\(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [226]

Optimal result

Integrand size = 28, antiderivative size = 218 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 \sqrt [4]{-1} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \tan ^{\frac {3}{2}}(c+d x)}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

2*(-1)^(1/4)*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)^(5/2)/d/(a+I*a*tan(d*x+c))^( 
5/2)+1/2*I*tan(d*x+c)^(3/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)+7/4*tan(d*x+c)^(1 
/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.74 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^2(c+d x) \left (80 \sqrt [4]{-1} \sqrt {a} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) (-i \cos (2 (c+d x))+\sin (2 (c+d x))) \sqrt {1+i \tan (c+d x)}-2 \sqrt {a} (-7+42 \cos (2 (c+d x))+40 i \sin (2 (c+d x))) \sqrt {\tan (c+d x)}-(5-5 i) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{40 a^{5/2} d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(Sec[c + d*x]^2*(80*(-1)^(1/4)*Sqrt[a]*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x 
]]]*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*Sqrt[1 + I*Tan[c + d*x]] - 
2*Sqrt[a]*(-7 + 42*Cos[2*(c + d*x)] + (40*I)*Sin[2*(c + d*x)])*Sqrt[Tan[c 
+ d*x]] - (5 - 5*I)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
I*a*Tan[c + d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[a + I*a*Ta 
n[c + d*x]]))/(40*a^(5/2)*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x 
]])
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.06, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)}}dx}{a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a^3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a^3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {2 i a^5 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {8 a^4 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {16 a^4 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-\frac {16 \sqrt [4]{-1} a^{7/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

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

Output:

-1/5*Tan[c + d*x]^(5/2)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I*a*Tan[c + d 
*x]^(3/2))/(d*(a + I*a*Tan[c + d*x])^(3/2)) - (((-16*(-1)^(1/4)*a^(7/2)*Ar 
cTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/ 
d - ((1 - I)*a^(7/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
 I*a*Tan[c + d*x]]])/d)/(2*a^2) - (7*a^2*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I 
*a*Tan[c + d*x]]))/(2*a^2))/(2*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 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. 957 vs. \(2 (170 ) = 340\).

Time = 1.64 (sec) , antiderivative size = 958, normalized size of antiderivative = 4.39

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(958\)
default	\(\text {Expression too large to display}\)	\(958\)

Input:

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

Output:

1/80/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(320*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)^3+30*I*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) 
*(I*a)^(1/2)*a*tan(d*x+c)^2-80*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 
)^4-320*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)-20*(I*a)^(1/2)*2^(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))*a*tan(d*x+c)^3-516*(-I*a)^(1/2)*(I*a)^(1/2) 
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+460*I*(-I*a)^(1/2)*(I* 
a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-5*I*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)*(I*a)^(1/2)*a*tan(d*x+c)^4+480*(-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)^2-5*I*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)*(I*a)^(1 
/2)*a+20*2^(1/2)*(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))*a*tan(d*x+c)-196*I 
*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x...
 

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. 621 vs. \(2 (160) = 320\).

Time = 0.15 (sec) , antiderivative size = 621, normalized size of antiderivative = 2.85 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/40*(10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(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)) - 10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c) 
*log(-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)) - 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e^( 
5*I*d*x + 5*I*c)*log(52/605*(4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s 
qrt((-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*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(-4* 
I/(a^5*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e 
^(5*I*d*x + 5*I*c)*log(52/605*(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^(3*I*d*x 
+ 3*I*c) + e^(I*d*x + I*c)) - (3*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(- 
4*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))* 
(41*e^(6*I*d*x + 6*I*c) + 34*e^(4*I*d*x + 4*I*c) - 6*e^(2*I*d*x + 2*I*c) + 
 1))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

integrate(tan(d*x+c)^(7/2)/(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. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int(tan(c + d*x)^(7/2)/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

int(tan(c + d*x)^(7/2)/(a + a*tan(c + d*x)*1i)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

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

Output:

( - int((sqrt(tan(c + d*x))*tan(c + d*x)**3)/(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)