Integrand size = 28, antiderivative size = 114 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {1}{5} a^2 d x^5+\frac {1}{6} a^2 e x^6+\frac {1}{7} a^2 f x^7+\frac {2}{9} a b d x^9+\frac {1}{5} a b e x^{10}+\frac {2}{11} a b f x^{11}+\frac {1}{13} b^2 d x^{13}+\frac {1}{14} b^2 e x^{14}+\frac {1}{15} b^2 f x^{15}+\frac {c \left (a+b x^4\right )^3}{12 b} \] Output:
1/5*a^2*d*x^5+1/6*a^2*e*x^6+1/7*a^2*f*x^7+2/9*a*b*d*x^9+1/5*a*b*e*x^10+2/1 1*a*b*f*x^11+1/13*b^2*d*x^13+1/14*b^2*e*x^14+1/15*b^2*f*x^15+1/12*c*(b*x^4 +a)^3/b
Time = 0.01 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.13 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {1}{4} a^2 c x^4+\frac {1}{5} a^2 d x^5+\frac {1}{6} a^2 e x^6+\frac {1}{7} a^2 f x^7+\frac {1}{4} a b c x^8+\frac {2}{9} a b d x^9+\frac {1}{5} a b e x^{10}+\frac {2}{11} a b f x^{11}+\frac {1}{12} b^2 c x^{12}+\frac {1}{13} b^2 d x^{13}+\frac {1}{14} b^2 e x^{14}+\frac {1}{15} b^2 f x^{15} \] Input:
Integrate[x^3*(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^2,x]
Output:
(a^2*c*x^4)/4 + (a^2*d*x^5)/5 + (a^2*e*x^6)/6 + (a^2*f*x^7)/7 + (a*b*c*x^8 )/4 + (2*a*b*d*x^9)/9 + (a*b*e*x^10)/5 + (2*a*b*f*x^11)/11 + (b^2*c*x^12)/ 12 + (b^2*d*x^13)/13 + (b^2*e*x^14)/14 + (b^2*f*x^15)/15
Time = 0.51 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2017, 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.
\(\displaystyle \int x^3 \left (a+b x^4\right )^2 \left (c+d x+e x^2+f x^3\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (b x^4+a\right )^2 \left (x^3 \left (f x^3+e x^2+d x+c\right )-c x^3\right )dx+\frac {c \left (a+b x^4\right )^3}{12 b}\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (b^2 f x^{14}+b^2 e x^{13}+b^2 d x^{12}+2 a b f x^{10}+2 a b e x^9+2 a b d x^8+a^2 f x^6+a^2 e x^5+a^2 d x^4\right )dx+\frac {c \left (a+b x^4\right )^3}{12 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} a^2 d x^5+\frac {1}{6} a^2 e x^6+\frac {1}{7} a^2 f x^7+\frac {c \left (a+b x^4\right )^3}{12 b}+\frac {2}{9} a b d x^9+\frac {1}{5} a b e x^{10}+\frac {2}{11} a b f x^{11}+\frac {1}{13} b^2 d x^{13}+\frac {1}{14} b^2 e x^{14}+\frac {1}{15} b^2 f x^{15}\) |
Input:
Int[x^3*(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^2,x]
Output:
(a^2*d*x^5)/5 + (a^2*e*x^6)/6 + (a^2*f*x^7)/7 + (2*a*b*d*x^9)/9 + (a*b*e*x ^10)/5 + (2*a*b*f*x^11)/11 + (b^2*d*x^13)/13 + (b^2*e*x^14)/14 + (b^2*f*x^ 15)/15 + (c*(a + b*x^4)^3)/(12*b)
Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coeff[Px, x, n - 1] *x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && IGtQ[p , 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] && !MatchQ[Px, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ [{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Coeff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]
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 = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.93
method | result | size |
gosper | \(\frac {1}{4} a^{2} c \,x^{4}+\frac {1}{5} a^{2} d \,x^{5}+\frac {1}{6} a^{2} e \,x^{6}+\frac {1}{7} a^{2} f \,x^{7}+\frac {1}{4} a b c \,x^{8}+\frac {2}{9} a b d \,x^{9}+\frac {1}{5} a b e \,x^{10}+\frac {2}{11} a b f \,x^{11}+\frac {1}{12} b^{2} c \,x^{12}+\frac {1}{13} b^{2} d \,x^{13}+\frac {1}{14} b^{2} e \,x^{14}+\frac {1}{15} b^{2} f \,x^{15}\) | \(106\) |
default | \(\frac {1}{4} a^{2} c \,x^{4}+\frac {1}{5} a^{2} d \,x^{5}+\frac {1}{6} a^{2} e \,x^{6}+\frac {1}{7} a^{2} f \,x^{7}+\frac {1}{4} a b c \,x^{8}+\frac {2}{9} a b d \,x^{9}+\frac {1}{5} a b e \,x^{10}+\frac {2}{11} a b f \,x^{11}+\frac {1}{12} b^{2} c \,x^{12}+\frac {1}{13} b^{2} d \,x^{13}+\frac {1}{14} b^{2} e \,x^{14}+\frac {1}{15} b^{2} f \,x^{15}\) | \(106\) |
norman | \(\frac {1}{4} a^{2} c \,x^{4}+\frac {1}{5} a^{2} d \,x^{5}+\frac {1}{6} a^{2} e \,x^{6}+\frac {1}{7} a^{2} f \,x^{7}+\frac {1}{4} a b c \,x^{8}+\frac {2}{9} a b d \,x^{9}+\frac {1}{5} a b e \,x^{10}+\frac {2}{11} a b f \,x^{11}+\frac {1}{12} b^{2} c \,x^{12}+\frac {1}{13} b^{2} d \,x^{13}+\frac {1}{14} b^{2} e \,x^{14}+\frac {1}{15} b^{2} f \,x^{15}\) | \(106\) |
risch | \(\frac {1}{4} a^{2} c \,x^{4}+\frac {1}{5} a^{2} d \,x^{5}+\frac {1}{6} a^{2} e \,x^{6}+\frac {1}{7} a^{2} f \,x^{7}+\frac {1}{4} a b c \,x^{8}+\frac {2}{9} a b d \,x^{9}+\frac {1}{5} a b e \,x^{10}+\frac {2}{11} a b f \,x^{11}+\frac {1}{12} b^{2} c \,x^{12}+\frac {1}{13} b^{2} d \,x^{13}+\frac {1}{14} b^{2} e \,x^{14}+\frac {1}{15} b^{2} f \,x^{15}\) | \(106\) |
parallelrisch | \(\frac {1}{4} a^{2} c \,x^{4}+\frac {1}{5} a^{2} d \,x^{5}+\frac {1}{6} a^{2} e \,x^{6}+\frac {1}{7} a^{2} f \,x^{7}+\frac {1}{4} a b c \,x^{8}+\frac {2}{9} a b d \,x^{9}+\frac {1}{5} a b e \,x^{10}+\frac {2}{11} a b f \,x^{11}+\frac {1}{12} b^{2} c \,x^{12}+\frac {1}{13} b^{2} d \,x^{13}+\frac {1}{14} b^{2} e \,x^{14}+\frac {1}{15} b^{2} f \,x^{15}\) | \(106\) |
orering | \(\frac {x^{4} \left (12012 b^{2} f \,x^{11}+12870 e \,b^{2} x^{10}+13860 b^{2} d \,x^{9}+15015 b^{2} c \,x^{8}+32760 x^{7} a b f +36036 a b e \,x^{6}+40040 x^{5} a d b +45045 a b c \,x^{4}+25740 a^{2} f \,x^{3}+30030 a^{2} e \,x^{2}+36036 a^{2} d x +45045 a^{2} c \right )}{180180}\) | \(106\) |
Input:
int(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^2,x,method=_RETURNVERBOSE)
Output:
1/4*a^2*c*x^4+1/5*a^2*d*x^5+1/6*a^2*e*x^6+1/7*a^2*f*x^7+1/4*a*b*c*x^8+2/9* a*b*d*x^9+1/5*a*b*e*x^10+2/11*a*b*f*x^11+1/12*b^2*c*x^12+1/13*b^2*d*x^13+1 /14*b^2*e*x^14+1/15*b^2*f*x^15
Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {1}{15} \, b^{2} f x^{15} + \frac {1}{14} \, b^{2} e x^{14} + \frac {1}{13} \, b^{2} d x^{13} + \frac {1}{12} \, b^{2} c x^{12} + \frac {2}{11} \, a b f x^{11} + \frac {1}{5} \, a b e x^{10} + \frac {2}{9} \, a b d x^{9} + \frac {1}{4} \, a b c x^{8} + \frac {1}{7} \, a^{2} f x^{7} + \frac {1}{6} \, a^{2} e x^{6} + \frac {1}{5} \, a^{2} d x^{5} + \frac {1}{4} \, a^{2} c x^{4} \] Input:
integrate(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^2,x, algorithm="fricas")
Output:
1/15*b^2*f*x^15 + 1/14*b^2*e*x^14 + 1/13*b^2*d*x^13 + 1/12*b^2*c*x^12 + 2/ 11*a*b*f*x^11 + 1/5*a*b*e*x^10 + 2/9*a*b*d*x^9 + 1/4*a*b*c*x^8 + 1/7*a^2*f *x^7 + 1/6*a^2*e*x^6 + 1/5*a^2*d*x^5 + 1/4*a^2*c*x^4
Time = 0.02 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.09 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {a^{2} c x^{4}}{4} + \frac {a^{2} d x^{5}}{5} + \frac {a^{2} e x^{6}}{6} + \frac {a^{2} f x^{7}}{7} + \frac {a b c x^{8}}{4} + \frac {2 a b d x^{9}}{9} + \frac {a b e x^{10}}{5} + \frac {2 a b f x^{11}}{11} + \frac {b^{2} c x^{12}}{12} + \frac {b^{2} d x^{13}}{13} + \frac {b^{2} e x^{14}}{14} + \frac {b^{2} f x^{15}}{15} \] Input:
integrate(x**3*(f*x**3+e*x**2+d*x+c)*(b*x**4+a)**2,x)
Output:
a**2*c*x**4/4 + a**2*d*x**5/5 + a**2*e*x**6/6 + a**2*f*x**7/7 + a*b*c*x**8 /4 + 2*a*b*d*x**9/9 + a*b*e*x**10/5 + 2*a*b*f*x**11/11 + b**2*c*x**12/12 + b**2*d*x**13/13 + b**2*e*x**14/14 + b**2*f*x**15/15
Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {1}{15} \, b^{2} f x^{15} + \frac {1}{14} \, b^{2} e x^{14} + \frac {1}{13} \, b^{2} d x^{13} + \frac {1}{12} \, b^{2} c x^{12} + \frac {2}{11} \, a b f x^{11} + \frac {1}{5} \, a b e x^{10} + \frac {2}{9} \, a b d x^{9} + \frac {1}{4} \, a b c x^{8} + \frac {1}{7} \, a^{2} f x^{7} + \frac {1}{6} \, a^{2} e x^{6} + \frac {1}{5} \, a^{2} d x^{5} + \frac {1}{4} \, a^{2} c x^{4} \] Input:
integrate(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^2,x, algorithm="maxima")
Output:
1/15*b^2*f*x^15 + 1/14*b^2*e*x^14 + 1/13*b^2*d*x^13 + 1/12*b^2*c*x^12 + 2/ 11*a*b*f*x^11 + 1/5*a*b*e*x^10 + 2/9*a*b*d*x^9 + 1/4*a*b*c*x^8 + 1/7*a^2*f *x^7 + 1/6*a^2*e*x^6 + 1/5*a^2*d*x^5 + 1/4*a^2*c*x^4
Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {1}{15} \, b^{2} f x^{15} + \frac {1}{14} \, b^{2} e x^{14} + \frac {1}{13} \, b^{2} d x^{13} + \frac {1}{12} \, b^{2} c x^{12} + \frac {2}{11} \, a b f x^{11} + \frac {1}{5} \, a b e x^{10} + \frac {2}{9} \, a b d x^{9} + \frac {1}{4} \, a b c x^{8} + \frac {1}{7} \, a^{2} f x^{7} + \frac {1}{6} \, a^{2} e x^{6} + \frac {1}{5} \, a^{2} d x^{5} + \frac {1}{4} \, a^{2} c x^{4} \] Input:
integrate(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^2,x, algorithm="giac")
Output:
1/15*b^2*f*x^15 + 1/14*b^2*e*x^14 + 1/13*b^2*d*x^13 + 1/12*b^2*c*x^12 + 2/ 11*a*b*f*x^11 + 1/5*a*b*e*x^10 + 2/9*a*b*d*x^9 + 1/4*a*b*c*x^8 + 1/7*a^2*f *x^7 + 1/6*a^2*e*x^6 + 1/5*a^2*d*x^5 + 1/4*a^2*c*x^4
Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {f\,a^2\,x^7}{7}+\frac {e\,a^2\,x^6}{6}+\frac {d\,a^2\,x^5}{5}+\frac {c\,a^2\,x^4}{4}+\frac {2\,f\,a\,b\,x^{11}}{11}+\frac {e\,a\,b\,x^{10}}{5}+\frac {2\,d\,a\,b\,x^9}{9}+\frac {c\,a\,b\,x^8}{4}+\frac {f\,b^2\,x^{15}}{15}+\frac {e\,b^2\,x^{14}}{14}+\frac {d\,b^2\,x^{13}}{13}+\frac {c\,b^2\,x^{12}}{12} \] Input:
int(x^3*(a + b*x^4)^2*(c + d*x + e*x^2 + f*x^3),x)
Output:
(a^2*c*x^4)/4 + (a^2*d*x^5)/5 + (b^2*c*x^12)/12 + (a^2*e*x^6)/6 + (b^2*d*x ^13)/13 + (a^2*f*x^7)/7 + (b^2*e*x^14)/14 + (b^2*f*x^15)/15 + (a*b*c*x^8)/ 4 + (2*a*b*d*x^9)/9 + (a*b*e*x^10)/5 + (2*a*b*f*x^11)/11
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx=\frac {x^{4} \left (12012 b^{2} f \,x^{11}+12870 b^{2} e \,x^{10}+13860 b^{2} d \,x^{9}+15015 b^{2} c \,x^{8}+32760 a b f \,x^{7}+36036 a b e \,x^{6}+40040 a b d \,x^{5}+45045 a b c \,x^{4}+25740 a^{2} f \,x^{3}+30030 a^{2} e \,x^{2}+36036 a^{2} d x +45045 a^{2} c \right )}{180180} \] Input:
int(x^3*(f*x^3+e*x^2+d*x+c)*(b*x^4+a)^2,x)
Output:
(x**4*(45045*a**2*c + 36036*a**2*d*x + 30030*a**2*e*x**2 + 25740*a**2*f*x* *3 + 45045*a*b*c*x**4 + 40040*a*b*d*x**5 + 36036*a*b*e*x**6 + 32760*a*b*f* x**7 + 15015*b**2*c*x**8 + 13860*b**2*d*x**9 + 12870*b**2*e*x**10 + 12012* b**2*f*x**11))/180180