^10,x]
Output:
-1/11*(a^3*(a + b*x)^11)/b^4 + (a^2*(a + b*x)^12)/(4*b^4) - (3*a*(a + b*x) ^13)/(13*b^4) + (a + b*x)^14/(14*b^4)
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.77
method | result | size |
gosper | \(\frac {1}{4} a^{10} x^{4}+2 a^{9} b \,x^{5}+\frac {15}{2} a^{8} b^{2} x^{6}+\frac {120}{7} a^{7} b^{3} x^{7}+\frac {105}{4} a^{6} b^{4} x^{8}+28 a^{5} b^{5} x^{9}+21 a^{4} b^{6} x^{10}+\frac {120}{11} a^{3} b^{7} x^{11}+\frac {15}{4} a^{2} b^{8} x^{12}+\frac {10}{13} a \,b^{9} x^{13}+\frac {1}{14} b^{10} x^{14}\) | \(113\) |
default | \(\frac {1}{4} a^{10} x^{4}+2 a^{9} b \,x^{5}+\frac {15}{2} a^{8} b^{2} x^{6}+\frac {120}{7} a^{7} b^{3} x^{7}+\frac {105}{4} a^{6} b^{4} x^{8}+28 a^{5} b^{5} x^{9}+21 a^{4} b^{6} x^{10}+\frac {120}{11} a^{3} b^{7} x^{11}+\frac {15}{4} a^{2} b^{8} x^{12}+\frac {10}{13} a \,b^{9} x^{13}+\frac {1}{14} b^{10} x^{14}\) | \(113\) |
norman | \(\frac {1}{4} a^{10} x^{4}+2 a^{9} b \,x^{5}+\frac {15}{2} a^{8} b^{2} x^{6}+\frac {120}{7} a^{7} b^{3} x^{7}+\frac {105}{4} a^{6} b^{4} x^{8}+28 a^{5} b^{5} x^{9}+21 a^{4} b^{6} x^{10}+\frac {120}{11} a^{3} b^{7} x^{11}+\frac {15}{4} a^{2} b^{8} x^{12}+\frac {10}{13} a \,b^{9} x^{13}+\frac {1}{14} b^{10} x^{14}\) | \(113\) |
risch | \(\frac {1}{4} a^{10} x^{4}+2 a^{9} b \,x^{5}+\frac {15}{2} a^{8} b^{2} x^{6}+\frac {120}{7} a^{7} b^{3} x^{7}+\frac {105}{4} a^{6} b^{4} x^{8}+28 a^{5} b^{5} x^{9}+21 a^{4} b^{6} x^{10}+\frac {120}{11} a^{3} b^{7} x^{11}+\frac {15}{4} a^{2} b^{8} x^{12}+\frac {10}{13} a \,b^{9} x^{13}+\frac {1}{14} b^{10} x^{14}\) | \(113\) |
parallelrisch | \(\frac {1}{4} a^{10} x^{4}+2 a^{9} b \,x^{5}+\frac {15}{2} a^{8} b^{2} x^{6}+\frac {120}{7} a^{7} b^{3} x^{7}+\frac {105}{4} a^{6} b^{4} x^{8}+28 a^{5} b^{5} x^{9}+21 a^{4} b^{6} x^{10}+\frac {120}{11} a^{3} b^{7} x^{11}+\frac {15}{4} a^{2} b^{8} x^{12}+\frac {10}{13} a \,b^{9} x^{13}+\frac {1}{14} b^{10} x^{14}\) | \(113\) |
orering | \(\frac {x^{4} \left (286 b^{10} x^{10}+3080 a \,b^{9} x^{9}+15015 a^{2} b^{8} x^{8}+43680 a^{3} b^{7} x^{7}+84084 a^{4} b^{6} x^{6}+112112 a^{5} b^{5} x^{5}+105105 a^{6} b^{4} x^{4}+68640 a^{7} b^{3} x^{3}+30030 a^{8} b^{2} x^{2}+8008 a^{9} b x +1001 a^{10}\right )}{4004}\) | \(113\) |
Input:
int(x^3*(b*x+a)^10,x,method=_RETURNVERBOSE)
Output:
1/4*a^10*x^4+2*a^9*b*x^5+15/2*a^8*b^2*x^6+120/7*a^7*b^3*x^7+105/4*a^6*b^4* x^8+28*a^5*b^5*x^9+21*a^4*b^6*x^10+120/11*a^3*b^7*x^11+15/4*a^2*b^8*x^12+1 0/13*a*b^9*x^13+1/14*b^10*x^14
Time = 0.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75 \[ \int x^3 (a+b x)^{10} \, dx=\frac {1}{14} \, b^{10} x^{14} + \frac {10}{13} \, a b^{9} x^{13} + \frac {15}{4} \, a^{2} b^{8} x^{12} + \frac {120}{11} \, a^{3} b^{7} x^{11} + 21 \, a^{4} b^{6} x^{10} + 28 \, a^{5} b^{5} x^{9} + \frac {105}{4} \, a^{6} b^{4} x^{8} + \frac {120}{7} \, a^{7} b^{3} x^{7} + \frac {15}{2} \, a^{8} b^{2} x^{6} + 2 \, a^{9} b x^{5} + \frac {1}{4} \, a^{10} x^{4} \] Input:
integrate(x^3*(b*x+a)^10,x, algorithm="fricas")
Output:
1/14*b^10*x^14 + 10/13*a*b^9*x^13 + 15/4*a^2*b^8*x^12 + 120/11*a^3*b^7*x^1 1 + 21*a^4*b^6*x^10 + 28*a^5*b^5*x^9 + 105/4*a^6*b^4*x^8 + 120/7*a^7*b^3*x ^7 + 15/2*a^8*b^2*x^6 + 2*a^9*b*x^5 + 1/4*a^10*x^4
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (56) = 112\).
Time = 0.03 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.02 \[ \int x^3 (a+b x)^{10} \, dx=\frac {a^{10} x^{4}}{4} + 2 a^{9} b x^{5} + \frac {15 a^{8} b^{2} x^{6}}{2} + \frac {120 a^{7} b^{3} x^{7}}{7} + \frac {105 a^{6} b^{4} x^{8}}{4} + 28 a^{5} b^{5} x^{9} + 21 a^{4} b^{6} x^{10} + \frac {120 a^{3} b^{7} x^{11}}{11} + \frac {15 a^{2} b^{8} x^{12}}{4} + \frac {10 a b^{9} x^{13}}{13} + \frac {b^{10} x^{14}}{14} \] Input:
integrate(x**3*(b*x+a)**10,x)
Output:
a**10*x**4/4 + 2*a**9*b*x**5 + 15*a**8*b**2*x**6/2 + 120*a**7*b**3*x**7/7 + 105*a**6*b**4*x**8/4 + 28*a**5*b**5*x**9 + 21*a**4*b**6*x**10 + 120*a**3 *b**7*x**11/11 + 15*a**2*b**8*x**12/4 + 10*a*b**9*x**13/13 + b**10*x**14/1 4
Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75 \[ \int x^3 (a+b x)^{10} \, dx=\frac {1}{14} \, b^{10} x^{14} + \frac {10}{13} \, a b^{9} x^{13} + \frac {15}{4} \, a^{2} b^{8} x^{12} + \frac {120}{11} \, a^{3} b^{7} x^{11} + 21 \, a^{4} b^{6} x^{10} + 28 \, a^{5} b^{5} x^{9} + \frac {105}{4} \, a^{6} b^{4} x^{8} + \frac {120}{7} \, a^{7} b^{3} x^{7} + \frac {15}{2} \, a^{8} b^{2} x^{6} + 2 \, a^{9} b x^{5} + \frac {1}{4} \, a^{10} x^{4} \] Input:
integrate(x^3*(b*x+a)^10,x, algorithm="maxima")
Output:
1/14*b^10*x^14 + 10/13*a*b^9*x^13 + 15/4*a^2*b^8*x^12 + 120/11*a^3*b^7*x^1 1 + 21*a^4*b^6*x^10 + 28*a^5*b^5*x^9 + 105/4*a^6*b^4*x^8 + 120/7*a^7*b^3*x ^7 + 15/2*a^8*b^2*x^6 + 2*a^9*b*x^5 + 1/4*a^10*x^4
Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75 \[ \int x^3 (a+b x)^{10} \, dx=\frac {1}{14} \, b^{10} x^{14} + \frac {10}{13} \, a b^{9} x^{13} + \frac {15}{4} \, a^{2} b^{8} x^{12} + \frac {120}{11} \, a^{3} b^{7} x^{11} + 21 \, a^{4} b^{6} x^{10} + 28 \, a^{5} b^{5} x^{9} + \frac {105}{4} \, a^{6} b^{4} x^{8} + \frac {120}{7} \, a^{7} b^{3} x^{7} + \frac {15}{2} \, a^{8} b^{2} x^{6} + 2 \, a^{9} b x^{5} + \frac {1}{4} \, a^{10} x^{4} \] Input:
integrate(x^3*(b*x+a)^10,x, algorithm="giac")
Output:
1/14*b^10*x^14 + 10/13*a*b^9*x^13 + 15/4*a^2*b^8*x^12 + 120/11*a^3*b^7*x^1 1 + 21*a^4*b^6*x^10 + 28*a^5*b^5*x^9 + 105/4*a^6*b^4*x^8 + 120/7*a^7*b^3*x ^7 + 15/2*a^8*b^2*x^6 + 2*a^9*b*x^5 + 1/4*a^10*x^4
Time = 0.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75 \[ \int x^3 (a+b x)^{10} \, dx=\frac {a^{10}\,x^4}{4}+2\,a^9\,b\,x^5+\frac {15\,a^8\,b^2\,x^6}{2}+\frac {120\,a^7\,b^3\,x^7}{7}+\frac {105\,a^6\,b^4\,x^8}{4}+28\,a^5\,b^5\,x^9+21\,a^4\,b^6\,x^{10}+\frac {120\,a^3\,b^7\,x^{11}}{11}+\frac {15\,a^2\,b^8\,x^{12}}{4}+\frac {10\,a\,b^9\,x^{13}}{13}+\frac {b^{10}\,x^{14}}{14} \] Input:
int(x^3*(a + b*x)^10,x)
Output:
(a^10*x^4)/4 + (b^10*x^14)/14 + 2*a^9*b*x^5 + (10*a*b^9*x^13)/13 + (15*a^8 *b^2*x^6)/2 + (120*a^7*b^3*x^7)/7 + (105*a^6*b^4*x^8)/4 + 28*a^5*b^5*x^9 + 21*a^4*b^6*x^10 + (120*a^3*b^7*x^11)/11 + (15*a^2*b^8*x^12)/4
Time = 0.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75 \[ \int x^3 (a+b x)^{10} \, dx=\frac {x^{4} \left (286 b^{10} x^{10}+3080 a \,b^{9} x^{9}+15015 a^{2} b^{8} x^{8}+43680 a^{3} b^{7} x^{7}+84084 a^{4} b^{6} x^{6}+112112 a^{5} b^{5} x^{5}+105105 a^{6} b^{4} x^{4}+68640 a^{7} b^{3} x^{3}+30030 a^{8} b^{2} x^{2}+8008 a^{9} b x +1001 a^{10}\right )}{4004} \] Input:
int(x^3*(b*x+a)^10,x)
Output:
(x**4*(1001*a**10 + 8008*a**9*b*x + 30030*a**8*b**2*x**2 + 68640*a**7*b**3 *x**3 + 105105*a**6*b**4*x**4 + 112112*a**5*b**5*x**5 + 84084*a**4*b**6*x* *6 + 43680*a**3*b**7*x**7 + 15015*a**2*b**8*x**8 + 3080*a*b**9*x**9 + 286* b**10*x**10))/4004