Integrand size = 17, antiderivative size = 177 \[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\frac {d^3 \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+c^3 x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+\frac {3}{2} c^2 d x^2 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )+c d^2 x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right ) \] Output:
1/4*d^3*(b*x^4+a)^(p+1)/b/(p+1)+c^3*x*(b*x^4+a)^p*hypergeom([1/4, -p],[5/4 ],-b*x^4/a)/((1+b*x^4/a)^p)+3/2*c^2*d*x^2*(b*x^4+a)^p*hypergeom([1/2, -p], [3/2],-b*x^4/a)/((1+b*x^4/a)^p)+c*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)
Time = 0.72 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.86 \[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\frac {\left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \left (4 b c^3 (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+d \left (6 b c^2 (1+p) x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )+d \left (d \left (a+b x^4\right ) \left (1+\frac {b x^4}{a}\right )^p+4 b c (1+p) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )\right )\right )\right )}{4 b (1+p)} \] Input:
Integrate[(c + d*x)^3*(a + b*x^4)^p,x]
Output:
((a + b*x^4)^p*(4*b*c^3*(1 + p)*x*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4 )/a)] + d*(6*b*c^2*(1 + p)*x^2*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a )] + d*(d*(a + b*x^4)*(1 + (b*x^4)/a)^p + 4*b*c*(1 + p)*x^3*Hypergeometric 2F1[3/4, -p, 7/4, -((b*x^4)/a)]))))/(4*b*(1 + p)*(1 + (b*x^4)/a)^p)
Time = 0.57 (sec) , antiderivative size = 177, 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.
| \(\displaystyle \int (c+d x)^3 \left (a+b x^4\right )^p \, dx\) |
| \(\Big \downarrow \) 2424 |
| \(\displaystyle \int \left (\left (c^3+3 c d^2 x^2\right ) \left (a+b x^4\right )^p+x \left (3 c^2 d+d^3 x^2\right ) \left (a+b x^4\right )^p\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle c^3 x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+\frac {3}{2} c^2 d x^2 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )+c d^2 x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )+\frac {d^3 \left (a+b x^4\right )^{p+1}}{4 b (p+1)}\) |
Input:
Int[(c + d*x)^3*(a + b*x^4)^p,x]
Output:
(d^3*(a + b*x^4)^(1 + p))/(4*b*(1 + p)) + (c^3*x*(a + b*x^4)^p*Hypergeomet ric2F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (3*c^2*d*x^2*(a + b*x^4)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a)])/(2*(1 + (b*x^4)/a) ^p) + (c*d^2*x^3*(a + b*x^4)^p*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^4)/a )])/(1 + (b*x^4)/a)^p
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)]
\[\int \left (d x +c \right )^{3} \left (b \,x^{4}+a \right )^{p}d x\]
Input:
int((d*x+c)^3*(b*x^4+a)^p,x)
Output:
int((d*x+c)^3*(b*x^4+a)^p,x)
\[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{4} + a\right )}^{p} \,d x } \] Input:
integrate((d*x+c)^3*(b*x^4+a)^p,x, algorithm="fricas")
Output:
integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(b*x^4 + a)^p, x)
Time = 24.64 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.88 \[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\frac {a^{p} c^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {3 a^{p} c^{2} d x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2} + \frac {3 a^{p} c d^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + d^{3} \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{4}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{4} \right )} & \text {otherwise} \end {cases}}{4 b} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((d*x+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**2*hyper((1/2, -p), (3/2,), b*x**4*exp_po lar(I*pi)/a)/2 + 3*a**p*c*d**2*x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b* x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + d**3*Piecewise((a**p*x**4/4, Eq(b , 0)), (Piecewise(((a + b*x**4)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + b*x **4), True))/(4*b), True))
\[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{4} + a\right )}^{p} \,d x } \] Input:
integrate((d*x+c)^3*(b*x^4+a)^p,x, algorithm="maxima")
Output:
integrate((d*x + c)^3*(b*x^4 + a)^p, x)
\[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{4} + a\right )}^{p} \,d x } \] Input:
integrate((d*x+c)^3*(b*x^4+a)^p,x, algorithm="giac")
Output:
integrate((d*x + c)^3*(b*x^4 + a)^p, x)
Timed out. \[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\int {\left (b\,x^4+a\right )}^p\,{\left (c+d\,x\right )}^3 \,d x \] Input:
int((a + b*x^4)^p*(c + d*x)^3,x)
Output:
int((a + b*x^4)^p*(c + d*x)^3, x)
\[ \int (c+d x)^3 \left (a+b x^4\right )^p \, dx=\text {too large to display} \] Input:
int((d*x+c)^3*(b*x^4+a)^p,x)
Output:
(32*(a + b*x**4)**p*a*d**3*p**3 + 48*(a + b*x**4)**p*a*d**3*p**2 + 22*(a + b*x**4)**p*a*d**3*p + 3*(a + b*x**4)**p*a*d**3 + 32*(a + b*x**4)**p*b*c** 3*p**3*x + 72*(a + b*x**4)**p*b*c**3*p**2*x + 52*(a + b*x**4)**p*b*c**3*p* x + 12*(a + b*x**4)**p*b*c**3*x + 96*(a + b*x**4)**p*b*c**2*d*p**3*x**2 + 192*(a + b*x**4)**p*b*c**2*d*p**2*x**2 + 114*(a + b*x**4)**p*b*c**2*d*p*x* *2 + 18*(a + b*x**4)**p*b*c**2*d*x**2 + 96*(a + b*x**4)**p*b*c*d**2*p**3*x **3 + 168*(a + b*x**4)**p*b*c*d**2*p**2*x**3 + 84*(a + b*x**4)**p*b*c*d**2 *p*x**3 + 12*(a + b*x**4)**p*b*c*d**2*x**3 + 32*(a + b*x**4)**p*b*d**3*p** 3*x**4 + 48*(a + b*x**4)**p*b*d**3*p**2*x**4 + 22*(a + b*x**4)**p*b*d**3*p *x**4 + 3*(a + b*x**4)**p*b*d**3*x**4 + 4096*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*b*c**3*p**7 + 15360*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*b*c**3*p**6 + 23296*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*b*c**3*p**5 + 18240*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*b*c**3*p**4 + 7744*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*b *c**3*p**3 + 1680*int((a + b*x**4)**p/(32*a*p**3 + 48*a*p**2 + 22*a*p +...