3.3.15 \(\int \frac {1}{(a+b x^4)^{7/4} (c+d x^4)^2} \, dx\) [215]

Optimal. Leaf size=390 \[ \frac {b (4 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}-\frac {b^{3/2} \left (8 b^2 c^2-32 a b c d+3 a^2 d^2\right ) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 a^{3/2} c (b c-a d)^3 \left (a+b x^4\right )^{3/4}}+\frac {d^2 (10 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^3}+\frac {d^2 (10 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)^3} \]

[Out]

1/12*b*(3*a*d+4*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^4+a)^(3/4)-1/4*d*x/c/(-a*d+b*c)/(b*x^4+a)^(3/4)/(d*x^4+c)-1/12*b^
(3/2)*(3*a^2*d^2-32*a*b*c*d+8*b^2*c^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)/co
s(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)/c/(-a*d+b*c
)^3/(b*x^4+a)^(3/4)+1/8*d^2*(-3*a*d+10*b*c)*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^2/(-a*d+b*c)^3+1/8*d^2*(-3*a*d+10*b*c)*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^2/(-a*d+
b*c)^3

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Rubi [A]
time = 0.35, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {425, 541, 543, 243, 342, 281, 237, 416, 418, 1232} \begin {gather*} -\frac {b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \left (3 a^2 d^2-32 a b c d+8 b^2 c^2\right ) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 a^{3/2} c \left (a+b x^4\right )^{3/4} (b c-a d)^3}+\frac {d^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (10 b c-3 a d) \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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3}+\frac {d^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (10 b c-3 a d) \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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3}+\frac {b x (3 a d+4 b c)}{12 a c \left (a+b x^4\right )^{3/4} (b c-a d)^2}-\frac {d x}{4 c \left (a+b x^4\right )^{3/4} \left (c+d x^4\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^4)^(7/4)*(c + d*x^4)^2),x]

[Out]

(b*(4*b*c + 3*a*d)*x)/(12*a*c*(b*c - a*d)^2*(a + b*x^4)^(3/4)) - (d*x)/(4*c*(b*c - a*d)*(a + b*x^4)^(3/4)*(c +
d*x^4)) - (b^(3/2)*(8*b^2*c^2 - 32*a*b*c*d + 3*a^2*d^2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x
^2)/Sqrt[a]]/2, 2])/(12*a^(3/2)*c*(b*c - a*d)^3*(a + b*x^4)^(3/4)) + (d^2*(10*b*c - 3*a*d)*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])/
(8*b^(1/4)*c^2*(b*c - a*d)^3) + (d^2*(10*b*c - 3*a*d)*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])/(8*b^(1/4)*c^2*(b*c - a*d)^3)

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 541

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 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*} \int \frac {1}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )^2} \, dx &=-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}+\frac {\int \frac {4 b c-3 a d-6 b d x^4}{\left (a+b x^4\right )^{7/4} \left (c+d x^4\right )} \, dx}{4 c (b c-a d)}\\ &=\frac {b (4 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}-\frac {\int \frac {-8 b^2 c^2+24 a b c d-9 a^2 d^2-2 b d (4 b c+3 a d) x^4}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx}{12 a c (b c-a d)^2}\\ &=\frac {b (4 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}+\frac {\left (d^2 (10 b c-3 a d)\right ) \int \frac {\sqrt [4]{a+b x^4}}{c+d x^4} \, dx}{4 c (b c-a d)^3}+\frac {\left (b \left (8 b^2 c^2-32 a b c d+3 a^2 d^2\right )\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{12 a c (b c-a d)^3}\\ &=\frac {b (4 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}+\frac {\left (b \left (8 b^2 c^2-32 a b c d+3 a^2 d^2\right ) \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}{12 a c (b c-a d)^3 \left (a+b x^4\right )^{3/4}}+\frac {\left (d^2 (10 b c-3 a 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 )}{4 c (b c-a d)^3}\\ &=\frac {b (4 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}-\frac {\left (b \left (8 b^2 c^2-32 a b c d+3 a^2 d^2\right ) \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 )}{12 a c (b c-a d)^3 \left (a+b x^4\right )^{3/4}}+\frac {\left (d^2 (10 b c-3 a 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 )}{8 c^2 (b c-a d)^3}+\frac {\left (d^2 (10 b c-3 a 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 )}{8 c^2 (b c-a d)^3}\\ &=\frac {b (4 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}+\frac {d^2 (10 b c-3 a 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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3}+\frac {d^2 (10 b c-3 a 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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3}-\frac {\left (b \left (8 b^2 c^2-32 a b c d+3 a^2 d^2\right ) \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 )}{24 a c (b c-a d)^3 \left (a+b x^4\right )^{3/4}}\\ &=\frac {b (4 b c+3 a d) x}{12 a c (b c-a d)^2 \left (a+b x^4\right )^{3/4}}-\frac {d x}{4 c (b c-a d) \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )}-\frac {b^{3/2} \left (8 b^2 c^2-32 a b c d+3 a^2 d^2\right ) \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 )}{12 a^{3/2} c (b c-a d)^3 \left (a+b x^4\right )^{3/4}}+\frac {d^2 (10 b c-3 a 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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3}+\frac {d^2 (10 b c-3 a 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 )}{8 \sqrt [4]{b} c^2 (b c-a d)^3}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.37, size = 387, normalized size = 0.99 \begin {gather*} \frac {x \left (2 b d (4 b c+3 a d) x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} F_1\left (\frac {5}{4};\frac {3}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+\frac {c \left (25 a c \left (12 a^2 d^2+3 a b d \left (-8 c+d x^4\right )+4 b^2 c \left (3 c+d x^4\right )\right ) F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-5 x^4 \left (3 a^2 d^2+3 a b d^2 x^4+4 b^2 c \left (c+d x^4\right )\right ) \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 )}{\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 )}\right )}{60 a c^2 (b c-a d)^2 \left (a+b x^4\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^4)^(7/4)*(c + d*x^4)^2),x]

[Out]

(x*(2*b*d*(4*b*c + 3*a*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*(25*a*c*(12*a^2*d^2 + 3*a*b*d*(-8*c + d*x^4) + 4*b^2*c*(3*c + d*x^4))*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a
), -((d*x^4)/c)] - 5*x^4*(3*a^2*d^2 + 3*a*b*d^2*x^4 + 4*b^2*c*(c + d*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)])))/((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)])))))/(60*a*c^2*(b*c - a*d)^2*(a
+ b*x^4)^(3/4))

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {7}{4}} \left (d \,x^{4}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^4+a)^(7/4)/(d*x^4+c)^2,x)

[Out]

int(1/(b*x^4+a)^(7/4)/(d*x^4+c)^2,x)

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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)^(7/4)/(d*x^4+c)^2,x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(7/4)*(d*x^4 + c)^2), x)

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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)^(7/4)/(d*x^4+c)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x^{4}\right )^{\frac {7}{4}} \left (c + d x^{4}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**4+a)**(7/4)/(d*x**4+c)**2,x)

[Out]

Integral(1/((a + b*x**4)**(7/4)*(c + d*x**4)**2), x)

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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)^(7/4)/(d*x^4+c)^2,x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(7/4)*(d*x^4 + c)^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,x^4+a\right )}^{7/4}\,{\left (d\,x^4+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^4)^(7/4)*(c + d*x^4)^2),x)

[Out]

int(1/((a + b*x^4)^(7/4)*(c + d*x^4)^2), x)

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