3.854 \(\int \cot ^{\frac{9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=310 \[ \frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}+\frac{2 b \left (49 a^2-3 b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{21 a d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}+\frac{i (-b+i a)^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{i (b+i a)^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

[Out]

(I*(I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt
[Tan[c + d*x]])/d + (I*(I*a + b)^(5/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sq
rt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d + (2*b*(49*a^2 - 3*b^2)*Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(2
1*a*d) + (2*(7*a^2 - 9*b^2)*Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(21*d) - (6*a*b*Cot[c + d*x]^(5/2)*Sq
rt[a + b*Tan[c + d*x]])/(7*d) - (2*a^2*Cot[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]])/(7*d)

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Rubi [A]  time = 1.34736, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4241, 3565, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}+\frac{2 b \left (49 a^2-3 b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{21 a d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}+\frac{i (-b+i a)^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{i (b+i a)^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(9/2)*(a + b*Tan[c + d*x])^(5/2),x]

[Out]

(I*(I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt
[Tan[c + d*x]])/d + (I*(I*a + b)^(5/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sq
rt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d + (2*b*(49*a^2 - 3*b^2)*Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(2
1*a*d) + (2*(7*a^2 - 9*b^2)*Cot[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(21*d) - (6*a*b*Cot[c + d*x]^(5/2)*Sq
rt[a + b*Tan[c + d*x]])/(7*d) - (2*a^2*Cot[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]])/(7*d)

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, 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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3616

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

Rule 3615

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 203

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

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])

Rubi steps

\begin{align*} \int \cot ^{\frac{9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}+\frac{1}{7} \left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{15 a^2 b}{2}-\frac{7}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac{1}{2} b \left (6 a^2-7 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{5}{4} a^2 \left (7 a^2-9 b^2\right )+\frac{35}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)+15 a^2 b^2 \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{35 a}\\ &=\frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}+\frac{\left (8 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{5}{8} a^2 b \left (49 a^2-3 b^2\right )+\frac{105}{8} a^3 \left (a^2-3 b^2\right ) \tan (c+d x)+\frac{5}{4} a^2 b \left (7 a^2-9 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{105 a^2}\\ &=\frac{2 b \left (49 a^2-3 b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{21 a d}+\frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{\left (16 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{105}{16} a^4 \left (a^2-3 b^2\right )-\frac{105}{16} a^3 b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{105 a^3}\\ &=\frac{2 b \left (49 a^2-3 b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{21 a d}+\frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}+\frac{1}{2} \left ((a-i b)^3 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} \left ((a+i b)^3 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{2 b \left (49 a^2-3 b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{21 a d}+\frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}+\frac{\left ((a-i b)^3 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left ((a+i b)^3 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{2 b \left (49 a^2-3 b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{21 a d}+\frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}+\frac{\left ((a-i b)^3 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\left ((a+i b)^3 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=\frac{i (i a-b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}+\frac{i (i a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{d}+\frac{2 b \left (49 a^2-3 b^2\right ) \sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}}{21 a d}+\frac{2 \left (7 a^2-9 b^2\right ) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{21 d}-\frac{6 a b \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}-\frac{2 a^2 \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}}{7 d}\\ \end{align*}

Mathematica [A]  time = 3.33589, size = 274, normalized size = 0.88 \[ \frac{\cot ^{\frac{7}{2}}(c+d x) \left (-\frac{1}{2} \sec ^3(c+d x) \sqrt{a+b \tan (c+d x)} \left (a \left (2 a^2+9 b^2\right ) \cos (c+d x)+\left (10 a^3-9 a b^2\right ) \cos (3 (c+d x))+2 b \sin (c+d x) \left (\left (58 a^2-3 b^2\right ) \cos (2 (c+d x))-40 a^2+3 b^2\right )\right )+21 (-1)^{3/4} a (-a-i b)^{5/2} \tan ^{\frac{7}{2}}(c+d x) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-21 (-1)^{3/4} a (a-i b)^{5/2} \tan ^{\frac{7}{2}}(c+d x) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )}{21 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(9/2)*(a + b*Tan[c + d*x])^(5/2),x]

[Out]

(Cot[c + d*x]^(7/2)*(21*(-1)^(3/4)*a*(-a - I*b)^(5/2)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/S
qrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(7/2) - 21*(-1)^(3/4)*a*(a - I*b)^(5/2)*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b
]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(7/2) - (Sec[c + d*x]^3*(a*(2*a^2 + 9*b^2)*Cos[c
+ d*x] + (10*a^3 - 9*a*b^2)*Cos[3*(c + d*x)] + 2*b*(-40*a^2 + 3*b^2 + (58*a^2 - 3*b^2)*Cos[2*(c + d*x)])*Sin[c
+ d*x])*Sqrt[a + b*Tan[c + d*x]])/2))/(21*a*d)

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Maple [C]  time = 1.242, size = 33802, normalized size = 109. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cot \left (d x + c\right )^{\frac{9}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(9/2), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(9/2)*(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out