\(\int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [615]

Optimal result

Integrand size = 36, antiderivative size = 116 \[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {B \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {B \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {B \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cot (c+d x)}}{1+\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 B}{3 d \cot ^{\frac {3}{2}}(c+d x)} \] Output:

1/2*B*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/d+1/2*B*arctan(1+2^(1/2) 
*cot(d*x+c)^(1/2))*2^(1/2)/d+1/2*B*arctanh(2^(1/2)*cot(d*x+c)^(1/2)/(1+cot 
(d*x+c)))*2^(1/2)/d+2/3*B/d/cot(d*x+c)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.71 \[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {B \left (-2+3 \arctan \left (\sqrt [4]{-\cot ^2(c+d x)}\right ) \left (-\cot ^2(c+d x)\right )^{3/4}+3 \text {arctanh}\left (\sqrt [4]{-\cot ^2(c+d x)}\right ) \left (-\cot ^2(c+d x)\right )^{3/4}\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

-1/3*(B*(-2 + 3*ArcTan[(-Cot[c + d*x]^2)^(1/4)]*(-Cot[c + d*x]^2)^(3/4) + 
3*ArcTanh[(-Cot[c + d*x]^2)^(1/4)]*(-Cot[c + d*x]^2)^(3/4)))/(d*Cot[c + d* 
x]^(3/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.35, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2011, 3042, 3955, 3042, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 2011

\(\displaystyle B \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle B \int \frac {1}{\left (-\tan \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 3955

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle B \left (\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}+\frac {2}{3 d \cot ^{\frac {3}{2}}(c+d x)}\right )\)

Input:

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

Output:

B*(2/(3*d*Cot[c + d*x]^(3/2)) + (2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x 
]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2])/2 + (-1/2*L 
og[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2 
]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(2*Sqrt[2]))/2))/d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

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

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3955

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x] 
)^(n + 1)/(b*d*(n + 1)), x] - Simp[1/b^2   Int[(b*Tan[c + d*x])^(n + 2), x] 
, x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {B \left (\frac {2}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}+1}{\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(102\)
default	\(\frac {B \left (\frac {2}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}+1}{\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(102\)

Input:

int((B*a+b*B*tan(d*x+c))/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (95) = 190\).

Time = 0.11 (sec) , antiderivative size = 429, normalized size of antiderivative = 3.70 \[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {6 \, {\left (\sqrt {2} B \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} B\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) + 6 \, {\left (\sqrt {2} B \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} B\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - \cos \left (2 \, d x + 2 \, c\right ) - 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) - 3 \, {\left (\sqrt {2} B \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} B\right )} \log \left (\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) + 3 \, {\left (\sqrt {2} B \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} B\right )} \log \left (-\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - \cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) - 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) + 8 \, {\left (B \cos \left (2 \, d x + 2 \, c\right ) - B\right )} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}}}{12 \, {\left (d \cos \left (2 \, d x + 2 \, c\right ) + d\right )}} \] Input:

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

Output:

-1/12*(6*(sqrt(2)*B*cos(2*d*x + 2*c) + sqrt(2)*B)*arctan((sqrt(2)*sqrt((co 
s(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + cos(2*d*x + 2*c) 
+ 1)/(cos(2*d*x + 2*c) + 1)) + 6*(sqrt(2)*B*cos(2*d*x + 2*c) + sqrt(2)*B)* 
arctan((sqrt(2)*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 
2*c) - cos(2*d*x + 2*c) - 1)/(cos(2*d*x + 2*c) + 1)) - 3*(sqrt(2)*B*cos(2* 
d*x + 2*c) + sqrt(2)*B)*log((sqrt(2)*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x 
 + 2*c))*sin(2*d*x + 2*c) + cos(2*d*x + 2*c) + sin(2*d*x + 2*c) + 1)/(cos( 
2*d*x + 2*c) + 1)) + 3*(sqrt(2)*B*cos(2*d*x + 2*c) + sqrt(2)*B)*log(-(sqrt 
(2)*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) - cos(2 
*d*x + 2*c) - sin(2*d*x + 2*c) - 1)/(cos(2*d*x + 2*c) + 1)) + 8*(B*cos(2*d 
*x + 2*c) - B)*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c)))/(d*cos(2*d*x 
 + 2*c) + d)
 

Sympy [F]

\[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=B \int \frac {1}{\cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:

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

Output:

B*Integral(cot(c + d*x)**(-5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.10 \[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {6 \, \sqrt {2} B \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 6 \, \sqrt {2} B \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 3 \, \sqrt {2} B \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 3 \, \sqrt {2} B \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + 8 \, B \tan \left (d x + c\right )^{\frac {3}{2}}}{12 \, d} \] Input:

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

Output:

1/12*(6*sqrt(2)*B*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 6 
*sqrt(2)*B*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + 3*sqrt( 
2)*B*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 3*sqrt(2)*B*lo 
g(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + 8*B*tan(d*x + c)^(3/ 
2))/d
 

Giac [F(-1)]

Timed out. \[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 6.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.56 \[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {2\,B}{3\,d\,{\left (\frac {1}{\mathrm {tan}\left (c+d\,x\right )}\right )}^{3/2}}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )\,1{}\mathrm {i}}{d}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )\,1{}\mathrm {i}}{d} \] Input:

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

Output:

(2*B)/(3*d*(1/tan(c + d*x))^(3/2)) - ((-1)^(1/4)*B*atan((-1)^(1/4)*(1/tan( 
c + d*x))^(1/2))*1i)/d - ((-1)^(1/4)*B*atanh((-1)^(1/4)*(1/tan(c + d*x))^( 
1/2))*1i)/d
 

Reduce [F]

\[ \int \frac {a B+b B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\left (\int \frac {\sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )^{3}}d x \right ) b \] Input:

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

Output:

int(sqrt(cot(c + d*x))/cot(c + d*x)**3,x)*b