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

Optimal. Leaf size=131 \[ -\frac{(2 A+i B) \cot ^2(c+d x)}{2 a d}+\frac{3 (-B+i A) \cot (c+d x)}{2 a d}-\frac{(2 A+i B) \log (\sin (c+d x))}{a d}+\frac{(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{3 x (-B+i A)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.212175, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3596, 3529, 3531, 3475} \[ -\frac{(2 A+i B) \cot ^2(c+d x)}{2 a d}+\frac{3 (-B+i A) \cot (c+d x)}{2 a d}-\frac{(2 A+i B) \log (\sin (c+d x))}{a d}+\frac{(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{3 x (-B+i A)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3529

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 + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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 ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx &=\frac{(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \cot ^3(c+d x) (2 a (2 A+i B)-3 a (i A-B) \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac{(2 A+i B) \cot ^2(c+d x)}{2 a d}+\frac{(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \cot ^2(c+d x) (-3 a (i A-B)-2 a (2 A+i B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{3 (i A-B) \cot (c+d x)}{2 a d}-\frac{(2 A+i B) \cot ^2(c+d x)}{2 a d}+\frac{(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \cot (c+d x) (-2 a (2 A+i B)+3 a (i A-B) \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{3 (i A-B) x}{2 a}+\frac{3 (i A-B) \cot (c+d x)}{2 a d}-\frac{(2 A+i B) \cot ^2(c+d x)}{2 a d}+\frac{(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{(2 A+i B) \int \cot (c+d x) \, dx}{a}\\ &=\frac{3 (i A-B) x}{2 a}+\frac{3 (i A-B) \cot (c+d x)}{2 a d}-\frac{(2 A+i B) \cot ^2(c+d x)}{2 a d}-\frac{(2 A+i B) \log (\sin (c+d x))}{a d}+\frac{(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 7.14485, size = 902, normalized size = 6.89 \[ \frac{\left (-\frac{1}{2} A \cos (c)-\frac{1}{2} i A \sin (c)\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \csc ^2(c+d x)}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) (\cos (d x)+i \sin (d x)) \left (\frac{1}{2} A \cos (c-d x)+\frac{1}{2} i B \cos (c-d x)-\frac{1}{2} A \cos (c+d x)-\frac{1}{2} i B \cos (c+d x)+\frac{1}{2} i A \sin (c-d x)-\frac{1}{2} B \sin (c-d x)-\frac{1}{2} i A \sin (c+d x)+\frac{1}{2} B \sin (c+d x)\right ) (A+B \tan (c+d x)) \csc (c+d x)}{2 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{\left (2 A \cos \left (\frac{c}{2}\right )+i B \cos \left (\frac{c}{2}\right )+2 i A \sin \left (\frac{c}{2}\right )-B \sin \left (\frac{c}{2}\right )\right ) \left (i \tan ^{-1}(\tan (d x)) \cos \left (\frac{c}{2}\right )-\tan ^{-1}(\tan (d x)) \sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{\left (2 A \cos \left (\frac{c}{2}\right )+i B \cos \left (\frac{c}{2}\right )+2 i A \sin \left (\frac{c}{2}\right )-B \sin \left (\frac{c}{2}\right )\right ) \left (-\frac{1}{2} \cos \left (\frac{c}{2}\right ) \log \left (\sin ^2(c+d x)\right )-\frac{1}{2} i \sin \left (\frac{c}{2}\right ) \log \left (\sin ^2(c+d x)\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{x (2 A \csc (c)+i B \csc (c)+(2 A+i B) \cot (c) (-\cos (c)-i \sin (c))) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{(A+i B) \cos (2 d x) \left (\frac{1}{4} i \sin (c)-\frac{\cos (c)}{4}\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{(A+i B) \left (\frac{3}{2} i d x \cos (c)-\frac{3}{2} d x \sin (c)\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac{(A+i B) \left (\frac{1}{4} i \cos (c)+\frac{\sin (c)}{4}\right ) (\cos (d x)+i \sin (d x)) \sin (2 d x) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*Cos[c/2] + I*B*Cos[c/2] + (2*I)*A*Sin[c/2] - B*Sin[c/2])*(I*ArcTan[Tan[d*x]]*Cos[c/2] - ArcTan[Tan[d*x]]
*Sin[c/2])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c +
d*x])) + ((2*A*Cos[c/2] + I*B*Cos[c/2] + (2*I)*A*Sin[c/2] - B*Sin[c/2])*(-(Cos[c/2]*Log[Sin[c + d*x]^2])/2 -
(I/2)*Log[Sin[c + d*x]^2]*Sin[c/2])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c
+ d*x])*(a + I*a*Tan[c + d*x])) + (x*(2*A*Csc[c] + I*B*Csc[c] + (2*A + I*B)*Cot[c]*(-Cos[c] - I*Sin[c]))*(Cos
[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A + I
*B)*Cos[2*d*x]*(-Cos[c]/4 + (I/4)*Sin[c])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B
*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + (Csc[c + d*x]^2*(-(A*Cos[c])/2 - (I/2)*A*Sin[c])*(Cos[d*x] + I*Sin[d*
x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A + I*B)*(((3*I)/2)
*d*x*Cos[c] - (3*d*x*Sin[c])/2)*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d
*x])*(a + I*a*Tan[c + d*x])) + ((A + I*B)*((I/4)*Cos[c] + Sin[c]/4)*(Cos[d*x] + I*Sin[d*x])*Sin[2*d*x]*(A + B*
Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + (Csc[c/2]*Csc[c + d*x]*Sec[c/2]*
(Cos[d*x] + I*Sin[d*x])*((A*Cos[c - d*x])/2 + (I/2)*B*Cos[c - d*x] - (A*Cos[c + d*x])/2 - (I/2)*B*Cos[c + d*x]
+ (I/2)*A*Sin[c - d*x] - (B*Sin[c - d*x])/2 - (I/2)*A*Sin[c + d*x] + (B*Sin[c + d*x])/2)*(A + B*Tan[c + d*x])
)/(2*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x]))

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Maple [A]  time = 0.111, size = 206, normalized size = 1.6 \begin{align*}{\frac{{\frac{i}{2}}A}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{B}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{4\,ad}}+{\frac{{\frac{5\,i}{4}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{ad}}+{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{4\,ad}}-{\frac{{\frac{i}{4}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{ad}}-{\frac{A}{2\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{iA}{ad\tan \left ( dx+c \right ) }}-{\frac{B}{ad\tan \left ( dx+c \right ) }}-{\frac{iB\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}-2\,{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I/d/a/(tan(d*x+c)-I)*A-1/2/d/a/(tan(d*x+c)-I)*B+7/4/d/a*ln(tan(d*x+c)-I)*A+5/4*I/d/a*ln(tan(d*x+c)-I)*B+1/
4/d/a*A*ln(tan(d*x+c)+I)-1/4*I/d/a*B*ln(tan(d*x+c)+I)-1/2/d/a*A/tan(d*x+c)^2+I/d/a/tan(d*x+c)*A-1/d/a/tan(d*x+
c)*B-I/d/a*B*ln(tan(d*x+c))-2/a/d*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)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.53613, size = 528, normalized size = 4.03 \begin{align*} \frac{{\left (14 i \, A - 10 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} +{\left ({\left (-28 i \, A + 20 \, B\right )} d x - A - 9 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left ({\left (14 i \, A - 10 \, B\right )} d x + 10 \, A + 10 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \,{\left ({\left (2 \, A + i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \,{\left (2 \, A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (2 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - A - i \, B}{4 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((14*I*A - 10*B)*d*x*e^(6*I*d*x + 6*I*c) + ((-28*I*A + 20*B)*d*x - A - 9*I*B)*e^(4*I*d*x + 4*I*c) + ((14*I
*A - 10*B)*d*x + 10*A + 10*I*B)*e^(2*I*d*x + 2*I*c) - 4*((2*A + I*B)*e^(6*I*d*x + 6*I*c) - 2*(2*A + I*B)*e^(4*
I*d*x + 4*I*c) + (2*A + I*B)*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - A - I*B)/(a*d*e^(6*I*d*x + 6*
I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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Sympy [A]  time = 7.9065, size = 197, normalized size = 1.5 \begin{align*} \frac{- \frac{2 i B e^{- 2 i c} e^{2 i d x}}{a d} + \frac{\left (2 A + 2 i B\right ) e^{- 4 i c}}{a d}}{e^{4 i d x} - 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \frac{\left (\begin{cases} 7 i A x e^{2 i c} - \frac{A e^{- 2 i d x}}{2 d} - 5 B x e^{2 i c} - \frac{i B e^{- 2 i d x}}{2 d} & \text{for}\: d \neq 0 \\x \left (7 i A e^{2 i c} + i A - 5 B e^{2 i c} - B\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i c}}{2 a} - \frac{\left (2 A + i B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*I*B*exp(-2*I*c)*exp(2*I*d*x)/(a*d) + (2*A + 2*I*B)*exp(-4*I*c)/(a*d))/(exp(4*I*d*x) - 2*exp(-2*I*c)*exp(2*
I*d*x) + exp(-4*I*c)) + Piecewise((7*I*A*x*exp(2*I*c) - A*exp(-2*I*d*x)/(2*d) - 5*B*x*exp(2*I*c) - I*B*exp(-2*
I*d*x)/(2*d), Ne(d, 0)), (x*(7*I*A*exp(2*I*c) + I*A - 5*B*exp(2*I*c) - B), True))*exp(-2*I*c)/(2*a) - (2*A + I
*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a*d)

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Giac [A]  time = 1.41393, size = 223, normalized size = 1.7 \begin{align*} -\frac{\frac{4 \,{\left (2 \, A + i \, B\right )} \log \left (-i \, \tan \left (d x + c\right )\right )}{a} - \frac{{\left (7 \, A + 5 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} - \frac{{\left (A - i \, B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac{7 \, A \tan \left (d x + c\right ) + 5 i \, B \tan \left (d x + c\right ) - 9 i \, A + 7 \, B}{a{\left (\tan \left (d x + c\right ) - i\right )}} - \frac{2 \,{\left (6 \, A \tan \left (d x + c\right )^{2} + 3 i \, B \tan \left (d x + c\right )^{2} + 2 i \, A \tan \left (d x + c\right ) - 2 \, B \tan \left (d x + c\right ) - A\right )}}{a \tan \left (d x + c\right )^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(2*A + I*B)*log(-I*tan(d*x + c))/a - (7*A + 5*I*B)*log(tan(d*x + c) - I)/a - (A - I*B)*log(-I*tan(d*x
+ c) + 1)/a + (7*A*tan(d*x + c) + 5*I*B*tan(d*x + c) - 9*I*A + 7*B)/(a*(tan(d*x + c) - I)) - 2*(6*A*tan(d*x +
c)^2 + 3*I*B*tan(d*x + c)^2 + 2*I*A*tan(d*x + c) - 2*B*tan(d*x + c) - A)/(a*tan(d*x + c)^2))/d