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

Optimal. Leaf size=80 \[ \frac{B+i A}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{x (A-i B)}{4 a^2}+\frac{-B+i A}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0615542, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3526, 3479, 8} \[ \frac{B+i A}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{x (A-i B)}{4 a^2}+\frac{-B+i A}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.520707, size = 94, normalized size = 1.18 \[ -\frac{\sec ^2(c+d x) ((4 i A d x+A+4 B d x+i B) \sin (2 (c+d x))+(A (4 d x+i)+B (-1-4 i d x)) \cos (2 (c+d x))+4 i A)}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.028, size = 162, normalized size = 2. \begin{align*}{\frac{A}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{4}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{4}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{B}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{{a}^{2}d}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{8\,{a}^{2}d}}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}}+{\frac{{\frac{i}{8}}A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.36364, size = 150, normalized size = 1.88 \begin{align*} \frac{{\left (4 \,{\left (A - i \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, A e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.80745, size = 163, normalized size = 2.04 \begin{align*} \begin{cases} \frac{\left (16 i A a^{2} d e^{4 i c} e^{- 2 i d x} + \left (4 i A a^{2} d e^{2 i c} - 4 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}\: 64 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac{A - i B}{4 a^{2}} + \frac{\left (A e^{4 i c} + 2 A e^{2 i c} + A - i B e^{4 i c} + i B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (A - i B\right )}{4 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34553, size = 149, normalized size = 1.86 \begin{align*} -\frac{\frac{2 \,{\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} - \frac{2 \,{\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{3 i \, A \tan \left (d x + c\right )^{2} + 3 \, B \tan \left (d x + c\right )^{2} + 10 \, A \tan \left (d x + c\right ) - 10 i \, B \tan \left (d x + c\right ) - 11 i \, A - 3 \, B}{a^{2}{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(-I*A - B)*log(tan(d*x + c) + I)/a^2 - 2*(-I*A - B)*log(tan(d*x + c) - I)/a^2 - (3*I*A*tan(d*x + c)^2
+ 3*B*tan(d*x + c)^2 + 10*A*tan(d*x + c) - 10*I*B*tan(d*x + c) - 11*I*A - 3*B)/(a^2*(tan(d*x + c) - I)^2))/d