∫ cot5 _2 (c+dx)___ (a+b tan(c+dx))5∕2 dx

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

Optimal. Leaf size=338 \[ \frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right )}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]

[Out]

arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/(I*a-b)^(5/2)/

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

2)/d+4/3*b^2*(4*a^4+15*a^2*b^2+8*b^4)/a^4/(a^2+b^2)^2/d/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)-2/3*cot(d*x+c)

^(3/2)/a/d/(a+b*tan(d*x+c))^(3/2)+2/3*b^2*(7*a^2+8*b^2)/a^3/(a^2+b^2)/d/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2

)+4*b*cot(d*x+c)^(1/2)/a^2/d/(a+b*tan(d*x+c))^(3/2)

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Rubi [A] time = 1.28, antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4241, 3569, 3649, 3650, 3616, 3615, 93, 203, 206} \[ \frac {4 b^2 \left (15 a^2 b^2+4 a^4+8 b^4\right )}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(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]])/((

I*a - b)^(5/2)*d) + (ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*S

qrt[Tan[c + d*x]])/((I*a + b)^(5/2)*d) + (2*b^2*(7*a^2 + 8*b^2))/(3*a^3*(a^2 + b^2)*d*Sqrt[Cot[c + d*x]]*(a +

b*Tan[c + d*x])^(3/2)) + (4*b*Sqrt[Cot[c + d*x]])/(a^2*d*(a + b*Tan[c + d*x])^(3/2)) - (2*Cot[c + d*x]^(3/2))/

(3*a*d*(a + b*Tan[c + d*x])^(3/2)) + (4*b^2*(4*a^4 + 15*a^2*b^2 + 8*b^4))/(3*a^4*(a^2 + b^2)^2*d*Sqrt[Cot[c +

d*x]]*Sqrt[a + b*Tan[c + d*x]])

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])

Rule 3569

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 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 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 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 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*t

an[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 + a^2*C)*(a + b*Tan[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)) - a*C*(b*c*(m + 1)

+ a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 2)*Tan[e + f*x]^2,

x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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]

Rubi steps

\begin {align*} \int \frac {\cot ^{\frac {5}{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 {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\\ &=-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}-\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {3 b+\frac {3}{2} a \tan (c+d x)+3 b \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx}{3 a}\\ &=\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {3}{4} \left (a^2-8 b^2\right )+6 b^2 \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{3 a^2}\\ &=\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {3}{8} \left (3 a^4-14 a^2 b^2-16 b^4\right )+\frac {9}{8} a^3 b \tan (c+d x)+\frac {3}{4} b^2 \left (7 a^2+8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{9 a^3 \left (a^2+b^2\right )}\\ &=\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right )}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left (16 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {9}{16} a^4 \left (a^2-b^2\right )+\frac {9}{8} a^5 b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{9 a^4 \left (a^2+b^2\right )^2}\\ &=\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right )}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\left ((a-i b)^2 \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}{2 \left (a^2+b^2\right )^2}-\frac {\left ((a+i b)^2 \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}{2 \left (a^2+b^2\right )^2}\\ &=\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right )}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\left ((a-i b)^2 \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 \left (a^2+b^2\right )^2 d}-\frac {\left ((a+i b)^2 \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 \left (a^2+b^2\right )^2 d}\\ &=\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right )}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\left ((a-i b)^2 \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 )}{\left (a^2+b^2\right )^2 d}-\frac {\left ((a+i b)^2 \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 )}{\left (a^2+b^2\right )^2 d}\\ &=\frac {\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)}}{(i a-b)^{5/2} d}+\frac {\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)}}{(i a+b)^{5/2} d}+\frac {2 b^2 \left (7 a^2+8 b^2\right )}{3 a^3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 b \sqrt {\cot (c+d x)}}{a^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 \cot ^{\frac {3}{2}}(c+d x)}{3 a d (a+b \tan (c+d x))^{3/2}}+\frac {4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right )}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}\\ \end {align*}

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Mathematica [A] time = 6.31, size = 504, normalized size = 1.49 \[ \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (-\frac {6 b}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {3 \left (\frac {32 b^2 \sqrt {\tan (c+d x)}}{3 a^2 \sqrt {a+b \tan (c+d x)}}-\frac {\frac {b (5 a-2 i b) \sqrt {\tan (c+d x)}}{(a-i b) \sqrt {a+b \tan (c+d x)}}-\frac {3 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}}{3 (b+i a)}+\frac {\frac {b (5 a+2 i b) \sqrt {\tan (c+d x)}}{(a+i b) \sqrt {a+b \tan (c+d x)}}-\frac {3 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{3/2}}}{3 (-b+i a)}+\frac {16 b^2 \sqrt {\tan (c+d x)}}{3 a (a+b \tan (c+d x))^{3/2}}+\frac {a b \sqrt {\tan (c+d x)}}{3 (-b+i a) (a+b \tan (c+d x))^{3/2}}-\frac {a b \sqrt {\tan (c+d x)}}{3 (b+i a) (a+b \tan (c+d x))^{3/2}}\right )}{2 a d}\right )}{3 a}\right ) \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-2/(3*a*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)) - (2*((-6*b)/(

a*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) - (3*((a*b*Sqrt[Tan[c + d*x]])/(3*(I*a - b)*(a + b*Tan[c +

d*x])^(3/2)) + (16*b^2*Sqrt[Tan[c + d*x]])/(3*a*(a + b*Tan[c + d*x])^(3/2)) - (a*b*Sqrt[Tan[c + d*x]])/(3*(I*a

+ b)*(a + b*Tan[c + d*x])^(3/2)) + (32*b^2*Sqrt[Tan[c + d*x]])/(3*a^2*Sqrt[a + b*Tan[c + d*x]]) - ((-3*(-1)^(

1/4)*a^2*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(-a + I*b)^(3/2) + (

(5*a - (2*I)*b)*b*Sqrt[Tan[c + d*x]])/((a - I*b)*Sqrt[a + b*Tan[c + d*x]]))/(3*(I*a + b)) + ((-3*(-1)^(1/4)*a^

2*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(a + I*b)^(3/2) + ((5*a + (2

*I)*b)*b*Sqrt[Tan[c + d*x]])/((a + I*b)*Sqrt[a + b*Tan[c + d*x]]))/(3*(I*a - b))))/(2*a*d)))/(3*a))

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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn

ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab

s(sin(d*x+c))]Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2

)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no

step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p

i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign

: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to chec

k sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable t

o check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Un

able to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_noste

p/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t

_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-

2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep

/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_

nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2

*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check si

gn: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to ch

eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable

to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)

Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nos

tep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi

/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>

(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nost

ep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/

t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign:

(2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check

sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to

check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unab

le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/

2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n

ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*

pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2