\(\int \frac {1}{(a+b x^4)^{3/4} (c+d x^4)} \, dx\) [112]

Optimal result

Integrand size = 21, antiderivative size = 259 \[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=-\frac {b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{\sqrt {a} (b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{2 \sqrt [4]{b} c (b c-a d)}-\frac {d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{2 \sqrt [4]{b} c (b c-a d)} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 10.07 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=-\frac {5 a c x \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right ) \left (-5 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+x^4 \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )} \] Input:

Integrate[1/((a + b*x^4)^(3/4)*(c + d*x^4)),x]
 

Output:

(-5*a*c*x*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)])/((a + b* 
x^4)^(3/4)*(c + d*x^4)*(-5*a*c*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -( 
(d*x^4)/c)] + x^4*(4*a*d*AppellF1[5/4, 3/4, 2, 9/4, -((b*x^4)/a), -((d*x^4 
)/c)] + 3*b*c*AppellF1[5/4, 7/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.87, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {926, 768, 858, 807, 229, 923, 925, 27, 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.

\(\displaystyle \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx\)

\(\Big \downarrow \) 926

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

\(\Big \downarrow \) 923

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

\(\Big \downarrow \) 925

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1542

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

Input:

Int[1/((a + b*x^4)^(3/4)*(c + d*x^4)),x]
 

Output:

-((b^(3/2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2 
)]/2, 2])/(Sqrt[a]*(b*c - a*d)*(a + b*x^4)^(3/4))) - (d*Sqrt[a/(a + b*x^4) 
]*Sqrt[a + b*x^4]*(EllipticPi[-(Sqrt[b*c - a*d]/(Sqrt[b]*Sqrt[c])), ArcSin 
[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1]/(2*b^(1/4)*c) + EllipticPi[Sqrt[b*c - 
 a*d]/(Sqrt[b]*Sqrt[c]), ArcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1]/(2*b^( 
1/4)*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 229

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

rule 768

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

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 923

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

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 926

Int[1/(((a_) + (b_.)*(x_)^4)^(3/4)*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Sim 
p[b/(b*c - a*d)   Int[1/(a + b*x^4)^(3/4), x], x] - Simp[d/(b*c - a*d)   In 
t[(a + b*x^4)^(1/4)/(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]

\[\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (d \,x^{4}+c \right )}d x\]

Input:

int(1/(b*x^4+a)^(3/4)/(d*x^4+c),x)
 

Output:

int(1/(b*x^4+a)^(3/4)/(d*x^4+c),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x^4+a)^(3/4)/(d*x^4+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (a + b x^{4}\right )^{\frac {3}{4}} \left (c + d x^{4}\right )}\, dx \] Input:

integrate(1/(b*x**4+a)**(3/4)/(d*x**4+c),x)
 

Output:

Integral(1/((a + b*x**4)**(3/4)*(c + d*x**4)), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} {\left (d x^{4} + c\right )}} \,d x } \] Input:

integrate(1/(b*x^4+a)^(3/4)/(d*x^4+c),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate(1/((b*x^4 + a)^(3/4)*(d*x^4 + c)), x)
 

Giac [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} {\left (d x^{4} + c\right )}} \,d x } \] Input:

integrate(1/(b*x^4+a)^(3/4)/(d*x^4+c),x, algorithm="giac")
 

Output:

integrate(1/((b*x^4 + a)^(3/4)*(d*x^4 + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{3/4}\,\left (d\,x^4+c\right )} \,d x \] Input:

int(1/((a + b*x^4)^(3/4)*(c + d*x^4)),x)
 

Output:

int(1/((a + b*x^4)^(3/4)*(c + d*x^4)), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} c +\left (b \,x^{4}+a \right )^{\frac {3}{4}} d \,x^{4}}d x \] Input:

int(1/(b*x^4+a)^(3/4)/(d*x^4+c),x)
 

Output:

int(1/((a + b*x**4)**(3/4)*c + (a + b*x**4)**(3/4)*d*x**4),x)