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

Optimal. Leaf size=439 \[ \frac {b x \left (-3 a^2 d^2-17 a b c d+5 b^2 c^2\right )}{12 a^2 c \sqrt {a-b x^4} (b c-a d)^3}+\frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} \left (-3 a^2 d^2-17 a b c d+5 b^2 c^2\right ) F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{12 a^{7/4} c \sqrt {a-b x^4} (b c-a d)^3}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} (13 b c-3 a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 \sqrt {a-b x^4} (b c-a d)^3}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} (13 b c-3 a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 \sqrt {a-b x^4} (b c-a d)^3}-\frac {d x}{4 c \left (a-b x^4\right )^{3/2} \left (c-d x^4\right ) (b c-a d)}+\frac {b x (3 a d+2 b c)}{12 a c \left (a-b x^4\right )^{3/2} (b c-a d)^2} \]

[Out]

1/12*b*(3*a*d+2*b*c)*x/a/c/(-a*d+b*c)^2/(-b*x^4+a)^(3/2)-1/4*d*x/c/(-a*d+b*c)/(-b*x^4+a)^(3/2)/(-d*x^4+c)+1/12
*b*(-3*a^2*d^2-17*a*b*c*d+5*b^2*c^2)*x/a^2/c/(-a*d+b*c)^3/(-b*x^4+a)^(1/2)+1/12*b^(3/4)*(-3*a^2*d^2-17*a*b*c*d
+5*b^2*c^2)*EllipticF(b^(1/4)*x/a^(1/4),I)*(1-b*x^4/a)^(1/2)/a^(7/4)/c/(-a*d+b*c)^3/(-b*x^4+a)^(1/2)+1/8*a^(1/
4)*d^2*(-3*a*d+13*b*c)*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^2/(-a*d+b*c)^3/(-b*x^4+a)^(1/2)+1/8*a^(1/4)*d^2*(-3*a*d+13*b*c)*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^2/(-a*d+b*c)^3/(-b*x^4+a)^(1/2)

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Rubi [A]  time = 0.54, antiderivative size = 439, normalized size of antiderivative = 1.00, number of steps used = 11, 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} \[ \frac {b x \left (-3 a^2 d^2-17 a b c d+5 b^2 c^2\right )}{12 a^2 c \sqrt {a-b x^4} (b c-a d)^3}+\frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} \left (-3 a^2 d^2-17 a b c d+5 b^2 c^2\right ) F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{12 a^{7/4} c \sqrt {a-b x^4} (b c-a d)^3}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} (13 b c-3 a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 \sqrt {a-b x^4} (b c-a d)^3}+\frac {\sqrt [4]{a} d^2 \sqrt {1-\frac {b x^4}{a}} (13 b c-3 a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 \sqrt {a-b x^4} (b c-a d)^3}-\frac {d x}{4 c \left (a-b x^4\right )^{3/2} \left (c-d x^4\right ) (b c-a d)}+\frac {b x (3 a d+2 b c)}{12 a c \left (a-b x^4\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(2*b*c + 3*a*d)*x)/(12*a*c*(b*c - a*d)^2*(a - b*x^4)^(3/2)) + (b*(5*b^2*c^2 - 17*a*b*c*d - 3*a^2*d^2)*x)/(1
2*a^2*c*(b*c - a*d)^3*Sqrt[a - b*x^4]) - (d*x)/(4*c*(b*c - a*d)*(a - b*x^4)^(3/2)*(c - d*x^4)) + (b^(3/4)*(5*b
^2*c^2 - 17*a*b*c*d - 3*a^2*d^2)*Sqrt[1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(12*a^(7/4)*c
*(b*c - a*d)^3*Sqrt[a - b*x^4]) + (a^(1/4)*d^2*(13*b*c - 3*a*d)*Sqrt[1 - (b*x^4)/a]*EllipticPi[-((Sqrt[a]*Sqrt
[d])/(Sqrt[b]*Sqrt[c])), ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(8*b^(1/4)*c^2*(b*c - a*d)^3*Sqrt[a - b*x^4]) + (a^
(1/4)*d^2*(13*b*c - 3*a*d)*Sqrt[1 - (b*x^4)/a]*EllipticPi[(Sqrt[a]*Sqrt[d])/(Sqrt[b]*Sqrt[c]), ArcSin[(b^(1/4)
*x)/a^(1/4)], -1])/(8*b^(1/4)*c^2*(b*c - a*d)^3*Sqrt[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 )^2} \, dx &=-\frac {d x}{4 c (b c-a d) \left (a-b x^4\right )^{3/2} \left (c-d x^4\right )}-\frac {\int \frac {-4 b c+3 a d-9 b d x^4}{\left (a-b x^4\right )^{5/2} \left (c-d x^4\right )} \, dx}{4 c (b c-a d)}\\ &=\frac {b (2 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a-b x^4\right )^{3/2}}-\frac {d x}{4 c (b c-a d) \left (a-b x^4\right )^{3/2} \left (c-d x^4\right )}-\frac {\int \frac {-2 \left (10 b^2 c^2-24 a b c d+9 a^2 d^2\right )+10 b d (2 b c+3 a d) x^4}{\left (a-b x^4\right )^{3/2} \left (c-d x^4\right )} \, dx}{24 a c (b c-a d)^2}\\ &=\frac {b (2 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a-b x^4\right )^{3/2}}+\frac {b \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) x}{12 a^2 c (b c-a d)^3 \sqrt {a-b x^4}}-\frac {d x}{4 c (b c-a d) \left (a-b x^4\right )^{3/2} \left (c-d x^4\right )}-\frac {\int \frac {-4 \left (5 b^3 c^3-17 a b^2 c^2 d+36 a^2 b c d^2-9 a^3 d^3\right )+4 b d \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) x^4}{\sqrt {a-b x^4} \left (c-d x^4\right )} \, dx}{48 a^2 c (b c-a d)^3}\\ &=\frac {b (2 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a-b x^4\right )^{3/2}}+\frac {b \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) x}{12 a^2 c (b c-a d)^3 \sqrt {a-b x^4}}-\frac {d x}{4 c (b c-a d) \left (a-b x^4\right )^{3/2} \left (c-d x^4\right )}+\frac {\left (d^2 (13 b c-3 a d)\right ) \int \frac {1}{\sqrt {a-b x^4} \left (c-d x^4\right )} \, dx}{4 c (b c-a d)^3}+\frac {\left (b \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a-b x^4}} \, dx}{12 a^2 c (b c-a d)^3}\\ &=\frac {b (2 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a-b x^4\right )^{3/2}}+\frac {b \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) x}{12 a^2 c (b c-a d)^3 \sqrt {a-b x^4}}-\frac {d x}{4 c (b c-a d) \left (a-b x^4\right )^{3/2} \left (c-d x^4\right )}+\frac {\left (d^2 (13 b c-3 a d)\right ) \int \frac {1}{\left (1-\frac {\sqrt {d} x^2}{\sqrt {c}}\right ) \sqrt {a-b x^4}} \, dx}{8 c^2 (b c-a d)^3}+\frac {\left (d^2 (13 b c-3 a d)\right ) \int \frac {1}{\left (1+\frac {\sqrt {d} x^2}{\sqrt {c}}\right ) \sqrt {a-b x^4}} \, dx}{8 c^2 (b c-a d)^3}+\frac {\left (b \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1}{\sqrt {1-\frac {b x^4}{a}}} \, dx}{12 a^2 c (b c-a d)^3 \sqrt {a-b x^4}}\\ &=\frac {b (2 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a-b x^4\right )^{3/2}}+\frac {b \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) x}{12 a^2 c (b c-a d)^3 \sqrt {a-b x^4}}-\frac {d x}{4 c (b c-a d) \left (a-b x^4\right )^{3/2} \left (c-d x^4\right )}+\frac {b^{3/4} \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) \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} c (b c-a d)^3 \sqrt {a-b x^4}}+\frac {\left (d^2 (13 b c-3 a d) \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}{8 c^2 (b c-a d)^3 \sqrt {a-b x^4}}+\frac {\left (d^2 (13 b c-3 a d) \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}{8 c^2 (b c-a d)^3 \sqrt {a-b x^4}}\\ &=\frac {b (2 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a-b x^4\right )^{3/2}}+\frac {b \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) x}{12 a^2 c (b c-a d)^3 \sqrt {a-b x^4}}-\frac {d x}{4 c (b c-a d) \left (a-b x^4\right )^{3/2} \left (c-d x^4\right )}+\frac {b^{3/4} \left (5 b^2 c^2-17 a b c d-3 a^2 d^2\right ) \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} c (b c-a d)^3 \sqrt {a-b x^4}}+\frac {\sqrt [4]{a} d^2 (13 b c-3 a d) \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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3 \sqrt {a-b x^4}}+\frac {\sqrt [4]{a} d^2 (13 b c-3 a d) \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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3 \sqrt {a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.88, size = 382, normalized size = 0.87 \[ \frac {x \left (\frac {b d x^4 \sqrt {1-\frac {b x^4}{a}} \left (3 a^2 d^2+17 a b c d-5 b^2 c^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )}{a^2 c^2}+5 \left (\frac {2 b^3 c}{a^2-a b x^4}+\frac {5 b^3 c}{a^2}+\frac {5 \left (-9 a^3 d^3+36 a^2 b c d^2-17 a b^2 c^2 d+5 b^3 c^3\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )}{a \left (c-d x^4\right ) \left (2 x^4 \left (2 a d F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )+b c F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )\right )+5 a c F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\frac {b x^4}{a},\frac {d x^4}{c}\right )\right )}-\frac {2 b^2 d}{a-b x^4}-\frac {17 b^2 d}{a}-\frac {3 a d^3}{c^2-c d x^4}+\frac {3 b d^3 x^4}{c^2-c d x^4}\right )\right )}{60 \sqrt {a-b x^4} (b c-a d)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*((b*d*(-5*b^2*c^2 + 17*a*b*c*d + 3*a^2*d^2)*x^4*Sqrt[1 - (b*x^4)/a]*AppellF1[5/4, 1/2, 1, 9/4, (b*x^4)/a, (
d*x^4)/c])/(a^2*c^2) + 5*((5*b^3*c)/a^2 - (17*b^2*d)/a - (2*b^2*d)/(a - b*x^4) + (2*b^3*c)/(a^2 - a*b*x^4) - (
3*a*d^3)/(c^2 - c*d*x^4) + (3*b*d^3*x^4)/(c^2 - c*d*x^4) + (5*(5*b^3*c^3 - 17*a*b^2*c^2*d + 36*a^2*b*c*d^2 - 9
*a^3*d^3)*AppellF1[1/4, 1/2, 1, 5/4, (b*x^4)/a, (d*x^4)/c])/(a*(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*(b*c - a*d)^3*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)^2,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.35, size = 484, normalized size = 1.10 \[ -\frac {\sqrt {-b \,x^{4}+a}\, b \,d^{3} x}{4 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (a d -b c \right ) \left (b d \,x^{4}-b c \right ) c}+\frac {\left (17 a d -5 b c \right ) b^{2} x}{12 \left (a d -b c \right )^{3} \sqrt {-\left (x^{4}-\frac {a}{b}\right ) b}\, a^{2}}-\frac {d \left (3 a d -13 b c \right ) \left (-\frac {2 \sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{3} d \EllipticPi \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , \frac {\RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{2} \sqrt {a}\, d}{\sqrt {b}\, c}, \frac {\sqrt {-\frac {\sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}}\right )}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, c}-\frac {\arctanh \left (\frac {-2 \RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{2} b \,x^{2}+2 a}{2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {-b \,x^{4}+a}}\right )}{\sqrt {\frac {a d -b c}{d}}}\right )}{32 c \left (a d -b c \right )^{3} \RootOf \left (d \,\textit {\_Z}^{4}-c \right )^{3}}+\frac {\left (\frac {b \,d^{2}}{4 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (a d -b c \right ) c}+\frac {\left (17 a d -5 b c \right ) b^{2}}{12 \left (a d -b c \right )^{3} a^{2}}\right ) \sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \EllipticF \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}+\frac {\sqrt {-b \,x^{4}+a}\, x}{6 \left (a d -b c \right )^{2} \left (x^{4}-\frac {a}{b}\right )^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*d^3/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a*d-b*c)/c*x*(-b*x^4+a)^(1/2)/(b*d*x^4-b*c)+1/6/(a*d-b*c)^2/a*x*(-b*x^
4+a)^(1/2)/(x^4-a/b)^2+1/12*b^2/a^2*x*(17*a*d-5*b*c)/(a*d-b*c)^3/(-(x^4-a/b)*b)^(1/2)+(1/4*b*d^2/(a^2*d^2-2*a*
b*c*d+b^2*c^2)/(a*d-b*c)/c+1/12/a^2*b^2*(17*a*d-5*b*c)/(a*d-b*c)^3)/(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)-1/3
2/c*d*sum((3*a*d-13*b*c)/(a*d-b*c)^3/_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 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {5}{2}} {\left (d x^{4} - c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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