\(\int \frac {(c+d x^2) (e+f x^2)}{(a+b x^2)^{5/4}} \, dx\) [560]

Optimal result

Integrand size = 26, antiderivative size = 190 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {2 (b c-a d) (b e-a f) x}{a b^2 \sqrt [4]{a+b x^2}}-\frac {2 \left (5 b^2 c e+12 a^2 d f-10 a b (d e+c f)\right ) x}{5 a b^2 \sqrt [4]{a+b x^2}}+\frac {2 d f x \left (a+b x^2\right )^{3/4}}{5 b^2}+\frac {2 \left (5 b^2 c e+12 a^2 d f-10 a b (d e+c f)\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 \sqrt {a} b^{5/2} \sqrt [4]{a+b x^2}} \] Output:

2*(-a*d+b*c)*(-a*f+b*e)*x/a/b^2/(b*x^2+a)^(1/4)-2/5*(5*b^2*c*e+12*a^2*d*f- 
10*a*b*(c*f+d*e))*x/a/b^2/(b*x^2+a)^(1/4)+2/5*d*f*x*(b*x^2+a)^(3/4)/b^2+2/ 
5*(5*b^2*c*e+12*a^2*d*f-10*a*b*(c*f+d*e))*(1+b*x^2/a)^(1/4)*EllipticE(sin( 
1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/a^(1/2)/b^(5/2)/(b*x^2+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.62 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {x \left (2 \left (5 b^2 c e+6 a^2 d f+a b \left (-5 d e-5 c f+d f x^2\right )\right )+\left (-5 b^2 c e-12 a^2 d f+10 a b (d e+c f)\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{5 a b^2 \sqrt [4]{a+b x^2}} \] Input:

Integrate[((c + d*x^2)*(e + f*x^2))/(a + b*x^2)^(5/4),x]
 

Output:

(x*(2*(5*b^2*c*e + 6*a^2*d*f + a*b*(-5*d*e - 5*c*f + d*f*x^2)) + (-5*b^2*c 
*e - 12*a^2*d*f + 10*a*b*(d*e + c*f))*(1 + (b*x^2)/a)^(1/4)*Hypergeometric 
2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(5*a*b^2*(a + b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {401, 27, 299, 227, 225, 212}

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.

\(\displaystyle \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 401

\(\displaystyle \frac {2 x \left (c+d x^2\right ) (b e-a f)}{a b \sqrt [4]{a+b x^2}}-\frac {2 \int \frac {d (5 b e-6 a f) x^2+c (b e-2 a f)}{2 \sqrt [4]{b x^2+a}}dx}{a b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 x \left (c+d x^2\right ) (b e-a f)}{a b \sqrt [4]{a+b x^2}}-\frac {\int \frac {d (5 b e-6 a f) x^2+c (b e-2 a f)}{\sqrt [4]{b x^2+a}}dx}{a b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {2 x \left (c+d x^2\right ) (b e-a f)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {2 d x \left (a+b x^2\right )^{3/4} (5 b e-6 a f)}{5 b}-\frac {(2 a d (5 b e-6 a f)-5 b c (b e-2 a f)) \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}}{a b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {2 x \left (c+d x^2\right ) (b e-a f)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {2 d x \left (a+b x^2\right )^{3/4} (5 b e-6 a f)}{5 b}-\frac {\sqrt [4]{\frac {b x^2}{a}+1} (2 a d (5 b e-6 a f)-5 b c (b e-2 a f)) \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}}{a b}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {2 x \left (c+d x^2\right ) (b e-a f)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {2 d x \left (a+b x^2\right )^{3/4} (5 b e-6 a f)}{5 b}-\frac {\sqrt [4]{\frac {b x^2}{a}+1} (2 a d (5 b e-6 a f)-5 b c (b e-2 a f)) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}}{a b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {2 x \left (c+d x^2\right ) (b e-a f)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {2 d x \left (a+b x^2\right )^{3/4} (5 b e-6 a f)}{5 b}-\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right ) (2 a d (5 b e-6 a f)-5 b c (b e-2 a f))}{5 b \sqrt [4]{a+b x^2}}}{a b}\)

Input:

Int[((c + d*x^2)*(e + f*x^2))/(a + b*x^2)^(5/4),x]
 

Output:

(2*(b*e - a*f)*x*(c + d*x^2))/(a*b*(a + b*x^2)^(1/4)) - ((2*d*(5*b*e - 6*a 
*f)*x*(a + b*x^2)^(3/4))/(5*b) - ((2*a*d*(5*b*e - 6*a*f) - 5*b*c*(b*e - 2* 
a*f))*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*Elli 
pticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(5*b*(a + b*x^2)^(1/4)) 
)/(a*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( 
a + b*x^2)^(1/4)   Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

rule 299

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

rule 401

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
q/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(a + b*x^2)^(p + 1)*( 
c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + 
(b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L 
tQ[p, -1] && GtQ[q, 0]
 

Maple [F]

\[\int \frac {\left (x^{2} d +c \right ) \left (f \,x^{2}+e \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

int((d*x^2+c)*(f*x^2+e)/(b*x^2+a)^(5/4),x)
 

Output:

int((d*x^2+c)*(f*x^2+e)/(b*x^2+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (f x^{2} + e\right )}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((d*x^2+c)*(f*x^2+e)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((d*f*x^4 + (d*e + c*f)*x^2 + c*e)*(b*x^2 + a)^(3/4)/(b^2*x^4 + 2* 
a*b*x^2 + a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.92 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.65 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c e x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}}} + \frac {c f x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{4}}} + \frac {d e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{4}}} + \frac {d f x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {5}{4}}} \] Input:

integrate((d*x**2+c)*(f*x**2+e)/(b*x**2+a)**(5/4),x)
 

Output:

c*e*x*hyper((1/2, 5/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(5/4) + c*f*x 
**3*hyper((5/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(5/4)) + d*e 
*x**3*hyper((5/4, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(5/4)) + d 
*f*x**5*hyper((5/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(5/4))
 

Maxima [F]

\[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (f x^{2} + e\right )}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((d*x^2+c)*(f*x^2+e)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*(f*x^2 + e)/(b*x^2 + a)^(5/4), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} {\left (f x^{2} + e\right )}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((d*x^2+c)*(f*x^2+e)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*(f*x^2 + e)/(b*x^2 + a)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {\left (d\,x^2+c\right )\,\left (f\,x^2+e\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:

int(((c + d*x^2)*(e + f*x^2))/(a + b*x^2)^(5/4),x)
 

Output:

int(((c + d*x^2)*(e + f*x^2))/(a + b*x^2)^(5/4), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d f +\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c f +\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d e +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c e \] Input:

int((d*x^2+c)*(f*x^2+e)/(b*x^2+a)^(5/4),x)
 

Output:

int(x**4/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*d*f + int 
(x**2/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*c*f + int(x* 
*2/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*d*e + int(1/((a 
 + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*c*e