\(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [53]

Optimal result

Integrand size = 34, antiderivative size = 124 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(A-i B) x}{8 a^3}+\frac {(i A-B) \tan ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {i A-7 B}{24 a d (a+i a \tan (c+d x))^2}+\frac {i A+17 B}{24 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:

-1/8*(A-I*B)*x/a^3+1/6*(I*A-B)*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^3+1/24*(I 
*A-7*B)/a/d/(a+I*a*tan(d*x+c))^2+1/24*(I*A+17*B)/d/(a^3+I*a^3*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.19 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) (-9 (A-i B) \cos (c+d x)+2 (A+i B-6 i A d x-6 B d x) \cos (3 (c+d x))-3 i A \sin (c+d x)-27 B \sin (c+d x)-2 i A \sin (3 (c+d x))+2 B \sin (3 (c+d x))+12 A d x \sin (3 (c+d x))-12 i B d x \sin (3 (c+d x)))}{96 a^3 d (-i+\tan (c+d x))^3} \] Input:

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

Output:

(Sec[c + d*x]^3*(-9*(A - I*B)*Cos[c + d*x] + 2*(A + I*B - (6*I)*A*d*x - 6* 
B*d*x)*Cos[3*(c + d*x)] - (3*I)*A*Sin[c + d*x] - 27*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*(-I + Tan[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3042, 4078, 3042, 4073, 3042, 4009, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x) (2 a (i A-B)-a (A-5 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x) (2 a (i A-B)-a (A-5 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}\)

\(\Big \downarrow \) 4073

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {-\frac {i \int \frac {a^2 (i A-7 B)-2 a^2 (A-5 i B) \tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {a (-7 B+i A)}{4 d (a+i a \tan (c+d x))^2}}{6 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {-\frac {i \int \frac {a^2 (i A-7 B)-2 a^2 (A-5 i B) \tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {a (-7 B+i A)}{4 d (a+i a \tan (c+d x))^2}}{6 a^2}\)

\(\Big \downarrow \) 4009

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {-\frac {i \left (\frac {3}{2} a (B+i A) \int 1dx+\frac {a^2 (A-17 i B)}{2 d (a+i a \tan (c+d x))}\right )}{2 a^2}-\frac {a (-7 B+i A)}{4 d (a+i a \tan (c+d x))^2}}{6 a^2}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {-\frac {i \left (\frac {a^2 (A-17 i B)}{2 d (a+i a \tan (c+d x))}+\frac {3}{2} a x (B+i A)\right )}{2 a^2}-\frac {a (-7 B+i A)}{4 d (a+i a \tan (c+d x))^2}}{6 a^2}\)

Input:

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

Output:

((I*A - B)*Tan[c + d*x]^2)/(6*d*(a + I*a*Tan[c + d*x])^3) - (-1/4*(a*(I*A 
- 7*B))/(d*(a + I*a*Tan[c + d*x])^2) - ((I/2)*((3*a*(I*A + B)*x)/2 + (a^2* 
(A - (17*I)*B))/(2*d*(a + I*a*Tan[c + d*x]))))/a^2)/(6*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 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]
 

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 = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {i x B}{8 a^{3}}-\frac {x A}{8 a^{3}}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}-\frac {3 \,{\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}\)	\(128\)
derivativedivides	\(-\frac {A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {7 i B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}+\frac {A}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {i B}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {3 i A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {5 B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}\)	\(158\)
default	\(-\frac {A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {7 i B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}+\frac {A}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {i B}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {3 i A}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {5 B}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}\)	\(158\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (12 \, {\left (A - i \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (-i \, A - 3 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (-i \, A + 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.08 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (\left (- 512 i A a^{6} d^{2} e^{6 i c} + 512 B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (768 i A a^{6} d^{2} e^{8 i c} - 2304 B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (1536 i A a^{6} d^{2} e^{10 i c} + 4608 B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- A + i B}{8 a^{3}} + \frac {\left (- A e^{6 i c} + A e^{4 i c} + A e^{2 i c} - A + i B e^{6 i c} - 3 i B e^{4 i c} + 3 i B e^{2 i c} - i B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- A + i B\right )}{8 a^{3}} \] Input:

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

Output:

Piecewise((((-512*I*A*a**6*d**2*exp(6*I*c) + 512*B*a**6*d**2*exp(6*I*c))*e 
xp(-6*I*d*x) + (768*I*A*a**6*d**2*exp(8*I*c) - 2304*B*a**6*d**2*exp(8*I*c) 
)*exp(-4*I*d*x) + (1536*I*A*a**6*d**2*exp(10*I*c) + 4608*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*(-(-A + I*B)/(8*a**3) + (-A*exp(6*I*c) + A*exp(4*I*c) + A*e 
xp(2*I*c) - A + I*B*exp(6*I*c) - 3*I*B*exp(4*I*c) + 3*I*B*exp(2*I*c) - I*B 
)*exp(-6*I*c)/(8*a**3)), True)) + x*(-A + I*B)/(8*a**3)
 

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{16 \, a^{3} d} - \frac {{\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{16 \, a^{3} d} - \frac {3 \, {\left (A + 7 i \, B\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (i \, A + 9 \, B\right )} \tan \left (d x + c\right ) + 2 \, A - 10 i \, B}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \] Input:

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

Output:

-1/16*(I*A + B)*log(tan(d*x + c) + I)/(a^3*d) - 1/16*(-I*A - B)*log(tan(d* 
x + c) - I)/(a^3*d) - 1/24*(3*(A + 7*I*B)*tan(d*x + c)^2 + 3*(I*A + 9*B)*t 
an(d*x + c) + 2*A - 10*I*B)/(a^3*d*(tan(d*x + c) - I)^3)
 

Mupad [B] (verification not implemented)

Time = 3.61 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {7\,B}{8\,a^3}+\frac {A\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {A\,1{}\mathrm {i}}{12\,a^3}+\frac {5\,B}{12\,a^3}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {A}{8\,a^3}-\frac {B\,9{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {x\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3} \] Input:

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

Output:

(tan(c + d*x)^2*((A*1i)/(8*a^3) - (7*B)/(8*a^3)) + (A*1i)/(12*a^3) + (5*B) 
/(12*a^3) - tan(c + d*x)*(A/(8*a^3) - (B*9i)/(8*a^3)))/(d*(tan(c + d*x)*3i 
 - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + (x*(A*1i + B)*1i)/(8*a^3)
 

Reduce [F]

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

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

Output:

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