Integrand size = 17, antiderivative size = 42 \[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right ) \] Output:
x*hypergeom([1/4, -p],[5/4],-b*x^4)-1/3*x^3*hypergeom([3/4, -p],[7/4],-b*x ^4)
Time = 0.55 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right ) \] Input:
Integrate[(1 - x^2)*(1 + b*x^4)^p,x]
Output:
x*Hypergeometric2F1[1/4, -p, 5/4, -(b*x^4)] - (x^3*Hypergeometric2F1[3/4, -p, 7/4, -(b*x^4)])/3
Time = 0.29 (sec) , antiderivative size = 42, 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 = {1516, 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.
\(\displaystyle \int \left (1-x^2\right ) \left (b x^4+1\right )^p \, dx\) |
\(\Big \downarrow \) 1516 |
\(\displaystyle \int \left (\left (b x^4+1\right )^p-x^2 \left (b x^4+1\right )^p\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )\) |
Input:
Int[(1 - x^2)*(1 + b*x^4)^p,x]
Output:
x*Hypergeometric2F1[1/4, -p, 5/4, -(b*x^4)] - (x^3*Hypergeometric2F1[3/4, -p, 7/4, -(b*x^4)])/3
Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[Expa ndIntegrand[(d + e*x^2)*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]
Time = 0.65 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88
method | result | size |
meijerg | \(x \operatorname {hypergeom}\left (\left [\frac {1}{4}, -p \right ], \left [\frac {5}{4}\right ], -b \,x^{4}\right )-\frac {x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, -p \right ], \left [\frac {7}{4}\right ], -b \,x^{4}\right )}{3}\) | \(37\) |
Input:
int((-x^2+1)*(b*x^4+1)^p,x,method=_RETURNVERBOSE)
Output:
x*hypergeom([1/4,-p],[5/4],-b*x^4)-1/3*x^3*hypergeom([3/4,-p],[7/4],-b*x^4 )
\[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:
integrate((-x^2+1)*(b*x^4+1)^p,x, algorithm="fricas")
Output:
integral(-(x^2 - 1)*(b*x^4 + 1)^p, x)
Result contains complex when optimal does not.
Time = 13.81 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.45 \[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=- \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((-x**2+1)*(b*x**4+1)**p,x)
Output:
-x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**4*exp_polar(I*pi))/(4*gamma (7/4)) + x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*pi))/(4* gamma(5/4))
\[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:
integrate((-x^2+1)*(b*x^4+1)^p,x, algorithm="maxima")
Output:
-integrate((x^2 - 1)*(b*x^4 + 1)^p, x)
\[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:
integrate((-x^2+1)*(b*x^4+1)^p,x, algorithm="giac")
Output:
integrate(-(x^2 - 1)*(b*x^4 + 1)^p, x)
Timed out. \[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=-\int \left (x^2-1\right )\,{\left (b\,x^4+1\right )}^p \,d x \] Input:
int(-(x^2 - 1)*(b*x^4 + 1)^p,x)
Output:
-int((x^2 - 1)*(b*x^4 + 1)^p, x)
\[ \int \left (1-x^2\right ) \left (1+b x^4\right )^p \, dx=\frac {-4 \left (b \,x^{4}+1\right )^{p} p \,x^{3}+4 \left (b \,x^{4}+1\right )^{p} p x -\left (b \,x^{4}+1\right )^{p} x^{3}+3 \left (b \,x^{4}+1\right )^{p} x +256 \left (\int \frac {\left (b \,x^{4}+1\right )^{p}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p^{4}+448 \left (\int \frac {\left (b \,x^{4}+1\right )^{p}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p^{3}+240 \left (\int \frac {\left (b \,x^{4}+1\right )^{p}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p^{2}+36 \left (\int \frac {\left (b \,x^{4}+1\right )^{p}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p -256 \left (\int \frac {\left (b \,x^{4}+1\right )^{p} x^{2}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p^{4}-320 \left (\int \frac {\left (b \,x^{4}+1\right )^{p} x^{2}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p^{3}-112 \left (\int \frac {\left (b \,x^{4}+1\right )^{p} x^{2}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p^{2}-12 \left (\int \frac {\left (b \,x^{4}+1\right )^{p} x^{2}}{16 b \,p^{2} x^{4}+16 b p \,x^{4}+3 b \,x^{4}+16 p^{2}+16 p +3}d x \right ) p}{16 p^{2}+16 p +3} \] Input:
int((-x^2+1)*(b*x^4+1)^p,x)
Output:
( - 4*(b*x**4 + 1)**p*p*x**3 + 4*(b*x**4 + 1)**p*p*x - (b*x**4 + 1)**p*x** 3 + 3*(b*x**4 + 1)**p*x + 256*int((b*x**4 + 1)**p/(16*b*p**2*x**4 + 16*b*p *x**4 + 3*b*x**4 + 16*p**2 + 16*p + 3),x)*p**4 + 448*int((b*x**4 + 1)**p/( 16*b*p**2*x**4 + 16*b*p*x**4 + 3*b*x**4 + 16*p**2 + 16*p + 3),x)*p**3 + 24 0*int((b*x**4 + 1)**p/(16*b*p**2*x**4 + 16*b*p*x**4 + 3*b*x**4 + 16*p**2 + 16*p + 3),x)*p**2 + 36*int((b*x**4 + 1)**p/(16*b*p**2*x**4 + 16*b*p*x**4 + 3*b*x**4 + 16*p**2 + 16*p + 3),x)*p - 256*int(((b*x**4 + 1)**p*x**2)/(16 *b*p**2*x**4 + 16*b*p*x**4 + 3*b*x**4 + 16*p**2 + 16*p + 3),x)*p**4 - 320* int(((b*x**4 + 1)**p*x**2)/(16*b*p**2*x**4 + 16*b*p*x**4 + 3*b*x**4 + 16*p **2 + 16*p + 3),x)*p**3 - 112*int(((b*x**4 + 1)**p*x**2)/(16*b*p**2*x**4 + 16*b*p*x**4 + 3*b*x**4 + 16*p**2 + 16*p + 3),x)*p**2 - 12*int(((b*x**4 + 1)**p*x**2)/(16*b*p**2*x**4 + 16*b*p*x**4 + 3*b*x**4 + 16*p**2 + 16*p + 3) ,x)*p)/(16*p**2 + 16*p + 3)