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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.105837, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {1582, 1657} \[ \frac{3}{5} a^2 b c x^5+\frac{1}{2} a^2 b d x^6+\frac{3}{7} a^2 b e x^7+a^3 c x+\frac{1}{2} a^3 d x^2+\frac{1}{3} a^3 e x^3+\frac{1}{3} a b^2 c x^9+\frac{3}{10} a b^2 d x^{10}+\frac{3}{11} a b^2 e x^{11}+\frac{f \left (a+b x^4\right )^4}{16 b}+\frac{1}{13} b^3 c x^{13}+\frac{1}{14} b^3 d x^{14}+\frac{1}{15} b^3 e x^{15} \]

Antiderivative was successfully verified.

[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 1582

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 1657

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*} \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^3 \, dx &=\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]  time = 0.0042748, size = 180, normalized size = 1.19 \[ \frac{3}{5} a^2 b c x^5+\frac{1}{2} a^2 b d x^6+\frac{3}{7} a^2 b e x^7+\frac{3}{8} a^2 b f x^8+a^3 c x+\frac{1}{2} a^3 d x^2+\frac{1}{3} a^3 e x^3+\frac{1}{4} a^3 f x^4+\frac{1}{3} a b^2 c x^9+\frac{3}{10} a b^2 d x^{10}+\frac{3}{11} a b^2 e x^{11}+\frac{1}{4} a b^2 f x^{12}+\frac{1}{13} b^3 c x^{13}+\frac{1}{14} b^3 d x^{14}+\frac{1}{15} b^3 e x^{15}+\frac{1}{16} b^3 f x^{16} \]

Antiderivative was successfully verified.

[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]  time = 0.001, size = 151, normalized size = 1. \begin{align*}{\frac{{b}^{3}f{x}^{16}}{16}}+{\frac{{b}^{3}e{x}^{15}}{15}}+{\frac{{b}^{3}d{x}^{14}}{14}}+{\frac{{b}^{3}c{x}^{13}}{13}}+{\frac{a{b}^{2}f{x}^{12}}{4}}+{\frac{3\,a{b}^{2}e{x}^{11}}{11}}+{\frac{3\,a{b}^{2}d{x}^{10}}{10}}+{\frac{a{b}^{2}c{x}^{9}}{3}}+{\frac{3\,fb{a}^{2}{x}^{8}}{8}}+{\frac{3\,{a}^{2}be{x}^{7}}{7}}+{\frac{{a}^{2}bd{x}^{6}}{2}}+{\frac{3\,{a}^{2}bc{x}^{5}}{5}}+{\frac{f{a}^{3}{x}^{4}}{4}}+{\frac{{a}^{3}e{x}^{3}}{3}}+{\frac{{a}^{3}d{x}^{2}}{2}}+{a}^{3}cx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*f*b*a^2*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*f*a^3*x^4+1/3*a^3*e
*x^3+1/2*a^3*d*x^2+a^3*c*x

________________________________________________________________________________________

Maxima [A]  time = 0.950192, size = 203, normalized size = 1.34 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 0.924338, size = 375, normalized size = 2.48 \begin{align*} \frac{1}{16} x^{16} f b^{3} + \frac{1}{15} x^{15} e b^{3} + \frac{1}{14} x^{14} d b^{3} + \frac{1}{13} x^{13} c b^{3} + \frac{1}{4} x^{12} f b^{2} a + \frac{3}{11} x^{11} e b^{2} a + \frac{3}{10} x^{10} d b^{2} a + \frac{1}{3} x^{9} c b^{2} a + \frac{3}{8} x^{8} f b a^{2} + \frac{3}{7} x^{7} e b a^{2} + \frac{1}{2} x^{6} d b a^{2} + \frac{3}{5} x^{5} c b a^{2} + \frac{1}{4} x^{4} f a^{3} + \frac{1}{3} x^{3} e a^{3} + \frac{1}{2} x^{2} d a^{3} + x c a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.084647, size = 180, normalized size = 1.19 \begin{align*} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Giac [A]  time = 1.05389, size = 208, normalized size = 1.38 \begin{align*} \frac{1}{16} \, b^{3} f x^{16} + \frac{1}{15} \, b^{3} x^{15} e + \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} x^{11} e + \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 x^{7} e + \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} x^{3} e + \frac{1}{2} \, a^{3} d x^{2} + a^{3} c x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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*x^15*e + 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*x^11*e +
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*x^7*e + 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*x^3*e + 1/2*a^3*d*x^2 + a^3*c*x