\(\int (c+d x+e x^2+f x^3) (a+b x^4)^3 \, dx\) [148]

Optimal result

Integrand size = 25, antiderivative size = 151 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {1}{3} a^3 e x^3+\frac {3}{5} a^2 b c x^5+\frac {1}{2} a^2 b d x^6+\frac {3}{7} a^2 b e x^7+\frac {1}{3} a b^2 c x^9+\frac {3}{10} a b^2 d x^{10}+\frac {3}{11} a b^2 e x^{11}+\frac {1}{13} b^3 c x^{13}+\frac {1}{14} b^3 d x^{14}+\frac {1}{15} b^3 e x^{15}+\frac {f \left (a+b x^4\right )^4}{16 b} \]

[Out]

a^3*c*x+1/2*a^3*d*x^2+1/3*a^3*e*x^3+3/5*a^2*b*c*x^5+1/2*a^2*b*d*x^6+3/7*a^2*b*e*x^7+1/3*a*b^2*c*x^9+3/10*a*b^2
*d*x^10+3/11*a*b^2*e*x^11+1/13*b^3*c*x^13+1/14*b^3*d*x^14+1/15*b^3*e*x^15+1/16*f*(b*x^4+a)^4/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1596, 1671} \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {1}{3} a^3 e x^3+\frac {3}{5} a^2 b c x^5+\frac {1}{2} a^2 b d x^6+\frac {3}{7} a^2 b e x^7+\frac {1}{3} a b^2 c x^9+\frac {3}{10} a b^2 d x^{10}+\frac {3}{11} a b^2 e x^{11}+\frac {f \left (a+b x^4\right )^4}{16 b}+\frac {1}{13} b^3 c x^{13}+\frac {1}{14} b^3 d x^{14}+\frac {1}{15} b^3 e x^{15} \]

[In]

Int[(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^3,x]

[Out]

a^3*c*x + (a^3*d*x^2)/2 + (a^3*e*x^3)/3 + (3*a^2*b*c*x^5)/5 + (a^2*b*d*x^6)/2 + (3*a^2*b*e*x^7)/7 + (a*b^2*c*x
^9)/3 + (3*a*b^2*d*x^10)/10 + (3*a*b^2*e*x^11)/11 + (b^3*c*x^13)/13 + (b^3*d*x^14)/14 + (b^3*e*x^15)/15 + (f*(
a + b*x^4)^4)/(16*b)

Rule 1596

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] && I
GtQ[p, 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] &&  !MatchQ[P
x, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ[{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Co
eff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \frac {f \left (a+b x^4\right )^4}{16 b}+\int \left (c+d x+e x^2\right ) \left (a+b x^4\right )^3 \, dx \\ & = \frac {f \left (a+b x^4\right )^4}{16 b}+\int \left (a^3 c+a^3 d x+a^3 e x^2+3 a^2 b c x^4+3 a^2 b d x^5+3 a^2 b e x^6+3 a b^2 c x^8+3 a b^2 d x^9+3 a b^2 e x^{10}+b^3 c x^{12}+b^3 d x^{13}+b^3 e x^{14}\right ) \, dx \\ & = a^3 c x+\frac {1}{2} a^3 d x^2+\frac {1}{3} a^3 e x^3+\frac {3}{5} a^2 b c x^5+\frac {1}{2} a^2 b d x^6+\frac {3}{7} a^2 b e x^7+\frac {1}{3} a b^2 c x^9+\frac {3}{10} a b^2 d x^{10}+\frac {3}{11} a b^2 e x^{11}+\frac {1}{13} b^3 c x^{13}+\frac {1}{14} b^3 d x^{14}+\frac {1}{15} b^3 e x^{15}+\frac {f \left (a+b x^4\right )^4}{16 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.19 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=a^3 c x+\frac {1}{2} a^3 d x^2+\frac {1}{3} a^3 e x^3+\frac {1}{4} a^3 f x^4+\frac {3}{5} a^2 b c x^5+\frac {1}{2} a^2 b d x^6+\frac {3}{7} a^2 b e x^7+\frac {3}{8} a^2 b f x^8+\frac {1}{3} a b^2 c x^9+\frac {3}{10} a b^2 d x^{10}+\frac {3}{11} a b^2 e x^{11}+\frac {1}{4} a b^2 f x^{12}+\frac {1}{13} b^3 c x^{13}+\frac {1}{14} b^3 d x^{14}+\frac {1}{15} b^3 e x^{15}+\frac {1}{16} b^3 f x^{16} \]

[In]

Integrate[(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^3,x]

[Out]

a^3*c*x + (a^3*d*x^2)/2 + (a^3*e*x^3)/3 + (a^3*f*x^4)/4 + (3*a^2*b*c*x^5)/5 + (a^2*b*d*x^6)/2 + (3*a^2*b*e*x^7
)/7 + (3*a^2*b*f*x^8)/8 + (a*b^2*c*x^9)/3 + (3*a*b^2*d*x^10)/10 + (3*a*b^2*e*x^11)/11 + (a*b^2*f*x^12)/4 + (b^
3*c*x^13)/13 + (b^3*d*x^14)/14 + (b^3*e*x^15)/15 + (b^3*f*x^16)/16

Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00

method	result	size
gosper	\(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{4} f \,a^{3} x^{4}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {3}{8} f \,a^{2} b \,x^{8}+\frac {1}{3} a \,b^{2} c \,x^{9}+\frac {3}{10} a \,b^{2} d \,x^{10}+\frac {3}{11} a \,b^{2} e \,x^{11}+\frac {1}{4} a \,b^{2} f \,x^{12}+\frac {1}{13} b^{3} c \,x^{13}+\frac {1}{14} b^{3} d \,x^{14}+\frac {1}{15} b^{3} e \,x^{15}+\frac {1}{16} b^{3} f \,x^{16}\)	\(151\)
default	\(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{4} f \,a^{3} x^{4}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {3}{8} f \,a^{2} b \,x^{8}+\frac {1}{3} a \,b^{2} c \,x^{9}+\frac {3}{10} a \,b^{2} d \,x^{10}+\frac {3}{11} a \,b^{2} e \,x^{11}+\frac {1}{4} a \,b^{2} f \,x^{12}+\frac {1}{13} b^{3} c \,x^{13}+\frac {1}{14} b^{3} d \,x^{14}+\frac {1}{15} b^{3} e \,x^{15}+\frac {1}{16} b^{3} f \,x^{16}\)	\(151\)
norman	\(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{4} f \,a^{3} x^{4}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {3}{8} f \,a^{2} b \,x^{8}+\frac {1}{3} a \,b^{2} c \,x^{9}+\frac {3}{10} a \,b^{2} d \,x^{10}+\frac {3}{11} a \,b^{2} e \,x^{11}+\frac {1}{4} a \,b^{2} f \,x^{12}+\frac {1}{13} b^{3} c \,x^{13}+\frac {1}{14} b^{3} d \,x^{14}+\frac {1}{15} b^{3} e \,x^{15}+\frac {1}{16} b^{3} f \,x^{16}\)	\(151\)
risch	\(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{4} f \,a^{3} x^{4}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {3}{8} f \,a^{2} b \,x^{8}+\frac {1}{3} a \,b^{2} c \,x^{9}+\frac {3}{10} a \,b^{2} d \,x^{10}+\frac {3}{11} a \,b^{2} e \,x^{11}+\frac {1}{4} a \,b^{2} f \,x^{12}+\frac {1}{13} b^{3} c \,x^{13}+\frac {1}{14} b^{3} d \,x^{14}+\frac {1}{15} b^{3} e \,x^{15}+\frac {1}{16} b^{3} f \,x^{16}\)	\(151\)
parallelrisch	\(a^{3} c x +\frac {1}{2} a^{3} d \,x^{2}+\frac {1}{3} a^{3} e \,x^{3}+\frac {1}{4} f \,a^{3} x^{4}+\frac {3}{5} a^{2} b c \,x^{5}+\frac {1}{2} a^{2} b d \,x^{6}+\frac {3}{7} a^{2} b e \,x^{7}+\frac {3}{8} f \,a^{2} b \,x^{8}+\frac {1}{3} a \,b^{2} c \,x^{9}+\frac {3}{10} a \,b^{2} d \,x^{10}+\frac {3}{11} a \,b^{2} e \,x^{11}+\frac {1}{4} a \,b^{2} f \,x^{12}+\frac {1}{13} b^{3} c \,x^{13}+\frac {1}{14} b^{3} d \,x^{14}+\frac {1}{15} b^{3} e \,x^{15}+\frac {1}{16} b^{3} f \,x^{16}\)	\(151\)

[In]

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^3,x,method=_RETURNVERBOSE)

[Out]

a^3*c*x+1/2*a^3*d*x^2+1/3*a^3*e*x^3+1/4*f*a^3*x^4+3/5*a^2*b*c*x^5+1/2*a^2*b*d*x^6+3/7*a^2*b*e*x^7+3/8*f*a^2*b*
x^8+1/3*a*b^2*c*x^9+3/10*a*b^2*d*x^10+3/11*a*b^2*e*x^11+1/4*a*b^2*f*x^12+1/13*b^3*c*x^13+1/14*b^3*d*x^14+1/15*
b^3*e*x^15+1/16*b^3*f*x^16

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=\frac {1}{16} \, b^{3} f x^{16} + \frac {1}{15} \, b^{3} e x^{15} + \frac {1}{14} \, b^{3} d x^{14} + \frac {1}{13} \, b^{3} c x^{13} + \frac {1}{4} \, a b^{2} f x^{12} + \frac {3}{11} \, a b^{2} e x^{11} + \frac {3}{10} \, a b^{2} d x^{10} + \frac {1}{3} \, a b^{2} c x^{9} + \frac {3}{8} \, a^{2} b f x^{8} + \frac {3}{7} \, a^{2} b e x^{7} + \frac {1}{2} \, a^{2} b d x^{6} + \frac {3}{5} \, a^{2} b c x^{5} + \frac {1}{4} \, a^{3} f x^{4} + \frac {1}{3} \, a^{3} e x^{3} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \]

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^3,x, algorithm="fricas")

[Out]

1/16*b^3*f*x^16 + 1/15*b^3*e*x^15 + 1/14*b^3*d*x^14 + 1/13*b^3*c*x^13 + 1/4*a*b^2*f*x^12 + 3/11*a*b^2*e*x^11 +
3/10*a*b^2*d*x^10 + 1/3*a*b^2*c*x^9 + 3/8*a^2*b*f*x^8 + 3/7*a^2*b*e*x^7 + 1/2*a^2*b*d*x^6 + 3/5*a^2*b*c*x^5 +
1/4*a^3*f*x^4 + 1/3*a^3*e*x^3 + 1/2*a^3*d*x^2 + a^3*c*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.19 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=a^{3} c x + \frac {a^{3} d x^{2}}{2} + \frac {a^{3} e x^{3}}{3} + \frac {a^{3} f x^{4}}{4} + \frac {3 a^{2} b c x^{5}}{5} + \frac {a^{2} b d x^{6}}{2} + \frac {3 a^{2} b e x^{7}}{7} + \frac {3 a^{2} b f x^{8}}{8} + \frac {a b^{2} c x^{9}}{3} + \frac {3 a b^{2} d x^{10}}{10} + \frac {3 a b^{2} e x^{11}}{11} + \frac {a b^{2} f x^{12}}{4} + \frac {b^{3} c x^{13}}{13} + \frac {b^{3} d x^{14}}{14} + \frac {b^{3} e x^{15}}{15} + \frac {b^{3} f x^{16}}{16} \]

[In]

integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**3,x)

[Out]

a**3*c*x + a**3*d*x**2/2 + a**3*e*x**3/3 + a**3*f*x**4/4 + 3*a**2*b*c*x**5/5 + a**2*b*d*x**6/2 + 3*a**2*b*e*x*
*7/7 + 3*a**2*b*f*x**8/8 + a*b**2*c*x**9/3 + 3*a*b**2*d*x**10/10 + 3*a*b**2*e*x**11/11 + a*b**2*f*x**12/4 + b*
*3*c*x**13/13 + b**3*d*x**14/14 + b**3*e*x**15/15 + b**3*f*x**16/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=\frac {1}{16} \, b^{3} f x^{16} + \frac {1}{15} \, b^{3} e x^{15} + \frac {1}{14} \, b^{3} d x^{14} + \frac {1}{13} \, b^{3} c x^{13} + \frac {1}{4} \, a b^{2} f x^{12} + \frac {3}{11} \, a b^{2} e x^{11} + \frac {3}{10} \, a b^{2} d x^{10} + \frac {1}{3} \, a b^{2} c x^{9} + \frac {3}{8} \, a^{2} b f x^{8} + \frac {3}{7} \, a^{2} b e x^{7} + \frac {1}{2} \, a^{2} b d x^{6} + \frac {3}{5} \, a^{2} b c x^{5} + \frac {1}{4} \, a^{3} f x^{4} + \frac {1}{3} \, a^{3} e x^{3} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \]

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^3,x, algorithm="maxima")

[Out]

1/16*b^3*f*x^16 + 1/15*b^3*e*x^15 + 1/14*b^3*d*x^14 + 1/13*b^3*c*x^13 + 1/4*a*b^2*f*x^12 + 3/11*a*b^2*e*x^11 +
3/10*a*b^2*d*x^10 + 1/3*a*b^2*c*x^9 + 3/8*a^2*b*f*x^8 + 3/7*a^2*b*e*x^7 + 1/2*a^2*b*d*x^6 + 3/5*a^2*b*c*x^5 +
1/4*a^3*f*x^4 + 1/3*a^3*e*x^3 + 1/2*a^3*d*x^2 + a^3*c*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=\frac {1}{16} \, b^{3} f x^{16} + \frac {1}{15} \, b^{3} e x^{15} + \frac {1}{14} \, b^{3} d x^{14} + \frac {1}{13} \, b^{3} c x^{13} + \frac {1}{4} \, a b^{2} f x^{12} + \frac {3}{11} \, a b^{2} e x^{11} + \frac {3}{10} \, a b^{2} d x^{10} + \frac {1}{3} \, a b^{2} c x^{9} + \frac {3}{8} \, a^{2} b f x^{8} + \frac {3}{7} \, a^{2} b e x^{7} + \frac {1}{2} \, a^{2} b d x^{6} + \frac {3}{5} \, a^{2} b c x^{5} + \frac {1}{4} \, a^{3} f x^{4} + \frac {1}{3} \, a^{3} e x^{3} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \]

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^3,x, algorithm="giac")

[Out]

1/16*b^3*f*x^16 + 1/15*b^3*e*x^15 + 1/14*b^3*d*x^14 + 1/13*b^3*c*x^13 + 1/4*a*b^2*f*x^12 + 3/11*a*b^2*e*x^11 +
3/10*a*b^2*d*x^10 + 1/3*a*b^2*c*x^9 + 3/8*a^2*b*f*x^8 + 3/7*a^2*b*e*x^7 + 1/2*a^2*b*d*x^6 + 3/5*a^2*b*c*x^5 +
1/4*a^3*f*x^4 + 1/3*a^3*e*x^3 + 1/2*a^3*d*x^2 + a^3*c*x

Mupad [B] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx=\frac {f\,a^3\,x^4}{4}+\frac {e\,a^3\,x^3}{3}+\frac {d\,a^3\,x^2}{2}+c\,a^3\,x+\frac {3\,f\,a^2\,b\,x^8}{8}+\frac {3\,e\,a^2\,b\,x^7}{7}+\frac {d\,a^2\,b\,x^6}{2}+\frac {3\,c\,a^2\,b\,x^5}{5}+\frac {f\,a\,b^2\,x^{12}}{4}+\frac {3\,e\,a\,b^2\,x^{11}}{11}+\frac {3\,d\,a\,b^2\,x^{10}}{10}+\frac {c\,a\,b^2\,x^9}{3}+\frac {f\,b^3\,x^{16}}{16}+\frac {e\,b^3\,x^{15}}{15}+\frac {d\,b^3\,x^{14}}{14}+\frac {c\,b^3\,x^{13}}{13} \]

[In]

int((a + b*x^4)^3*(c + d*x + e*x^2 + f*x^3),x)

[Out]

(a^3*d*x^2)/2 + (b^3*c*x^13)/13 + (a^3*e*x^3)/3 + (b^3*d*x^14)/14 + (a^3*f*x^4)/4 + (b^3*e*x^15)/15 + (b^3*f*x
^16)/16 + a^3*c*x + (3*a^2*b*c*x^5)/5 + (a*b^2*c*x^9)/3 + (a^2*b*d*x^6)/2 + (3*a*b^2*d*x^10)/10 + (3*a^2*b*e*x
^7)/7 + (3*a*b^2*e*x^11)/11 + (3*a^2*b*f*x^8)/8 + (a*b^2*f*x^12)/4