3.9.25 \(\int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [825]

Optimal. Leaf size=250 \[ -\frac {(a+b) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a+b) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 a^{5/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}+\frac {2}{b d \sqrt {\cot (c+d x)}}-\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]

[Out]

2*a^(5/2)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/b^(1/2))/b^(3/2)/(a^2+b^2)/d+1/2*(a+b)*arctan(-1+2^(1/2)*cot(d*x+c)^
(1/2))/(a^2+b^2)/d*2^(1/2)+1/2*(a+b)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)-1/4*(a-b)*ln(1+cot
(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)+1/4*(a-b)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(a^2
+b^2)/d*2^(1/2)+2/b/d/cot(d*x+c)^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3754, 3650, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {(a+b) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {(a+b) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )}-\frac {(a-b) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}+\frac {(a-b) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}+\frac {2 a^{5/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d \left (a^2+b^2\right )}+\frac {2}{b d \sqrt {\cot (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]

[Out]

-(((a + b)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((a + b)*ArcTan[1 + Sqrt[2]*Sqrt
[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) + (2*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(b^(3/2)*(
a^2 + b^2)*d) + 2/(b*d*Sqrt[Cot[c + d*x]]) - ((a - b)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*S
qrt[2]*(a^2 + b^2)*d) + ((a - b)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d)

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 210

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

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rule 3754

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))} \, dx\\ &=\frac {2}{b d \sqrt {\cot (c+d x)}}+\frac {2 \int \frac {-\frac {a}{2}-\frac {1}{2} b \cot (c+d x)-\frac {1}{2} a \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{b}\\ &=\frac {2}{b d \sqrt {\cot (c+d x)}}+\frac {2 \int \frac {-\frac {a b}{2}-\frac {1}{2} b^2 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{b \left (a^2+b^2\right )}-\frac {a^3 \int \frac {1+\cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac {2}{b d \sqrt {\cot (c+d x)}}+\frac {4 \text {Subst}\left (\int \frac {\frac {a b}{2}+\frac {b^2 x^2}{2}}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{b \left (a^2+b^2\right ) d}-\frac {a^3 \text {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=\frac {2}{b d \sqrt {\cot (c+d x)}}+\frac {(a-b) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}+\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{b \left (a^2+b^2\right ) d}+\frac {(a+b) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}+\frac {2}{b d \sqrt {\cot (c+d x)}}-\frac {(a-b) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a-b) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}+\frac {2}{b d \sqrt {\cot (c+d x)}}-\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a+b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}\\ &=-\frac {(a+b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a+b) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}+\frac {2}{b d \sqrt {\cot (c+d x)}}-\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.41, size = 193, normalized size = 0.77 \begin {gather*} \frac {8 a^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {a \cot (c+d x)}{b}\right )+b \left (8 b \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right )+\sqrt {2} a \sqrt {\cot (c+d x)} \left (-2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )+2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )-\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )+\log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right )\right )}{4 b \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]

[Out]

(8*a^2*Hypergeometric2F1[-1/2, 1, 1/2, -((a*Cot[c + d*x])/b)] + b*(8*b*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c
+ d*x]^2] + Sqrt[2]*a*Sqrt[Cot[c + d*x]]*(-2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] + 2*ArcTan[1 + Sqrt[2]*Sqr
t[Cot[c + d*x]]] - Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + C
ot[c + d*x]])))/(4*b*(a^2 + b^2)*d*Sqrt[Cot[c + d*x]])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 55.60, size = 5450, normalized size = 21.80

method	result	size
default	\(\text {Expression too large to display}\)	\(5450\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [A]
time = 0.53, size = 189, normalized size = 0.76 \begin {gather*} \frac {\frac {8 \, a^{3} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{2} b + b^{3}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (a - b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (a - b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{2} + b^{2}} + \frac {8 \, \sqrt {\tan \left (d x + c\right )}}{b}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/4*(8*a^3*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^2*b + b^3)*sqrt(a*b)) + (2*sqrt(2)*(a + b)*arctan(1/2*
sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a + b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x +
c)))) + sqrt(2)*(a - b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - sqrt(2)*(a - b)*log(-sqrt(2)/sq
rt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/(a^2 + b^2) + 8*sqrt(tan(d*x + c))/b)/d

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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/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),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 \tan {\left (c + d x \right )}\right ) \cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)**(5/2)/(a+b*tan(d*x+c)),x)

[Out]

Integral(1/((a + b*tan(c + d*x))*cot(c + d*x)**(5/2)), 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/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((b*tan(d*x + c) + a)*cot(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cot(c + d*x)^(5/2)*(a + b*tan(c + d*x))),x)

[Out]

int(1/(cot(c + d*x)^(5/2)*(a + b*tan(c + d*x))), x)

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