\(\int \frac {(c+d x^2)^2}{\sqrt [4]{a-b x^2}} \, dx\) [471]

Optimal result

Integrand size = 22, antiderivative size = 136 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt [4]{a-b x^2}} \, dx=-\frac {4 d (3 b c+a d) x \left (a-b x^2\right )^{3/4}}{15 b^2}-\frac {2 d^2 x^3 \left (a-b x^2\right )^{3/4}}{9 b}+\frac {2 \sqrt {a} \left (15 b^2 c^2+4 a d (3 b c+a d)\right ) \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a-b x^2}} \] Output:

-4/15*d*(a*d+3*b*c)*x*(-b*x^2+a)^(3/4)/b^2-2/9*d^2*x^3*(-b*x^2+a)^(3/4)/b+ 
2/15*a^(1/2)*(15*b^2*c^2+4*a*d*(a*d+3*b*c))*(1-b*x^2/a)^(1/4)*EllipticE(si 
n(1/2*arcsin(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(5/2)/(-b*x^2+a)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 10.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.16 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt [4]{a-b x^2}} \, dx=\frac {x \sqrt [4]{1-\frac {b x^2}{a}} \left (7 a \left (15 c^2+10 c d x^2+3 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {7}{2},\frac {b x^2}{a}\right )+2 b x^2 \left (2 c^2+3 c d x^2+d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {3}{2},\frac {9}{2},\frac {b x^2}{a}\right )+b x^2 \left (c+d x^2\right )^2 \, _3F_2\left (\frac {5}{4},\frac {3}{2},2;1,\frac {9}{2};\frac {b x^2}{a}\right )\right )}{105 a \sqrt [4]{a-b x^2}} \] Input:

Integrate[(c + d*x^2)^2/(a - b*x^2)^(1/4),x]
 

Output:

(x*(1 - (b*x^2)/a)^(1/4)*(7*a*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*Hypergeome 
tric2F1[1/4, 1/2, 7/2, (b*x^2)/a] + 2*b*x^2*(2*c^2 + 3*c*d*x^2 + d^2*x^4)* 
Hypergeometric2F1[5/4, 3/2, 9/2, (b*x^2)/a] + b*x^2*(c + d*x^2)^2*Hypergeo 
metricPFQ[{5/4, 3/2, 2}, {1, 9/2}, (b*x^2)/a]))/(105*a*(a - b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {318, 27, 299, 227, 226}

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.

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

\(\Big \downarrow \) 318

\(\displaystyle -\frac {2 \int -\frac {d (13 b c+6 a d) x^2+c (9 b c+2 a d)}{2 \sqrt [4]{a-b x^2}}dx}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )}{9 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {d (13 b c+6 a d) x^2+c (9 b c+2 a d)}{\sqrt [4]{a-b x^2}}dx}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )}{9 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {3 \left (4 a^2 d^2+12 a b c d+15 b^2 c^2\right ) \int \frac {1}{\sqrt [4]{a-b x^2}}dx}{5 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} (6 a d+13 b c)}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )}{9 b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {\frac {3 \sqrt [4]{1-\frac {b x^2}{a}} \left (4 a^2 d^2+12 a b c d+15 b^2 c^2\right ) \int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}}dx}{5 b \sqrt [4]{a-b x^2}}-\frac {2 d x \left (a-b x^2\right )^{3/4} (6 a d+13 b c)}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )}{9 b}\)

\(\Big \downarrow \) 226

\(\displaystyle \frac {\frac {6 \sqrt {a} \sqrt [4]{1-\frac {b x^2}{a}} \left (4 a^2 d^2+12 a b c d+15 b^2 c^2\right ) E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {2 d x \left (a-b x^2\right )^{3/4} (6 a d+13 b c)}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )}{9 b}\)

Input:

Int[(c + d*x^2)^2/(a - b*x^2)^(1/4),x]
 

Output:

(-2*d*x*(a - b*x^2)^(3/4)*(c + d*x^2))/(9*b) + ((-2*d*(13*b*c + 6*a*d)*x*( 
a - b*x^2)^(3/4))/(5*b) + (6*Sqrt[a]*(15*b^2*c^2 + 12*a*b*c*d + 4*a^2*d^2) 
*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^( 
3/2)*(a - b*x^2)^(1/4)))/(9*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 226

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] 
))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[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 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

Maple [F]

\[\int \frac {\left (x^{2} d +c \right )^{2}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((d*x^2+c)^2/(-b*x^2+a)^(1/4),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c+d x^2\right )^2}{\sqrt [4]{a-b x^2}} \, dx=\frac {c^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{\sqrt [4]{a}} + \frac {2 c d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{3 \sqrt [4]{a}} + \frac {d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{5 \sqrt [4]{a}} \] Input:

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

Output:

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

Maxima [F]

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

integrate((d*x^2+c)^2/(-b*x^2+a)^(1/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^2/(-b*x^2 + a)^(1/4), x)
 

Giac [F]

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

integrate((d*x^2+c)^2/(-b*x^2+a)^(1/4),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x^2 + c)^2/(-b*x^2 + a)^(1/4), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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