\(\int \frac {\sqrt [4]{a+b x^4}}{(c+d x^4)^2} \, dx\) [213]

Optimal result

Integrand size = 21, antiderivative size = 308 \[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}-\frac {\sqrt {a} b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{4 c (b c-a d) \left (a+b x^4\right )^{3/4}}+\frac {(2 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)}+\frac {(2 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)} \]

[Out]

1/4*x*(b*x^4+a)^(1/4)/c/(d*x^4+c)-1/4*b^(3/2)*(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))*a^(1/2)/c/(-
a*d+b*c)/(b*x^4+a)^(3/4)+1/8*(-3*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)+1/8*(-3*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)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {423, 543, 243, 342, 281, 237, 416, 418, 1232} \[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-3 a d) \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)}+\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-3 a d) \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)}-\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{4 c \left (a+b x^4\right )^{3/4} (b c-a d)}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )} \]

[In]

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

[Out]

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

Rule 237

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 243

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 281

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 342

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 416

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 418

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 423

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

Rule 543

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 1232

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]

Rubi steps \begin{align*} \text {integral}& = \frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}-\frac {\int \frac {-3 a-2 b x^4}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx}{4 c} \\ & = \frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}+\frac {(a b) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{4 c (b c-a d)}+\frac {(2 b c-3 a d) \int \frac {\sqrt [4]{a+b x^4}}{c+d x^4} \, dx}{4 c (b c-a d)} \\ & = \frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}+\frac {\left (a b \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) \left (a+b x^4\right )^{3/4}}+\frac {\left ((2 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \text {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)} \\ & = \frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}-\frac {\left (a b \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {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) \left (a+b x^4\right )^{3/4}}+\frac {\left ((2 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \text {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)}+\frac {\left ((2 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \text {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)} \\ & = \frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}+\frac {(2 b c-3 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)}+\frac {(2 b c-3 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)}-\frac {\left (a b \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {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) \left (a+b x^4\right )^{3/4}} \\ & = \frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}-\frac {\sqrt {a} b^{3/2} \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 c (b c-a d) \left (a+b x^4\right )^{3/4}}+\frac {(2 b c-3 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)}+\frac {(2 b c-3 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)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 10.22 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \left (\frac {2 b x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2}+\frac {5 \left (\frac {a+b x^4}{c}-\frac {15 a^2 \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{-5 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+x^4 \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}\right )}{c+d x^4}\right )}{20 \left (a+b x^4\right )^{3/4}} \]

[In]

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

[Out]

(x*((2*b*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)])/c^2 + (5*((a + b*x^
4)/c - (15*a^2*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)])/(-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)]))))/(c + d*x^4)))/(20*(a + b*x^4)^(3/4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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