\(\int \frac {1}{(a+b x^4)^{7/4} (c+d x^4)} \, dx\) [203]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (warning: unable to verify)
   Maple [F]
   Fricas [F(-1)]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 21, antiderivative size = 304 \[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=\frac {b x}{3 a (b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {b^{3/2} (2 b c-5 a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{3 a^{3/2} (b c-a d)^2 \left (a+b x^4\right )^{3/4}}+\frac {d^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{2 \sqrt [4]{b} c (b c-a d)^2}+\frac {d^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{2 \sqrt [4]{b} c (b c-a d)^2} \]

[Out]

1/3*b*x/a/(-a*d+b*c)/(b*x^4+a)^(3/4)-1/3*b^(3/2)*(-5*a*d+2*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))/a^(3/2)/(-a*d+b*c)^2/(b*x^4+a)^(3/4)+1/2*d^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)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1/2)/b^(1/4)/c/(-a*d+b*c)^2+1/2*d^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)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1/2)/b^(1/4)/c/(-a*d+b*c)^2

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 304, 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 = {425, 543, 243, 342, 281, 237, 416, 418, 1232} \[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=-\frac {b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (2 b c-5 a d) \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4} (b c-a d)^2}+\frac {d^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c (b c-a d)^2}+\frac {d^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c (b c-a d)^2}+\frac {b x}{3 a \left (a+b x^4\right )^{3/4} (b c-a d)} \]

[In]

Int[1/((a + b*x^4)^(7/4)*(c + d*x^4)),x]

[Out]

(b*x)/(3*a*(b*c - a*d)*(a + b*x^4)^(3/4)) - (b^(3/2)*(2*b*c - 5*a*d)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCo
t[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(3*a^(3/2)*(b*c - a*d)^2*(a + b*x^4)^(3/4)) + (d^2*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*(b*c - a*d)^2) + (d^2*Sqrt[a/(a + b*x^4)]*Sqrt[a + b*x^4]*EllipticPi[Sqrt[b*c - a*d]/(Sqrt[b]*Sqrt[c]), A
rcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1])/(2*b^(1/4)*c*(b*c - a*d)^2)

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 425

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]) && IntBinomi
alQ[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 {b x}{3 a (b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {\int \frac {-2 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}{3 a (b c-a d)} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^4\right )^{3/4}}+\frac {d^2 \int \frac {\sqrt [4]{a+b x^4}}{c+d x^4} \, dx}{(b c-a d)^2}+\frac {(b (2 b c-5 a d)) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{3 a (b c-a d)^2} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^4\right )^{3/4}}+\frac {\left (b (2 b c-5 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}{3 a (b c-a d)^2 \left (a+b x^4\right )^{3/4}}+\frac {\left (d^2 \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)^2} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {\left (b (2 b c-5 a d) \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 )}{3 a (b c-a d)^2 \left (a+b x^4\right )^{3/4}}+\frac {\left (d^2 \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)^2}+\frac {\left (d^2 \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)^2} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^4\right )^{3/4}}+\frac {d^2 \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)^2}+\frac {d^2 \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)^2}-\frac {\left (b (2 b c-5 a d) \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 )}{6 a (b c-a d)^2 \left (a+b x^4\right )^{3/4}} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^4\right )^{3/4}}-\frac {b^{3/2} (2 b c-5 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 )}{3 a^{3/2} (b c-a d)^2 \left (a+b x^4\right )^{3/4}}+\frac {d^2 \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)^2}+\frac {d^2 \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)^2} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 10.27 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=\frac {x \left (-\frac {2 b d x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c}+\frac {5 \left (5 a c \left (3 a d-b \left (3 c+d x^4\right )\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b x^4 \left (c+d x^4\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}{\left (c+d x^4\right ) \left (5 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-x^4 \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}\right )}{15 a (-b c+a d) \left (a+b x^4\right )^{3/4}} \]

[In]

Integrate[1/((a + b*x^4)^(7/4)*(c + d*x^4)),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)])/c + (5*(5*a*c*(3
*a*d - b*(3*c + d*x^4))*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + b*x^4*(c + d*x^4)*(4*a*d*Appe
llF1[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)*(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)])))))/(1
5*a*(-(b*c) + a*d)*(a + b*x^4)^(3/4))

Maple [F]

\[\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {7}{4}} \left (d \,x^{4}+c \right )}d x\]

[In]

int(1/(b*x^4+a)^(7/4)/(d*x^4+c),x)

[Out]

int(1/(b*x^4+a)^(7/4)/(d*x^4+c),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x^4+a)^(7/4)/(d*x^4+c),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (a + b x^{4}\right )^{\frac {7}{4}} \left (c + d x^{4}\right )}\, dx \]

[In]

integrate(1/(b*x**4+a)**(7/4)/(d*x**4+c),x)

[Out]

Integral(1/((a + b*x**4)**(7/4)*(c + d*x**4)), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {7}{4}} {\left (d x^{4} + c\right )}} \,d x } \]

[In]

integrate(1/(b*x^4+a)^(7/4)/(d*x^4+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(7/4)*(d*x^4 + c)), x)

Giac [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {7}{4}} {\left (d x^{4} + c\right )}} \,d x } \]

[In]

integrate(1/(b*x^4+a)^(7/4)/(d*x^4+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(7/4)*(d*x^4 + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{7/4}\,\left (d\,x^4+c\right )} \,d x \]

[In]

int(1/((a + b*x^4)^(7/4)*(c + d*x^4)),x)

[Out]

int(1/((a + b*x^4)^(7/4)*(c + d*x^4)), x)