\(\int \frac {\cot (x)}{(a+b \tan ^4(x))^{5/2}} \, dx\) [405]
Optimal result
Integrand size = 15, antiderivative size = 183 \[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^4(x)}}{\sqrt {a}}\right )}{2 a^{5/2}}+\frac {1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac {1}{2 a^2 \sqrt {a+b \tan ^4(x)}}-\frac {3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tan ^4(x)}}
\]
[Out]
-1/2*arctanh((a+b*tan(x)^4)^(1/2)/a^(1/2))/a^(5/2)+1/2*arctanh((a-b*tan(x)^2)/(a+b)^(1/2)/(a+b*tan(x)^4)^(1/2)
)/(a+b)^(5/2)+1/2/a^2/(a+b*tan(x)^4)^(1/2)+1/6*(-3*a^2-b*(5*a+2*b)*tan(x)^2)/a^2/(a+b)^2/(a+b*tan(x)^4)^(1/2)+
1/6/a/(a+b*tan(x)^4)^(3/2)+1/6*(-a-b*tan(x)^2)/a/(a+b)/(a+b*tan(x)^4)^(3/2)
Rubi [A] (verified)
Time = 0.33 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of
steps used = 14, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3751, 1266, 975, 755, 837,
12, 739, 212, 272, 53, 65, 214} \[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^4(x)}}{\sqrt {a}}\right )}{2 a^{5/2}}+\frac {1}{2 a^2 \sqrt {a+b \tan ^4(x)}}-\frac {3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tan ^4(x)}}+\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}}+\frac {1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}
\]
[In]
Int[Cot[x]/(a + b*Tan[x]^4)^(5/2),x]
[Out]
ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*(a + b)^(5/2)) - ArcTanh[Sqrt[a + b*Tan[x]^4]/
Sqrt[a]]/(2*a^(5/2)) + 1/(6*a*(a + b*Tan[x]^4)^(3/2)) - (a + b*Tan[x]^2)/(6*a*(a + b)*(a + b*Tan[x]^4)^(3/2))
+ 1/(2*a^2*Sqrt[a + b*Tan[x]^4]) - (3*a^2 + b*(5*a + 2*b)*Tan[x]^2)/(6*a^2*(a + b)^2*Sqrt[a + b*Tan[x]^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 53
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
Rule 65
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
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 272
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 739
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]
Rule 755
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]
Rule 837
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Rule 975
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) && !(IGtQ[m, 0] || IGtQ[n, 0])
Rule 1266
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
Rule 3751
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))
Rubi steps \begin{align*}
\text {integral}& = \text {Subst}\left (\int \frac {1}{x \left (1+x^2\right ) \left (a+b x^4\right )^{5/2}} \, dx,x,\tan (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x (1+x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan ^2(x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{(-1-x) \left (a+b x^2\right )^{5/2}}+\frac {1}{x \left (a+b x^2\right )^{5/2}}\right ) \, dx,x,\tan ^2(x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(-1-x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan ^2(x)\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \left (a+b x^2\right )^{5/2}} \, dx,x,\tan ^2(x)\right ) \\ & = -\frac {a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\tan ^4(x)\right )-\frac {\text {Subst}\left (\int \frac {-3 a-2 b-2 b x}{(-1-x) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan ^2(x)\right )}{6 a (a+b)} \\ & = \frac {1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tan ^4(x)}}+\frac {\text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\tan ^4(x)\right )}{4 a}+\frac {\text {Subst}\left (\int \frac {3 a^2 b}{(-1-x) \sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right )}{6 a^2 b (a+b)^2} \\ & = \frac {1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac {1}{2 a^2 \sqrt {a+b \tan ^4(x)}}-\frac {3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tan ^4(x)}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\tan ^4(x)\right )}{4 a^2}+\frac {\text {Subst}\left (\int \frac {1}{(-1-x) \sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right )}{2 (a+b)^2} \\ & = \frac {1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac {1}{2 a^2 \sqrt {a+b \tan ^4(x)}}-\frac {3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tan ^4(x)}}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^4(x)}\right )}{2 a^2 b}-\frac {\text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\frac {-a+b \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{2 (a+b)^2} \\ & = \frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^4(x)}}{\sqrt {a}}\right )}{2 a^{5/2}}+\frac {1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac {1}{2 a^2 \sqrt {a+b \tan ^4(x)}}-\frac {3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt {a+b \tan ^4(x)}} \\
\end{align*}
Mathematica [C] (verified)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.86 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.81
\[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=\frac {1}{6} \left (\frac {3 \text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{(a+b)^{5/2}}+\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \tan ^4(x)}{a}\right )}{a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac {a^2 (4 a+b)+3 a b (2 a+b) \tan ^2(x)+3 a^2 b \tan ^4(x)+b^2 (5 a+2 b) \tan ^6(x)}{a^2 (a+b)^2 \left (a+b \tan ^4(x)\right )^{3/2}}\right )
\]
[In]
Integrate[Cot[x]/(a + b*Tan[x]^4)^(5/2),x]
[Out]
((3*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])])/(a + b)^(5/2) + Hypergeometric2F1[-3/2, 1, -
1/2, 1 + (b*Tan[x]^4)/a]/(a*(a + b*Tan[x]^4)^(3/2)) - (a^2*(4*a + b) + 3*a*b*(2*a + b)*Tan[x]^2 + 3*a^2*b*Tan[
x]^4 + b^2*(5*a + 2*b)*Tan[x]^6)/(a^2*(a + b)^2*(a + b*Tan[x]^4)^(3/2)))/6
Maple [F]
\[\int \frac {\cot \left (x \right )}{\left (a +b \tan \left (x \right )^{4}\right )^{\frac {5}{2}}}d x\]
[In]
int(cot(x)/(a+b*tan(x)^4)^(5/2),x)
[Out]
int(cot(x)/(a+b*tan(x)^4)^(5/2),x)
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (153) = 306\).
Time = 0.68 (sec) , antiderivative size = 1749, normalized size of antiderivative = 9.56
\[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=\text {Too large to display}
\]
[In]
integrate(cot(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="fricas")
[Out]
[1/12*(3*(a^3*b^2*tan(x)^8 + 2*a^4*b*tan(x)^4 + a^5)*sqrt(a + b)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2
- 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) + 3*((a^3*b^
2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*tan(x)^8 + a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b
^3 + a*b^4)*tan(x)^4)*sqrt(a)*log(-(b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*sqrt(a) + 2*a)/tan(x)^4) - 2*((5*a^3*b
^2 + 7*a^2*b^3 + 2*a*b^4)*tan(x)^6 - 7*a^4*b - 11*a^3*b^2 - 4*a^2*b^3 - 3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*tan(
x)^4 + 3*(2*a^4*b + 3*a^3*b^2 + a^2*b^3)*tan(x)^2)*sqrt(b*tan(x)^4 + a))/((a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a
^3*b^5)*tan(x)^8 + a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*tan(x)^4)
, 1/12*(6*((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*tan(x)^8 + a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 + 2*(a^4*b + 3
*a^3*b^2 + 3*a^2*b^3 + a*b^4)*tan(x)^4)*sqrt(-a)*arctan(sqrt(b*tan(x)^4 + a)*sqrt(-a)/a) + 3*(a^3*b^2*tan(x)^8
+ 2*a^4*b*tan(x)^4 + a^5)*sqrt(a + b)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(
b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) - 2*((5*a^3*b^2 + 7*a^2*b^3 + 2*a*b^4)
*tan(x)^6 - 7*a^4*b - 11*a^3*b^2 - 4*a^2*b^3 - 3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*tan(x)^4 + 3*(2*a^4*b + 3*a^3
*b^2 + a^2*b^3)*tan(x)^2)*sqrt(b*tan(x)^4 + a))/((a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*tan(x)^8 + a^8 +
3*a^7*b + 3*a^6*b^2 + a^5*b^3 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*tan(x)^4), 1/12*(6*(a^3*b^2*tan(x)
^8 + 2*a^4*b*tan(x)^4 + a^5)*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^
2)*tan(x)^4 + a^2 + a*b)) + 3*((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*tan(x)^8 + a^5 + 3*a^4*b + 3*a^3*b^2 + a^
2*b^3 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*tan(x)^4)*sqrt(a)*log(-(b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*
sqrt(a) + 2*a)/tan(x)^4) - 2*((5*a^3*b^2 + 7*a^2*b^3 + 2*a*b^4)*tan(x)^6 - 7*a^4*b - 11*a^3*b^2 - 4*a^2*b^3 -
3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*tan(x)^4 + 3*(2*a^4*b + 3*a^3*b^2 + a^2*b^3)*tan(x)^2)*sqrt(b*tan(x)^4 + a))
/((a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*tan(x)^8 + a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3 + 2*(a^7*b + 3*a^
6*b^2 + 3*a^5*b^3 + a^4*b^4)*tan(x)^4), 1/6*(3*(a^3*b^2*tan(x)^8 + 2*a^4*b*tan(x)^4 + a^5)*sqrt(-a - b)*arctan
(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2)*tan(x)^4 + a^2 + a*b)) + 3*((a^3*b^2 + 3*a^2*
b^3 + 3*a*b^4 + b^5)*tan(x)^8 + a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4
)*tan(x)^4)*sqrt(-a)*arctan(sqrt(b*tan(x)^4 + a)*sqrt(-a)/a) - ((5*a^3*b^2 + 7*a^2*b^3 + 2*a*b^4)*tan(x)^6 - 7
*a^4*b - 11*a^3*b^2 - 4*a^2*b^3 - 3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*tan(x)^4 + 3*(2*a^4*b + 3*a^3*b^2 + a^2*b^
3)*tan(x)^2)*sqrt(b*tan(x)^4 + a))/((a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*tan(x)^8 + a^8 + 3*a^7*b + 3*a
^6*b^2 + a^5*b^3 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*tan(x)^4)]
Sympy [F]
\[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=\int \frac {\cot {\left (x \right )}}{\left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx
\]
[In]
integrate(cot(x)/(a+b*tan(x)**4)**(5/2),x)
[Out]
Integral(cot(x)/(a + b*tan(x)**4)**(5/2), x)
Maxima [F(-2)]
Exception generated. \[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError}
\]
[In]
integrate(cot(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.
Giac [F(-2)]
Exception generated. \[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError}
\]
[In]
integrate(cot(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type
Mupad [F(-1)]
Timed out. \[
\int \frac {\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx=\text {Hanged}
\]
[In]
int(cot(x)/(a + b*tan(x)^4)^(5/2),x)
[Out]
\text{Hanged}