\(\int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [84]

Optimal result

Integrand size = 26, antiderivative size = 168 \[ \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {i \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {62 i (a+i a \tan (c+d x))^{3/2}}{105 a d} \] Output:

-I*2^(1/2)*a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d 
+8/35*I*(a+I*a*tan(d*x+c))^(1/2)/d-2/35*I*tan(d*x+c)^2*(a+I*a*tan(d*x+c))^ 
(1/2)/d+2/7*tan(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d+62/105*I*(a+I*a*tan(d* 
x+c))^(3/2)/a/d
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.62 \[ \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {-105 i \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 \sqrt {a+i a \tan (c+d x)} \left (43 i-31 \tan (c+d x)-3 i \tan ^2(c+d x)+15 \tan ^3(c+d x)\right )}{105 d} \] Input:

Integrate[Tan[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

((-105*I)*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt 
[a])] + 2*Sqrt[a + I*a*Tan[c + d*x]]*(43*I - 31*Tan[c + d*x] - (3*I)*Tan[c 
 + d*x]^2 + 15*Tan[c + d*x]^3))/(105*d)
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4043, 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.

\(\displaystyle \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4043

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4080

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 \int -\frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 i a^2-31 a^2 \tan (c+d x)\right )dx}{5 a}+\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{7 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 i a^2-31 a^2 \tan (c+d x)\right )dx}{5 a}}{7 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 i a^2-31 a^2 \tan (c+d x)\right )dx}{5 a}}{7 a}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (4 i \tan (c+d x) a^2+31 a^2\right )dx+\frac {62 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left (4 i \tan (c+d x) a^2+31 a^2\right )dx+\frac {62 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}\)

\(\Big \downarrow \) 4010

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {35 a^2 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {8 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {62 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {35 a^2 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {8 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {62 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {70 i a^3 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {8 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {62 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {\frac {2 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {35 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {62 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{7 a}\)

Input:

Int[Tan[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

(2*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(7*d) - ((((2*I)/5)*a*Tan[c 
+ d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d - (((-35*I)*Sqrt[2]*a^(5/2)*ArcTanh 
[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((8*I)*a^2*Sqrt[a + I* 
a*Tan[c + d*x]])/d + (((62*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/d)/(5*a)) 
/(7*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 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 4043

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*(m + n - 1))), x] - Simp[1/(a*(m + n - 1))   Int[(a + b 
*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) 
 - a*c^2*(m + n - 1) + d*(b*d*m - a*c*(m + 2*n - 2))*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] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] 
 || 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 = 1.56 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.56

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{3}}\)	\(94\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{3}}\)	\(94\)

Input:

int(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2*I/d/a^3*(1/7*(a+I*a*tan(d*x+c))^(7/2)-2/5*a*(a+I*a*tan(d*x+c))^(5/2)+2/3 
*a^2*(a+I*a*tan(d*x+c))^(3/2)-1/2*a^(7/2)*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. 330 vs. \(2 (125) = 250\).

Time = 0.09 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.96 \[ \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {105 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 105 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 16 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-23 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 28 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 35 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{210 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/210*(105*sqrt(2)*(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(-a/d^2)*log(4*((I*d*e^(2*I*d*x + 2*I*c) + I* 
d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + a*e^(I*d*x + I*c))*e^( 
-I*d*x - I*c)) - 105*sqrt(2)*(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(-a/d^2)*log(4*((-I*d*e^(2*I*d*x + 
2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + a*e^(I*d*x 
+ I*c))*e^(-I*d*x - I*c)) - 16*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*( 
-23*I*e^(7*I*d*x + 7*I*c) - 28*I*e^(5*I*d*x + 5*I*c) - 35*I*e^(3*I*d*x + 3 
*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]

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

integrate(tan(d*x+c)**4*(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (105 \, \sqrt {2} a^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 60 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 168 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 280 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4}\right )}}{210 \, a^{5} d} \] Input:

integrate(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/210*I*(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))) + 60*(I*a*tan(d*x 
+ c) + a)^(7/2)*a^2 - 168*(I*a*tan(d*x + c) + a)^(5/2)*a^3 + 280*(I*a*tan( 
d*x + c) + a)^(3/2)*a^4)/(a^5*d)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^4*(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 = 0.35 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.65 \[ \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,4{}\mathrm {i}}{5\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,2{}\mathrm {i}}{7\,a^3\,d}-\frac {\sqrt {2}\,\sqrt {-a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{d} \] Input:

int(tan(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(1/2),x)
 

Output:

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

Reduce [F]

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

int(tan(d*x+c)^4*(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)**4,x)