Integrand size = 26, antiderivative size = 57 \[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=-\frac {x^{-2 p} \left (1+b x^2\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-d) x^2}{c \left (1+b x^2\right )}\right )}{2 c p} \] 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))
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02 \[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=-\frac {x^{-2 p} \left (1+\frac {d x^2}{c}\right )^p \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {(-b c+d) x^2}{c+d x^2}\right )}{2 c p} \] 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))
Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {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.
\(\displaystyle \int \frac {x^{-2 p} \left (b x^2+1\right )^p}{c x+d x^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^{-2 p-1} \left (b x^2+1\right )^p}{c+d x^2}dx\) |
\(\Big \downarrow \) 393 |
\(\displaystyle \frac {1}{2} x^{-2 (p+1)} \left (x^2\right )^{p+1} \int \frac {\left (x^2\right )^{-p-1} \left (b x^2+1\right )^p}{d x^2+c}dx^2\) |
\(\Big \downarrow \) 141 |
\(\displaystyle -\frac {x^{2-2 (p+1)} \left (b x^2+1\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-d) x^2}{c \left (b x^2+1\right )}\right )}{2 c p}\) |
Input:
Int[(1 + b*x^2)^p/(x^(2*p)*(c*x + d*x^3)),x]
Output:
-1/2*(x^(2 - 2*(1 + p))*(1 + b*x^2)^p*Hypergeometric2F1[1, -p, 1 - p, ((b* c - d)*x^2)/(c*(1 + b*x^2))])/(c*p)
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]
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]
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]
\[\int \frac {\left (b \,x^{2}+1\right )^{p} x^{-2 p}}{d \,x^{3}+c x}d x\]
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)
\[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int { \frac {{\left (b x^{2} + 1\right )}^{p}}{{\left (d x^{3} + c x\right )} x^{2 \, p}} \,d x } \] 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)
Timed out. \[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\text {Timed out} \] Input:
integrate((b*x**2+1)**p/(x**(2*p))/(d*x**3+c*x),x)
Output:
Timed out
\[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int { \frac {{\left (b x^{2} + 1\right )}^{p}}{{\left (d x^{3} + c x\right )} x^{2 \, p}} \,d x } \] 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)
\[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int { \frac {{\left (b x^{2} + 1\right )}^{p}}{{\left (d x^{3} + c x\right )} x^{2 \, p}} \,d x } \] 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)
Timed out. \[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\int \frac {{\left (b\,x^2+1\right )}^p}{x^{2\,p}\,\left (d\,x^3+c\,x\right )} \,d x \] Input:
int((b*x^2 + 1)^p/(x^(2*p)*(c*x + d*x^3)),x)
Output:
int((b*x^2 + 1)^p/(x^(2*p)*(c*x + d*x^3)), x)
\[ \int \frac {x^{-2 p} \left (1+b x^2\right )^p}{c x+d x^3} \, dx=\frac {-\left (b \,x^{2}+1\right )^{p}+2 x^{2 p} \left (\int \frac {\left (b \,x^{2}+1\right )^{p} x}{x^{2 p} b c \,x^{2}+x^{2 p} b d \,x^{4}+x^{2 p} c +x^{2 p} d \,x^{2}}d x \right ) b c p -2 x^{2 p} \left (\int \frac {\left (b \,x^{2}+1\right )^{p} x}{x^{2 p} b c \,x^{2}+x^{2 p} b d \,x^{4}+x^{2 p} c +x^{2 p} d \,x^{2}}d x \right ) d p}{2 x^{2 p} c p} \] 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)