\(\int (a+b x^4)^p (c+d x^4)^3 \, dx\) [129]

Optimal result

Integrand size = 19, antiderivative size = 282 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right )^3 \, dx=\frac {3 d \left (15 a^2 d^2-5 a b c d (13+4 p)+b^2 c^2 \left (117+88 p+16 p^2\right )\right ) x \left (a+b x^4\right )^{1+p}}{b^3 (5+4 p) (9+4 p) (13+4 p)}-\frac {3 d^2 (3 a d-b c (13+4 p)) x^5 \left (a+b x^4\right )^{1+p}}{b^2 (9+4 p) (13+4 p)}+\frac {d^3 x^9 \left (a+b x^4\right )^{1+p}}{b (13+4 p)}+\frac {\left (b^3 c^3 \left (117+88 p+16 p^2\right )-\frac {3 a d \left (15 a^2 d^2-5 a b c d (13+4 p)+b^2 c^2 \left (117+88 p+16 p^2\right )\right )}{5+4 p}\right ) x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )}{b^3 (9+4 p) (13+4 p)} \] Output:

3*d*(15*a^2*d^2-5*a*b*c*d*(13+4*p)+b^2*c^2*(16*p^2+88*p+117))*x*(b*x^4+a)^ 
(p+1)/b^3/(5+4*p)/(9+4*p)/(13+4*p)-3*d^2*(3*a*d-b*c*(13+4*p))*x^5*(b*x^4+a 
)^(p+1)/b^2/(9+4*p)/(13+4*p)+d^3*x^9*(b*x^4+a)^(p+1)/b/(13+4*p)+(b^3*c^3*( 
16*p^2+88*p+117)-3*a*d*(15*a^2*d^2-5*a*b*c*d*(13+4*p)+b^2*c^2*(16*p^2+88*p 
+117))/(5+4*p))*x*(b*x^4+a)^p*hypergeom([1/4, -p],[5/4],-b*x^4/a)/b^3/(9+4 
*p)/(13+4*p)/((1+b*x^4/a)^p)
 

Mathematica [A] (verified)

Time = 5.44 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.49 \[ \int \left (a+b x^4\right )^p \left (c+d x^4\right )^3 \, dx=\frac {1}{195} x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (195 c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+d x^4 \left (117 c^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )+5 d x^4 \left (13 c \operatorname {Hypergeometric2F1}\left (\frac {9}{4},-p,\frac {13}{4},-\frac {b x^4}{a}\right )+3 d x^4 \operatorname {Hypergeometric2F1}\left (\frac {13}{4},-p,\frac {17}{4},-\frac {b x^4}{a}\right )\right )\right )\right ) \] Input:

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

Output:

(x*(a + b*x^4)^p*(195*c^3*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + 
d*x^4*(117*c^2*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)] + 5*d*x^4*(13 
*c*Hypergeometric2F1[9/4, -p, 13/4, -((b*x^4)/a)] + 3*d*x^4*Hypergeometric 
2F1[13/4, -p, 17/4, -((b*x^4)/a)]))))/(195*(1 + (b*x^4)/a)^p)
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {933, 25, 1025, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\left (b x^4+a\right )^p \left (d x^4+c\right ) \left (d (9 a d-b c (4 p+21)) x^4+c (a d-b c (4 p+13))\right )dx}{b (4 p+13)}+\frac {d x \left (c+d x^4\right )^2 \left (a+b x^4\right )^{p+1}}{b (4 p+13)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^4\right )^2 \left (a+b x^4\right )^{p+1}}{b (4 p+13)}-\frac {\int \left (b x^4+a\right )^p \left (d x^4+c\right ) \left (d (9 a d-b c (4 p+21)) x^4+c (a d-b c (4 p+13))\right )dx}{b (4 p+13)}\)

\(\Big \downarrow \) 1025

\(\displaystyle \frac {d x \left (c+d x^4\right )^2 \left (a+b x^4\right )^{p+1}}{b (4 p+13)}-\frac {\frac {\int -\left (b x^4+a\right )^p \left (d \left (b^2 \left (16 p^2+104 p+201\right ) c^2-6 a b d (4 p+25) c+45 a^2 d^2\right ) x^4+c \left (b^2 \left (16 p^2+88 p+117\right ) c^2-2 a b d (4 p+15) c+9 a^2 d^2\right )\right )dx}{b (4 p+9)}+\frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1} (9 a d-b c (4 p+21))}{b (4 p+9)}}{b (4 p+13)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^4\right )^2 \left (a+b x^4\right )^{p+1}}{b (4 p+13)}-\frac {\frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1} (9 a d-b c (4 p+21))}{b (4 p+9)}-\frac {\int \left (b x^4+a\right )^p \left (d \left (b^2 \left (16 p^2+104 p+201\right ) c^2-6 a b d (4 p+25) c+45 a^2 d^2\right ) x^4+c \left (b^2 \left (16 p^2+88 p+117\right ) c^2-2 a b d (4 p+15) c+9 a^2 d^2\right )\right )dx}{b (4 p+9)}}{b (4 p+13)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d x \left (c+d x^4\right )^2 \left (a+b x^4\right )^{p+1}}{b (4 p+13)}-\frac {\frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1} (9 a d-b c (4 p+21))}{b (4 p+9)}-\frac {\frac {d x \left (a+b x^4\right )^{p+1} \left (45 a^2 d^2-6 a b c d (4 p+25)+b^2 c^2 \left (16 p^2+104 p+201\right )\right )}{b (4 p+5)}-\frac {\left (45 a^3 d^3-15 a^2 b c d^2 (4 p+13)+3 a b^2 c^2 d \left (16 p^2+88 p+117\right )-b^3 c^3 \left (64 p^3+432 p^2+908 p+585\right )\right ) \int \left (b x^4+a\right )^pdx}{b (4 p+5)}}{b (4 p+9)}}{b (4 p+13)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {d x \left (c+d x^4\right )^2 \left (a+b x^4\right )^{p+1}}{b (4 p+13)}-\frac {\frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1} (9 a d-b c (4 p+21))}{b (4 p+9)}-\frac {\frac {d x \left (a+b x^4\right )^{p+1} \left (45 a^2 d^2-6 a b c d (4 p+25)+b^2 c^2 \left (16 p^2+104 p+201\right )\right )}{b (4 p+5)}-\frac {\left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (45 a^3 d^3-15 a^2 b c d^2 (4 p+13)+3 a b^2 c^2 d \left (16 p^2+88 p+117\right )-b^3 c^3 \left (64 p^3+432 p^2+908 p+585\right )\right ) \int \left (\frac {b x^4}{a}+1\right )^pdx}{b (4 p+5)}}{b (4 p+9)}}{b (4 p+13)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {d x \left (c+d x^4\right )^2 \left (a+b x^4\right )^{p+1}}{b (4 p+13)}-\frac {\frac {d x \left (c+d x^4\right ) \left (a+b x^4\right )^{p+1} (9 a d-b c (4 p+21))}{b (4 p+9)}-\frac {\frac {d x \left (a+b x^4\right )^{p+1} \left (45 a^2 d^2-6 a b c d (4 p+25)+b^2 c^2 \left (16 p^2+104 p+201\right )\right )}{b (4 p+5)}-\frac {x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (45 a^3 d^3-15 a^2 b c d^2 (4 p+13)+3 a b^2 c^2 d \left (16 p^2+88 p+117\right )-b^3 c^3 \left (64 p^3+432 p^2+908 p+585\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )}{b (4 p+5)}}{b (4 p+9)}}{b (4 p+13)}\)

Input:

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

Output:

(d*x*(a + b*x^4)^(1 + p)*(c + d*x^4)^2)/(b*(13 + 4*p)) - ((d*(9*a*d - b*c* 
(21 + 4*p))*x*(a + b*x^4)^(1 + p)*(c + d*x^4))/(b*(9 + 4*p)) - ((d*(45*a^2 
*d^2 - 6*a*b*c*d*(25 + 4*p) + b^2*c^2*(201 + 104*p + 16*p^2))*x*(a + b*x^4 
)^(1 + p))/(b*(5 + 4*p)) - ((45*a^3*d^3 - 15*a^2*b*c*d^2*(13 + 4*p) + 3*a* 
b^2*c^2*d*(117 + 88*p + 16*p^2) - b^3*c^3*(585 + 908*p + 432*p^2 + 64*p^3) 
)*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)))/(b*(13 + 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]
 

rule 1025

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^12 + 3*c*d^2*x^8 + 3*c^2*d*x^4 + c^3)*(b*x^4 + a)^p, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(180*(a + b*x**4)**p*a**3*d**3*p*x - 240*(a + b*x**4)**p*a**2*b*c*d**2*p** 
2*x - 780*(a + b*x**4)**p*a**2*b*c*d**2*p*x - 144*(a + b*x**4)**p*a**2*b*d 
**3*p**2*x**5 - 36*(a + b*x**4)**p*a**2*b*d**3*p*x**5 + 192*(a + b*x**4)** 
p*a*b**2*c**2*d*p**3*x + 1056*(a + b*x**4)**p*a*b**2*c**2*d*p**2*x + 1404* 
(a + b*x**4)**p*a*b**2*c**2*d*p*x + 192*(a + b*x**4)**p*a*b**2*c*d**2*p**3 
*x**5 + 672*(a + b*x**4)**p*a*b**2*c*d**2*p**2*x**5 + 156*(a + b*x**4)**p* 
a*b**2*c*d**2*p*x**5 + 64*(a + b*x**4)**p*a*b**2*d**3*p**3*x**9 + 96*(a + 
b*x**4)**p*a*b**2*d**3*p**2*x**9 + 20*(a + b*x**4)**p*a*b**2*d**3*p*x**9 + 
 64*(a + b*x**4)**p*b**3*c**3*p**3*x + 432*(a + b*x**4)**p*b**3*c**3*p**2* 
x + 908*(a + b*x**4)**p*b**3*c**3*p*x + 585*(a + b*x**4)**p*b**3*c**3*x + 
192*(a + b*x**4)**p*b**3*c**2*d*p**3*x**5 + 1104*(a + b*x**4)**p*b**3*c**2 
*d*p**2*x**5 + 1668*(a + b*x**4)**p*b**3*c**2*d*p*x**5 + 351*(a + b*x**4)* 
*p*b**3*c**2*d*x**5 + 192*(a + b*x**4)**p*b**3*c*d**2*p**3*x**9 + 912*(a + 
 b*x**4)**p*b**3*c*d**2*p**2*x**9 + 996*(a + b*x**4)**p*b**3*c*d**2*p*x**9 
 + 195*(a + b*x**4)**p*b**3*c*d**2*x**9 + 64*(a + b*x**4)**p*b**3*d**3*p** 
3*x**13 + 240*(a + b*x**4)**p*b**3*d**3*p**2*x**13 + 236*(a + b*x**4)**p*b 
**3*d**3*p*x**13 + 45*(a + b*x**4)**p*b**3*d**3*x**13 - 46080*int((a + b*x 
**4)**p/(256*a*p**4 + 1792*a*p**3 + 4064*a*p**2 + 3248*a*p + 585*a + 256*b 
*p**4*x**4 + 1792*b*p**3*x**4 + 4064*b*p**2*x**4 + 3248*b*p*x**4 + 585*b*x 
**4),x)*a**4*d**3*p**5 - 322560*int((a + b*x**4)**p/(256*a*p**4 + 1792*...