\(\int x^3 (c+d x+e x^2+f x^3) (a+b x^4)^p \, dx\) [113]

Optimal result

Integrand size = 28, antiderivative size = 175 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\frac {c \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+\frac {1}{5} d x^5 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )+\frac {1}{6} e x^6 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^4}{a}\right )+\frac {1}{7} f x^7 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^4}{a}\right ) \] Output:

1/4*c*(b*x^4+a)^(p+1)/b/(p+1)+1/5*d*x^5*(b*x^4+a)^p*hypergeom([5/4, -p],[9 
/4],-b*x^4/a)/((1+b*x^4/a)^p)+1/6*e*x^6*(b*x^4+a)^p*hypergeom([3/2, -p],[5 
/2],-b*x^4/a)/((1+b*x^4/a)^p)+1/7*f*x^7*(b*x^4+a)^p*hypergeom([7/4, -p],[1 
1/4],-b*x^4/a)/((1+b*x^4/a)^p)
 

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.83 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\frac {\left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (105 c \left (a+b x^4\right ) \left (1+\frac {b x^4}{a}\right )^p+84 b d (1+p) x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )+70 b e (1+p) x^6 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^4}{a}\right )+60 b f (1+p) x^7 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^4}{a}\right )\right )}{420 b (1+p)} \] Input:

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

Output:

((a + b*x^4)^p*(105*c*(a + b*x^4)*(1 + (b*x^4)/a)^p + 84*b*d*(1 + p)*x^5*H 
ypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)] + 70*b*e*(1 + p)*x^6*Hypergeo 
metric2F1[3/2, -p, 5/2, -((b*x^4)/a)] + 60*b*f*(1 + p)*x^7*Hypergeometric2 
F1[7/4, -p, 11/4, -((b*x^4)/a)]))/(420*b*(1 + p)*(1 + (b*x^4)/a)^p)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2372, 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.

\(\displaystyle \int x^3 \left (a+b x^4\right )^p \left (c+d x+e x^2+f x^3\right ) \, dx\)

\(\Big \downarrow \) 2372

\(\displaystyle \int \left (x^3 \left (c+e x^2\right ) \left (a+b x^4\right )^p+x^4 \left (d+f x^2\right ) \left (a+b x^4\right )^p\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {c \left (a+b x^4\right )^{p+1}}{4 b (p+1)}+\frac {1}{5} d x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )+\frac {1}{6} e x^6 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^4}{a}\right )+\frac {1}{7} f x^7 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^4}{a}\right )\)

Input:

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

Output:

(c*(a + b*x^4)^(1 + p))/(4*b*(1 + p)) + (d*x^5*(a + b*x^4)^p*Hypergeometri 
c2F1[5/4, -p, 9/4, -((b*x^4)/a)])/(5*(1 + (b*x^4)/a)^p) + (e*x^6*(a + b*x^ 
4)^p*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^4)/a)])/(6*(1 + (b*x^4)/a)^p) 
+ (f*x^7*(a + b*x^4)^p*Hypergeometric2F1[7/4, -p, 11/4, -((b*x^4)/a)])/(7* 
(1 + (b*x^4)/a)^p)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2372

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Mo 
dule[{q = Expon[Pq, x], j, k}, Int[Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, 
 j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, 
 n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0 
] &&  !PolyQ[Pq, x^(n/2)]
 

Maple [F]

\[\int x^{3} \left (f \,x^{3}+e \,x^{2}+d x +c \right ) \left (b \,x^{4}+a \right )^{p}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int { {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p} x^{3} \,d x } \] Input:

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

Output:

integral((f*x^6 + e*x^5 + d*x^4 + c*x^3)*(b*x^4 + a)^p, x)
 

Sympy [A] (verification not implemented)

Time = 45.65 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.82 \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\frac {a^{p} d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {a^{p} e x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6} + \frac {a^{p} f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, - p \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + c \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{4}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{4} \right )} & \text {otherwise} \end {cases}}{4 b} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**p*d*x**5*gamma(5/4)*hyper((5/4, -p), (9/4,), b*x**4*exp_polar(I*pi)/a)/ 
(4*gamma(9/4)) + a**p*e*x**6*hyper((3/2, -p), (5/2,), b*x**4*exp_polar(I*p 
i)/a)/6 + a**p*f*x**7*gamma(7/4)*hyper((7/4, -p), (11/4,), b*x**4*exp_pola 
r(I*pi)/a)/(4*gamma(11/4)) + c*Piecewise((a**p*x**4/4, Eq(b, 0)), (Piecewi 
se(((a + b*x**4)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + b*x**4), True))/(4 
*b), True))
 

Maxima [F]

\[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int { {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p} x^{3} \,d x } \] Input:

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

Output:

1/4*(b*x^4 + a)^(p + 1)*c/(b*(p + 1)) + integrate((f*x^6 + e*x^5 + d*x^4)* 
(b*x^4 + a)^p, x)
 

Giac [F]

\[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int { {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p} x^{3} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int x^3\,{\left (b\,x^4+a\right )}^p\,\left (f\,x^3+e\,x^2+d\,x+c\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^3 \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\text {too large to display} \] Input:

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

Output:

(1024*(a + b*x**4)**p*a*c*p**6 + 6144*(a + b*x**4)**p*a*c*p**5 + 14464*(a 
+ b*x**4)**p*a*c*p**4 + 16896*(a + b*x**4)**p*a*c*p**3 + 10180*(a + b*x**4 
)**p*a*c*p**2 + 2952*(a + b*x**4)**p*a*c*p + 315*(a + b*x**4)**p*a*c + 102 
4*(a + b*x**4)**p*a*d*p**6*x + 5632*(a + b*x**4)**p*a*d*p**5*x + 11840*(a 
+ b*x**4)**p*a*d*p**4*x + 11840*(a + b*x**4)**p*a*d*p**3*x + 5616*(a + b*x 
**4)**p*a*d*p**2*x + 1008*(a + b*x**4)**p*a*d*p*x + 1024*(a + b*x**4)**p*a 
*e*p**6*x**2 + 5120*(a + b*x**4)**p*a*e*p**5*x**2 + 9600*(a + b*x**4)**p*a 
*e*p**4*x**2 + 8320*(a + b*x**4)**p*a*e*p**3*x**2 + 3236*(a + b*x**4)**p*a 
*e*p**2*x**2 + 420*(a + b*x**4)**p*a*e*p*x**2 + 1024*(a + b*x**4)**p*a*f*p 
**6*x**3 + 4608*(a + b*x**4)**p*a*f*p**5*x**3 + 7744*(a + b*x**4)**p*a*f*p 
**4*x**3 + 5952*(a + b*x**4)**p*a*f*p**3*x**3 + 2032*(a + b*x**4)**p*a*f*p 
**2*x**3 + 240*(a + b*x**4)**p*a*f*p*x**3 + 1024*(a + b*x**4)**p*b*c*p**6* 
x**4 + 6144*(a + b*x**4)**p*b*c*p**5*x**4 + 14464*(a + b*x**4)**p*b*c*p**4 
*x**4 + 16896*(a + b*x**4)**p*b*c*p**3*x**4 + 10180*(a + b*x**4)**p*b*c*p* 
*2*x**4 + 2952*(a + b*x**4)**p*b*c*p*x**4 + 315*(a + b*x**4)**p*b*c*x**4 + 
 1024*(a + b*x**4)**p*b*d*p**6*x**5 + 5888*(a + b*x**4)**p*b*d*p**5*x**5 + 
 13248*(a + b*x**4)**p*b*d*p**4*x**5 + 14800*(a + b*x**4)**p*b*d*p**3*x**5 
 + 8576*(a + b*x**4)**p*b*d*p**2*x**5 + 2412*(a + b*x**4)**p*b*d*p*x**5 + 
252*(a + b*x**4)**p*b*d*x**5 + 1024*(a + b*x**4)**p*b*e*p**6*x**6 + 5632*( 
a + b*x**4)**p*b*e*p**5*x**6 + 12160*(a + b*x**4)**p*b*e*p**4*x**6 + 13...