\(\int (c+d x) (a+b x^4)^p \, dx\) [219]

Optimal result

Integrand size = 15, antiderivative size = 96 \[ \int (c+d x) \left (a+b x^4\right )^p \, dx=c x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+\frac {1}{2} d x^2 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int (c+d x) \left (a+b x^4\right )^p \, dx=\frac {1}{2} x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (2 c \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+d x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )\right ) \] Input:

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

Output:

(x*(a + b*x^4)^p*(2*c*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + d*x* 
Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a)]))/(2*(1 + (b*x^4)/a)^p)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2424, 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.

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

\(\Big \downarrow \) 2424

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c*x*(a + b*x^4)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b* 
x^4)/a)^p + (d*x^2*(a + b*x^4)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4) 
/a)])/(2*(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 2424

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, 
 x], j, k}, Int[Sum[x^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, p}, 
 x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]
 

Maple [F]

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

Input:

int((d*x+c)*(b*x^4+a)^p,x)
 

Output:

int((d*x+c)*(b*x^4+a)^p,x)
 

Fricas [F]

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

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

Output:

integral((d*x + c)*(b*x^4 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 14.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int (c+d x) \left (a+b x^4\right )^p \, dx=\frac {a^{p} c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} d x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2} \] Input:

integrate((d*x+c)*(b*x**4+a)**p,x)
 

Output:

a**p*c*x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4* 
gamma(5/4)) + a**p*d*x**2*hyper((1/2, -p), (3/2,), b*x**4*exp_polar(I*pi)/ 
a)/2
 

Maxima [F]

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

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

Output:

integrate((d*x + c)*(b*x^4 + a)^p, x)
 

Giac [F]

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

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

Output:

integrate((d*x + c)*(b*x^4 + a)^p, x)
 

Mupad [F(-1)]

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

int((a + b*x^4)^p*(c + d*x),x)
 

Output:

int((a + b*x^4)^p*(c + d*x), x)
 

Reduce [F]

\[ \int (c+d x) \left (a+b x^4\right )^p \, dx=\frac {4 \left (b \,x^{4}+a \right )^{p} c p x +2 \left (b \,x^{4}+a \right )^{p} c x +4 \left (b \,x^{4}+a \right )^{p} d p \,x^{2}+\left (b \,x^{4}+a \right )^{p} d \,x^{2}+128 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a c \,p^{4}+160 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a c \,p^{3}+64 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a c \,p^{2}+8 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a c p +128 \left (\int \frac {\left (b \,x^{4}+a \right )^{p} x}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a d \,p^{4}+128 \left (\int \frac {\left (b \,x^{4}+a \right )^{p} x}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a d \,p^{3}+40 \left (\int \frac {\left (b \,x^{4}+a \right )^{p} x}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a d \,p^{2}+4 \left (\int \frac {\left (b \,x^{4}+a \right )^{p} x}{8 b \,p^{2} x^{4}+6 b p \,x^{4}+b \,x^{4}+8 a \,p^{2}+6 a p +a}d x \right ) a d p}{16 p^{2}+12 p +2} \] Input:

int((d*x+c)*(b*x^4+a)^p,x)
 

Output:

(4*(a + b*x**4)**p*c*p*x + 2*(a + b*x**4)**p*c*x + 4*(a + b*x**4)**p*d*p*x 
**2 + (a + b*x**4)**p*d*x**2 + 128*int((a + b*x**4)**p/(8*a*p**2 + 6*a*p + 
 a + 8*b*p**2*x**4 + 6*b*p*x**4 + b*x**4),x)*a*c*p**4 + 160*int((a + b*x** 
4)**p/(8*a*p**2 + 6*a*p + a + 8*b*p**2*x**4 + 6*b*p*x**4 + b*x**4),x)*a*c* 
p**3 + 64*int((a + b*x**4)**p/(8*a*p**2 + 6*a*p + a + 8*b*p**2*x**4 + 6*b* 
p*x**4 + b*x**4),x)*a*c*p**2 + 8*int((a + b*x**4)**p/(8*a*p**2 + 6*a*p + a 
 + 8*b*p**2*x**4 + 6*b*p*x**4 + b*x**4),x)*a*c*p + 128*int(((a + b*x**4)** 
p*x)/(8*a*p**2 + 6*a*p + a + 8*b*p**2*x**4 + 6*b*p*x**4 + b*x**4),x)*a*d*p 
**4 + 128*int(((a + b*x**4)**p*x)/(8*a*p**2 + 6*a*p + a + 8*b*p**2*x**4 + 
6*b*p*x**4 + b*x**4),x)*a*d*p**3 + 40*int(((a + b*x**4)**p*x)/(8*a*p**2 + 
6*a*p + a + 8*b*p**2*x**4 + 6*b*p*x**4 + b*x**4),x)*a*d*p**2 + 4*int(((a + 
 b*x**4)**p*x)/(8*a*p**2 + 6*a*p + a + 8*b*p**2*x**4 + 6*b*p*x**4 + b*x**4 
),x)*a*d*p)/(2*(8*p**2 + 6*p + 1))