\(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [97]

Optimal result

Integrand size = 36, antiderivative size = 209 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d} \] Output:

1/4*(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/ 
a^(3/2)/d+1/3*(I*A-B)*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)+1/2*(3*A+5*I 
*B)*tan(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^(1/2)-2*(3*A+5*I*B)*(a+I*a*tan(d*x 
+c))^(1/2)/a^2/d+1/6*(11*A+21*I*B)*(a+I*a*tan(d*x+c))^(3/2)/a^3/d
 

Mathematica [A] (verified)

Time = 1.83 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {25 i A-39 B-3 (13 A+19 i B) \tan (c+d x)+12 (-i A+B) \tan ^2(c+d x)-4 i B \tan ^3(c+d x)}{6 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[ 
2]*a^(3/2)*d) + ((25*I)*A - 39*B - 3*(13*A + (19*I)*B)*Tan[c + d*x] + 12*( 
(-I)*A + B)*Tan[c + d*x]^2 - (4*I)*B*Tan[c + d*x]^3)/(6*a*d*(-I + Tan[c + 
d*x])*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 4078, 27, 3042, 4078, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {\int -\frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^2 (3 A+5 i B)-a^2 (11 i A-21 B) \tan (c+d x)\right )dx}{a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^2 (3 A+5 i B)-a^2 (11 i A-21 B) \tan (c+d x)\right )dx}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^2 (3 A+5 i B)-a^2 (11 i A-21 B) \tan (c+d x)\right )dx}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \sqrt {i \tan (c+d x) a+a} \left ((11 i A-21 B) a^2+4 (3 A+5 i B) \tan (c+d x) a^2\right )dx-\frac {2 a (11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \sqrt {i \tan (c+d x) a+a} \left ((11 i A-21 B) a^2+4 (3 A+5 i B) \tan (c+d x) a^2\right )dx-\frac {2 a (11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 4010

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-\left (a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx\right )+\frac {8 a^2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-\left (a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx\right )+\frac {8 a^2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {2 i a^3 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {8 a^2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 a^2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{2 a^2}-\frac {a (3 A+5 i B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\)

Input:

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

Output:

((I*A - B)*Tan[c + d*x]^3)/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) - (-((a*(3*A 
 + (5*I)*B)*Tan[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]])) + ((I*Sqrt[2]* 
a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d 
 + (8*a^2*(3*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/d - (2*a*(11*A + (21 
*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d))/(2*a^2))/(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 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 4078

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(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 - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a 
*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] 
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 i a B \sqrt {a +i a \tan \left (d x +c \right )}-2 a A \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{2} \left (7 i B +5 A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{3} \left (i B +A \right )}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{a^{3} d}\)	\(154\)
default	\(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 i a B \sqrt {a +i a \tan \left (d x +c \right )}-2 a A \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{2} \left (7 i B +5 A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{3} \left (i B +A \right )}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{a^{3} d}\)	\(154\)
parts	\(\frac {2 A \left (-\sqrt {a +i a \tan \left (d x +c \right )}+\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {5 a}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{2}}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{2}}+\frac {2 i B \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {7 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{3}}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}}\)	\(208\)

Input:

int(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETUR 
NVERBOSE)
 

Output:

2/d/a^3*(1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)-2*I*a*B*(a+I*a*tan(d*x+c))^(1/2) 
-a*A*(a+I*a*tan(d*x+c))^(1/2)+1/8*a^(3/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I 
*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-1/4*a^2*(5*A+7*I*B)/(a+I*a*tan(d*x+c 
))^(1/2)+1/6*a^3*(A+I*B)/(a+I*a*tan(d*x+c))^(3/2))
 

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. 440 vs. \(2 (162) = 324\).

Time = 0.13 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.11 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left (2 \, {\left (19 \, A + 26 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (17 \, A + 29 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (2 \, A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \] Input:

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algori 
thm="fricas")
 

Output:

-1/12*(3*sqrt(1/2)*(a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c)) 
*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(I*a^2*d* 
e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 
 - 2*I*A*B - B^2)/(a^3*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I 
*c)/(I*A + B)) - 3*sqrt(1/2)*(a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x 
 + 3*I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2) 
*(-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) 
)*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^ 
(-I*d*x - I*c)/(I*A + B)) + sqrt(2)*(2*(19*A + 26*I*B)*e^(6*I*d*x + 6*I*c) 
 + 3*(17*A + 29*I*B)*e^(4*I*d*x + 4*I*c) + 6*(2*A + 3*I*B)*e^(2*I*d*x + 2* 
I*c) - A - I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(a^2*d*e^(5*I*d*x + 5*I 
*c) + a^2*d*e^(3*I*d*x + 3*I*c))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 16 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 48 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 2 i \, B\right )} a^{2} + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (5 \, A + 7 i \, B\right )} a^{3} - 2 \, {\left (A + i \, B\right )} a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a^{4} d} \] Input:

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algori 
thm="maxima")
 

Output:

-1/24*(3*sqrt(2)*(A - I*B)*a^(5/2)*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))) - 16*I*(I*a*t 
an(d*x + c) + a)^(3/2)*B*a + 48*sqrt(I*a*tan(d*x + c) + a)*(A + 2*I*B)*a^2 
 + 4*(3*(I*a*tan(d*x + c) + a)*(5*A + 7*I*B)*a^3 - 2*(A + I*B)*a^4)/(I*a*t 
an(d*x + c) + a)^(3/2))/(a^4*d)
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2),x, algori 
thm="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 [B] (verification not implemented)

Time = 4.01 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.11 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\frac {B\,1{}\mathrm {i}}{3\,d}-\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\frac {A\,a}{3}-\frac {5\,A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}-\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d}-\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d}+\frac {\sqrt {2}\,A\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(a)*( - 4*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*a*i - 2*sqrt(tan(c + 
d*x)*i + 1)*a + 5*int(sqrt(tan(c + d*x)*i + 1)/(tan(c + d*x)**3*i + tan(c 
+ d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*a*d*i + 5*int(sqrt(tan( 
c + d*x)*i + 1)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1) 
,x)*a*d*i - 2*int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**5)/(tan(c + d*x) 
**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*b*d*i - 2 
*int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**5)/(tan(c + d*x)**3*i + tan(c 
 + d*x)**2 + tan(c + d*x)*i + 1),x)*b*d*i - 2*int((sqrt(tan(c + d*x)*i + 1 
)*tan(c + d*x)**4)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 
 1),x)*tan(c + d*x)**2*a*d*i + 2*int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x 
)**4)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c 
+ d*x)**2*b*d - 2*int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**4)/(tan(c + 
d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*a*d*i + 2*int((sqrt(t 
an(c + d*x)*i + 1)*tan(c + d*x)**4)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + 
 tan(c + d*x)*i + 1),x)*b*d + 3*int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x) 
**2)/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + 
 d*x)**2*a*d*i + 3*int((sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2)/(tan(c + 
 d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*a*d*i))/(2*a**2*d*(t 
an(c + d*x)**2 + 1))