\(\int \frac {(c+d x^2)^2}{(a+b x^2)^{9/4}} \, dx\) [446]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 134 \[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 (b c-a d)^2 x}{5 a b^2 \left (a+b x^2\right )^{5/4}}+\frac {2 d^2 x}{b^2 \sqrt [4]{a+b x^2}}+\frac {2 \left (3 b^2 c^2+4 a b c d-12 a^2 d^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} b^{5/2} \sqrt [4]{a+b x^2}} \] Output:

2/5*(-a*d+b*c)^2*x/a/b^2/(b*x^2+a)^(5/4)+2*d^2*x/b^2/(b*x^2+a)^(1/4)+2/5*( 
-12*a^2*d^2+4*a*b*c*d+3*b^2*c^2)*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arcta 
n(b^(1/2)*x/a^(1/2))),2^(1/2))/a^(3/2)/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 = 15.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=-\frac {x \left (-2 (b c-a d) \left (6 a^2 d+3 b^2 c x^2+a b \left (4 c+7 d x^2\right )\right )+\left (3 b^2 c^2+4 a b c d-12 a^2 d^2\right ) \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{5 a^2 b^2 \left (a+b x^2\right )^{5/4}} \] Input:

Integrate[(c + d*x^2)^2/(a + b*x^2)^(9/4),x]
 

Output:

-1/5*(x*(-2*(b*c - a*d)*(6*a^2*d + 3*b^2*c*x^2 + a*b*(4*c + 7*d*x^2)) + (3 
*b^2*c^2 + 4*a*b*c*d - 12*a^2*d^2)*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hyper 
geometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(a^2*b^2*(a + b*x^2)^(5/4))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.37, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {315, 27, 298, 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.

\(\displaystyle \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {6 x (b c-a d) (2 a d+b c)}{a b \sqrt [4]{a+b x^2}}-\left (\frac {3 b c^2}{a}-\frac {12 a d^2}{b}+4 c d\right ) \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 a b}+\frac {2 x \left (c+d x^2\right ) (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {\frac {6 x (b c-a d) (2 a d+b c)}{a b \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {3 b c^2}{a}-\frac {12 a d^2}{b}+4 c d\right ) \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{\sqrt [4]{a+b x^2}}}{5 a b}+\frac {2 x \left (c+d x^2\right ) (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {\frac {6 x (b c-a d) (2 a d+b c)}{a b \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {3 b c^2}{a}-\frac {12 a d^2}{b}+4 c d\right ) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{\sqrt [4]{a+b x^2}}}{5 a b}+\frac {2 x \left (c+d x^2\right ) (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {\frac {6 x (b c-a d) (2 a d+b c)}{a b \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {3 b c^2}{a}-\frac {12 a d^2}{b}+4 c d\right ) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{\sqrt [4]{a+b x^2}}}{5 a b}+\frac {2 x \left (c+d x^2\right ) (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}\)

Input:

Int[(c + d*x^2)^2/(a + b*x^2)^(9/4),x]
 

Output:

(2*(b*c - a*d)*x*(c + d*x^2))/(5*a*b*(a + b*x^2)^(5/4)) + ((6*(b*c - a*d)* 
(b*c + 2*a*d)*x)/(a*b*(a + b*x^2)^(1/4)) - (((3*b*c^2)/a + 4*c*d - (12*a*d 
^2)/b)*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*Ell 
ipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(a + b*x^2)^(1/4))/(5* 
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 298
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 315
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 
Maple [F]

\[\int \frac {\left (x^{2} d +c \right )^{2}}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]

Input:

int((d*x^2+c)^2/(b*x^2+a)^(9/4),x)
 

Output:

int((d*x^2+c)^2/(b*x^2+a)^(9/4),x)
 

Fricas [F]

\[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{2}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x^2+c)^2/(b*x^2+a)^(9/4),x, algorithm="fricas")
 

Output:

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*(b*x^2 + a)^(3/4)/(b^3*x^6 + 3*a*b^2* 
x^4 + 3*a^2*b*x^2 + a^3), x)
 

Sympy [F]

\[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {\left (c + d x^{2}\right )^{2}}{\left (a + b x^{2}\right )^{\frac {9}{4}}}\, dx \] Input:

integrate((d*x**2+c)**2/(b*x**2+a)**(9/4),x)
 

Output:

Integral((c + d*x**2)**2/(a + b*x**2)**(9/4), x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{2}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x^2+c)^2/(b*x^2+a)^(9/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^2/(b*x^2 + a)^(9/4), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{2}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x^2+c)^2/(b*x^2+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^2/(b*x^2 + a)^(9/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {{\left (d\,x^2+c\right )}^2}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \] Input:

int((c + d*x^2)^2/(a + b*x^2)^(9/4),x)
 

Output:

int((c + d*x^2)^2/(a + b*x^2)^(9/4), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{9/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) d^{2}+2 \left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) c^{2} \] Input:

int((d*x^2+c)^2/(b*x^2+a)^(9/4),x)
                                                                                    
                                                                                    
 

Output:

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