\(\int (a+b x^4)^p (c+d x^4) \, dx\) [131]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right ) \, dx=\frac {d x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\left (c-\frac {a d}{5 b+4 b p}\right ) x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right ) \, dx=\frac {x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (d \left (a+b x^4\right ) \left (1+\frac {b x^4}{a}\right )^p+(-a d+b c (5+4 p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{b (5+4 p)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {913, 779, 778}

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.

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

\(\Big \downarrow \) 913

\(\displaystyle \left (c-\frac {a d}{4 b p+5 b}\right ) \int \left (b x^4+a\right )^pdx+\frac {d x \left (a+b x^4\right )^{p+1}}{b (4 p+5)}\)

\(\Big \downarrow \) 779

\(\displaystyle \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (c-\frac {a d}{4 b p+5 b}\right ) \int \left (\frac {b x^4}{a}+1\right )^pdx+\frac {d x \left (a+b x^4\right )^{p+1}}{b (4 p+5)}\)

\(\Big \downarrow \) 778

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x 
^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(1 + b*(x^n/a))^p, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 28.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right ) \, dx=\frac {a^{p} c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} \] Input:

integrate((b*x**4+a)**p*(d*x**4+c),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**5*gamma(5/4)*hyper((5/4, -p), (9/4,), b*x**4*exp_p 
olar(I*pi)/a)/(4*gamma(9/4))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b x^4\right )^p \left (c+d x^4\right ) \, dx=\frac {4 \left (b \,x^{4}+a \right )^{p} a d p x +4 \left (b \,x^{4}+a \right )^{p} b c p x +5 \left (b \,x^{4}+a \right )^{p} b c x +4 \left (b \,x^{4}+a \right )^{p} b d p \,x^{5}+\left (b \,x^{4}+a \right )^{p} b d \,x^{5}-64 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{16 b \,p^{2} x^{4}+24 b p \,x^{4}+5 b \,x^{4}+16 a \,p^{2}+24 a p +5 a}d x \right ) a^{2} d \,p^{3}-96 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{16 b \,p^{2} x^{4}+24 b p \,x^{4}+5 b \,x^{4}+16 a \,p^{2}+24 a p +5 a}d x \right ) a^{2} d \,p^{2}-20 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{16 b \,p^{2} x^{4}+24 b p \,x^{4}+5 b \,x^{4}+16 a \,p^{2}+24 a p +5 a}d x \right ) a^{2} d p +256 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{16 b \,p^{2} x^{4}+24 b p \,x^{4}+5 b \,x^{4}+16 a \,p^{2}+24 a p +5 a}d x \right ) a b c \,p^{4}+704 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{16 b \,p^{2} x^{4}+24 b p \,x^{4}+5 b \,x^{4}+16 a \,p^{2}+24 a p +5 a}d x \right ) a b c \,p^{3}+560 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{16 b \,p^{2} x^{4}+24 b p \,x^{4}+5 b \,x^{4}+16 a \,p^{2}+24 a p +5 a}d x \right ) a b c \,p^{2}+100 \left (\int \frac {\left (b \,x^{4}+a \right )^{p}}{16 b \,p^{2} x^{4}+24 b p \,x^{4}+5 b \,x^{4}+16 a \,p^{2}+24 a p +5 a}d x \right ) a b c p}{b \left (16 p^{2}+24 p +5\right )} \] Input:

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

Output:

(4*(a + b*x**4)**p*a*d*p*x + 4*(a + b*x**4)**p*b*c*p*x + 5*(a + b*x**4)**p 
*b*c*x + 4*(a + b*x**4)**p*b*d*p*x**5 + (a + b*x**4)**p*b*d*x**5 - 64*int( 
(a + b*x**4)**p/(16*a*p**2 + 24*a*p + 5*a + 16*b*p**2*x**4 + 24*b*p*x**4 + 
 5*b*x**4),x)*a**2*d*p**3 - 96*int((a + b*x**4)**p/(16*a*p**2 + 24*a*p + 5 
*a + 16*b*p**2*x**4 + 24*b*p*x**4 + 5*b*x**4),x)*a**2*d*p**2 - 20*int((a + 
 b*x**4)**p/(16*a*p**2 + 24*a*p + 5*a + 16*b*p**2*x**4 + 24*b*p*x**4 + 5*b 
*x**4),x)*a**2*d*p + 256*int((a + b*x**4)**p/(16*a*p**2 + 24*a*p + 5*a + 1 
6*b*p**2*x**4 + 24*b*p*x**4 + 5*b*x**4),x)*a*b*c*p**4 + 704*int((a + b*x** 
4)**p/(16*a*p**2 + 24*a*p + 5*a + 16*b*p**2*x**4 + 24*b*p*x**4 + 5*b*x**4) 
,x)*a*b*c*p**3 + 560*int((a + b*x**4)**p/(16*a*p**2 + 24*a*p + 5*a + 16*b* 
p**2*x**4 + 24*b*p*x**4 + 5*b*x**4),x)*a*b*c*p**2 + 100*int((a + b*x**4)** 
p/(16*a*p**2 + 24*a*p + 5*a + 16*b*p**2*x**4 + 24*b*p*x**4 + 5*b*x**4),x)* 
a*b*c*p)/(b*(16*p**2 + 24*p + 5))