\(\int (c+d x)^2 (a+b x^4)^p \, dx\) [218]

Optimal result

Integrand size = 17, antiderivative size = 148 \[ \int (c+d x)^2 \left (a+b x^4\right )^p \, dx=c^2 x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+c d x^2 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )+\frac {1}{3} d^2 x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.68 \[ \int (c+d x)^2 \left (a+b x^4\right )^p \, dx=\frac {1}{3} x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (3 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+d x \left (3 c \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )+d x \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )\right )\right ) \] Input:

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

Output:

(x*(a + b*x^4)^p*(3*c^2*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + d* 
x*(3*c*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a)] + d*x*Hypergeometric2 
F1[3/4, -p, 7/4, -((b*x^4)/a)])))/(3*(1 + (b*x^4)/a)^p)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, 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.

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

\(\Big \downarrow \) 2424

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c^2*x*(a + b*x^4)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + ( 
b*x^4)/a)^p + (c*d*x^2*(a + b*x^4)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b* 
x^4)/a)])/(1 + (b*x^4)/a)^p + (d^2*x^3*(a + b*x^4)^p*Hypergeometric2F1[3/4 
, -p, 7/4, -((b*x^4)/a)])/(3*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^2*x^2 + 2*c*d*x + c^2)*(b*x^4 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 22.77 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.72 \[ \int (c+d x)^2 \left (a+b x^4\right )^p \, dx=\frac {a^{p} c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + a^{p} c d x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )} + \frac {a^{p} d^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

a**p*c**2*x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*pi)/a)/ 
(4*gamma(5/4)) + a**p*c*d*x**2*hyper((1/2, -p), (3/2,), b*x**4*exp_polar(I 
*pi)/a) + a**p*d**2*x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**4*exp_po 
lar(I*pi)/a)/(4*gamma(7/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(8*(a + b*x**4)**p*c**2*p**2*x + 10*(a + b*x**4)**p*c**2*p*x + 3*(a + b*x* 
*4)**p*c**2*x + 16*(a + b*x**4)**p*c*d*p**2*x**2 + 16*(a + b*x**4)**p*c*d* 
p*x**2 + 3*(a + b*x**4)**p*c*d*x**2 + 8*(a + b*x**4)**p*d**2*p**2*x**3 + 6 
*(a + b*x**4)**p*d**2*p*x**3 + (a + b*x**4)**p*d**2*x**3 + 1024*int((a + b 
*x**4)**p/(32*a*p**3 + 48*a*p**2 + 22*a*p + 3*a + 32*b*p**3*x**4 + 48*b*p* 
*2*x**4 + 22*b*p*x**4 + 3*b*x**4),x)*a*c**2*p**6 + 2816*int((a + b*x**4)** 
p/(32*a*p**3 + 48*a*p**2 + 22*a*p + 3*a + 32*b*p**3*x**4 + 48*b*p**2*x**4 
+ 22*b*p*x**4 + 3*b*x**4),x)*a*c**2*p**5 + 3008*int((a + b*x**4)**p/(32*a* 
p**3 + 48*a*p**2 + 22*a*p + 3*a + 32*b*p**3*x**4 + 48*b*p**2*x**4 + 22*b*p 
*x**4 + 3*b*x**4),x)*a*c**2*p**4 + 1552*int((a + b*x**4)**p/(32*a*p**3 + 4 
8*a*p**2 + 22*a*p + 3*a + 32*b*p**3*x**4 + 48*b*p**2*x**4 + 22*b*p*x**4 + 
3*b*x**4),x)*a*c**2*p**3 + 384*int((a + b*x**4)**p/(32*a*p**3 + 48*a*p**2 
+ 22*a*p + 3*a + 32*b*p**3*x**4 + 48*b*p**2*x**4 + 22*b*p*x**4 + 3*b*x**4) 
,x)*a*c**2*p**2 + 36*int((a + b*x**4)**p/(32*a*p**3 + 48*a*p**2 + 22*a*p + 
 3*a + 32*b*p**3*x**4 + 48*b*p**2*x**4 + 22*b*p*x**4 + 3*b*x**4),x)*a*c**2 
*p + 1024*int(((a + b*x**4)**p*x**2)/(32*a*p**3 + 48*a*p**2 + 22*a*p + 3*a 
 + 32*b*p**3*x**4 + 48*b*p**2*x**4 + 22*b*p*x**4 + 3*b*x**4),x)*a*d**2*p** 
6 + 2304*int(((a + b*x**4)**p*x**2)/(32*a*p**3 + 48*a*p**2 + 22*a*p + 3*a 
+ 32*b*p**3*x**4 + 48*b*p**2*x**4 + 22*b*p*x**4 + 3*b*x**4),x)*a*d**2*p**5 
 + 1984*int(((a + b*x**4)**p*x**2)/(32*a*p**3 + 48*a*p**2 + 22*a*p + 3*...