3.214 \(\int \frac {1}{(a+b x^4)^{3/4} (c+d x^4)^2} \, dx\)

Optimal. Leaf size=330 \[ -\frac {b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (4 b c-a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 \sqrt {a} c \left (a+b x^4\right )^{3/4} (b c-a d)^2}-\frac {3 d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-a d) \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}-\frac {3 d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-a d) \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}-\frac {d x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right ) (b c-a d)} \]

[Out]

-1/4*d*x*(b*x^4+a)^(1/4)/c/(-a*d+b*c)/(d*x^4+c)-1/4*b^(3/2)*(-a*d+4*b*c)*(1+a/b/x^4)^(3/4)*x^3*(cos(1/2*arccot
(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x^2*b^(1/2)/a^(1
/2))),2^(1/2))/c/(-a*d+b*c)^2/(b*x^4+a)^(3/4)/a^(1/2)-3/8*d*(-a*d+2*b*c)*EllipticPi(b^(1/4)*x/(b*x^4+a)^(1/4),
-(-a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1/2)/b^(1/4)/c^2/(-a*d+b*c)^2-3/8*d*(-a*d+
2*b*c)*EllipticPi(b^(1/4)*x/(b*x^4+a)^(1/4),(-a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^
(1/2)/b^(1/4)/c^2/(-a*d+b*c)^2

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {414, 529, 237, 335, 275, 231, 407, 409, 1218} \[ -\frac {b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (4 b c-a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 \sqrt {a} c \left (a+b x^4\right )^{3/4} (b c-a d)^2}-\frac {3 d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-a d) \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}-\frac {3 d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-a d) \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}-\frac {d x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^4)^(3/4)*(c + d*x^4)^2),x]

[Out]

-(d*x*(a + b*x^4)^(1/4))/(4*c*(b*c - a*d)*(c + d*x^4)) - (b^(3/2)*(4*b*c - a*d)*(1 + a/(b*x^4))^(3/4)*x^3*Elli
pticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*Sqrt[a]*c*(b*c - a*d)^2*(a + b*x^4)^(3/4)) - (3*d*(2*b*c - a*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])/(8*b^(1/4)*c^2*(b*c - a*d)^2) - (3*d*(2*b*c - a*d)*Sqrt[a/(a + b*x^4)]*Sqrt[a + b*x^4]*Elli
pticPi[Sqrt[b*c - a*d]/(Sqrt[b]*Sqrt[c]), ArcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1])/(8*b^(1/4)*c^2*(b*c - a*
d)^2)

Rule 231

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

Rule 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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 335

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] && IntegerQ[m]

Rule 407

Int[((a_) + (b_.)*(x_)^4)^(1/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[Sqrt[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 409

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-(d/c), 2]*x^2)), x], x] + Dist[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 414

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

Rule 529

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

Rule 1218

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

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )^2} \, dx &=-\frac {d x \sqrt [4]{a+b x^4}}{4 c (b c-a d) \left (c+d x^4\right )}+\frac {\int \frac {4 b c-3 a d-2 b d x^4}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx}{4 c (b c-a d)}\\ &=-\frac {d x \sqrt [4]{a+b x^4}}{4 c (b c-a d) \left (c+d x^4\right )}-\frac {(3 d (2 b c-a d)) \int \frac {\sqrt [4]{a+b x^4}}{c+d x^4} \, dx}{4 c (b c-a d)^2}+\frac {(b (4 b c-a d)) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{4 c (b c-a d)^2}\\ &=-\frac {d x \sqrt [4]{a+b x^4}}{4 c (b c-a d) \left (c+d x^4\right )}+\frac {\left (b (4 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{4 c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {\left (3 d (2 b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-b x^4} \left (c-(b c-a d) x^4\right )} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4 c (b c-a d)^2}\\ &=-\frac {d x \sqrt [4]{a+b x^4}}{4 c (b c-a d) \left (c+d x^4\right )}-\frac {\left (b (4 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{4 c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {\left (3 d (2 b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {b c-a d} x^2}{\sqrt {c}}\right ) \sqrt {1-b x^4}} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 c^2 (b c-a d)^2}-\frac {\left (3 d (2 b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {b c-a d} x^2}{\sqrt {c}}\right ) \sqrt {1-b x^4}} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 c^2 (b c-a d)^2}\\ &=-\frac {d x \sqrt [4]{a+b x^4}}{4 c (b c-a d) \left (c+d x^4\right )}-\frac {3 d (2 b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}-\frac {3 d (2 b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}-\frac {\left (b (4 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{8 c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac {d x \sqrt [4]{a+b x^4}}{4 c (b c-a d) \left (c+d x^4\right )}-\frac {b^{3/2} (4 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 \sqrt {a} c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {3 d (2 b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}-\frac {3 d (2 b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.34, size = 337, normalized size = 1.02 \[ \frac {x \left (\frac {2 b d x^4 \left (\frac {b x^4}{a}+1\right )^{3/4} F_1\left (\frac {5}{4};\frac {3}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a d-b c}+\frac {c \left (25 a c \left (4 a d-4 b c+b d x^4\right ) F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-5 d x^4 \left (a+b x^4\right ) \left (4 a d F_1\left (\frac {5}{4};\frac {3}{4},2;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c F_1\left (\frac {5}{4};\frac {7}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}{\left (c+d x^4\right ) (b c-a d) \left (x^4 \left (4 a d F_1\left (\frac {5}{4};\frac {3}{4},2;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c F_1\left (\frac {5}{4};\frac {7}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )-5 a c F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}\right )}{20 c^2 \left (a+b x^4\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^4)^(3/4)*(c + d*x^4)^2),x]

[Out]

(x*((2*b*d*x^4*(1 + (b*x^4)/a)^(3/4)*AppellF1[5/4, 3/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])/(-(b*c) + a*d) +
(c*(25*a*c*(-4*b*c + 4*a*d + b*d*x^4)*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] - 5*d*x^4*(a + b*
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)])))/((b*c - a*d)*(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)])))))/(20*c^2*(a + b*x^4)^(3/4))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(3/4)/(d*x^4+c)^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} {\left (d x^{4} + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(3/4)/(d*x^4+c)^2,x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*(d*x^4 + c)^2), x)

________________________________________________________________________________________

maple [F]  time = 0.68, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (d \,x^{4}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^4+a)^(3/4)/(d*x^4+c)^2,x)

[Out]

int(1/(b*x^4+a)^(3/4)/(d*x^4+c)^2,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} {\left (d x^{4} + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(3/4)/(d*x^4+c)^2,x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*(d*x^4 + c)^2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,x^4+a\right )}^{3/4}\,{\left (d\,x^4+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^4)^(3/4)*(c + d*x^4)^2),x)

[Out]

int(1/((a + b*x^4)^(3/4)*(c + d*x^4)^2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{4}\right )^{\frac {3}{4}} \left (c + d x^{4}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**4+a)**(3/4)/(d*x**4+c)**2,x)

[Out]

Integral(1/((a + b*x**4)**(3/4)*(c + d*x**4)**2), x)

________________________________________________________________________________________