3.405 \(\int \frac{\cot (x)}{(a+b \tan ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ -\frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}+\frac{1}{2 a^2 \sqrt{a+b \tan ^4(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^4(x)}}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}} \]

[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])

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Rubi [A]  time = 0.302311, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {3670, 1252, 961, 741, 823, 12, 725, 206, 266, 51, 63, 208} \[ -\frac{3 a^2+b (5 a+2 b) \tan ^2(x)}{6 a^2 (a+b)^2 \sqrt{a+b \tan ^4(x)}}+\frac{1}{2 a^2 \sqrt{a+b \tan ^4(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^4(x)}}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{a+b \tan ^2(x)}{6 a (a+b) \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{1}{6 a \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[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 3670

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]))

Rule 1252

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 961

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 741

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 823

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] + Di
st[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 12

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

Rule 725

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 266

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 51

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 63

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 208

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

Rubi steps

\begin{align*} \int \frac{\cot (x)}{\left (a+b \tan ^4(x)\right )^{5/2}} \, dx &=\operatorname{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} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{(-1-x) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan ^2(x)\right )+\frac{1}{2} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,\tan ^4(x)\right )-\frac{\operatorname{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{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\tan ^4(x)\right )}{4 a}+\frac{\operatorname{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{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan ^4(x)\right )}{4 a^2}+\frac{\operatorname{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{\operatorname{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{\operatorname{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{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 (a+b)^{5/2}}-\frac{\tanh ^{-1}\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]  time = 1.40658, size = 149, normalized size = 0.81 \[ \frac{1}{6} \left (\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{b \tan ^4(x)}{a}+1\right )}{a \left (a+b \tan ^4(x)\right )^{3/2}}-\frac{3 a^2 b \tan ^4(x)+a^2 (4 a+b)+b^2 (5 a+2 b) \tan ^6(x)+3 a b (2 a+b) \tan ^2(x)}{a^2 (a+b)^2 \left (a+b \tan ^4(x)\right )^{3/2}}+\frac{3 \tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{(a+b)^{5/2}}\right ) \]

Antiderivative was successfully verified.

[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

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Maple [F]  time = 0.128, size = 0, normalized size = 0. \begin{align*} \int{\cot \left ( x \right ) \left ( a+b \left ( \tan \left ( x \right ) \right ) ^{4} \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)/(a+b*tan(x)^4)^(5/2),x)

[Out]

int(cot(x)/(a+b*tan(x)^4)^(5/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 7.16085, size = 3970, normalized size = 21.69 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (x \right )}}{\left (a + b \tan ^{4}{\left (x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*tan(x)**4)**(5/2),x)

[Out]

Integral(cot(x)/(a + b*tan(x)**4)**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*tan(x)^4)^(5/2),x, algorithm="giac")

[Out]

integrate(cot(x)/(b*tan(x)^4 + a)^(5/2), x)