\(\int (c+d x^2)^3 (a+b x^4)^p \, dx\) [481]

Optimal result

Integrand size = 19, antiderivative size = 189 \[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\frac {3 c d^2 x \left (a+b x^4\right )^{1+p}}{b (5+4 p)}+\frac {d^3 x^3 \left (a+b x^4\right )^{1+p}}{b (7+4 p)}+c \left (c^2-\frac {3 a d^2}{5 b+4 b p}\right ) x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+d \left (c^2-\frac {a d^2}{7 b+4 b p}\right ) x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right ) \] Output:

3*c*d^2*x*(b*x^4+a)^(p+1)/b/(5+4*p)+d^3*x^3*(b*x^4+a)^(p+1)/b/(7+4*p)+c*(c 
^2-3*a*d^2/(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)+d*(c^2-a*d^2/(4*b*p+7*b))*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 = 0.58 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.72 \[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\frac {1}{35} x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (35 c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+d x^2 \left (35 c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )+d x^2 \left (21 c \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )+5 d x^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-\frac {b x^4}{a}\right )\right )\right )\right ) \] Input:

Integrate[(c + d*x^2)^3*(a + b*x^4)^p,x]
 

Output:

(x*(a + b*x^4)^p*(35*c^3*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + d 
*x^2*(35*c^2*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^4)/a)] + d*x^2*(21*c*H 
ypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)] + 5*d*x^2*Hypergeometric2F1[7 
/4, -p, 11/4, -((b*x^4)/a)]))))/(35*(1 + (b*x^4)/a)^p)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1519, 2432, 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.

\(\displaystyle \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx\)

\(\Big \downarrow \) 1519

\(\displaystyle \frac {\int \left (b x^4+a\right )^p \left (3 b c d^2 (4 p+7) x^4-3 d \left (a d^2-b c^2 (4 p+7)\right ) x^2+b c^3 (4 p+7)\right )dx}{b (4 p+7)}+\frac {d^3 x^3 \left (a+b x^4\right )^{p+1}}{b (4 p+7)}\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {\int \left (3 b c d^2 (4 p+7) x^4 \left (b x^4+a\right )^p+3 d \left (b c^2 (4 p+7)-a d^2\right ) x^2 \left (b x^4+a\right )^p+b c^3 (4 p+7) \left (b x^4+a\right )^p\right )dx}{b (4 p+7)}+\frac {d^3 x^3 \left (a+b x^4\right )^{p+1}}{b (4 p+7)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b c^3 (4 p+7) x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )-d x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (a d^2-b c^2 (4 p+7)\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )+\frac {3}{5} b c d^2 (4 p+7) x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^4}{a}\right )}{b (4 p+7)}+\frac {d^3 x^3 \left (a+b x^4\right )^{p+1}}{b (4 p+7)}\)

Input:

Int[(c + d*x^2)^3*(a + b*x^4)^p,x]
 

Output:

(d^3*x^3*(a + b*x^4)^(1 + p))/(b*(7 + 4*p)) + ((b*c^3*(7 + 4*p)*x*(a + b*x 
^4)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p - ( 
d*(a*d^2 - b*c^2*(7 + 4*p))*x^3*(a + b*x^4)^p*Hypergeometric2F1[3/4, -p, 7 
/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (3*b*c*d^2*(7 + 4*p)*x^5*(a + b*x^4 
)^p*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)])/(5*(1 + (b*x^4)/a)^p))/ 
(b*(7 + 4*p))
 

Defintions of rubi rules used

rule 1519

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Sim 
p[e^q*x^(2*q - 3)*((a + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Simp[1/(c 
*(4*p + 2*q + 1))   Int[(a + c*x^4)^p*ExpandToSum[c*(4*p + 2*q + 1)*(d + e* 
x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x 
], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[q, 1]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

Maple [F]

\[\int \left (d \,x^{2}+c \right )^{3} \left (b \,x^{4}+a \right )^{p}d x\]

Input:

int((d*x^2+c)^3*(b*x^4+a)^p,x)
 

Output:

int((d*x^2+c)^3*(b*x^4+a)^p,x)
 

Fricas [F]

\[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\int { {\left (d x^{2} + c\right )}^{3} {\left (b x^{4} + a\right )}^{p} \,d x } \] Input:

integrate((d*x^2+c)^3*(b*x^4+a)^p,x, algorithm="fricas")
 

Output:

integral((d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3)*(b*x^4 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 58.62 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.88 \[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\frac {a^{p} c^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {3 a^{p} c^{2} d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {3 a^{p} c d^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {a^{p} d^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, - p \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate((d*x**2+c)**3*(b*x**4+a)**p,x)
 

Output:

a**p*c**3*x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*pi)/a)/ 
(4*gamma(5/4)) + 3*a**p*c**2*d*x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b* 
x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + 3*a**p*c*d**2*x**5*gamma(5/4)*hyp 
er((5/4, -p), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4)) + a**p*d**3 
*x**7*gamma(7/4)*hyper((7/4, -p), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*ga 
mma(11/4))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\int { {\left (d x^{2} + c\right )}^{3} {\left (b x^{4} + a\right )}^{p} \,d x } \] Input:

integrate((d*x^2+c)^3*(b*x^4+a)^p,x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^3*(b*x^4 + a)^p, x)
 

Giac [F]

\[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\int { {\left (d x^{2} + c\right )}^{3} {\left (b x^{4} + a\right )}^{p} \,d x } \] Input:

integrate((d*x^2+c)^3*(b*x^4+a)^p,x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^3*(b*x^4 + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\int {\left (b\,x^4+a\right )}^p\,{\left (d\,x^2+c\right )}^3 \,d x \] Input:

int((a + b*x^4)^p*(c + d*x^2)^3,x)
 

Output:

int((a + b*x^4)^p*(c + d*x^2)^3, x)
 

Reduce [F]

\[ \int \left (c+d x^2\right )^3 \left (a+b x^4\right )^p \, dx=\text {too large to display} \] Input:

int((d*x^2+c)^3*(b*x^4+a)^p,x)
 

Output:

(192*(a + b*x**4)**p*a*c*d**2*p**3*x + 480*(a + b*x**4)**p*a*c*d**2*p**2*x 
 + 252*(a + b*x**4)**p*a*c*d**2*p*x + 64*(a + b*x**4)**p*a*d**3*p**3*x**3 
+ 96*(a + b*x**4)**p*a*d**3*p**2*x**3 + 20*(a + b*x**4)**p*a*d**3*p*x**3 + 
 64*(a + b*x**4)**p*b*c**3*p**3*x + 240*(a + b*x**4)**p*b*c**3*p**2*x + 28 
4*(a + b*x**4)**p*b*c**3*p*x + 105*(a + b*x**4)**p*b*c**3*x + 192*(a + b*x 
**4)**p*b*c**2*d*p**3*x**3 + 624*(a + b*x**4)**p*b*c**2*d*p**2*x**3 + 564* 
(a + b*x**4)**p*b*c**2*d*p*x**3 + 105*(a + b*x**4)**p*b*c**2*d*x**3 + 192* 
(a + b*x**4)**p*b*c*d**2*p**3*x**5 + 528*(a + b*x**4)**p*b*c*d**2*p**2*x** 
5 + 372*(a + b*x**4)**p*b*c*d**2*p*x**5 + 63*(a + b*x**4)**p*b*c*d**2*x**5 
 + 64*(a + b*x**4)**p*b*d**3*p**3*x**7 + 144*(a + b*x**4)**p*b*d**3*p**2*x 
**7 + 92*(a + b*x**4)**p*b*d**3*p*x**7 + 15*(a + b*x**4)**p*b*d**3*x**7 - 
49152*int((a + b*x**4)**p/(256*a*p**4 + 1024*a*p**3 + 1376*a*p**2 + 704*a* 
p + 105*a + 256*b*p**4*x**4 + 1024*b*p**3*x**4 + 1376*b*p**2*x**4 + 704*b* 
p*x**4 + 105*b*x**4),x)*a**2*c*d**2*p**7 - 319488*int((a + b*x**4)**p/(256 
*a*p**4 + 1024*a*p**3 + 1376*a*p**2 + 704*a*p + 105*a + 256*b*p**4*x**4 + 
1024*b*p**3*x**4 + 1376*b*p**2*x**4 + 704*b*p*x**4 + 105*b*x**4),x)*a**2*c 
*d**2*p**6 - 820224*int((a + b*x**4)**p/(256*a*p**4 + 1024*a*p**3 + 1376*a 
*p**2 + 704*a*p + 105*a + 256*b*p**4*x**4 + 1024*b*p**3*x**4 + 1376*b*p**2 
*x**4 + 704*b*p*x**4 + 105*b*x**4),x)*a**2*c*d**2*p**5 - 1053696*int((a + 
b*x**4)**p/(256*a*p**4 + 1024*a*p**3 + 1376*a*p**2 + 704*a*p + 105*a + ...