\(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [54]

Optimal result

Integrand size = 32, antiderivative size = 110 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(i A+B) x}{8 a^3}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.35 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {(\cos (3 (c+d x))-i \sin (3 (c+d x))) (3 (A+3 i B) \cos (c+d x)-2 (A+6 i A d x+B (i+6 d x)) \cos (3 (c+d x))+9 i A \sin (c+d x)-3 B \sin (c+d x)+2 i A \sin (3 (c+d x))-2 B \sin (3 (c+d x))+12 A d x \sin (3 (c+d x))-12 i B d x \sin (3 (c+d x)))}{96 a^3 d} \] Input:

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

Output:

((Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)])*(3*(A + (3*I)*B)*Cos[c + d*x] - 2 
*(A + (6*I)*A*d*x + B*(I + 6*d*x))*Cos[3*(c + d*x)] + (9*I)*A*Sin[c + d*x] 
 - 3*B*Sin[c + d*x] + (2*I)*A*Sin[3*(c + d*x)] - 2*B*Sin[3*(c + d*x)] + 12 
*A*d*x*Sin[3*(c + d*x)] - (12*I)*B*d*x*Sin[3*(c + d*x)]))/(96*a^3*d)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4073, 3042, 4009, 3042, 3960, 24}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4073

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4009

\(\displaystyle -\frac {i \left (\frac {1}{2} (A-i B) \int \frac {1}{i \tan (c+d x) a+a}dx+\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {i \left (\frac {1}{2} (A-i B) \int \frac {1}{i \tan (c+d x) a+a}dx+\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 3960

\(\displaystyle -\frac {i \left (\frac {1}{2} (A-i B) \left (\frac {\int 1dx}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}\right )+\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {i \left (\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}+\frac {1}{2} (A-i B) \left (\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}\right )\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\)

Input:

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

Output:

-1/6*(A + I*B)/(d*(a + I*a*Tan[c + d*x])^3) - ((I/2)*((a*(I*A - 3*B))/(4*d 
*(a + I*a*Tan[c + d*x])^2) + ((A - I*B)*(x/(2*a) + (I/2)/(d*(a + I*a*Tan[c 
 + d*x]))))/2))/a^2
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3960

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

rule 4009

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

rule 4073

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {x B}{8 a^{3}}-\frac {i x A}{8 a^{3}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}\)	\(128\)
derivativedivides	\(-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {i A}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {B}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {3 i B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}\)	\(158\)
default	\(-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {i A}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {B}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {3 i B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}\)	\(158\)

Input:

int(tan(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBO 
SE)
 

Output:

-1/8*x/a^3*B-1/8*I*x/a^3*A+1/16*I/a^3/d*exp(-2*I*(d*x+c))*B+1/16/a^3/d*exp 
(-2*I*(d*x+c))*A+1/32*I/a^3/d*exp(-4*I*(d*x+c))*B-1/32/a^3/d*exp(-4*I*(d*x 
+c))*A-1/48*I/a^3/d*exp(-6*I*(d*x+c))*B-1/48/a^3/d*exp(-6*I*(d*x+c))*A
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.67 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (12 \, {\left (i \, A + B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, {\left (A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, A + 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \] Input:

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="f 
ricas")
 

Output:

-1/96*(12*(I*A + B)*d*x*e^(6*I*d*x + 6*I*c) - 6*(A + I*B)*e^(4*I*d*x + 4*I 
*c) + 3*(A - I*B)*e^(2*I*d*x + 2*I*c) + 2*A + 2*I*B)*e^(-6*I*d*x - 6*I*c)/ 
(a^3*d)
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.36 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (\left (- 512 A a^{6} d^{2} e^{6 i c} - 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 768 A a^{6} d^{2} e^{8 i c} + 768 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (1536 A a^{6} d^{2} e^{10 i c} + 1536 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- i A - B}{8 a^{3}} + \frac {\left (- i A e^{6 i c} - i A e^{4 i c} + i A e^{2 i c} + i A - B e^{6 i c} + B e^{4 i c} + B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- i A - B\right )}{8 a^{3}} \] Input:

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

Output:

Piecewise((((-512*A*a**6*d**2*exp(6*I*c) - 512*I*B*a**6*d**2*exp(6*I*c))*e 
xp(-6*I*d*x) + (-768*A*a**6*d**2*exp(8*I*c) + 768*I*B*a**6*d**2*exp(8*I*c) 
)*exp(-4*I*d*x) + (1536*A*a**6*d**2*exp(10*I*c) + 1536*I*B*a**6*d**2*exp(1 
0*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12 
*I*c), 0)), (x*(-(-I*A - B)/(8*a**3) + (-I*A*exp(6*I*c) - I*A*exp(4*I*c) + 
 I*A*exp(2*I*c) + I*A - B*exp(6*I*c) + B*exp(4*I*c) + B*exp(2*I*c) - B)*ex 
p(-6*I*c)/(8*a**3)), True)) + x*(-I*A - B)/(8*a**3)
 

Maxima [F(-2)]

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

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="m 
axima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{16 \, a^{3} d} - \frac {{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{16 \, a^{3} d} - \frac {3 \, {\left (i \, A + B\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + i \, B\right )} \tan \left (d x + c\right ) - 2 i \, A + 2 \, B}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \] Input:

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="g 
iac")
 

Output:

1/16*(A - I*B)*log(tan(d*x + c) + I)/(a^3*d) - 1/16*(A - I*B)*log(tan(d*x 
+ c) - I)/(a^3*d) - 1/24*(3*(I*A + B)*tan(d*x + c)^2 + 3*(3*A + I*B)*tan(d 
*x + c) - 2*I*A + 2*B)/(a^3*d*(tan(d*x + c) - I)^3)
 

Mupad [B] (verification not implemented)

Time = 3.61 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.34 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A}{8\,a^3}-\frac {B\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,1{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{8\,a^3}+\frac {A\,3{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{16\,a^3\,d} \] Input:

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

Output:

(A/(12*a^3) - tan(c + d*x)^2*(A/(8*a^3) - (B*1i)/(8*a^3)) + (B*1i)/(12*a^3 
) + tan(c + d*x)*((A*3i)/(8*a^3) - B/(8*a^3)))/(d*(tan(c + d*x)*3i - 3*tan 
(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + (log(tan(c + d*x) - 1i)*(A*1i + B) 
*1i)/(16*a^3*d) + (log(tan(c + d*x) + 1i)*(A - B*1i))/(16*a^3*d)
 

Reduce [F]

\[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {-\left (\int \frac {\tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) b -\left (\int \frac {\tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) a}{a^{3}} \] Input:

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

Output:

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