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

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

[Out]

(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])

________________________________________________________________________________________

Rubi [A]  time = 0.399788, antiderivative size = 334, normalized size of antiderivative = 1., 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 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 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 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 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 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 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 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]

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]

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*}

Mathematica [C]  time = 0.72477, size = 422, normalized size = 1.26 \[ \frac{x \left (\frac{b d x^4 \sqrt{1-\frac{b x^4}{a}} (11 a d-5 b c) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\frac{b x^4}{a},\frac{d x^4}{c}\right )}{c}-\frac{5 \left (5 a c \left (a^2 b d \left (d x^4-24 c\right )+12 a^3 d^2+a b^2 \left (12 c^2+15 c d x^4-11 d^2 x^8\right )+5 b^3 c x^4 \left (d x^4-2 c\right )\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\frac{b x^4}{a},\frac{d x^4}{c}\right )+2 b x^4 \left (d x^4-c\right ) \left (13 a^2 d-a b \left (7 c+11 d x^4\right )+5 b^2 c x^4\right ) \left (2 a d F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};\frac{b x^4}{a},\frac{d x^4}{c}\right )+b c F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};\frac{b x^4}{a},\frac{d x^4}{c}\right )\right )\right )}{\left (a-b x^4\right ) \left (d x^4-c\right ) \left (2 x^4 \left (2 a d F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};\frac{b x^4}{a},\frac{d x^4}{c}\right )+b c F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};\frac{b x^4}{a},\frac{d x^4}{c}\right )\right )+5 a c F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\frac{b x^4}{a},\frac{d x^4}{c}\right )\right )}\right )}{60 a^2 \sqrt{a-b x^4} (b c-a d)^2} \]

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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Maple [C]  time = 0.032, size = 361, normalized size = 1.1 \begin{align*} -{\frac{x}{6\,ab \left ( ad-bc \right ) }\sqrt{-b{x}^{4}+a} \left ({x}^{4}-{\frac{a}{b}} \right ) ^{-2}}-{\frac{bx \left ( 11\,ad-5\,bc \right ) }{12\,{a}^{2} \left ( ad-bc \right ) ^{2}}{\frac{1}{\sqrt{- \left ({x}^{4}-{\frac{a}{b}} \right ) b}}}}-{\frac{b \left ( 11\,ad-5\,bc \right ) }{12\,{a}^{2} \left ( ad-bc \right ) ^{2}}\sqrt{1-{{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{-b{x}^{4}+a}}}}-{\frac{d}{8}\sum _{{\it \_alpha}={\it RootOf} \left ({{\it \_Z}}^{4}d-c \right ) }{\frac{1}{ \left ( ad-bc \right ) ^{2}{{\it \_alpha}}^{3}} \left ( -{{\it Artanh} \left ({\frac{-2\,{{\it \_alpha}}^{2}b{x}^{2}+2\,a}{2}{\frac{1}{\sqrt{{\frac{ad-bc}{d}}}}}{\frac{1}{\sqrt{-b{x}^{4}+a}}}} \right ){\frac{1}{\sqrt{{\frac{ad-bc}{d}}}}}}-2\,{\frac{{{\it \_alpha}}^{3}d}{c\sqrt{-b{x}^{4}+a}}\sqrt{1-{\frac{{x}^{2}\sqrt{b}}{\sqrt{a}}}}\sqrt{1+{\frac{{x}^{2}\sqrt{b}}{\sqrt{a}}}}{\it EllipticPi} \left ( x\sqrt{{\frac{\sqrt{b}}{\sqrt{a}}}},{\frac{\sqrt{a}{{\it \_alpha}}^{2}d}{c\sqrt{b}}},{\sqrt{-{\frac{\sqrt{b}}{\sqrt{a}}}}{\frac{1}{\sqrt{{\frac{\sqrt{b}}{\sqrt{a}}}}}}} \right ){\frac{1}{\sqrt{{\frac{\sqrt{b}}{\sqrt{a}}}}}}} \right ) }} \end{align*}

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*x/a/b/(a*d-b*c)*(-b*x^4+a)^(1/2)/(x^4-1/b*a)^2-1/12*b*x/a^2*(11*a*d-5*b*c)/(a*d-b*c)^2/(-(x^4-1/b*a)*b)^(
1/2)-1/12*b/a^2*(11*a*d-5*b*c)/(a*d-b*c)^2/(1/a^(1/2)*b^(1/2))^(1/2)*(1-x^2*b^(1/2)/a^(1/2))^(1/2)*(1+x^2*b^(1
/2)/a^(1/2))^(1/2)/(-b*x^4+a)^(1/2)*EllipticF(x*(1/a^(1/2)*b^(1/2))^(1/2),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-x^2*b^(1/2)/a^(1/2))^(1/2)*(1+x^2*b^(1/2)/a^(1/2))^(1/2)/(-b*x^4+a)^(1/2)*Elli
pticPi(x*(1/a^(1/2)*b^(1/2))^(1/2),a^(1/2)/b^(1/2)*_alpha^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., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{{\left (-b x^{4} + a\right )}^{\frac{5}{2}}{\left (d x^{4} - c\right )}}\,{d x} \end{align*}

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

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{1}{{\left (-b x^{4} + a\right )}^{\frac{5}{2}}{\left (d x^{4} - c\right )}}\,{d x} \end{align*}

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)