3.61 \(\int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=59 \[ \frac {1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {i x}{4 a^2}-\frac {1}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3526, 3479, 8} \[ \frac {1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {i x}{4 a^2}-\frac {1}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[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 3526

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]

Rubi steps

\begin {align*} \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {1}{4 d (a+i a \tan (c+d x))^2}-\frac {i \int \frac {1}{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {i \int 1 \, dx}{4 a^2}\\ &=-\frac {i x}{4 a^2}-\frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {1}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 66, normalized size = 1.12 \[ \frac {\sec ^2(c+d x) ((1+4 i d x) \cos (2 (c+d x))-(4 d x+i) \sin (2 (c+d x)))}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*((1 + (4*I)*d*x)*Cos[2*(c + d*x)] - (I + 4*d*x)*Sin[2*(c + d*x)]))/(16*a^2*d*(-I + Tan[c + d*x
])^2)

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 32, normalized size = 0.54 \[ \frac {{\left (-4 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.96, size = 70, normalized size = 1.19 \[ -\frac {\frac {\log \left (\tan \left (2 \, d x + 2 \, c\right ) - i\right )}{a^{2}} - \frac {\log \left (-i \, \tan \left (2 \, d x + 2 \, c\right ) + 1\right )}{a^{2}} - \frac {\tan \left (2 \, d x + 2 \, c\right ) + i}{a^{2} {\left (\tan \left (2 \, d x + 2 \, c\right ) - i\right )}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.12, size = 77, normalized size = 1.31 \[ \frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\ln \left (\tan \left (d x +c \right )-i\right )}{8 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a^2*ln(tan(d*x+c)+I)-1/4*I/d/a^2/(tan(d*x+c)-I)+1/4/d/a^2/(tan(d*x+c)-I)^2-1/8/d/a^2*ln(tan(d*x+c)-I)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

mupad [B]  time = 3.97, size = 46, normalized size = 0.78 \[ -\frac {x\,1{}\mathrm {i}}{4\,a^2}+\frac {\mathrm {tan}\left (c+d\,x\right )}{4\,a^2\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.25, size = 75, normalized size = 1.27 \[ \begin {cases} - \frac {e^{- 4 i c} e^{- 4 i d x}}{16 a^{2} d} & \text {for}\: 16 a^{2} d e^{4 i c} \neq 0 \\x \left (\frac {\left (- i e^{4 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} + \frac {i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {i x}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________