\(\int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [855]

Optimal result

Integrand size = 25, antiderivative size = 259 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=-\frac {(i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \] Output:

-(I*a-b)^(5/2)*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)/d+(I*a+b)^(5/2)*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 
)/d+2/15*(15*a^2-23*b^2)*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d-22/15*a 
*b*cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2)/d-2/5*a^2*cot(d*x+c)^(5/2)*(a+b 
*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.83 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {\cot ^{\frac {5}{2}}(c+d x) \left (15 \sqrt [4]{-1} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {5}{2}}(c+d x)+15 \sqrt [4]{-1} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {5}{2}}(c+d x)+2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2-11 a b \tan (c+d x)+\left (15 a^2-23 b^2\right ) \tan ^2(c+d x)\right )\right )}{15 d} \] Input:

Integrate[Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

(Cot[c + d*x]^(5/2)*(15*(-1)^(1/4)*(-a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqr 
t[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(5/ 
2) + 15*(-1)^(1/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[T 
an[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(5/2) + 2*Sqrt[a + b* 
Tan[c + d*x]]*(-3*a^2 - 11*a*b*Tan[c + d*x] + (15*a^2 - 23*b^2)*Tan[c + d* 
x]^2)))/(15*d)
 

Rubi [A] (verified)

Time = 1.79 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 4729, 3042, 4048, 27, 3042, 4132, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

\(\displaystyle \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cot (c+d x)^{7/2} (a+b \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan (c+d x)^{7/2}}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2}{5} \int \frac {11 b a^2-5 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (4 a^2-5 b^2\right ) \tan ^2(c+d x)}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \int \frac {11 b a^2-5 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (4 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \int \frac {11 b a^2-5 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (4 a^2-5 b^2\right ) \tan (c+d x)^2}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4132

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \int \frac {22 b^2 \tan ^2(c+d x) a^2+\left (15 a^2-23 b^2\right ) a^2+15 b \left (3 a^2-b^2\right ) \tan (c+d x) a}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {\int \frac {22 b^2 \tan ^2(c+d x) a^2+\left (15 a^2-23 b^2\right ) a^2+15 b \left (3 a^2-b^2\right ) \tan (c+d x) a}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {\int \frac {22 b^2 \tan (c+d x)^2 a^2+\left (15 a^2-23 b^2\right ) a^2+15 b \left (3 a^2-b^2\right ) \tan (c+d x) a}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4132

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {-\frac {2 \int -\frac {15 \left (a^2 b \left (3 a^2-b^2\right )-a^3 \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {\frac {15 \int \frac {a^2 b \left (3 a^2-b^2\right )-a^3 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {\frac {15 \int \frac {a^2 b \left (3 a^2-b^2\right )-a^3 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4099

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^2 (-b+i a)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^2 (b+i a)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^2 (-b+i a)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^2 (b+i a)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\right )\)

\(\Big \downarrow \) 4098

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 (-b+i a)^3 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a^2 (b+i a)^3 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}}{3 a}\right )\right )\)

\(\Big \downarrow \) 104

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 (-b+i a)^3 \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^2 (b+i a)^3 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}\right )\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {a^2 (b+i a)^3 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {22 a b \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 a \left (15 a^2-23 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {a^2 (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )}{a}}{3 a}\right )\right )\)

Input:

Int[Cot[c + d*x]^(7/2)*(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*a^2*Sqrt[a + b*Tan[c + d*x]])/( 
5*d*Tan[c + d*x]^(5/2)) + ((-22*a*b*Sqrt[a + b*Tan[c + d*x]])/(3*d*Tan[c + 
 d*x]^(3/2)) - ((15*((a^2*(I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c 
 + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (a^2*(I*a + b)^(5/2)*ArcTanh[(Sqr 
t[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d))/a - (2*a*(15 
*a^2 - 23*b^2)*Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/5)
 

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 216

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

rule 219

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

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 4098

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[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 4099

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(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 4132

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[(A*b^2 - a*(b*B - a*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] + Simp[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)*Ta 
n[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] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 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. \(2626\) vs. \(2(213)=426\).

Time = 3.89 (sec) , antiderivative size = 2627, normalized size of antiderivative = 10.14

method	result	size
default	\(\text {Expression too large to display}\)	\(2627\)

Input:

int(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/60/d*2^(1/2)/(-b+(a^2+b^2)^(1/2))^(1/2)*((30*cos(d*x+c)-30)*sin(d*x+c)^2 
*(a^2+b^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b*ln 
(-(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)-a*cot(d*x+c)*cos(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)))+(15*cos(d*x+c)-15)*sin(d*x+c)^2*( 
b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*a^2*ln(-(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)-a*cot(d*x+c)*cos(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)))+(-45*cos(d*x+c)+45)*sin(d*x+c)^2*(b+(a^2+b^2)^(1/ 
2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b^2*ln(-(2*2^(1/2)*((a*cos(d*x+c)+b*s 
in(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*si 
n(d*x+c)-a*cot(d*x+c)*cos(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)))+(-30*cos(d*x+c)+30)*sin(d*x+c)^2*(a^2+b^2)^(1/2)*(b+(a^2+b^2)^(1 
/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*b*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+c...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5165 vs. \(2 (209) = 418\).

Time = 0.68 (sec) , antiderivative size = 5165, normalized size of antiderivative = 19.94 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**(7/2)*(a+b*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(7/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,14,5]%%%}+%%%{6,[0,12,5]%%%}+%%%{15,[0,10,5]%%%}+ 
%%%{20,[0
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int(cot(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(5/2),x)
 

Output:

int(cot(c + d*x)^(7/2)*(a + b*tan(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\int \cot \left (d x +c \right )^{\frac {7}{2}} \left (a +\tan \left (d x +c \right ) b \right )^{\frac {5}{2}}d x \] Input:

int(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(5/2),x)
 

Output:

int(cot(d*x+c)^(7/2)*(a+b*tan(d*x+c))^(5/2),x)