\(\int \frac {(a+b x^2)^p (c+d x^2)}{x^2} \, dx\) [517]

Optimal result

Integrand size = 20, antiderivative size = 80 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{x^2} \, dx=-\frac {c \left (a+b x^2\right )^{1+p}}{a x}+\frac {(a d+b (c+2 c p)) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{x^2} \, dx=\frac {\left (a+b x^2\right )^p \left (-c \left (a+b x^2\right )+(a d+b (c+2 c p)) x^2 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{a x} \] Input:

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

Output:

((a + b*x^2)^p*(-(c*(a + b*x^2)) + ((a*d + b*(c + 2*c*p))*x^2*Hypergeometr 
ic2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/(a*x)
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {359, 238, 237}

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.

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

\(\displaystyle \frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} (a d+b (2 c p+c)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {c \left (a+b x^2\right )^{p+1}}{a x}\)

Input:

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

Output:

-((c*(a + b*x^2)^(1 + p))/(a*x)) + ((a*d + b*(c + 2*c*p))*x*(a + b*x^2)^p* 
Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(a*(1 + (b*x^2)/a)^p)
 

Defintions of rubi rules used

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.59 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.64 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{x^2} \, dx=- \frac {a^{p} c {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + a^{p} d x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \] Input:

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

Output:

-a**p*c*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x + a**p*d*x*h 
yper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a)
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{x^2} \, dx=\frac {d\,x\,{\left (b\,x^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^p}+\frac {c\,{\left (b\,x^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2}-p,-p;\ \frac {3}{2}-p;\ -\frac {a}{b\,x^2}\right )}{x\,\left (2\,p-1\right )\,{\left (\frac {a}{b\,x^2}+1\right )}^p} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )}{x^2} \, dx=\frac {2 \left (b \,x^{2}+a \right )^{p} a d p +2 \left (b \,x^{2}+a \right )^{p} b c p +\left (b \,x^{2}+a \right )^{p} b c +2 \left (b \,x^{2}+a \right )^{p} b d p \,x^{2}-\left (b \,x^{2}+a \right )^{p} b d \,x^{2}+8 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{4}+4 a \,p^{2} x^{2}-b \,x^{4}-a \,x^{2}}d x \right ) a^{2} d \,p^{3} x -2 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{4}+4 a \,p^{2} x^{2}-b \,x^{4}-a \,x^{2}}d x \right ) a^{2} d p x +16 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{4}+4 a \,p^{2} x^{2}-b \,x^{4}-a \,x^{2}}d x \right ) a b c \,p^{4} x +8 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{4}+4 a \,p^{2} x^{2}-b \,x^{4}-a \,x^{2}}d x \right ) a b c \,p^{3} x -4 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{4}+4 a \,p^{2} x^{2}-b \,x^{4}-a \,x^{2}}d x \right ) a b c \,p^{2} x -2 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{4}+4 a \,p^{2} x^{2}-b \,x^{4}-a \,x^{2}}d x \right ) a b c p x}{b x \left (4 p^{2}-1\right )} \] Input:

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

Output:

(2*(a + b*x**2)**p*a*d*p + 2*(a + b*x**2)**p*b*c*p + (a + b*x**2)**p*b*c + 
 2*(a + b*x**2)**p*b*d*p*x**2 - (a + b*x**2)**p*b*d*x**2 + 8*int((a + b*x* 
*2)**p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a**2*d*p**3*x 
- 2*int((a + b*x**2)**p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4), 
x)*a**2*d*p*x + 16*int((a + b*x**2)**p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2* 
x**4 - b*x**4),x)*a*b*c*p**4*x + 8*int((a + b*x**2)**p/(4*a*p**2*x**2 - a* 
x**2 + 4*b*p**2*x**4 - b*x**4),x)*a*b*c*p**3*x - 4*int((a + b*x**2)**p/(4* 
a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a*b*c*p**2*x - 2*int((a 
+ b*x**2)**p/(4*a*p**2*x**2 - a*x**2 + 4*b*p**2*x**4 - b*x**4),x)*a*b*c*p* 
x)/(b*x*(4*p**2 - 1))