\(\int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [192]

Optimal result

Integrand size = 28, antiderivative size = 154 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {26 \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}} \] Output:

(-1-I)*a^(1/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^( 
1/2))/d-2/5*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-2/15*I*(a+I*a*tan( 
d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)+26/15*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+ 
c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {-15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \tan ^2(c+d x) \sqrt {i a \tan (c+d x)}+2 \sqrt {a+i a \tan (c+d x)} \left (-3-i \tan (c+d x)+13 \tan ^2(c+d x)\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \] Input:

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

Output:

(-15*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + 
 d*x]]]*Tan[c + d*x]^2*Sqrt[I*a*Tan[c + d*x]] + 2*Sqrt[a + I*a*Tan[c + d*x 
]]*(-3 - I*Tan[c + d*x] + 13*Tan[c + d*x]^2))/(15*d*Tan[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4044, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4044

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

\(\displaystyle \frac {-\frac {\frac {2 \int \frac {15 i a^3 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {26 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {15 i a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {15 i a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {-\frac {\frac {30 a^4 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {26 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {-\frac {\frac {(15+15 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {26 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\)

Input:

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

Output:

(-2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2)) + ((((-2*I)/3)*a* 
Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) - (((15 + 15*I)*a^(5/2) 
*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]) 
/d - (26*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/(5 
*a)
 

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 4044

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 358 vs. \(2 (123 ) = 246\).

Time = 1.75 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.33

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-15 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+52 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3}-16 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-56 i \tan \left (d x +c \right )^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}+12 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{30 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )}\)	\(359\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-15 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+52 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3}-16 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-56 i \tan \left (d x +c \right )^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}+12 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{30 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )}\)	\(359\)

Input:

int((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-1/30/d*(a*(1+I*tan(d*x+c)))^(1/2)/tan(d*x+c)^(5/2)*(15*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))*a*tan(d*x+c)^3-15*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+52*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan 
(d*x+c)^3-16*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2) 
-56*I*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)+12*I 
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*ta 
n(d*x+c)))^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)
 

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. 424 vs. \(2 (114) = 228\).

Time = 0.09 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.75 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-17 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 3 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 5 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, e^{\left (i \, d x + i \, c\right )}\right )} - 15 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{30 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:

integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

-1/30*(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))*(-17*I*e^(7*I*d*x + 7*I*c) + 3*I*e^( 
5*I*d*x + 5*I*c) + 5*I*e^(3*I*d*x + 3*I*c) - 15*I*e^(I*d*x + I*c)) - 15*(d 
*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - 
 d)*sqrt(2*I*a/d^2)*log((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) + I*d*sqrt(2*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 15*(d*e^( 
6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)* 
sqrt(2*I*a/d^2)*log((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 
) - I*d*sqrt(2*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(d*e^(6*I*d*x 
+ 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [F]

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

integrate((a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(7/2),x)
 

Output:

Integral(sqrt(I*a*(tan(c + d*x) - I))/tan(c + d*x)**(7/2), x)
 

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. 1151 vs. \(2 (114) = 228\).

Time = 0.28 (sec) , antiderivative size = 1151, normalized size of antiderivative = 7.47 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

1/30*(3*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) 
+ 1)*(((10*I - 10)*cos(3*d*x + 3*c) - (13*I - 13)*cos(d*x + c) - (10*I + 1 
0)*sin(3*d*x + 3*c) + (13*I + 13)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x 
+ 2*c), -cos(2*d*x + 2*c) + 1)) + (-(10*I + 10)*cos(3*d*x + 3*c) + (13*I + 
 13)*cos(d*x + c) - (10*I - 10)*sin(3*d*x + 3*c) + (13*I - 13)*sin(d*x + c 
))*sin(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + 15 
*(2*(-(I - 1)*cos(2*d*x + 2*c)^2 - (I - 1)*sin(2*d*x + 2*c)^2 + (2*I - 2)* 
cos(2*d*x + 2*c) - I + 1)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 
 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2* 
d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 
- 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d 
*x + 2*c) + 1)) - sin(d*x + c)) + ((I + 1)*cos(2*d*x + 2*c)^2 + (I + 1)*si 
n(2*d*x + 2*c)^2 - (2*I + 2)*cos(2*d*x + 2*c) + I + 1)*log(cos(d*x + c)^2 
+ sin(d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2* 
d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) 
^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos 
(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos( 
1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d 
*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos( 
2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqr...
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {a}\, \left (-4 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -6 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}-15 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{3} d \right )}{15 \tan \left (d x +c \right )^{3} d} \] Input:

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

Output:

(sqrt(a)*( - 4*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 
 - 2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - 6*sqrt(t 
an(c + d*x))*sqrt(tan(c + d*x)*i + 1) - 15*int((sqrt(tan(c + d*x))*sqrt(ta 
n(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)**3*d))/(15*tan(c + d*x) 
**3*d)