\(\int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [225]

Optimal result

Integrand size = 28, antiderivative size = 257 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {5 (-1)^{3/4} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \] Output:

5*(-1)^(3/4)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)) 
^(1/2))/a^(5/2)/d-(1/8+1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I* 
a*tan(d*x+c))^(1/2))/a^(5/2)/d-1/5*tan(d*x+c)^(7/2)/d/(a+I*a*tan(d*x+c))^( 
5/2)+19/30*I*tan(d*x+c)^(5/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)+41/12*tan(d*x+c 
)^(3/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+21/4*I*tan(d*x+c)^(1/2)*(a+I*a*tan( 
d*x+c))^(1/2)/a^3/d
 

Mathematica [A] (verified)

Time = 3.96 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.05 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-\sqrt {-1+e^{2 i (c+d x)}} \left (3-28 e^{2 i (c+d x)}+252 e^{4 i (c+d x)}+403 e^{6 i (c+d x)}\right )+15 e^{5 i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \text {arctanh}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+300 \sqrt {2} e^{5 i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \text {arctanh}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\tan (c+d x)}}{60 \sqrt {2} a^3 d \sqrt {-1+e^{2 i (c+d x)}}} \] Input:

Integrate[Tan[c + d*x]^(9/2)/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

((-1/60*I)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-(Sqrt 
[-1 + E^((2*I)*(c + d*x))]*(3 - 28*E^((2*I)*(c + d*x)) + 252*E^((4*I)*(c + 
 d*x)) + 403*E^((6*I)*(c + d*x)))) + 15*E^((5*I)*(c + d*x))*(1 + E^((2*I)* 
(c + d*x)))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] + 300* 
Sqrt[2]*E^((5*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))*ArcTanh[(Sqrt[2]*E^( 
I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]])/(Sqrt[2 
]*a^3*d*E^((6*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))])
 

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4080, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}

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.

\(\displaystyle \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{9/2}}{(a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\tan ^{\frac {5}{2}}(c+d x) (7 a-12 i a \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) (7 a-12 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^{5/2} (7 a-12 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {5 \tan ^{\frac {3}{2}}(c+d x) \left (22 \tan (c+d x) a^2+19 i a^2\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (22 \tan (c+d x) a^2+19 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \int \frac {\tan (c+d x)^{3/2} \left (22 \tan (c+d x) a^2+19 i a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (-\frac {\int -\frac {3}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (41 a^3-42 i a^3 \tan (c+d x)\right )dx}{a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (41 a^3-42 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (41 a^3-42 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (20 \tan (c+d x) a^4+21 i a^4\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (20 \tan (c+d x) a^4+21 i a^4\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {i a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {2 a^6 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\frac {20 i a^5 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\frac {40 i a^5 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {19 i a \tan ^{\frac {5}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {5 \left (\frac {3 \left (\frac {\frac {(1+i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {40 (-1)^{3/4} a^{9/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {42 i a^3 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}-\frac {41 a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}}{10 a^2}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

Int[Tan[c + d*x]^(9/2)/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

-1/5*Tan[c + d*x]^(7/2)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((((19*I)/3)*a* 
Tan[c + d*x]^(5/2))/(d*(a + I*a*Tan[c + d*x])^(3/2)) - (5*((-41*a^2*Tan[c 
+ d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (3*(((-40*(-1)^(3/4)*a^(9/2 
)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x] 
]])/d + ((1 + I)*a^(9/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt 
[a + I*a*Tan[c + d*x]]])/d)/a - ((42*I)*a^3*Sqrt[Tan[c + d*x]]*Sqrt[a + I* 
a*Tan[c + d*x]])/d))/(2*a^2)))/(6*a^2))/(10*a^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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 4027

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4041

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* 
((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m)   Int[(a + 
b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n 
 - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta 
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] 
 && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In 
tegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4078

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), 
 x] + Simp[1/(2*a^2*m)   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* 
x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a 
*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] 
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
 

rule 4080

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 
1/(a*(m + n))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim 
p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T 
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
 

rule 4082

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, 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] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b + a*B)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1006 vs. \(2 (201 ) = 402\).

Time = 1.63 (sec) , antiderivative size = 1007, normalized size of antiderivative = 3.92

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1007\)
default	\(\text {Expression too large to display}\)	\(1007\)

Input:

int(tan(d*x+c)^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(2228*(a*tan(d*x+c 
)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3-60*I*(I*a) 
^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+ 
I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)+1260*I*(I*a)^(1/2 
)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)+3600*I*ln(1/2*(2*I*a* 
tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1 
/2))*(-I*a)^(1/2)*a*tan(d*x+c)^2-15*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/ 
2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I 
))*2^(1/2)*a*tan(d*x+c)^4-600*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*( 
1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+ 
c)^4-4948*I*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2) 
*tan(d*x+c)^2-600*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+ 
c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a+90*ln(-(-2*2^(1/2)*( 
-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan 
(d*x+c)+I))*(I*a)^(1/2)*2^(1/2)*a*tan(d*x+c)^2-2400*(-I*a)^(1/2)*ln(1/2*(2 
*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I* 
a)^(1/2))*a*tan(d*x+c)^3+60*I*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a* 
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^( 
1/2)*a*tan(d*x+c)^3+240*I*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2 
)*(-I*a)^(1/2)*tan(d*x+c)^4-15*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 623 vs. \(2 (189) = 378\).

Time = 0.16 (sec) , antiderivative size = 623, normalized size of antiderivative = 2.42 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/120*(30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(I*a^3*d*sqrt 
(1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c 
) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2 
*I*d*x + 2*I*c) + 1)) - 30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c) 
*log(-I*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/( 
e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 
2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)) - 30*a^3*d*sqrt(25*I/(a^5*d^2))*e^ 
(5*I*d*x + 5*I*c)*log(104/3025*(10*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1 
))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(3*I*d* 
x + 3*I*c) + e^(I*d*x + I*c)) - (3*I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)* 
sqrt(25*I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + 30*a^3*d*sqrt(25*I/(a^5 
*d^2))*e^(5*I*d*x + 5*I*c)*log(104/3025*(10*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2 
*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*( 
e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c)) - (-3*I*a^3*d*e^(2*I*d*x + 2*I*c) + 
 I*a^3*d)*sqrt(25*I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt(2)*sqrt( 
a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x 
 + 2*I*c) + 1))*(403*I*e^(6*I*d*x + 6*I*c) + 252*I*e^(4*I*d*x + 4*I*c) - 2 
8*I*e^(2*I*d*x + 2*I*c) + 3*I))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(9/2)/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+I*a*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:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{9/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int(tan(c + d*x)^(9/2)/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

int(tan(c + d*x)^(9/2)/(a + a*tan(c + d*x)*1i)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

int(tan(d*x+c)^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int((sqrt(tan(c + d*x))*tan(c + d*x)**4)/(sqrt(tan(c + d*x)*i + 1)*tan 
(c + d*x)**2 - 2*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d* 
x)*i + 1)),x))/(sqrt(a)*a**2)