\(\int \frac {(c+d x^4) (a+b x^8)^p}{x^2} \, dx\) [10]

Optimal result

Integrand size = 20, antiderivative size = 99 \[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=-\frac {c \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},-p,\frac {7}{8},-\frac {b x^8}{a}\right )}{x}+\frac {1}{3} d x^3 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right ) \] Output:

-c*(b*x^8+a)^p*hypergeom([-1/8, -p],[7/8],-b*x^8/a)/x/((1+b*x^8/a)^p)+1/3* 
d*x^3*(b*x^8+a)^p*hypergeom([3/8, -p],[11/8],-b*x^8/a)/((1+b*x^8/a)^p)
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (-3 c \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},-p,\frac {7}{8},-\frac {b x^8}{a}\right )+d x^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )\right )}{3 x} \] Input:

Integrate[((c + d*x^4)*(a + b*x^8)^p)/x^2,x]
 

Output:

((a + b*x^8)^p*(-3*c*Hypergeometric2F1[-1/8, -p, 7/8, -((b*x^8)/a)] + d*x^ 
4*Hypergeometric2F1[3/8, -p, 11/8, -((b*x^8)/a)]))/(3*x*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1865, 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.

\(\displaystyle \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx\)

\(\Big \downarrow \) 1865

\(\displaystyle \int \left (\frac {c \left (a+b x^8\right )^p}{x^2}+d x^2 \left (a+b x^8\right )^p\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} d x^3 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )-\frac {c \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},-p,\frac {7}{8},-\frac {b x^8}{a}\right )}{x}\)

Input:

Int[((c + d*x^4)*(a + b*x^8)^p)/x^2,x]
 

Output:

-((c*(a + b*x^8)^p*Hypergeometric2F1[-1/8, -p, 7/8, -((b*x^8)/a)])/(x*(1 + 
 (b*x^8)/a)^p)) + (d*x^3*(a + b*x^8)^p*Hypergeometric2F1[3/8, -p, 11/8, -( 
(b*x^8)/a)])/(3*(1 + (b*x^8)/a)^p)
 

Defintions of rubi rules used

rule 1865

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ 
(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2*n))^p, (f*x)^m*(d 
 + e*x^n)^q, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n] && I 
GtQ[n, 0] && IGtQ[q, 0]
 

rule 2009

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

Maple [F]

\[\int \frac {\left (x^{4} d +c \right ) \left (b \,x^{8}+a \right )^{p}}{x^{2}}d x\]

Input:

int((d*x^4+c)*(b*x^8+a)^p/x^2,x)
 

Output:

int((d*x^4+c)*(b*x^8+a)^p/x^2,x)
 

Fricas [F]

\[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=\int { \frac {{\left (d x^{4} + c\right )} {\left (b x^{8} + a\right )}^{p}}{x^{2}} \,d x } \] Input:

integrate((d*x^4+c)*(b*x^8+a)^p/x^2,x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 123.98 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.79 \[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=\frac {a^{p} c \Gamma \left (- \frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{8}, - p \\ \frac {7}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 x \Gamma \left (\frac {7}{8}\right )} + \frac {a^{p} d x^{3} \Gamma \left (\frac {3}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{8}, - p \\ \frac {11}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {11}{8}\right )} \] Input:

integrate((d*x**4+c)*(b*x**8+a)**p/x**2,x)
 

Output:

a**p*c*gamma(-1/8)*hyper((-1/8, -p), (7/8,), b*x**8*exp_polar(I*pi)/a)/(8* 
x*gamma(7/8)) + a**p*d*x**3*gamma(3/8)*hyper((3/8, -p), (11/8,), b*x**8*ex 
p_polar(I*pi)/a)/(8*gamma(11/8))
 

Maxima [F]

\[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=\int { \frac {{\left (d x^{4} + c\right )} {\left (b x^{8} + a\right )}^{p}}{x^{2}} \,d x } \] Input:

integrate((d*x^4+c)*(b*x^8+a)^p/x^2,x, algorithm="maxima")
 

Output:

integrate((d*x^4 + c)*(b*x^8 + a)^p/x^2, x)
 

Giac [F]

\[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=\int { \frac {{\left (d x^{4} + c\right )} {\left (b x^{8} + a\right )}^{p}}{x^{2}} \,d x } \] Input:

integrate((d*x^4+c)*(b*x^8+a)^p/x^2,x, algorithm="giac")
 

Output:

integrate((d*x^4 + c)*(b*x^8 + a)^p/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=\int \frac {{\left (b\,x^8+a\right )}^p\,\left (d\,x^4+c\right )}{x^2} \,d x \] Input:

int(((a + b*x^8)^p*(c + d*x^4))/x^2,x)
 

Output:

int(((a + b*x^8)^p*(c + d*x^4))/x^2, x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x^2} \, dx=\frac {8 \left (b \,x^{8}+a \right )^{p} c p +3 \left (b \,x^{8}+a \right )^{p} c +8 \left (b \,x^{8}+a \right )^{p} d p \,x^{4}-\left (b \,x^{8}+a \right )^{p} d \,x^{4}+4096 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{10}+16 b p \,x^{10}-3 b \,x^{10}+64 a \,p^{2} x^{2}+16 a p \,x^{2}-3 a \,x^{2}}d x \right ) a c \,p^{4} x +2560 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{10}+16 b p \,x^{10}-3 b \,x^{10}+64 a \,p^{2} x^{2}+16 a p \,x^{2}-3 a \,x^{2}}d x \right ) a c \,p^{3} x +192 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{10}+16 b p \,x^{10}-3 b \,x^{10}+64 a \,p^{2} x^{2}+16 a p \,x^{2}-3 a \,x^{2}}d x \right ) a c \,p^{2} x -72 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{10}+16 b p \,x^{10}-3 b \,x^{10}+64 a \,p^{2} x^{2}+16 a p \,x^{2}-3 a \,x^{2}}d x \right ) a c p x +4096 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+16 b p \,x^{8}-3 b \,x^{8}+64 a \,p^{2}+16 a p -3 a}d x \right ) a d \,p^{4} x +512 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+16 b p \,x^{8}-3 b \,x^{8}+64 a \,p^{2}+16 a p -3 a}d x \right ) a d \,p^{3} x -320 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+16 b p \,x^{8}-3 b \,x^{8}+64 a \,p^{2}+16 a p -3 a}d x \right ) a d \,p^{2} x +24 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+16 b p \,x^{8}-3 b \,x^{8}+64 a \,p^{2}+16 a p -3 a}d x \right ) a d p x}{x \left (64 p^{2}+16 p -3\right )} \] Input:

int((d*x^4+c)*(b*x^8+a)^p/x^2,x)
 

Output:

(8*(a + b*x**8)**p*c*p + 3*(a + b*x**8)**p*c + 8*(a + b*x**8)**p*d*p*x**4 
- (a + b*x**8)**p*d*x**4 + 4096*int((a + b*x**8)**p/(64*a*p**2*x**2 + 16*a 
*p*x**2 - 3*a*x**2 + 64*b*p**2*x**10 + 16*b*p*x**10 - 3*b*x**10),x)*a*c*p* 
*4*x + 2560*int((a + b*x**8)**p/(64*a*p**2*x**2 + 16*a*p*x**2 - 3*a*x**2 + 
 64*b*p**2*x**10 + 16*b*p*x**10 - 3*b*x**10),x)*a*c*p**3*x + 192*int((a + 
b*x**8)**p/(64*a*p**2*x**2 + 16*a*p*x**2 - 3*a*x**2 + 64*b*p**2*x**10 + 16 
*b*p*x**10 - 3*b*x**10),x)*a*c*p**2*x - 72*int((a + b*x**8)**p/(64*a*p**2* 
x**2 + 16*a*p*x**2 - 3*a*x**2 + 64*b*p**2*x**10 + 16*b*p*x**10 - 3*b*x**10 
),x)*a*c*p*x + 4096*int(((a + b*x**8)**p*x**2)/(64*a*p**2 + 16*a*p - 3*a + 
 64*b*p**2*x**8 + 16*b*p*x**8 - 3*b*x**8),x)*a*d*p**4*x + 512*int(((a + b* 
x**8)**p*x**2)/(64*a*p**2 + 16*a*p - 3*a + 64*b*p**2*x**8 + 16*b*p*x**8 - 
3*b*x**8),x)*a*d*p**3*x - 320*int(((a + b*x**8)**p*x**2)/(64*a*p**2 + 16*a 
*p - 3*a + 64*b*p**2*x**8 + 16*b*p*x**8 - 3*b*x**8),x)*a*d*p**2*x + 24*int 
(((a + b*x**8)**p*x**2)/(64*a*p**2 + 16*a*p - 3*a + 64*b*p**2*x**8 + 16*b* 
p*x**8 - 3*b*x**8),x)*a*d*p*x)/(x*(64*p**2 + 16*p - 3))