Integrand size = 18, antiderivative size = 101 \[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\frac {1}{2} c x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+\frac {1}{6} d x^6 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^8}{a}\right ) \] Output:
1/2*c*x^2*(b*x^8+a)^p*hypergeom([1/4, -p],[5/4],-b*x^8/a)/((1+b*x^8/a)^p)+ 1/6*d*x^6*(b*x^8+a)^p*hypergeom([3/4, -p],[7/4],-b*x^8/a)/((1+b*x^8/a)^p)
Time = 0.56 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\frac {1}{6} x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (3 c \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+d x^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^8}{a}\right )\right ) \] Input:
Integrate[x*(c + d*x^4)*(a + b*x^8)^p,x]
Output:
(x^2*(a + b*x^8)^p*(3*c*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8)/a)] + d* x^4*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^8)/a)]))/(6*(1 + (b*x^8)/a)^p)
Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1815, 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.
\(\displaystyle \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx\) |
\(\Big \downarrow \) 1815 |
\(\displaystyle \frac {1}{2} \int \left (d x^4+c\right ) \left (b x^8+a\right )^pdx^2\) |
\(\Big \downarrow \) 1516 |
\(\displaystyle \frac {1}{2} \int \left (d x^4 \left (b x^8+a\right )^p+c \left (b x^8+a\right )^p\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (c x^2 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+\frac {1}{3} d x^6 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^8}{a}\right )\right )\) |
Input:
Int[x*(c + d*x^4)*(a + b*x^8)^p,x]
Output:
((c*x^2*(a + b*x^8)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8)/a)])/(1 + (b*x^8)/a)^p + (d*x^6*(a + b*x^8)^p*Hypergeometric2F1[3/4, -p, 7/4, -((b*x ^8)/a)])/(3*(1 + (b*x^8)/a)^p))/2
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]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_ .), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/ k - 1)*(d + e*x^(n/k))^q*(a + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IntegerQ[m ]
\[\int x \left (x^{4} d +c \right ) \left (b \,x^{8}+a \right )^{p}d x\]
Input:
int(x*(d*x^4+c)*(b*x^8+a)^p,x)
Output:
int(x*(d*x^4+c)*(b*x^8+a)^p,x)
\[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{4} + c\right )} {\left (b x^{8} + a\right )}^{p} x \,d x } \] Input:
integrate(x*(d*x^4+c)*(b*x^8+a)^p,x, algorithm="fricas")
Output:
integral((d*x^5 + c*x)*(b*x^8 + a)^p, x)
Result contains complex when optimal does not.
Time = 91.71 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75 \[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\frac {a^{p} c x^{2} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} d x^{6} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate(x*(d*x**4+c)*(b*x**8+a)**p,x)
Output:
a**p*c*x**2*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**8*exp_polar(I*pi)/a)/ (8*gamma(5/4)) + a**p*d*x**6*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**8*ex p_polar(I*pi)/a)/(8*gamma(7/4))
\[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{4} + c\right )} {\left (b x^{8} + a\right )}^{p} x \,d x } \] Input:
integrate(x*(d*x^4+c)*(b*x^8+a)^p,x, algorithm="maxima")
Output:
integrate((d*x^4 + c)*(b*x^8 + a)^p*x, x)
\[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{4} + c\right )} {\left (b x^{8} + a\right )}^{p} x \,d x } \] Input:
integrate(x*(d*x^4+c)*(b*x^8+a)^p,x, algorithm="giac")
Output:
integrate((d*x^4 + c)*(b*x^8 + a)^p*x, x)
Timed out. \[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\int x\,{\left (b\,x^8+a\right )}^p\,\left (d\,x^4+c\right ) \,d x \] Input:
int(x*(a + b*x^8)^p*(c + d*x^4),x)
Output:
int(x*(a + b*x^8)^p*(c + d*x^4), x)
\[ \int x \left (c+d x^4\right ) \left (a+b x^8\right )^p \, dx=\frac {4 \left (b \,x^{8}+a \right )^{p} c p \,x^{2}+3 \left (b \,x^{8}+a \right )^{p} c \,x^{2}+4 \left (b \,x^{8}+a \right )^{p} d p \,x^{6}+\left (b \,x^{8}+a \right )^{p} d \,x^{6}+512 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{5}}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a d \,p^{4}+640 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{5}}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a d \,p^{3}+224 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{5}}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a d \,p^{2}+24 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{5}}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a d p +512 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a c \,p^{4}+896 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a c \,p^{3}+480 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a c \,p^{2}+72 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{16 b \,p^{2} x^{8}+16 b p \,x^{8}+3 b \,x^{8}+16 a \,p^{2}+16 a p +3 a}d x \right ) a c p}{32 p^{2}+32 p +6} \] Input:
int(x*(d*x^4+c)*(b*x^8+a)^p,x)
Output:
(4*(a + b*x**8)**p*c*p*x**2 + 3*(a + b*x**8)**p*c*x**2 + 4*(a + b*x**8)**p *d*p*x**6 + (a + b*x**8)**p*d*x**6 + 512*int(((a + b*x**8)**p*x**5)/(16*a* p**2 + 16*a*p + 3*a + 16*b*p**2*x**8 + 16*b*p*x**8 + 3*b*x**8),x)*a*d*p**4 + 640*int(((a + b*x**8)**p*x**5)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x* *8 + 16*b*p*x**8 + 3*b*x**8),x)*a*d*p**3 + 224*int(((a + b*x**8)**p*x**5)/ (16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**8 + 16*b*p*x**8 + 3*b*x**8),x)*a* d*p**2 + 24*int(((a + b*x**8)**p*x**5)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p* *2*x**8 + 16*b*p*x**8 + 3*b*x**8),x)*a*d*p + 512*int(((a + b*x**8)**p*x)/( 16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**8 + 16*b*p*x**8 + 3*b*x**8),x)*a*c *p**4 + 896*int(((a + b*x**8)**p*x)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2* x**8 + 16*b*p*x**8 + 3*b*x**8),x)*a*c*p**3 + 480*int(((a + b*x**8)**p*x)/( 16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x**8 + 16*b*p*x**8 + 3*b*x**8),x)*a*c *p**2 + 72*int(((a + b*x**8)**p*x)/(16*a*p**2 + 16*a*p + 3*a + 16*b*p**2*x **8 + 16*b*p*x**8 + 3*b*x**8),x)*a*c*p)/(2*(16*p**2 + 16*p + 3))