\(\int (a+b x^4)^p (c+d x^4)^2 \, dx\) [130]

Optimal result

Integrand size = 19, antiderivative size = 173 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right )^2 \, dx=-\frac {d (5 a d-2 b c (9+4 p)) x \left (a+b x^4\right )^{1+p}}{b^2 (5+4 p) (9+4 p)}+\frac {d^2 x^5 \left (a+b x^4\right )^{1+p}}{b (9+4 p)}+\frac {\left (5 a^2 d^2-2 a b c d (9+4 p)+b^2 c^2 \left (45+56 p+16 p^2\right )\right ) x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )}{b^2 (5+4 p) (9+4 p)} \] Output:

-d*(5*a*d-2*b*c*(9+4*p))*x*(b*x^4+a)^(p+1)/b^2/(5+4*p)/(9+4*p)+d^2*x^5*(b* 
x^4+a)^(p+1)/b/(9+4*p)+(5*a^2*d^2-2*a*b*c*d*(9+4*p)+b^2*c^2*(16*p^2+56*p+4 
5))*x*(b*x^4+a)^p*hypergeom([1/4, -p],[5/4],-b*x^4/a)/b^2/(5+4*p)/(9+4*p)/ 
((1+b*x^4/a)^p)
 

Mathematica [A] (verified)

Time = 5.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.61 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right )^2 \, dx=\frac {1}{45} x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (45 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+d x^4 \left (18 c \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )+5 d x^4 \operatorname {Hypergeometric2F1}\left (\frac {9}{4},-p,\frac {13}{4},-\frac {b x^4}{a}\right )\right )\right ) \] Input:

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

Output:

(x*(a + b*x^4)^p*(45*c^2*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + d 
*x^4*(18*c*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)] + 5*d*x^4*Hyperge 
ometric2F1[9/4, -p, 13/4, -((b*x^4)/a)])))/(45*(1 + (b*x^4)/a)^p)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {933, 25, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\left (b x^4+a\right )^p \left (d (5 a d-b c (4 p+13)) x^4+c (a d-b c (4 p+9))\right )dx}{b (4 p+9)}+\frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1}}{b (4 p+9)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1}}{b (4 p+9)}-\frac {\int \left (b x^4+a\right )^p \left (d (5 a d-b c (4 p+13)) x^4+c (a d-b c (4 p+9))\right )dx}{b (4 p+9)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1}}{b (4 p+9)}-\frac {\frac {d x \left (a+b x^4\right )^{p+1} (5 a d-b c (4 p+13))}{b (4 p+5)}-\frac {\left (5 a^2 d^2-2 a b c d (4 p+9)+b^2 c^2 \left (16 p^2+56 p+45\right )\right ) \int \left (b x^4+a\right )^pdx}{b (4 p+5)}}{b (4 p+9)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1}}{b (4 p+9)}-\frac {\frac {d x \left (a+b x^4\right )^{p+1} (5 a d-b c (4 p+13))}{b (4 p+5)}-\frac {\left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (5 a^2 d^2-2 a b c d (4 p+9)+b^2 c^2 \left (16 p^2+56 p+45\right )\right ) \int \left (\frac {b x^4}{a}+1\right )^pdx}{b (4 p+5)}}{b (4 p+9)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1}}{b (4 p+9)}-\frac {\frac {d x \left (a+b x^4\right )^{p+1} (5 a d-b c (4 p+13))}{b (4 p+5)}-\frac {x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (5 a^2 d^2-2 a b c d (4 p+9)+b^2 c^2 \left (16 p^2+56 p+45\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )}{b (4 p+5)}}{b (4 p+9)}\)

Input:

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

Output:

(d*x*(a + b*x^4)^(1 + p)*(c + d*x^4))/(b*(9 + 4*p)) - ((d*(5*a*d - b*c*(13 
 + 4*p))*x*(a + b*x^4)^(1 + p))/(b*(5 + 4*p)) - ((5*a^2*d^2 - 2*a*b*c*d*(9 
 + 4*p) + b^2*c^2*(45 + 56*p + 16*p^2))*x*(a + b*x^4)^p*Hypergeometric2F1[ 
1/4, -p, 5/4, -((b*x^4)/a)])/(b*(5 + 4*p)*(1 + (b*x^4)/a)^p))/(b*(9 + 4*p) 
)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]
 

rule 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^2*x^8 + 2*c*d*x^4 + c^2)*(b*x^4 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 81.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.69 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right )^2 \, dx=\frac {a^{p} c^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} c d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} + \frac {a^{p} d^{2} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, - p \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \] Input:

integrate((b*x**4+a)**p*(d*x**4+c)**2,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**5*gamma(5/4)*hyper((5/4, -p), (9/4,), b*x**4* 
exp_polar(I*pi)/a)/(2*gamma(9/4)) + a**p*d**2*x**9*gamma(9/4)*hyper((9/4, 
-p), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 20*(a + b*x**4)**p*a**2*d**2*p*x + 32*(a + b*x**4)**p*a*b*c*d*p**2*x + 
 72*(a + b*x**4)**p*a*b*c*d*p*x + 16*(a + b*x**4)**p*a*b*d**2*p**2*x**5 + 
4*(a + b*x**4)**p*a*b*d**2*p*x**5 + 16*(a + b*x**4)**p*b**2*c**2*p**2*x + 
56*(a + b*x**4)**p*b**2*c**2*p*x + 45*(a + b*x**4)**p*b**2*c**2*x + 32*(a 
+ b*x**4)**p*b**2*c*d*p**2*x**5 + 80*(a + b*x**4)**p*b**2*c*d*p*x**5 + 18* 
(a + b*x**4)**p*b**2*c*d*x**5 + 16*(a + b*x**4)**p*b**2*d**2*p**2*x**9 + 2 
4*(a + b*x**4)**p*b**2*d**2*p*x**9 + 5*(a + b*x**4)**p*b**2*d**2*x**9 + 12 
80*int((a + b*x**4)**p/(64*a*p**3 + 240*a*p**2 + 236*a*p + 45*a + 64*b*p** 
3*x**4 + 240*b*p**2*x**4 + 236*b*p*x**4 + 45*b*x**4),x)*a**3*d**2*p**4 + 4 
800*int((a + b*x**4)**p/(64*a*p**3 + 240*a*p**2 + 236*a*p + 45*a + 64*b*p* 
*3*x**4 + 240*b*p**2*x**4 + 236*b*p*x**4 + 45*b*x**4),x)*a**3*d**2*p**3 + 
4720*int((a + b*x**4)**p/(64*a*p**3 + 240*a*p**2 + 236*a*p + 45*a + 64*b*p 
**3*x**4 + 240*b*p**2*x**4 + 236*b*p*x**4 + 45*b*x**4),x)*a**3*d**2*p**2 + 
 900*int((a + b*x**4)**p/(64*a*p**3 + 240*a*p**2 + 236*a*p + 45*a + 64*b*p 
**3*x**4 + 240*b*p**2*x**4 + 236*b*p*x**4 + 45*b*x**4),x)*a**3*d**2*p - 20 
48*int((a + b*x**4)**p/(64*a*p**3 + 240*a*p**2 + 236*a*p + 45*a + 64*b*p** 
3*x**4 + 240*b*p**2*x**4 + 236*b*p*x**4 + 45*b*x**4),x)*a**2*b*c*d*p**5 - 
12288*int((a + b*x**4)**p/(64*a*p**3 + 240*a*p**2 + 236*a*p + 45*a + 64*b* 
p**3*x**4 + 240*b*p**2*x**4 + 236*b*p*x**4 + 45*b*x**4),x)*a**2*b*c*d*p**4 
 - 24832*int((a + b*x**4)**p/(64*a*p**3 + 240*a*p**2 + 236*a*p + 45*a +...