\(\int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [108]

Optimal result

Integrand size = 26, antiderivative size = 201 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {188 \sqrt {a+i a \tan (c+d x)}}{35 a d}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}+\frac {223 (a+i a \tan (c+d x))^{3/2}}{105 a^2 d} \] Output:

-1/2*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(1/2) 
/d-tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^(1/2)-188/35*(a+I*a*tan(d*x+c))^(1/2) 
/a/d+47/35*tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/a/d-9/7*I*tan(d*x+c)^3*(a 
+I*a*tan(d*x+c))^(1/2)/a/d+223/105*(a+I*a*tan(d*x+c))^(3/2)/a^2/d
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.56 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {-341-118 i \tan (c+d x)-82 \tan ^2(c+d x)+6 i \tan ^3(c+d x)+30 \tan ^4(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Tan[c + d*x]^5/Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

-(ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(Sqrt[2]*Sqrt[a]*d 
)) + (-341 - (118*I)*Tan[c + d*x] - 82*Tan[c + d*x]^2 + (6*I)*Tan[c + d*x] 
^3 + 30*Tan[c + d*x]^4)/(105*d*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 4041, 27, 3042, 4080, 27, 3042, 4080, 27, 3042, 4075, 3042, 4010, 3042, 3961, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {1}{2} \tan ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (8 a-9 i a \tan (c+d x))dx}{a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \tan ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (8 a-9 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \tan (c+d x)^3 \sqrt {i \tan (c+d x) a+a} (8 a-9 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (47 \tan (c+d x) a^2+54 i a^2\right )dx}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (47 \tan (c+d x) a^2+54 i a^2\right )dx}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a} \left (47 \tan (c+d x) a^2+54 i a^2\right )dx}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {\frac {\frac {2 \int -\frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (188 a^3-223 i a^3 \tan (c+d x)\right )dx}{5 a}+\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (188 a^3-223 i a^3 \tan (c+d x)\right )dx}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (188 a^3-223 i a^3 \tan (c+d x)\right )dx}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (188 \tan (c+d x) a^3+223 i a^3\right )dx-\frac {446 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (188 \tan (c+d x) a^3+223 i a^3\right )dx-\frac {446 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4010

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {35 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {376 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {446 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {35 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {376 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {446 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\frac {70 a^4 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {376 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {446 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {94 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\frac {35 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {376 a^3 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {446 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}-\frac {18 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}}{2 a^2}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

Input:

Int[Tan[c + d*x]^5/Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

-(Tan[c + d*x]^4/(d*Sqrt[a + I*a*Tan[c + d*x]])) + ((((-18*I)/7)*a*Tan[c + 
 d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + ((94*a^2*Tan[c + d*x]^2*Sqrt[a + I 
*a*Tan[c + d*x]])/(5*d) - ((35*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c 
+ d*x]]/(Sqrt[2]*Sqrt[a])])/d + (376*a^3*Sqrt[a + I*a*Tan[c + d*x]])/d - ( 
446*a^2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d))/(5*a))/(7*a))/(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 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 3961

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

rule 4010

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Simp 
[(b*c + a*d)/b   Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e 
, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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 4075

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

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]
 

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}-\frac {a^{4}}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{4} d}\)	\(131\)
default	\(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}-\frac {a^{4}}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{4} d}\)	\(131\)

Input:

int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

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

Time = 0.09 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.94 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {105 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 105 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (353 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1708 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2030 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1260 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 105\right )}}{420 \, {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \] Input:

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

-1/420*(105*sqrt(2)*(a*d*e^(7*I*d*x + 7*I*c) + 3*a*d*e^(5*I*d*x + 5*I*c) + 
 3*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(4*(( 
a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a 
*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 105*sqrt(2)*(a*d*e^(7*I*d* 
x + 7*I*c) + 3*a*d*e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) + a*d*e 
^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(-4*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sq 
rt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - a*e^(I*d*x + I*c))*e^(-I 
*d*x - I*c)) + 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(353*e^(8*I*d*x 
 + 8*I*c) + 1708*e^(6*I*d*x + 6*I*c) + 2030*e^(4*I*d*x + 4*I*c) + 1260*e^( 
2*I*d*x + 2*I*c) + 105))/(a*d*e^(7*I*d*x + 7*I*c) + 3*a*d*e^(5*I*d*x + 5*I 
*c) + 3*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))
 

Sympy [F]

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

integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Integral(tan(c + d*x)**5/sqrt(I*a*(tan(c + d*x) - I)), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.78 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {105 \, \sqrt {2} a^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 120 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 504 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 1120 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 1680 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5} - \frac {420 \, a^{6}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{420 \, a^{6} d} \] Input:

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/420*(105*sqrt(2)*a^(11/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) 
+ a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 120*(I*a*tan(d*x + 
 c) + a)^(7/2)*a^2 - 504*(I*a*tan(d*x + c) + a)^(5/2)*a^3 + 1120*(I*a*tan( 
d*x + c) + a)^(3/2)*a^4 - 1680*sqrt(I*a*tan(d*x + c) + a)*a^5 - 420*a^6/sq 
rt(I*a*tan(d*x + c) + a))/(a^6*d)
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),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 TypeError: Bad Argument TypeError: Bad Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {1}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {4\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}+\frac {8\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}-\frac {6\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^3\,d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{7\,a^4\,d}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \] Input:

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

Output:

(8*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a^2*d) - (4*(a + a*tan(c + d*x)*1i)^( 
1/2))/(a*d) - 1/(d*(a + a*tan(c + d*x)*1i)^(1/2)) - (6*(a + a*tan(c + d*x) 
*1i)^(5/2))/(5*a^3*d) + (2*(a + a*tan(c + d*x)*1i)^(7/2))/(7*a^4*d) + (2^( 
1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(2*a 
^(1/2)*d)
 

Reduce [F]

\[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{6}}{\tan \left (d x +c \right )^{2}+1}d x \right ) i +\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{5}}{\tan \left (d x +c \right )^{2}+1}d x \right )}{a} \] Input:

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

Output:

(sqrt(a)*( - int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**6)/(tan(c + d*x)* 
*2 + 1),x)*i + int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**5)/(tan(c + d*x 
)**2 + 1),x)))/a