\(\int \frac {1}{(a+b x^4)^{13/4} (c+d x^4)} \, dx\) [198]

Optimal result

Integrand size = 21, antiderivative size = 233 \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}-\frac {d^3 \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}} \]

[Out]

1/9*b*x/a/(-a*d+b*c)/(b*x^4+a)^(9/4)+1/45*b*(-17*a*d+8*b*c)*x/a^2/(-a*d+b*c)^2/(b*x^4+a)^(5/4)+1/45*b*(113*a^2
*d^2-100*a*b*c*d+32*b^2*c^2)*x/a^3/(-a*d+b*c)^3/(b*x^4+a)^(1/4)-1/2*d^3*arctan((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x
^4+a)^(1/4))/c^(3/4)/(-a*d+b*c)^(13/4)-1/2*d^3*arctanh((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(3/4)/(-a
*d+b*c)^(13/4)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {425, 541, 12, 385, 218, 214, 211} \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\frac {b x (8 b c-17 a d)}{45 a^2 \left (a+b x^4\right )^{5/4} (b c-a d)^2}+\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{45 a^3 \sqrt [4]{a+b x^4} (b c-a d)^3}-\frac {d^3 \arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}-\frac {d^3 \text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)} \]

[In]

Int[1/((a + b*x^4)^(13/4)*(c + d*x^4)),x]

[Out]

(b*x)/(9*a*(b*c - a*d)*(a + b*x^4)^(9/4)) + (b*(8*b*c - 17*a*d)*x)/(45*a^2*(b*c - a*d)^2*(a + b*x^4)^(5/4)) +
(b*(32*b^2*c^2 - 100*a*b*c*d + 113*a^2*d^2)*x)/(45*a^3*(b*c - a*d)^3*(a + b*x^4)^(1/4)) - (d^3*ArcTan[((b*c -
a*d)^(1/4)*x)/(c^(1/4)*(a + b*x^4)^(1/4))])/(2*c^(3/4)*(b*c - a*d)^(13/4)) - (d^3*ArcTanh[((b*c - a*d)^(1/4)*x
)/(c^(1/4)*(a + b*x^4)^(1/4))])/(2*c^(3/4)*(b*c - a*d)^(13/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}-\frac {\int \frac {-8 b c+9 a d-8 b d x^4}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx}{9 a (b c-a d)} \\ & = \frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {\int \frac {32 b^2 c^2-68 a b c d+45 a^2 d^2+4 b d (8 b c-17 a d) x^4}{\left (a+b x^4\right )^{5/4} \left (c+d x^4\right )} \, dx}{45 a^2 (b c-a d)^2} \\ & = \frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {\int \frac {45 a^3 d^3}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )} \, dx}{45 a^3 (b c-a d)^3} \\ & = \frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )} \, dx}{(b c-a d)^3} \\ & = \frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \text {Subst}\left (\int \frac {1}{c-(b c-a d) x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{(b c-a d)^3} \\ & = \frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \text {Subst}\left (\int \frac {1}{\sqrt {c}-\sqrt {b c-a d} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt {c} (b c-a d)^3}-\frac {d^3 \text {Subst}\left (\int \frac {1}{\sqrt {c}+\sqrt {b c-a d} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt {c} (b c-a d)^3} \\ & = \frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \tan ^{-1}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}-\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 14.50 (sec) , antiderivative size = 1172, normalized size of antiderivative = 5.03 \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=-\frac {-16575 c^5 (b c-a d)^2 x^8 \left (a+b x^4\right )^2-39780 c^4 d (b c-a d)^2 x^{12} \left (a+b x^4\right )^2-35360 c^3 d^2 (b c-a d)^2 x^{16} \left (a+b x^4\right )^2-10880 c^2 d^3 (b c-a d)^2 x^{20} \left (a+b x^4\right )^2-29835 c^6 (b c-a d) x^4 \left (a+b x^4\right )^3-71604 c^5 d (b c-a d) x^8 \left (a+b x^4\right )^3-63648 c^4 d^2 (b c-a d) x^{12} \left (a+b x^4\right )^3-19584 c^3 d^3 (b c-a d) x^{16} \left (a+b x^4\right )^3-149175 c^7 \left (a+b x^4\right )^4-358020 c^6 d x^4 \left (a+b x^4\right )^4-318240 c^5 d^2 x^8 \left (a+b x^4\right )^4-97920 c^4 d^3 x^{12} \left (a+b x^4\right )^4+149175 c^7 \left (a+b x^4\right )^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+358020 c^6 d x^4 \left (a+b x^4\right )^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+318240 c^5 d^2 x^8 \left (a+b x^4\right )^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+97920 c^4 d^3 x^{12} \left (a+b x^4\right )^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+13620 c^3 (b c-a d)^4 x^{16} \operatorname {Hypergeometric2F1}\left (2,\frac {17}{4},\frac {21}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+36900 c^2 d (b c-a d)^4 x^{20} \operatorname {Hypergeometric2F1}\left (2,\frac {17}{4},\frac {21}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+33840 c d^2 (b c-a d)^4 x^{24} \operatorname {Hypergeometric2F1}\left (2,\frac {17}{4},\frac {21}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+10560 d^3 (b c-a d)^4 x^{28} \operatorname {Hypergeometric2F1}\left (2,\frac {17}{4},\frac {21}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+6480 c^3 (b c-a d)^4 x^{16} \, _3F_2\left (2,2,\frac {17}{4};1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+18720 c^2 d (b c-a d)^4 x^{20} \, _3F_2\left (2,2,\frac {17}{4};1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+18000 c d^2 (b c-a d)^4 x^{24} \, _3F_2\left (2,2,\frac {17}{4};1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+5760 d^3 (b c-a d)^4 x^{28} \, _3F_2\left (2,2,\frac {17}{4};1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+960 c^3 (b c-a d)^4 x^{16} \, _4F_3\left (2,2,2,\frac {17}{4};1,1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+2880 c^2 d (b c-a d)^4 x^{20} \, _4F_3\left (2,2,2,\frac {17}{4};1,1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+2880 c d^2 (b c-a d)^4 x^{24} \, _4F_3\left (2,2,2,\frac {17}{4};1,1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+960 d^3 (b c-a d)^4 x^{28} \, _4F_3\left (2,2,2,\frac {17}{4};1,1,\frac {21}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )}{11475 c^5 (-b c+a d)^3 x^{11} \left (a+b x^4\right )^{17/4}} \]

[In]

Integrate[1/((a + b*x^4)^(13/4)*(c + d*x^4)),x]

[Out]

-1/11475*(-16575*c^5*(b*c - a*d)^2*x^8*(a + b*x^4)^2 - 39780*c^4*d*(b*c - a*d)^2*x^12*(a + b*x^4)^2 - 35360*c^
3*d^2*(b*c - a*d)^2*x^16*(a + b*x^4)^2 - 10880*c^2*d^3*(b*c - a*d)^2*x^20*(a + b*x^4)^2 - 29835*c^6*(b*c - a*d
)*x^4*(a + b*x^4)^3 - 71604*c^5*d*(b*c - a*d)*x^8*(a + b*x^4)^3 - 63648*c^4*d^2*(b*c - a*d)*x^12*(a + b*x^4)^3
- 19584*c^3*d^3*(b*c - a*d)*x^16*(a + b*x^4)^3 - 149175*c^7*(a + b*x^4)^4 - 358020*c^6*d*x^4*(a + b*x^4)^4 -
318240*c^5*d^2*x^8*(a + b*x^4)^4 - 97920*c^4*d^3*x^12*(a + b*x^4)^4 + 149175*c^7*(a + b*x^4)^4*Hypergeometric2
F1[1/4, 1, 5/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 358020*c^6*d*x^4*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1,
5/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 318240*c^5*d^2*x^8*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ((b*
c - a*d)*x^4)/(c*(a + b*x^4))] + 97920*c^4*d^3*x^12*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ((b*c - a*d)*
x^4)/(c*(a + b*x^4))] + 13620*c^3*(b*c - a*d)^4*x^16*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a
+ b*x^4))] + 36900*c^2*d*(b*c - a*d)^4*x^20*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))
] + 33840*c*d^2*(b*c - a*d)^4*x^24*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 10560
*d^3*(b*c - a*d)^4*x^28*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 6480*c^3*(b*c -
a*d)^4*x^16*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 18720*c^2*d*(b*c -
a*d)^4*x^20*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 18000*c*d^2*(b*c
- a*d)^4*x^24*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 5760*d^3*(b*c -
a*d)^4*x^28*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 960*c^3*(b*c - a*d
)^4*x^16*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 2880*c^2*d*(b*c
- a*d)^4*x^20*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 2880*c*d^
2*(b*c - a*d)^4*x^24*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 960
*d^3*(b*c - a*d)^4*x^28*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))])/(
c^5*(-(b*c) + a*d)^3*x^11*(a + b*x^4)^(17/4))

Maple [A] (verified)

Time = 4.44 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.64

method	result	size
pseudoelliptic	\(-\frac {3 \left (\frac {a^{3} d^{3} \left (b \,x^{4}+a \right )^{\frac {9}{4}} \left (\ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )-2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x -\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x +\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}\right )\right ) \sqrt {2}}{24}+x \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} b \left (a^{4} d^{2}-b \left (-\frac {9 d \,x^{4}}{5}+c \right ) d \,a^{3}+\frac {b^{2} \left (\frac {113}{45} d^{2} x^{8}-5 c d \,x^{4}+c^{2}\right ) a^{2}}{3}+\frac {8 x^{4} b^{3} \left (-\frac {25 d \,x^{4}}{18}+c \right ) c a}{15}+\frac {32 b^{4} c^{2} x^{8}}{135}\right ) c \right )}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {9}{4}} \left (a d -b c \right )^{3} c \,a^{3}}\)	\(381\)

[In]

int(1/(b*x^4+a)^(13/4)/(d*x^4+c),x,method=_RETURNVERBOSE)

[Out]

-3/((a*d-b*c)/c)^(1/4)*(1/24*a^3*d^3*(b*x^4+a)^(9/4)*(ln((-((a*d-b*c)/c)^(1/4)*(b*x^4+a)^(1/4)*2^(1/2)*x+((a*d
-b*c)/c)^(1/2)*x^2+(b*x^4+a)^(1/2))/(((a*d-b*c)/c)^(1/4)*(b*x^4+a)^(1/4)*2^(1/2)*x+((a*d-b*c)/c)^(1/2)*x^2+(b*
x^4+a)^(1/2)))-2*arctan((((a*d-b*c)/c)^(1/4)*x-2^(1/2)*(b*x^4+a)^(1/4))/((a*d-b*c)/c)^(1/4)/x)+2*arctan((((a*d
-b*c)/c)^(1/4)*x+2^(1/2)*(b*x^4+a)^(1/4))/((a*d-b*c)/c)^(1/4)/x))*2^(1/2)+x*((a*d-b*c)/c)^(1/4)*b*(a^4*d^2-b*(
-9/5*d*x^4+c)*d*a^3+1/3*b^2*(113/45*d^2*x^8-5*c*d*x^4+c^2)*a^2+8/15*x^4*b^3*(-25/18*d*x^4+c)*c*a+32/135*b^4*c^
2*x^8)*c)/(b*x^4+a)^(9/4)/(a*d-b*c)^3/c/a^3

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (a + b x^{4}\right )^{\frac {13}{4}} \left (c + d x^{4}\right )}\, dx \]

[In]

integrate(1/(b*x**4+a)**(13/4)/(d*x**4+c),x)

[Out]

Integral(1/((a + b*x**4)**(13/4)*(c + d*x**4)), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} {\left (d x^{4} + c\right )}} \,d x } \]

[In]

integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(13/4)*(d*x^4 + c)), x)

Giac [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} {\left (d x^{4} + c\right )}} \,d x } \]

[In]

integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(13/4)*(d*x^4 + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{13/4}\,\left (d\,x^4+c\right )} \,d x \]

[In]

int(1/((a + b*x^4)^(13/4)*(c + d*x^4)),x)

[Out]

int(1/((a + b*x^4)^(13/4)*(c + d*x^4)), x)