3.3.8 \(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\) [208]

3.3.8.1 Optimal result
3.3.8.2 Mathematica [B] (verified)
3.3.8.3 Rubi [A] (verified)
3.3.8.4 Maple [B] (verified)
3.3.8.5 Fricas [B] (verification not implemented)
3.3.8.6 Sympy [F(-1)]
3.3.8.7 Maxima [B] (verification not implemented)
3.3.8.8 Giac [F(-2)]
3.3.8.9 Mupad [F(-1)]

3.3.8.1 Optimal result

Integrand size = 28, antiderivative size = 202 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(4-4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}} \]

output
(4-4*I)*a^(5/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^ 
(1/2))/d+104/21*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/7*a^2* 
(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(7/2)-6/7*I*a^2*(a+I*a*tan(d*x+c))^( 
1/2)/d/tan(d*x+c)^(5/2)+32/21*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3 
/2)
 
3.3.8.2 Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(479\) vs. \(2(202)=404\).

Time = 6.93 (sec) , antiderivative size = 479, normalized size of antiderivative = 2.37 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {4 i \sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {4 i a^{5/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {5 (-1)^{3/4} a^2 \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)}}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}}+\frac {i a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}} \]

input
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(9/2),x]
 
output
((-4*I)*Sqrt[2]*a^2*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a* 
Tan[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) + ((4*I)*a^( 
5/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt 
[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (5 
*(-1)^(3/4)*a^2*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c 
+ d*x]])/(d*Sqrt[1 + I*Tan[c + d*x]]) - (2*a^2*Sqrt[a + I*a*Tan[c + d*x]]) 
/(7*d*Tan[c + d*x]^(7/2)) - (((6*I)/7)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d* 
Tan[c + d*x]^(5/2)) + (32*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(21*d*Tan[c + d* 
x]^(3/2)) + (((104*I)/21)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + 
d*x]]) + (I*a^(3/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[I*a*Tan[c 
 + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[ 
c + d*x]])
 
3.3.8.3 Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {3042, 4036, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4027, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4036

\(\displaystyle -\frac {2}{7} \int -\frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{2 \tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{\tan (c+d x)^{7/2}}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{7} \left (\frac {2 \int -\frac {10 \sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (-\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (-\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan (c+d x)^{5/2}}dx}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\frac {2 \int -\frac {21 a^5 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {-21 a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {-21 a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\frac {42 i a^6 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{7} \left (-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 \left (\frac {-\frac {(21-21 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\)

input
Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(9/2),x]
 
output
(-2*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) + (((-6*I)*a^ 
2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(5/2)) - (4*((-8*a^3*Sqrt[a 
+ I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + (((-21 + 21*I)*a^(9/2)*Arc 
Tanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - 
 ((26*I)*a^4*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a)))/a 
)/7
 

3.3.8.3.1 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 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 4036
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] 
)^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si 
mp[a/(d*(b*c + a*d)*(n + 1))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ 
e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) 
 + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[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] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4081
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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && LtQ[n, -1]
 
3.3.8.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (163 ) = 326\).

Time = 0.94 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.26

method result size
derivativedivides \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (21 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )-21 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+84 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+32 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+104 i \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-18 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-6 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{21 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) \(457\)
default \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (21 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )-21 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+84 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+32 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+104 i \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-18 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-6 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{21 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) \(457\)

input
int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(9/2),x,method=_RETURNVERBOSE)
 
output
1/21/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(7/2)*(21*I*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))*(I*a)^(1/2)*a*tan(d*x+c)^4-21*(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)^4+84*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*ta 
n(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)^4+32*(-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+104*I*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^ 
(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)-18*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+ 
c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-6*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+ 
c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1 
/2)/(-I*a)^(1/2)
 
3.3.8.5 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. 509 vs. \(2 (152) = 304\).

Time = 0.25 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.52 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {8 \, \sqrt {2} {\left (40 \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} - 37 \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 7 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 49 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 21 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 21 \, \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 21 \, \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{42 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

input
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(9/2),x, algorithm="fricas")
 
output
-1/42*(8*sqrt(2)*(40*a^2*e^(9*I*d*x + 9*I*c) - 37*a^2*e^(7*I*d*x + 7*I*c) 
- 7*a^2*e^(5*I*d*x + 5*I*c) + 49*a^2*e^(3*I*d*x + 3*I*c) - 21*a^2*e^(I*d*x 
 + I*c))*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)) - 21*sqrt(-32*I*a^5/d^2)*(d*e^(8*I*d*x + 8*I 
*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 
 2*I*c) + d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^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)) + sqrt(-32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) 
+ 21*sqrt(-32*I*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) 
+ 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt(2 
)*(a^2*e^(2*I*d*x + 2*I*c) + a^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)) - sqrt(-32*I*a^5/d^ 
2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2))/(d*e^(8*I*d*x + 8*I*c) - 4*d* 
e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + 
d)
 
3.3.8.6 Sympy [F(-1)]

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

input
integrate((a+I*a*tan(d*x+c))**(5/2)/tan(d*x+c)**(9/2),x)
 
output
Timed out
 
3.3.8.7 Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3186 vs. \(2 (152) = 304\).

Time = 0.84 (sec) , antiderivative size = 3186, normalized size of antiderivative = 15.77 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]

input
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(9/2),x, algorithm="maxima")
 
output
-2/105*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c 
) + 1)*(3*((35*I + 35)*a^2*cos(7*d*x + 7*c) - (35*I + 35)*a^2*cos(5*d*x + 
5*c) + (21*I + 21)*a^2*cos(3*d*x + 3*c) - (I + 1)*a^2*cos(d*x + c) + (35*I 
 - 35)*a^2*sin(7*d*x + 7*c) - (35*I - 35)*a^2*sin(5*d*x + 5*c) + (21*I - 2 
1)*a^2*sin(3*d*x + 3*c) - (I - 1)*a^2*sin(d*x + c))*cos(7/2*arctan2(sin(2* 
d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + 5*(13*(-(I + 1)*a^2*cos(d*x + c) - ( 
I - 1)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (13*I + 13)*a^2*cos(d*x + c) 
 + 13*(-(I + 1)*a^2*cos(d*x + c) - (I - 1)*a^2*sin(d*x + c))*sin(2*d*x + 2 
*c)^2 - (13*I - 13)*a^2*sin(d*x + c) + 21*((I + 1)*a^2*cos(2*d*x + 2*c)^2 
+ (I + 1)*a^2*sin(2*d*x + 2*c)^2 - (2*I + 2)*a^2*cos(2*d*x + 2*c) + (I + 1 
)*a^2)*cos(3*d*x + 3*c) + 26*((I + 1)*a^2*cos(d*x + c) + (I - 1)*a^2*sin(d 
*x + c))*cos(2*d*x + 2*c) + 21*((I - 1)*a^2*cos(2*d*x + 2*c)^2 + (I - 1)*a 
^2*sin(2*d*x + 2*c)^2 - (2*I - 2)*a^2*cos(2*d*x + 2*c) + (I - 1)*a^2)*sin( 
3*d*x + 3*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + 
3*((35*I - 35)*a^2*cos(7*d*x + 7*c) - (35*I - 35)*a^2*cos(5*d*x + 5*c) + ( 
21*I - 21)*a^2*cos(3*d*x + 3*c) - (I - 1)*a^2*cos(d*x + c) - (35*I + 35)*a 
^2*sin(7*d*x + 7*c) + (35*I + 35)*a^2*sin(5*d*x + 5*c) - (21*I + 21)*a^2*s 
in(3*d*x + 3*c) + (I + 1)*a^2*sin(d*x + c))*sin(7/2*arctan2(sin(2*d*x + 2* 
c), -cos(2*d*x + 2*c) + 1)) + 5*(13*(-(I - 1)*a^2*cos(d*x + c) + (I + 1)*a 
^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (13*I - 13)*a^2*cos(d*x + c) + 13...
 
3.3.8.8 Giac [F(-2)]

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

input
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(9/2),x, algorithm="giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0] was discarded 
 and replaced randomly by 0=[48]Warning, replacing 48 by 64, a substitutio 
n variabl
 
3.3.8.9 Mupad [F(-1)]

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

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