\(\int \frac {1}{\sqrt [4]{a+b x^2} (c+d x^2)^2} \, dx\) [1538]

Optimal result

Integrand size = 21, antiderivative size = 336 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^2} \, dx=\frac {b x}{2 c (b c-a d) \sqrt [4]{a+b x^2}}-\frac {d x \left (a+b x^2\right )^{3/4}}{2 c (b c-a d) \left (c+d x^2\right )}-\frac {\sqrt {a} \sqrt {b} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c (b c-a d) \sqrt [4]{a+b x^2}}-\frac {\sqrt [4]{a} (3 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c \sqrt {d} (-b c+a d)^{3/2} x}+\frac {\sqrt [4]{a} (3 b c-2 a d) \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{4 c \sqrt {d} (-b c+a d)^{3/2} x} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 0.25 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^2} \, dx=\frac {-6 a c x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right ) \left (-6 c \left (-2 b c+2 a d+b d x^2\right )+b d x^2 \sqrt [4]{1+\frac {b x^2}{a}} \left (c+d x^2\right ) \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )-d x^3 \left (6 c \left (a+b x^2\right )-b x^2 \sqrt [4]{1+\frac {b x^2}{a}} \left (c+d x^2\right ) \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{12 c^2 (b c-a d) \sqrt [4]{a+b x^2} \left (c+d x^2\right ) \left (-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )} \] Input:

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

Output:

(-6*a*c*x*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)]*(-6*c*(-2 
*b*c + 2*a*d + b*d*x^2) + b*d*x^2*(1 + (b*x^2)/a)^(1/4)*(c + d*x^2)*Appell 
F1[3/2, 1/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]) - d*x^3*(6*c*(a + b*x^2) 
 - b*x^2*(1 + (b*x^2)/a)^(1/4)*(c + d*x^2)*AppellF1[3/2, 1/4, 1, 5/2, -((b 
*x^2)/a), -((d*x^2)/c)])*(4*a*d*AppellF1[3/2, 1/4, 2, 5/2, -((b*x^2)/a), - 
((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]) 
)/(12*c^2*(b*c - a*d)*(a + b*x^2)^(1/4)*(c + d*x^2)*(-6*a*c*AppellF1[1/2, 
1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 1/4, 2 
, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x 
^2)/a), -((d*x^2)/c)])))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {316, 27, 405, 227, 225, 212, 310, 993, 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.

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 405

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

\(\Big \downarrow \) 227

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

\(\Big \downarrow \) 225

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

\(\Big \downarrow \) 212

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

\(\Big \downarrow \) 310

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

\(\Big \downarrow \) 993

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

\(\Big \downarrow \) 1542

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

Input:

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

Output:

-1/2*(d*x*(a + b*x^2)^(3/4))/(c*(b*c - a*d)*(c + d*x^2)) + ((b*(1 + (b*x^2 
)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqr 
t[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(a + b*x^2)^(1/4) + (2*(3*b*c - 2*a*d)*S 
qrt[-((b*x^2)/a)]*((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*Sqrt[d]*Sqrt[-(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*Sqrt[d]*Sqrt[-(b*c) + a*d])))/x)/(4*c*(b*c 
 - a*d))
 

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 310

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

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[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 993

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> 
With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2* 
b)   Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b)   Int[1/((r 
 - s*x^2)*Sqrt[c + d*x^4]), 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]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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