\(\int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\) [873]

Optimal result

Integrand size = 25, antiderivative size = 310 \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{b^{5/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}-\frac {2 a^2}{b \left (a^2+b^2\right ) d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {\left (3 a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)}} \] Output:

I*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)^(3/2)/d-3*a*arctanh(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)/b^(5/2)/d+I*a 
rctanh((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)^(3/2)/d-2*a^2/b/(a^2+b^2)/d/cot(d*x+c)^(3/2) 
/(a+b*tan(d*x+c))^(1/2)+(3*a^2+b^2)*(a+b*tan(d*x+c))^(1/2)/b^2/(a^2+b^2)/d 
/cot(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.52 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {5 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}-\frac {5 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{3/2}}-\frac {5 \sqrt {\tan (c+d x)}}{(a-i b) \sqrt {a+b \tan (c+d x)}}-\frac {5 \sqrt {\tan (c+d x)}}{(a+i b) \sqrt {a+b \tan (c+d x)}}+\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {5}{2}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{a \sqrt {a+b \tan (c+d x)}}\right )}{5 d} \] Input:

Integrate[1/(Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(3/2)),x]
 

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((5*(-1)^(3/4)*ArcTan[((-1)^(1/4)*S 
qrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(-a + I*b)^(3 
/2) - (5*(-1)^(3/4)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/S 
qrt[a + b*Tan[c + d*x]]])/(a + I*b)^(3/2) - (5*Sqrt[Tan[c + d*x]])/((a - I 
*b)*Sqrt[a + b*Tan[c + d*x]]) - (5*Sqrt[Tan[c + d*x]])/((a + I*b)*Sqrt[a + 
 b*Tan[c + d*x]]) + (2*Hypergeometric2F1[3/2, 5/2, 7/2, -((b*Tan[c + d*x]) 
/a)]*Tan[c + d*x]^(5/2)*Sqrt[1 + (b*Tan[c + d*x])/a])/(a*Sqrt[a + b*Tan[c 
+ d*x]])))/(5*d)
 

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4729, 3042, 4048, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cot (c+d x)^{7/2} (a+b \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{7/2}}{(a+b \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 \int \frac {\sqrt {\tan (c+d x)} \left (3 a^2-b \tan (c+d x) a+\left (3 a^2+b^2\right ) \tan ^2(c+d x)\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {\sqrt {\tan (c+d x)} \left (3 a^2-b \tan (c+d x) a+\left (3 a^2+b^2\right ) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {\sqrt {\tan (c+d x)} \left (3 a^2-b \tan (c+d x) a+\left (3 a^2+b^2\right ) \tan (c+d x)^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 4130

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int -\frac {2 \tan (c+d x) b^3+3 a \left (a^2+b^2\right ) \tan ^2(c+d x)+a \left (3 a^2+b^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b}+\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {2 \tan (c+d x) b^3+3 a \left (a^2+b^2\right ) \tan ^2(c+d x)+a \left (3 a^2+b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {2 \tan (c+d x) b^3+3 a \left (a^2+b^2\right ) \tan (c+d x)^2+a \left (3 a^2+b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 4138

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {2 \tan (c+d x) b^3+3 a \left (a^2+b^2\right ) \tan ^2(c+d x)+a \left (3 a^2+b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 b d}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 2035

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {2 \tan (c+d x) b^3+3 a \left (a^2+b^2\right ) \tan ^2(c+d x)+a \left (3 a^2+b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 2257

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \left (\frac {3 a \left (a^2+b^2\right )}{\sqrt {a+b \tan (c+d x)}}-\frac {2 \left (a b^2-b^3 \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {3 a \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-\frac {b^2 (a-i b) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}-\frac {b^2 (a+i b) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}}{b d}}{b \left (a^2+b^2\right )}\right )\)

Input:

Int[1/(Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(3/2)),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*a^2*Tan[c + d*x]^(3/2))/(b*(a^2 
 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) + (-((-(((a - I*b)*b^2*ArcTan[(Sqrt[I* 
a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a - b]) + (3* 
a*(a^2 + b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x] 
]])/Sqrt[b] - ((a + I*b)*b^2*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sq 
rt[a + b*Tan[c + d*x]]])/Sqrt[I*a + b])/(b*d)) + ((3*a^2 + b^2)*Sqrt[Tan[c 
 + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(b*(a^2 + b^2)))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] - Simp[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] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4130

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

rule 4138

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

rule 4729

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2692\) vs. \(2(260)=520\).

Time = 5.42 (sec) , antiderivative size = 2693, normalized size of antiderivative = 8.69

method	result	size
default	\(\text {Expression too large to display}\)	\(2693\)

Input:

int(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*2^(1/2)/(a^2+b^2)^(3/2)/(-b+(a^2+b^2)^(1/2))^(1/2)/a/b^(5/2)*(b^(7/2 
)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(-b+(a^2 
+b^2)^(1/2))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*(a^2+b^2)^(1/2)*ln(-(a*cot(d* 
x+c)*cos(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x 
+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*a*cot(d*x+c)+2*(a^2 
+b^2)^(1/2)*cos(d*x+c)+2*cos(d*x+c)*b-sin(d*x+c)*a+a*csc(d*x+c)-2*(a^2+b^2 
)^(1/2)-2*b)/(-1+cos(d*x+c)))*(-cos(d*x+c)^2-cos(d*x+c))+b^(9/2)*((a*cos(d 
*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(-b+(a^2+b^2)^(1/2) 
)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*ln(-(a*cot(d*x+c)*cos(d*x+c)+2*2^(1/2)*( 
(a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2 
)^(1/2))^(1/2)*sin(d*x+c)-2*a*cot(d*x+c)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*co 
s(d*x+c)*b-sin(d*x+c)*a+a*csc(d*x+c)-2*(a^2+b^2)^(1/2)-2*b)/(-1+cos(d*x+c) 
))*(cos(d*x+c)^2+cos(d*x+c))+b^(5/2)*a^2*((a*cos(d*x+c)+b*sin(d*x+c))*sin( 
d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*(b+(a^2+b^2)^(1/ 
2))^(1/2)*ln(-(a*cot(d*x+c)*cos(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+ 
c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c 
)-2*a*cot(d*x+c)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*cos(d*x+c)*b-sin(d*x+c)*a+ 
a*csc(d*x+c)-2*(a^2+b^2)^(1/2)-2*b)/(-1+cos(d*x+c)))*(-cos(d*x+c)^2-cos(d* 
x+c))+b^(7/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1 
/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*(a^2+b^2)^(1/2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7939 vs. \(2 (254) = 508\).

Time = 2.20 (sec) , antiderivative size = 15911, normalized size of antiderivative = 51.33 \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/cot(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )^{4} \tan \left (d x +c \right )^{2} b^{2}+2 \cot \left (d x +c \right )^{4} \tan \left (d x +c \right ) a b +\cot \left (d x +c \right )^{4} a^{2}}d x \] Input:

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

Output:

int((sqrt(tan(c + d*x)*b + a)*sqrt(cot(c + d*x)))/(cot(c + d*x)**4*tan(c + 
 d*x)**2*b**2 + 2*cot(c + d*x)**4*tan(c + d*x)*a*b + cot(c + d*x)**4*a**2) 
,x)