\(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\) [46]

Optimal result

Integrand size = 32, antiderivative size = 76 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {(i A+B) x}{4 a^2}+\frac {A+3 i B}{4 a^2 d (1+i \tan (c+d x))}-\frac {A+i B}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-1/4*(I*A+B)*x/a^2+1/4*(A+3*I*B)/a^2/d/(1+I*tan(d*x+c))+1/4*(-A-I*B)/d/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3671, 3607, 8} \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {A+3 i B}{4 a^2 d (1+i \tan (c+d x))}-\frac {x (B+i A)}{4 a^2}-\frac {A+i B}{4 d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

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

Rule 8

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

Rule 3607

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] + Dist[(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 3671

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] + Dis
t[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {A+i B}{4 d (a+i a \tan (c+d x))^2}-\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{2 a^2} \\ & = \frac {A+3 i B}{4 a^2 d (1+i \tan (c+d x))}-\frac {A+i B}{4 d (a+i a \tan (c+d x))^2}-\frac {(i A+B) \int 1 \, dx}{4 a^2} \\ & = -\frac {(i A+B) x}{4 a^2}+\frac {A+3 i B}{4 a^2 d (1+i \tan (c+d x))}-\frac {A+i B}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\sec ^2(c+d x) (-4 i B+(A+4 i A d x+B (i+4 d x)) \cos (2 (c+d x))+(-i A+B-4 A d x+4 i B d x) \sin (2 (c+d x)))}{16 a^2 d (-i+\tan (c+d x))^2} \]

[In]

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

[Out]

(Sec[c + d*x]^2*((-4*I)*B + (A + (4*I)*A*d*x + B*(I + 4*d*x))*Cos[2*(c + d*x)] + ((-I)*A + B - 4*A*d*x + (4*I)
*B*d*x)*Sin[2*(c + d*x)]))/(16*a^2*d*(-I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {x B}{4 a^{2}}-\frac {i x A}{4 a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} B}{16 a^{2} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} A}{16 a^{2} d}\)	\(73\)
derivativedivides	\(-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {3 B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\)	\(117\)
default	\(-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {3 B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\)	\(117\)

[In]

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

[Out]

-1/4*x/a^2*B-1/4*I*x/a^2*A+1/4*I*B/a^2/d*exp(-2*I*(d*x+c))-1/16*I/a^2/d*exp(-4*I*(d*x+c))*B-1/16/a^2/d*exp(-4*
I*(d*x+c))*A

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.68 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {{\left (4 \, {\left (i \, A + B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, B e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

[In]

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

[Out]

-1/16*(4*(I*A + B)*d*x*e^(4*I*d*x + 4*I*c) - 4*I*B*e^(2*I*d*x + 2*I*c) + A + I*B)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.20 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 i B a^{2} d e^{4 i c} e^{- 2 i d x} + \left (- 4 A a^{2} d e^{2 i c} - 4 i B a^{2} d e^{2 i c}\right ) e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac {- i A - B}{4 a^{2}} + \frac {\left (- i A e^{4 i c} + i A - B e^{4 i c} + 2 B e^{2 i c} - B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- i A - B\right )}{4 a^{2}} \]

[In]

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

[Out]

Piecewise(((16*I*B*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + (-4*A*a**2*d*exp(2*I*c) - 4*I*B*a**2*d*exp(2*I*c))*exp(-4
*I*d*x))*exp(-6*I*c)/(64*a**4*d**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*(-(-I*A - B)/(4*a**2) + (-I*A*exp(4*I*c)
+ I*A - B*exp(4*I*c) + 2*B*exp(2*I*c) - B)*exp(-4*I*c)/(4*a**2)), True)) + x*(-I*A - B)/(4*a**2)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.38 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} - \frac {2 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {3 \, A \tan \left (d x + c\right )^{2} - 3 i \, B \tan \left (d x + c\right )^{2} - 10 i \, A \tan \left (d x + c\right ) + 6 \, B \tan \left (d x + c\right ) - 3 \, A - 5 i \, B}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

1/16*(2*(A - I*B)*log(tan(d*x + c) + I)/a^2 - 2*(A - I*B)*log(tan(d*x + c) - I)/a^2 + (3*A*tan(d*x + c)^2 - 3*
I*B*tan(d*x + c)^2 - 10*I*A*tan(d*x + c) + 6*B*tan(d*x + c) - 3*A - 5*I*B)/(a^2*(tan(d*x + c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 8.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.39 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {B}{2\,a^2}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {A}{4\,a^2}+\frac {B\,3{}\mathrm {i}}{4\,a^2}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{8\,a^2\,d} \]

[In]

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

[Out]

(B/(2*a^2) + tan(c + d*x)*(A/(4*a^2) + (B*3i)/(4*a^2)))/(d*(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i)) + (log(t
an(c + d*x) - 1i)*(A*1i + B)*1i)/(8*a^2*d) + (log(tan(c + d*x) + 1i)*(A - B*1i))/(8*a^2*d)