3.178 \(\int \frac {1}{(a-b x^4)^{5/2} (c-d x^4)} \, dx\)

Optimal. Leaf size=334 \[ \frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} (5 b c-11 a d) F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{12 a^{7/4} \sqrt {a-b x^4} (b c-a d)^2}+\frac {b x (5 b c-11 a d)}{12 a^2 \sqrt {a-b x^4} (b c-a d)^2}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c \sqrt {a-b x^4} (b c-a d)^2}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c \sqrt {a-b x^4} (b c-a d)^2}+\frac {b x}{6 a \left (a-b x^4\right )^{3/2} (b c-a d)} \]

[Out]

1/6*b*x/a/(-a*d+b*c)/(-b*x^4+a)^(3/2)+1/12*b*(-11*a*d+5*b*c)*x/a^2/(-a*d+b*c)^2/(-b*x^4+a)^(1/2)+1/12*b^(3/4)*
(-11*a*d+5*b*c)*EllipticF(b^(1/4)*x/a^(1/4),I)*(1-b*x^4/a)^(1/2)/a^(7/4)/(-a*d+b*c)^2/(-b*x^4+a)^(1/2)+1/2*a^(
1/4)*d^2*EllipticPi(b^(1/4)*x/a^(1/4),-a^(1/2)*d^(1/2)/b^(1/2)/c^(1/2),I)*(1-b*x^4/a)^(1/2)/b^(1/4)/c/(-a*d+b*
c)^2/(-b*x^4+a)^(1/2)+1/2*a^(1/4)*d^2*EllipticPi(b^(1/4)*x/a^(1/4),a^(1/2)*d^(1/2)/b^(1/2)/c^(1/2),I)*(1-b*x^4
/a)^(1/2)/b^(1/4)/c/(-a*d+b*c)^2/(-b*x^4+a)^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {414, 527, 523, 224, 221, 409, 1219, 1218} \[ \frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} (5 b c-11 a d) F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{12 a^{7/4} \sqrt {a-b x^4} (b c-a d)^2}+\frac {b x (5 b c-11 a d)}{12 a^2 \sqrt {a-b x^4} (b c-a d)^2}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c \sqrt {a-b x^4} (b c-a d)^2}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c \sqrt {a-b x^4} (b c-a d)^2}+\frac {b x}{6 a \left (a-b x^4\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a - b*x^4)^(5/2)*(c - d*x^4)),x]

[Out]

(b*x)/(6*a*(b*c - a*d)*(a - b*x^4)^(3/2)) + (b*(5*b*c - 11*a*d)*x)/(12*a^2*(b*c - a*d)^2*Sqrt[a - b*x^4]) + (b
^(3/4)*(5*b*c - 11*a*d)*Sqrt[1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(12*a^(7/4)*(b*c - a*d
)^2*Sqrt[a - b*x^4]) + (a^(1/4)*d^2*Sqrt[1 - (b*x^4)/a]*EllipticPi[-((Sqrt[a]*Sqrt[d])/(Sqrt[b]*Sqrt[c])), Arc
Sin[(b^(1/4)*x)/a^(1/4)], -1])/(2*b^(1/4)*c*(b*c - a*d)^2*Sqrt[a - b*x^4]) + (a^(1/4)*d^2*Sqrt[1 - (b*x^4)/a]*
EllipticPi[(Sqrt[a]*Sqrt[d])/(Sqrt[b]*Sqrt[c]), ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(2*b^(1/4)*c*(b*c - a*d)^2*S
qrt[a - b*x^4])

Rule 221

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

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 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 523

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

Rule 527

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

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]

Rule 1219

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4]
, Int[1/((d + e*x^2)*Sqrt[1 + (c*x^4)/a]), x], 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 )^{5/2} \left (c-d x^4\right )} \, dx &=\frac {b x}{6 a (b c-a d) \left (a-b x^4\right )^{3/2}}+\frac {\int \frac {5 b c-6 a d-5 b d x^4}{\left (a-b x^4\right )^{3/2} \left (c-d x^4\right )} \, dx}{6 a (b c-a d)}\\ &=\frac {b x}{6 a (b c-a d) \left (a-b x^4\right )^{3/2}}+\frac {b (5 b c-11 a d) x}{12 a^2 (b c-a d)^2 \sqrt {a-b x^4}}+\frac {\int \frac {5 b^2 c^2-11 a b c d+12 a^2 d^2-b d (5 b c-11 a d) x^4}{\sqrt {a-b x^4} \left (c-d x^4\right )} \, dx}{12 a^2 (b c-a d)^2}\\ &=\frac {b x}{6 a (b c-a d) \left (a-b x^4\right )^{3/2}}+\frac {b (5 b c-11 a d) x}{12 a^2 (b c-a d)^2 \sqrt {a-b x^4}}+\frac {d^2 \int \frac {1}{\sqrt {a-b x^4} \left (c-d x^4\right )} \, dx}{(b c-a d)^2}+\frac {(b (5 b c-11 a d)) \int \frac {1}{\sqrt {a-b x^4}} \, dx}{12 a^2 (b c-a d)^2}\\ &=\frac {b x}{6 a (b c-a d) \left (a-b x^4\right )^{3/2}}+\frac {b (5 b c-11 a d) x}{12 a^2 (b c-a d)^2 \sqrt {a-b x^4}}+\frac {d^2 \int \frac {1}{\left (1-\frac {\sqrt {d} x^2}{\sqrt {c}}\right ) \sqrt {a-b x^4}} \, dx}{2 c (b c-a d)^2}+\frac {d^2 \int \frac {1}{\left (1+\frac {\sqrt {d} x^2}{\sqrt {c}}\right ) \sqrt {a-b x^4}} \, dx}{2 c (b c-a d)^2}+\frac {\left (b (5 b c-11 a d) \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1}{\sqrt {1-\frac {b x^4}{a}}} \, dx}{12 a^2 (b c-a d)^2 \sqrt {a-b x^4}}\\ &=\frac {b x}{6 a (b c-a d) \left (a-b x^4\right )^{3/2}}+\frac {b (5 b c-11 a d) x}{12 a^2 (b c-a d)^2 \sqrt {a-b x^4}}+\frac {b^{3/4} (5 b c-11 a d) \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{12 a^{7/4} (b c-a d)^2 \sqrt {a-b x^4}}+\frac {\left (d^2 \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1}{\left (1-\frac {\sqrt {d} x^2}{\sqrt {c}}\right ) \sqrt {1-\frac {b x^4}{a}}} \, dx}{2 c (b c-a d)^2 \sqrt {a-b x^4}}+\frac {\left (d^2 \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1}{\left (1+\frac {\sqrt {d} x^2}{\sqrt {c}}\right ) \sqrt {1-\frac {b x^4}{a}}} \, dx}{2 c (b c-a d)^2 \sqrt {a-b x^4}}\\ &=\frac {b x}{6 a (b c-a d) \left (a-b x^4\right )^{3/2}}+\frac {b (5 b c-11 a d) x}{12 a^2 (b c-a d)^2 \sqrt {a-b x^4}}+\frac {b^{3/4} (5 b c-11 a d) \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{12 a^{7/4} (b c-a d)^2 \sqrt {a-b x^4}}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c (b c-a d)^2 \sqrt {a-b x^4}}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c (b c-a d)^2 \sqrt {a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.85, size = 422, normalized size = 1.26 \[ \frac {x \left (\frac {b d x^4 \sqrt {1-\frac {b x^4}{a}} (11 a d-5 b c) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )}{c}-\frac {5 \left (2 b x^4 \left (d x^4-c\right ) \left (13 a^2 d-a b \left (7 c+11 d x^4\right )+5 b^2 c x^4\right ) \left (2 a d F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )+b c F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )\right )+5 a c \left (12 a^3 d^2+a^2 b d \left (d x^4-24 c\right )+a b^2 \left (12 c^2+15 c d x^4-11 d^2 x^8\right )+5 b^3 c x^4 \left (d x^4-2 c\right )\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )\right )}{\left (a-b x^4\right ) \left (d x^4-c\right ) \left (2 x^4 \left (2 a d F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )+b c F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )\right )+5 a c F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )\right )}\right )}{60 a^2 \sqrt {a-b x^4} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a - b*x^4)^(5/2)*(c - d*x^4)),x]

[Out]

(x*((b*d*(-5*b*c + 11*a*d)*x^4*Sqrt[1 - (b*x^4)/a]*AppellF1[5/4, 1/2, 1, 9/4, (b*x^4)/a, (d*x^4)/c])/c - (5*(5
*a*c*(12*a^3*d^2 + a^2*b*d*(-24*c + d*x^4) + 5*b^3*c*x^4*(-2*c + d*x^4) + a*b^2*(12*c^2 + 15*c*d*x^4 - 11*d^2*
x^8))*AppellF1[1/4, 1/2, 1, 5/4, (b*x^4)/a, (d*x^4)/c] + 2*b*x^4*(-c + d*x^4)*(13*a^2*d + 5*b^2*c*x^4 - a*b*(7
*c + 11*d*x^4))*(2*a*d*AppellF1[5/4, 1/2, 2, 9/4, (b*x^4)/a, (d*x^4)/c] + b*c*AppellF1[5/4, 3/2, 1, 9/4, (b*x^
4)/a, (d*x^4)/c])))/((a - b*x^4)*(-c + d*x^4)*(5*a*c*AppellF1[1/4, 1/2, 1, 5/4, (b*x^4)/a, (d*x^4)/c] + 2*x^4*
(2*a*d*AppellF1[5/4, 1/2, 2, 9/4, (b*x^4)/a, (d*x^4)/c] + b*c*AppellF1[5/4, 3/2, 1, 9/4, (b*x^4)/a, (d*x^4)/c]
)))))/(60*a^2*(b*c - a*d)^2*Sqrt[a - b*x^4])

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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)^(5/2)/(-d*x^4+c),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {1}{{\left (-b x^{4} + a\right )}^{\frac {5}{2}} {\left (d x^{4} - c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^4+a)^(5/2)/(-d*x^4+c),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.32, size = 361, normalized size = 1.08 \[ -\frac {d \left (-\frac {2 \sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{3} d \EllipticPi \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , \frac {\RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{2} \sqrt {a}\, d}{\sqrt {b}\, c}, \frac {\sqrt {-\frac {\sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, c}-\frac {\arctanh \left (\frac {-2 \RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{2} b \,x^{2}+2 a}{2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {-b \,x^{4}+a}}\right )}{\sqrt {\frac {a d -b c}{d}}}\right )}{8 \left (a d -b c \right )^{2} \RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{3}}-\frac {\left (11 a d -5 b c \right ) b x}{12 \left (a d -b c \right )^{2} \sqrt {-\left (x^{4}-\frac {a}{b}\right ) b}\, a^{2}}-\frac {\left (11 a d -5 b c \right ) \sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b \EllipticF \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )}{12 \left (a d -b c \right )^{2} \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, a^{2}}-\frac {\sqrt {-b \,x^{4}+a}\, x}{6 \left (a d -b c \right ) \left (x^{4}-\frac {a}{b}\right )^{2} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/a*x/b/(a*d-b*c)*(-b*x^4+a)^(1/2)/(x^4-a/b)^2-1/12*b/a^2*x*(11*a*d-5*b*c)/(a*d-b*c)^2/(-(x^4-a/b)*b)^(1/2)
-1/12/a^2*b*(11*a*d-5*b*c)/(a*d-b*c)^2/(1/a^(1/2)*b^(1/2))^(1/2)*(-1/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*b
^(1/2)*x^2+1)^(1/2)/(-b*x^4+a)^(1/2)*EllipticF((1/a^(1/2)*b^(1/2))^(1/2)*x,I)-1/8*d*sum(1/(a*d-b*c)^2/_alpha^3
*(-1/((a*d-b*c)/d)^(1/2)*arctanh(1/2*(-2*_alpha^2*b*x^2+2*a)/((a*d-b*c)/d)^(1/2)/(-b*x^4+a)^(1/2))-2/(1/a^(1/2
)*b^(1/2))^(1/2)*_alpha^3*d/c*(-1/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(-b*x^4+a)^(1/2
)*EllipticPi((1/a^(1/2)*b^(1/2))^(1/2)*x,_alpha^2*a^(1/2)/b^(1/2)/c*d,(-1/a^(1/2)*b^(1/2))^(1/2)/(1/a^(1/2)*b^
(1/2))^(1/2))),_alpha=RootOf(_Z^4*d-c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {5}{2}} {\left (d x^{4} - c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^4+a)^(5/2)/(-d*x^4+c),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a-b\,x^4\right )}^{5/2}\,\left (c-d\,x^4\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a - b*x^4)^(5/2)*(c - d*x^4)),x)

[Out]

int(1/((a - b*x^4)^(5/2)*(c - d*x^4)), x)

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sympy [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)**(5/2)/(-d*x**4+c),x)

[Out]

Timed out

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