\(\int \frac {\sqrt [4]{a-b x^2}}{(c+d x^2)^2} \, dx\) [454]

Optimal result

Integrand size = 22, antiderivative size = 277 \[ \int \frac {\sqrt [4]{a-b x^2}}{\left (c+d x^2\right )^2} \, dx=\frac {x \sqrt [4]{a-b x^2}}{2 c \left (c+d x^2\right )}-\frac {\sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{2 c d \left (a-b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} (b c+2 a d) \sqrt {\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c+a d}},\arcsin \left (\frac {\sqrt [4]{a-b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c d (b c+a d) x}-\frac {\sqrt [4]{a} (b c+2 a d) \sqrt {\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c+a d}},\arcsin \left (\frac {\sqrt [4]{a-b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c d (b c+a d) x} \] Output:

1/2*x*(-b*x^2+a)^(1/4)/c/(d*x^2+c)-1/2*a^(1/2)*b^(1/2)*(1-b*x^2/a)^(3/4)*I 
nverseJacobiAM(1/2*arcsin(b^(1/2)*x/a^(1/2)),2^(1/2))/c/d/(-b*x^2+a)^(3/4) 
-1/4*a^(1/4)*(2*a*d+b*c)*(b*x^2/a)^(1/2)*EllipticPi((-b*x^2+a)^(1/4)/a^(1/ 
4),-a^(1/2)*d^(1/2)/(a*d+b*c)^(1/2),I)/c/d/(a*d+b*c)/x-1/4*a^(1/4)*(2*a*d+ 
b*c)*(b*x^2/a)^(1/2)*EllipticPi((-b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/( 
a*d+b*c)^(1/2),I)/c/d/(a*d+b*c)/x
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 10.26 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt [4]{a-b x^2}}{\left (c+d x^2\right )^2} \, dx=\frac {x \left (\frac {6 a-6 b x^2}{c^2+c d x^2}-\frac {b x^2 \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c^2}+\frac {36 a^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\left (c+d x^2\right ) \left (6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (-4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{12 \left (a-b x^2\right )^{3/4}} \] Input:

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

Output:

(x*((6*a - 6*b*x^2)/(c^2 + c*d*x^2) - (b*x^2*(1 - (b*x^2)/a)^(3/4)*AppellF 
1[3/2, 3/4, 1, 5/2, (b*x^2)/a, -((d*x^2)/c)])/c^2 + (36*a^2*AppellF1[1/2, 
3/4, 1, 3/2, (b*x^2)/a, -((d*x^2)/c)])/((c + d*x^2)*(6*a*c*AppellF1[1/2, 3 
/4, 1, 3/2, (b*x^2)/a, -((d*x^2)/c)] + x^2*(-4*a*d*AppellF1[3/2, 3/4, 2, 5 
/2, (b*x^2)/a, -((d*x^2)/c)] + 3*b*c*AppellF1[3/2, 7/4, 1, 5/2, (b*x^2)/a, 
 -((d*x^2)/c)])))))/(12*(a - b*x^2)^(3/4))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.91, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {314, 27, 405, 231, 230, 312, 118, 925, 1542}

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.

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

\(\Big \downarrow \) 314

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 405

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

\(\Big \downarrow \) 231

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

\(\Big \downarrow \) 230

\(\displaystyle \frac {\frac {(2 a d+b c) \int \frac {1}{\left (a-b x^2\right )^{3/4} \left (d x^2+c\right )}dx}{d}-\frac {2 \sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a-b x^2\right )^{3/4}}}{4 c}+\frac {x \sqrt [4]{a-b x^2}}{2 c \left (c+d x^2\right )}\)

\(\Big \downarrow \) 312

\(\displaystyle \frac {\frac {\sqrt {\frac {b x^2}{a}} (2 a d+b c) \int \frac {1}{\sqrt {\frac {b x^2}{a}} \left (a-b x^2\right )^{3/4} \left (d x^2+c\right )}dx^2}{2 d x}-\frac {2 \sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a-b x^2\right )^{3/4}}}{4 c}+\frac {x \sqrt [4]{a-b x^2}}{2 c \left (c+d x^2\right )}\)

\(\Big \downarrow \) 118

\(\displaystyle \frac {-\frac {2 \sqrt {\frac {b x^2}{a}} (2 a d+b c) \int \frac {1}{\sqrt {1-\frac {x^8}{a}} \left (-d x^8+b c+a d\right )}d\sqrt [4]{a-b x^2}}{d x}-\frac {2 \sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a-b x^2\right )^{3/4}}}{4 c}+\frac {x \sqrt [4]{a-b x^2}}{2 c \left (c+d x^2\right )}\)

\(\Big \downarrow \) 925

\(\displaystyle \frac {-\frac {2 \sqrt {\frac {b x^2}{a}} (2 a d+b c) \left (\frac {\int \frac {1}{\left (1-\frac {\sqrt {d} x^4}{\sqrt {b c+a d}}\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{a-b x^2}}{2 (a d+b c)}+\frac {\int \frac {1}{\left (\frac {\sqrt {d} x^4}{\sqrt {b c+a d}}+1\right ) \sqrt {1-\frac {x^8}{a}}}d\sqrt [4]{a-b x^2}}{2 (a d+b c)}\right )}{d x}-\frac {2 \sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a-b x^2\right )^{3/4}}}{4 c}+\frac {x \sqrt [4]{a-b x^2}}{2 c \left (c+d x^2\right )}\)

\(\Big \downarrow \) 1542

\(\displaystyle \frac {-\frac {2 \sqrt {\frac {b x^2}{a}} (2 a d+b c) \left (\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c+a d}},\arcsin \left (\frac {\sqrt [4]{a-b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{2 (a d+b c)}+\frac {\sqrt [4]{a} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c+a d}},\arcsin \left (\frac {\sqrt [4]{a-b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{2 (a d+b c)}\right )}{d x}-\frac {2 \sqrt {a} \sqrt {b} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{d \left (a-b x^2\right )^{3/4}}}{4 c}+\frac {x \sqrt [4]{a-b x^2}}{2 c \left (c+d x^2\right )}\)

Input:

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

Output:

(x*(a - b*x^2)^(1/4))/(2*c*(c + d*x^2)) + ((-2*Sqrt[a]*Sqrt[b]*(1 - (b*x^2 
)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(d*(a - b*x^2)^(3/ 
4)) - (2*(b*c + 2*a*d)*Sqrt[(b*x^2)/a]*((a^(1/4)*EllipticPi[-((Sqrt[a]*Sqr 
t[d])/Sqrt[b*c + a*d]), ArcSin[(a - b*x^2)^(1/4)/a^(1/4)], -1])/(2*(b*c + 
a*d)) + (a^(1/4)*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[b*c + a*d], ArcSin[(a - 
 b*x^2)^(1/4)/a^(1/4)], -1])/(2*(b*c + a*d))))/(d*x))/(4*c)
 

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 118

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^( 
3/4)), x_] :> Simp[-4   Subst[Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + 
 d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e, f}, x] & 
& GtQ[-f/(d*e - c*f), 0]
 

rule 230

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] 
))*EllipticF[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ 
[a, 0] && NegQ[b/a]
 

rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( 
a + b*x^2)^(3/4)   Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

rule 312

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim 
p[Sqrt[(-b)*(x^2/a)]/(2*x)   Subst[Int[1/(Sqrt[(-b)*(x/a)]*(a + b*x)^(3/4)* 
(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
 

rule 314

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

rule 405

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

rule 925

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Simp[ 
1/(2*c)   Int[1/(Sqrt[a + b*x^4]*(1 - Rt[-d/c, 2]*x^2)), x], x] + Simp[1/(2 
*c)   Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a, b, 
 c, d}, x] && NeQ[b*c - a*d, 0]
 

rule 1542

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x 
], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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