Integrand size = 30, antiderivative size = 88 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {3}{4} a^2 b c x^4+\frac {3}{5} a^2 b d x^5+\frac {3}{7} a b^2 c x^7+\frac {3}{8} a b^2 d x^8+\frac {1}{10} b^3 c x^{10}+\frac {1}{11} b^3 d x^{11} \] Output:
a^3*c*x+1/2*a^3*d*x^2+3/4*a^2*b*c*x^4+3/5*a^2*b*d*x^5+3/7*a*b^2*c*x^7+3/8* a*b^2*d*x^8+1/10*b^3*c*x^10+1/11*b^3*d*x^11
Time = 0.00 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {3}{4} a^2 b c x^4+\frac {3}{5} a^2 b d x^5+\frac {3}{7} a b^2 c x^7+\frac {3}{8} a b^2 d x^8+\frac {1}{10} b^3 c x^{10}+\frac {1}{11} b^3 d x^{11} \] Input:
Integrate[(a + b*x^3)^2*(a*c + a*d*x + b*c*x^3 + b*d*x^4),x]
Output:
a^3*c*x + (a^3*d*x^2)/2 + (3*a^2*b*c*x^4)/4 + (3*a^2*b*d*x^5)/5 + (3*a*b^2 *c*x^7)/7 + (3*a*b^2*d*x^8)/8 + (b^3*c*x^10)/10 + (b^3*d*x^11)/11
Time = 0.43 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2389, 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.
\(\displaystyle \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (a^3 c+a^3 d x+3 a^2 b c x^3+3 a^2 b d x^4+3 a b^2 c x^6+3 a b^2 d x^7+b^3 c x^9+b^3 d x^{10}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^3 c x+\frac {1}{2} a^3 d x^2+\frac {3}{4} a^2 b c x^4+\frac {3}{5} a^2 b d x^5+\frac {3}{7} a b^2 c x^7+\frac {3}{8} a b^2 d x^8+\frac {1}{10} b^3 c x^{10}+\frac {1}{11} b^3 d x^{11}\) |
Input:
Int[(a + b*x^3)^2*(a*c + a*d*x + b*c*x^3 + b*d*x^4),x]
Output:
a^3*c*x + (a^3*d*x^2)/2 + (3*a^2*b*c*x^4)/4 + (3*a^2*b*d*x^5)/5 + (3*a*b^2 *c*x^7)/7 + (3*a*b^2*d*x^8)/8 + (b^3*c*x^10)/10 + (b^3*d*x^11)/11
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.85
method | result | size |
default | \(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {3}{5} a^{2} b d \,x^{5}+\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {3}{8} a \,b^{2} d \,x^{8}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{11} b^{3} d \,x^{11}\) | \(75\) |
norman | \(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {3}{5} a^{2} b d \,x^{5}+\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {3}{8} a \,b^{2} d \,x^{8}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{11} b^{3} d \,x^{11}\) | \(75\) |
risch | \(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {3}{5} a^{2} b d \,x^{5}+\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {3}{8} a \,b^{2} d \,x^{8}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{11} b^{3} d \,x^{11}\) | \(75\) |
parallelrisch | \(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {3}{4} a^{2} b c \,x^{4}+\frac {3}{5} a^{2} b d \,x^{5}+\frac {3}{7} a \,b^{2} c \,x^{7}+\frac {3}{8} a \,b^{2} d \,x^{8}+\frac {1}{10} b^{3} c \,x^{10}+\frac {1}{11} b^{3} d \,x^{11}\) | \(75\) |
gosper | \(\frac {x \left (280 b^{3} d \,x^{10}+308 b^{3} c \,x^{9}+1155 a \,b^{2} d \,x^{7}+1320 a \,b^{2} c \,x^{6}+1848 a^{2} b d \,x^{4}+2310 a^{2} b c \,x^{3}+1540 a^{3} d x +3080 c \,a^{3}\right )}{3080}\) | \(76\) |
orering | \(\frac {x \left (280 b^{3} d \,x^{10}+308 b^{3} c \,x^{9}+1155 a \,b^{2} d \,x^{7}+1320 a \,b^{2} c \,x^{6}+1848 a^{2} b d \,x^{4}+2310 a^{2} b c \,x^{3}+1540 a^{3} d x +3080 c \,a^{3}\right ) \left (d b \,x^{4}+b c \,x^{3}+a d x +a c \right )}{3080 \left (d x +c \right ) \left (b \,x^{3}+a \right )}\) | \(112\) |
Input:
int((b*x^3+a)^2*(b*d*x^4+b*c*x^3+a*d*x+a*c),x,method=_RETURNVERBOSE)
Output:
a^3*c*x+1/2*a^3*d*x^2+3/4*a^2*b*c*x^4+3/5*a^2*b*d*x^5+3/7*a*b^2*c*x^7+3/8* a*b^2*d*x^8+1/10*b^3*c*x^10+1/11*b^3*d*x^11
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=\frac {1}{11} \, b^{3} d x^{11} + \frac {1}{10} \, b^{3} c x^{10} + \frac {3}{8} \, a b^{2} d x^{8} + \frac {3}{7} \, a b^{2} c x^{7} + \frac {3}{5} \, a^{2} b d x^{5} + \frac {3}{4} \, a^{2} b c x^{4} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \] Input:
integrate((b*x^3+a)^2*(b*d*x^4+b*c*x^3+a*d*x+a*c),x, algorithm="fricas")
Output:
1/11*b^3*d*x^11 + 1/10*b^3*c*x^10 + 3/8*a*b^2*d*x^8 + 3/7*a*b^2*c*x^7 + 3/ 5*a^2*b*d*x^5 + 3/4*a^2*b*c*x^4 + 1/2*a^3*d*x^2 + a^3*c*x
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=a^{3} c x + \frac {a^{3} d x^{2}}{2} + \frac {3 a^{2} b c x^{4}}{4} + \frac {3 a^{2} b d x^{5}}{5} + \frac {3 a b^{2} c x^{7}}{7} + \frac {3 a b^{2} d x^{8}}{8} + \frac {b^{3} c x^{10}}{10} + \frac {b^{3} d x^{11}}{11} \] Input:
integrate((b*x**3+a)**2*(b*d*x**4+b*c*x**3+a*d*x+a*c),x)
Output:
a**3*c*x + a**3*d*x**2/2 + 3*a**2*b*c*x**4/4 + 3*a**2*b*d*x**5/5 + 3*a*b** 2*c*x**7/7 + 3*a*b**2*d*x**8/8 + b**3*c*x**10/10 + b**3*d*x**11/11
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=\frac {1}{11} \, b^{3} d x^{11} + \frac {1}{10} \, b^{3} c x^{10} + \frac {3}{8} \, a b^{2} d x^{8} + \frac {3}{7} \, a b^{2} c x^{7} + \frac {3}{5} \, a^{2} b d x^{5} + \frac {3}{4} \, a^{2} b c x^{4} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \] Input:
integrate((b*x^3+a)^2*(b*d*x^4+b*c*x^3+a*d*x+a*c),x, algorithm="maxima")
Output:
1/11*b^3*d*x^11 + 1/10*b^3*c*x^10 + 3/8*a*b^2*d*x^8 + 3/7*a*b^2*c*x^7 + 3/ 5*a^2*b*d*x^5 + 3/4*a^2*b*c*x^4 + 1/2*a^3*d*x^2 + a^3*c*x
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=\frac {1}{11} \, b^{3} d x^{11} + \frac {1}{10} \, b^{3} c x^{10} + \frac {3}{8} \, a b^{2} d x^{8} + \frac {3}{7} \, a b^{2} c x^{7} + \frac {3}{5} \, a^{2} b d x^{5} + \frac {3}{4} \, a^{2} b c x^{4} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \] Input:
integrate((b*x^3+a)^2*(b*d*x^4+b*c*x^3+a*d*x+a*c),x, algorithm="giac")
Output:
1/11*b^3*d*x^11 + 1/10*b^3*c*x^10 + 3/8*a*b^2*d*x^8 + 3/7*a*b^2*c*x^7 + 3/ 5*a^2*b*d*x^5 + 3/4*a^2*b*c*x^4 + 1/2*a^3*d*x^2 + a^3*c*x
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=\frac {d\,a^3\,x^2}{2}+c\,a^3\,x+\frac {3\,d\,a^2\,b\,x^5}{5}+\frac {3\,c\,a^2\,b\,x^4}{4}+\frac {3\,d\,a\,b^2\,x^8}{8}+\frac {3\,c\,a\,b^2\,x^7}{7}+\frac {d\,b^3\,x^{11}}{11}+\frac {c\,b^3\,x^{10}}{10} \] Input:
int((a + b*x^3)^2*(a*c + a*d*x + b*c*x^3 + b*d*x^4),x)
Output:
(a^3*d*x^2)/2 + (b^3*c*x^10)/10 + (b^3*d*x^11)/11 + a^3*c*x + (3*a^2*b*c*x ^4)/4 + (3*a*b^2*c*x^7)/7 + (3*a^2*b*d*x^5)/5 + (3*a*b^2*d*x^8)/8
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.85 \[ \int \left (a+b x^3\right )^2 \left (a c+a d x+b c x^3+b d x^4\right ) \, dx=\frac {x \left (280 b^{3} d \,x^{10}+308 b^{3} c \,x^{9}+1155 a \,b^{2} d \,x^{7}+1320 a \,b^{2} c \,x^{6}+1848 a^{2} b d \,x^{4}+2310 a^{2} b c \,x^{3}+1540 a^{3} d x +3080 a^{3} c \right )}{3080} \] Input:
int((b*x^3+a)^2*(b*d*x^4+b*c*x^3+a*d*x+a*c),x)
Output:
(x*(3080*a**3*c + 1540*a**3*d*x + 2310*a**2*b*c*x**3 + 1848*a**2*b*d*x**4 + 1320*a*b**2*c*x**6 + 1155*a*b**2*d*x**7 + 308*b**3*c*x**9 + 280*b**3*d*x **10))/3080