\(\int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [610]

Optimal result

Integrand size = 36, antiderivative size = 117 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {B \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {B \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cot (c+d x)}}{1+\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {2 B \cot ^{\frac {3}{2}}(c+d x)}{3 d} \] 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*cot(d*x+c)^(3/2)/d
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \left (-3 \arctan \left (\sqrt [4]{-\cot ^2(c+d x)}\right ) \sqrt [4]{-\cot (c+d x)}+3 \text {arctanh}\left (\sqrt [4]{-\cot ^2(c+d x)}\right ) \sqrt [4]{-\cot (c+d x)}+2 \cot ^{\frac {7}{4}}(c+d x)\right )}{3 d \sqrt [4]{\cot (c+d x)}} \] Input:

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

Output:

-1/3*(B*(-3*ArcTan[(-Cot[c + d*x]^2)^(1/4)]*(-Cot[c + d*x])^(1/4) + 3*ArcT 
anh[(-Cot[c + d*x]^2)^(1/4)]*(-Cot[c + d*x])^(1/4) + 2*Cot[c + d*x]^(7/4)) 
)/(d*Cot[c + d*x]^(1/4))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2011, 3042, 3954, 3042, 3957, 266, 826, 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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

B*((-2*Cot[c + d*x]^(3/2))/(3*d) + (2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + 
d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2])/2 + (Log 
[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(2*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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[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 = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.87

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

Input:

int(cot(d*x+c)^(5/2)*(B*a+b*B*tan(d*x+c))/(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. 398 vs. \(2 (95) = 190\).

Time = 0.09 (sec) , antiderivative size = 398, normalized size of antiderivative = 3.40 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {6 \, \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) + 6 \, \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - \cos \left (2 \, d x + 2 \, c\right ) - 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) + 3 \, \sqrt {2} B \log \left (\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) - 3 \, \sqrt {2} B \log \left (-\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - \cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) - 1}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \sin \left (2 \, d x + 2 \, c\right ) + 8 \, {\left (B \cos \left (2 \, d x + 2 \, c\right ) + B\right )} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}}}{12 \, d \sin \left (2 \, d x + 2 \, c\right )} \] Input:

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

Output:

-1/12*(6*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))*si 
n(2*d*x + 2*c) + 6*sqrt(2)*B*arctan((sqrt(2)*sqrt((cos(2*d*x + 2*c) + 1)/s 
in(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))*sin(2*d*x + 2*c) + 3*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))*sin(2*d*x + 2*c) - 3*sqrt(2)*B*log(-(sqr 
t(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))*sin(2*d*x + 2 
*c) + 8*(B*cos(2*d*x + 2*c) + B)*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2 
*c)))/(d*sin(2*d*x + 2*c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {3 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} B - \frac {8 \, B}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 6.41 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.55 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )}{d}-\frac {2\,B\,{\left (\frac {1}{\mathrm {tan}\left (c+d\,x\right )}\right )}^{3/2}}{3\,d}-\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )}{d} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cot(c + d*x))*cot(c + d*x)**2,x)*b