\(\int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [99]

Optimal result

Integrand size = 26, antiderivative size = 204 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {4 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {368 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {92 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{105 d}+\frac {38 i a^2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{63 d}-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}-\frac {472 a (a+i a \tan (c+d x))^{3/2}}{315 d} \] Output:

4*2^(1/2)*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d- 
368/105*a^2*(a+I*a*tan(d*x+c))^(1/2)/d+92/105*a^2*tan(d*x+c)^2*(a+I*a*tan( 
d*x+c))^(1/2)/d+38/63*I*a^2*tan(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d-2/9*a^ 
2*tan(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2)/d-472/315*a*(a+I*a*tan(d*x+c))^(3/ 
2)/d
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.57 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {4 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)} \left (788+236 i \tan (c+d x)-138 \tan ^2(c+d x)-95 i \tan ^3(c+d x)+35 \tan ^4(c+d x)\right )}{315 d} \] Input:

Integrate[Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(4*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/ 
d - (2*a^2*Sqrt[a + I*a*Tan[c + d*x]]*(788 + (236*I)*Tan[c + d*x] - 138*Ta 
n[c + d*x]^2 - (95*I)*Tan[c + d*x]^3 + 35*Tan[c + d*x]^4))/(315*d)
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {3042, 4039, 27, 3042, 4080, 27, 3042, 4080, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

\(\displaystyle \frac {2}{9} a \int \frac {1}{2} \tan ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (19 i \tan (c+d x) a+17 a)dx-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} a \int \tan ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (19 i \tan (c+d x) a+17 a)dx-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} a \int \tan (c+d x)^3 \sqrt {i \tan (c+d x) a+a} (19 i \tan (c+d x) a+17 a)dx-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {1}{9} a \left (\frac {2 \int -3 \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (19 i a^2-23 a^2 \tan (c+d x)\right )dx}{7 a}+\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \int \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (19 i a^2-23 a^2 \tan (c+d x)\right )dx}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \int \tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a} \left (19 i a^2-23 a^2 \tan (c+d x)\right )dx}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (59 i \tan (c+d x) a^3+46 a^3\right )dx}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (59 i \tan (c+d x) a^3+46 a^3\right )dx}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \left (\int \sqrt {i \tan (c+d x) a+a} \left (46 a^3 \tan (c+d x)-59 i a^3\right )dx+\frac {118 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \left (\int \sqrt {i \tan (c+d x) a+a} \left (46 a^3 \tan (c+d x)-59 i a^3\right )dx+\frac {118 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 4010

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \left (-105 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {92 a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {118 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \left (-105 i a^3 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {92 a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {118 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \left (-\frac {210 a^4 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {92 a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {118 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{9} a \left (\frac {38 i a \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {6 \left (\frac {2 \left (-\frac {105 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {92 a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {118 a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}-\frac {46 a^2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )}{7 a}\right )-\frac {2 a^2 \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)}}{9 d}\)

Input:

Int[Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(-2*a^2*Tan[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]])/(9*d) + (a*((((38*I)/7) 
*a*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (6*((-46*a^2*Tan[c + d*x 
]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) + (2*((-105*Sqrt[2]*a^(7/2)*ArcTanh[ 
Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (92*a^3*Sqrt[a + I*a*Ta 
n[c + d*x]])/d + (118*a^2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d)))/(5*a)))/(7 
*a)))/9
 

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 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[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 = 1.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}+4 a^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{2} d}\)	\(131\)
default	\(\frac {-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}+4 a^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{2} d}\)	\(131\)

Input:

int(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/d/a^2*(-1/9*(a+I*a*tan(d*x+c))^(9/2)+1/7*a*(a+I*a*tan(d*x+c))^(7/2)-1/5* 
a^2*(a+I*a*tan(d*x+c))^(5/2)-1/3*a^3*(a+I*a*tan(d*x+c))^(3/2)-2*a^4*(a+I*a 
*tan(d*x+c))^(1/2)+2*a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)* 
2^(1/2)/a^(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. 412 vs. \(2 (161) = 322\).

Time = 0.09 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.02 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 \, {\left (315 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 315 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 2 \, \sqrt {2} {\left (646 \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} + 1647 \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 2331 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 1365 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 315 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

2/315*(315*sqrt(2)*sqrt(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(4*(a^ 
3*e^(I*d*x + I*c) + sqrt(a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2 
*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 315*sqrt(2)*sqrt(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(4*(a^3*e^(I*d*x + I*c) - sqrt(a^5/d^2)* 
(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - 
 I*c)/a^2) - 2*sqrt(2)*(646*a^2*e^(9*I*d*x + 9*I*c) + 1647*a^2*e^(7*I*d*x 
+ 7*I*c) + 2331*a^2*e^(5*I*d*x + 5*I*c) + 1365*a^2*e^(3*I*d*x + 3*I*c) + 3 
15*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(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)
 

Sympy [F]

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

integrate(tan(d*x+c)**3*(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**(5/2)*tan(c + d*x)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.76 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (315 \, \sqrt {2} a^{\frac {13}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 45 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 63 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} + 105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 630 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{6}\right )}}{315 \, a^{4} d} \] Input:

integrate(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

-2/315*(315*sqrt(2)*a^(13/2)*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))) + 35*(I*a*tan(d*x + 
 c) + a)^(9/2)*a^2 - 45*(I*a*tan(d*x + c) + a)^(7/2)*a^3 + 63*(I*a*tan(d*x 
 + c) + a)^(5/2)*a^4 + 105*(I*a*tan(d*x + c) + a)^(3/2)*a^5 + 630*sqrt(I*a 
*tan(d*x + c) + a)*a^6)/(a^4*d)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^3*(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 TypeError: Bad Argument TypeError: Bad Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 1.28 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.70 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}-\frac {4\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{7\,a\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{9/2}}{9\,a^2\,d}-\frac {2\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}-\frac {\sqrt {2}\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \] Input:

int(tan(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

(2*(a + a*tan(c + d*x)*1i)^(7/2))/(7*a*d) - (4*a^2*(a + a*tan(c + d*x)*1i) 
^(1/2))/d - (2*(a + a*tan(c + d*x)*1i)^(5/2))/(5*d) - (2*(a + a*tan(c + d* 
x)*1i)^(9/2))/(9*a^2*d) - (2*a*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d) - (2^( 
1/2)*a^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))* 
4i)/d
 

Reduce [F]

\[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{5}d x \right )+2 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{4}d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}d x \right ) \] Input:

int(tan(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

sqrt(a)*a**2*( - int(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**5,x) + 2*int(s 
qrt(tan(c + d*x)*i + 1)*tan(c + d*x)**4,x)*i + int(sqrt(tan(c + d*x)*i + 1 
)*tan(c + d*x)**3,x))