3.3.2 \(\int \frac {1}{(a+b x^4)^{3/4} (c+d x^4)} \, dx\) [202]

Optimal. Leaf size=259 \[ -\frac {b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} (b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{2 \sqrt [4]{b} c (b c-a d)}-\frac {d \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{2 \sqrt [4]{b} c (b c-a d)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {419, 243, 342, 281, 237, 416, 418, 1232} \begin {gather*} -\frac {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 .\text {ArcSin}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\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} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c (b c-a d)}-\frac {b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \left (a+b x^4\right )^{3/4} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^(3/2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/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])), ArcSi
n[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1])/(2*b^(1/4)*c*(b*c - a*d)) - (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])/(2*b^(1/4)*c*(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 419

Int[1/(((a_) + (b_.)*(x_)^4)^(3/4)*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^4)^
(3/4), x], x] - Dist[d/(b*c - a*d), Int[(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 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*} \int \frac {1}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx &=\frac {b \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{b c-a d}-\frac {d \int \frac {\sqrt [4]{a+b x^4}}{c+d x^4} \, dx}{b c-a d}\\ &=\frac {\left (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}{(b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {\left (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 )}{b c-a d}\\ &=-\frac {\left (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 )}{(b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {\left (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 )}{2 c (b c-a d)}-\frac {\left (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 )}{2 c (b c-a d)}\\ &=-\frac {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 )}{2 \sqrt [4]{b} c (b c-a d)}-\frac {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 )}{2 \sqrt [4]{b} c (b c-a d)}-\frac {\left (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 )}{2 (b c-a d) \left (a+b x^4\right )^{3/4}}\\ &=-\frac {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 )}{\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} \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 )}{2 \sqrt [4]{b} c (b c-a d)}-\frac {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 )}{2 \sqrt [4]{b} c (b c-a d)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.04, size = 161, normalized size = 0.62 \begin {gather*} -\frac {5 a c x F_1\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 F_1\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 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 )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-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*Appel
lF1[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)])))

________________________________________________________________________________________

Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} \left (d \,x^{4}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________