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

Optimal. Leaf size=95 \[ \frac{3 A+i B}{4 a^2 d (1+i \tan (c+d x))}-\frac{x (-B+3 i A)}{4 a^2}+\frac{A \log (\sin (c+d x))}{a^2 d}+\frac{A+i B}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.228687, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3596, 3531, 3475} \[ \frac{3 A+i B}{4 a^2 d (1+i \tan (c+d x))}-\frac{x (-B+3 i A)}{4 a^2}+\frac{A \log (\sin (c+d x))}{a^2 d}+\frac{A+i B}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*I)*A - B)*x)/(4*a^2) + (A*Log[Sin[c + d*x]])/(a^2*d) + (3*A + 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 3596

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

Rule 3531

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\frac{A+i B}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\cot (c+d x) (4 a A-2 a (i A-B) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{3 A+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{\int \cot (c+d x) \left (8 a^2 A-2 a^2 (3 i A-B) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{(3 i A-B) x}{4 a^2}+\frac{3 A+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{A \int \cot (c+d x) \, dx}{a^2}\\ &=-\frac{(3 i A-B) x}{4 a^2}+\frac{A \log (\sin (c+d x))}{a^2 d}+\frac{3 A+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]  time = 0.965453, size = 184, normalized size = 1.94 \[ -\frac{i \sec ^2(c+d x) \left (\cos (2 (c+d x)) \left (-8 i A \log \left (\sin ^2(c+d x)\right )+4 A d x-i A-4 i B d x+B\right )+4 i A d x \sin (2 (c+d x))-A \sin (2 (c+d x))+8 A \sin (2 (c+d x)) \log \left (\sin ^2(c+d x)\right )-16 A \tan ^{-1}(\tan (d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-8 i A-i B \sin (2 (c+d x))+4 B d x \sin (2 (c+d x))+4 B\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/16)*Sec[c + d*x]^2*((-8*I)*A + 4*B + Cos[2*(c + d*x)]*((-I)*A + B + 4*A*d*x - (4*I)*B*d*x - (8*I)*A*Log[S
in[c + d*x]^2]) - 16*A*ArcTan[Tan[d*x]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - A*Sin[2*(c + d*x)] - I*B*Sin
[2*(c + d*x)] + (4*I)*A*d*x*Sin[2*(c + d*x)] + 4*B*d*x*Sin[2*(c + d*x)] + 8*A*Log[Sin[c + d*x]^2]*Sin[2*(c + d
*x)]))/(a^2*d*(-I + Tan[c + d*x])^2)

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Maple [B]  time = 0.112, size = 177, normalized size = 1.9 \begin{align*}{\frac{B}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{3\,i}{4}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{8\,{a}^{2}d}}-{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{{a}^{2}d}}-{\frac{A}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{4}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}}+{\frac{{\frac{i}{8}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{2}d}}+{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(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)*B-3/4*I/d/a^2/(tan(d*x+c)-I)*A-7/8/d/a^2*ln(tan(d*x+c)-I)*A-1/8*I/d/a^2*ln(tan(d*x+c)
-I)*B-1/4/d/a^2/(tan(d*x+c)-I)^2*A-1/4*I/d/a^2/(tan(d*x+c)-I)^2*B-1/8/d/a^2*A*ln(tan(d*x+c)+I)+1/8*I/d/a^2*B*l
n(tan(d*x+c)+I)+1/d/a^2*A*ln(tan(d*x+c))

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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(cot(d*x+c)*(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 = 1.5515, size = 242, normalized size = 2.55 \begin{align*} \frac{{\left ({\left (-28 i \, A + 4 \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, A e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 4 \,{\left (2 \, A + i \, B\right )} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((-28*I*A + 4*B)*d*x*e^(4*I*d*x + 4*I*c) + 16*A*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) + 4*(2*A
+ I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [A]  time = 3.75569, size = 221, normalized size = 2.33 \begin{align*} \frac{A \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} + \begin{cases} \frac{\left (\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} + \left (32 A a^{2} d e^{4 i c} + 16 i B a^{2} d e^{4 i c}\right ) e^{- 2 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{7 i A - B}{4 a^{2}} - \frac{\left (7 i A e^{4 i c} + 4 i A e^{2 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 (- 7 i A + B\right )}{4 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d) + Piecewise((((4*A*a**2*d*exp(2*I*c) + 4*I*B*a**2*d*exp(2*I*c))*exp
(-4*I*d*x) + (32*A*a**2*d*exp(4*I*c) + 16*I*B*a**2*d*exp(4*I*c))*exp(-2*I*d*x))*exp(-6*I*c)/(64*a**4*d**2), Ne
(64*a**4*d**2*exp(6*I*c), 0)), (x*((7*I*A - B)/(4*a**2) - (7*I*A*exp(4*I*c) + 4*I*A*exp(2*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*(-7*I*A + B)/(4*a**2)

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Giac [A]  time = 1.29025, size = 165, normalized size = 1.74 \begin{align*} -\frac{\frac{2 \,{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{2 \,{\left (7 \, A + i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{16 \, A \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac{21 \, A \tan \left (d x + c\right )^{2} + 3 i \, B \tan \left (d x + c\right )^{2} - 54 i \, A \tan \left (d x + c\right ) + 10 \, B \tan \left (d x + c\right ) - 37 \, A - 11 i \, 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(cot(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*(7*A + I*B)*log(tan(d*x + c) - I)/a^2 - 16*A*log(abs(tan(d*x
+ c)))/a^2 - (21*A*tan(d*x + c)^2 + 3*I*B*tan(d*x + c)^2 - 54*I*A*tan(d*x + c) + 10*B*tan(d*x + c) - 37*A - 11
*I*B)/(a^2*(tan(d*x + c) - I)^2))/d