\(\int \frac {(x^2)^{-p} (1+b x^2)^p}{c x+d x^3} \, dx\) [515]

Optimal result

Integrand size = 28, antiderivative size = 59 \[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=-\frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-d) x^2}{c \left (1+b x^2\right )}\right )}{2 c p} \] Output:

-1/2*(b*x^2+1)^p*hypergeom([1, -p],[1-p],(b*c-d)*x^2/c/(b*x^2+1))/c/p/((x^ 
2)^p)
 

Mathematica [A] (warning: unable to verify)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.02 \[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=-\frac {\left (x^2\right )^{-p} \left (1+\frac {d x^2}{c}\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {(-b c+d) x^2}{c+d x^2}\right )}{2 c p} \] Input:

Integrate[(1 + b*x^2)^p/((x^2)^p*(c*x + d*x^3)),x]
 

Output:

-1/2*((1 + (d*x^2)/c)^p*Hypergeometric2F1[-p, -p, 1 - p, ((-(b*c) + d)*x^2 
)/(c + d*x^2)])/(c*p*(x^2)^p)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {34, 9, 393, 141}

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.

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

\(\Big \downarrow \) 34

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 393

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

\(\Big \downarrow \) 141

\(\displaystyle -\frac {x^{2 p-2 (p+1)+2} \left (x^2\right )^{-p} \left (b x^2+1\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-d) x^2}{c \left (b x^2+1\right )}\right )}{2 c p}\)

Input:

Int[(1 + b*x^2)^p/((x^2)^p*(c*x + d*x^3)),x]
 

Output:

-1/2*(x^(2 + 2*p - 2*(1 + p))*(1 + b*x^2)^p*Hypergeometric2F1[1, -p, 1 - p 
, ((b*c - d)*x^2)/(c*(1 + b*x^2))])/(c*p*(x^2)^p)
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

rule 34

Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F 
racPart[p]/x^(m*FracPart[p]))   Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x 
] &&  !IntegerQ[p]
 

rule 141

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( 
n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f 
))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, 
p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !Su 
mSimplerQ[p, 1]) &&  !ILtQ[m, 0]
 

rule 393

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q 
_.), x_Symbol] :> Simp[(e*x)^m/(2*x*(x^2)^(Simplify[(m + 1)/2] - 1))   Subs 
t[Int[x^(Simplify[(m + 1)/2] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] 
/; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ[Simp 
lify[m + 2*p]] &&  !IntegerQ[m]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+1\right )^{p} \left (x^{2}\right )^{-p}}{d \,x^{3}+c x}d x\]

Input:

int((b*x^2+1)^p/((x^2)^p)/(d*x^3+c*x),x)
 

Output:

int((b*x^2+1)^p/((x^2)^p)/(d*x^3+c*x),x)
 

Fricas [F]

\[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int { \frac {{\left (b x^{2} + 1\right )}^{p}}{{\left (d x^{3} + c x\right )} {\left (x^{2}\right )}^{p}} \,d x } \] Input:

integrate((b*x^2+1)^p/((x^2)^p)/(d*x^3+c*x),x, algorithm="fricas")
 

Output:

integral((b*x^2 + 1)^p/((d*x^3 + c*x)*(x^2)^p), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+1)**p/((x**2)**p)/(d*x**3+c*x),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int { \frac {{\left (b x^{2} + 1\right )}^{p}}{{\left (d x^{3} + c x\right )} {\left (x^{2}\right )}^{p}} \,d x } \] Input:

integrate((b*x^2+1)^p/((x^2)^p)/(d*x^3+c*x),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + 1)^p/((d*x^3 + c*x)*(x^2)^p), x)
 

Giac [F]

\[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int { \frac {{\left (b x^{2} + 1\right )}^{p}}{{\left (d x^{3} + c x\right )} {\left (x^{2}\right )}^{p}} \,d x } \] Input:

integrate((b*x^2+1)^p/((x^2)^p)/(d*x^3+c*x),x, algorithm="giac")
 

Output:

integrate((b*x^2 + 1)^p/((d*x^3 + c*x)*(x^2)^p), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int \frac {{\left (b\,x^2+1\right )}^p}{\left (d\,x^3+c\,x\right )\,{\left (x^2\right )}^p} \,d x \] Input:

int((b*x^2 + 1)^p/((c*x + d*x^3)*(x^2)^p),x)
 

Output:

int((b*x^2 + 1)^p/((c*x + d*x^3)*(x^2)^p), x)
 

Reduce [F]

\[ \int \frac {\left (x^2\right )^{-p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\frac {-\left (b \,x^{2}+1\right )^{p}+2 x^{2 p} \left (\int \frac {\left (b \,x^{2}+1\right )^{p} x}{x^{2 p} b c \,x^{2}+x^{2 p} b d \,x^{4}+x^{2 p} c +x^{2 p} d \,x^{2}}d x \right ) b c p -2 x^{2 p} \left (\int \frac {\left (b \,x^{2}+1\right )^{p} x}{x^{2 p} b c \,x^{2}+x^{2 p} b d \,x^{4}+x^{2 p} c +x^{2 p} d \,x^{2}}d x \right ) d p}{2 x^{2 p} c p} \] Input:

int((b*x^2+1)^p/((x^2)^p)/(d*x^3+c*x),x)
 

Output:

( - (b*x**2 + 1)**p + 2*x**(2*p)*int(((b*x**2 + 1)**p*x)/(x**(2*p)*b*c*x** 
2 + x**(2*p)*b*d*x**4 + x**(2*p)*c + x**(2*p)*d*x**2),x)*b*c*p - 2*x**(2*p 
)*int(((b*x**2 + 1)**p*x)/(x**(2*p)*b*c*x**2 + x**(2*p)*b*d*x**4 + x**(2*p 
)*c + x**(2*p)*d*x**2),x)*d*p)/(2*x**(2*p)*c*p)