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

Optimal. Leaf size=131 \[ \frac{7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{x (-B+7 i A)}{8 a^3}+\frac{A \log (\sin (c+d x))}{a^3 d}+\frac{A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac{3 A+i B}{8 a d (a+i a \tan (c+d x))^2} \]

[Out]

-(((7*I)*A - B)*x)/(8*a^3) + (A*Log[Sin[c + d*x]])/(a^3*d) + (A + I*B)/(6*d*(a + I*a*Tan[c + d*x])^3) + (3*A +
I*B)/(8*a*d*(a + I*a*Tan[c + d*x])^2) + (7*A + I*B)/(8*d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.360603, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3596, 3531, 3475} \[ \frac{7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{x (-B+7 i A)}{8 a^3}+\frac{A \log (\sin (c+d x))}{a^3 d}+\frac{A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac{3 A+i B}{8 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((7*I)*A - B)*x)/(8*a^3) + (A*Log[Sin[c + d*x]])/(a^3*d) + (A + I*B)/(6*d*(a + I*a*Tan[c + d*x])^3) + (3*A +
I*B)/(8*a*d*(a + I*a*Tan[c + d*x])^2) + (7*A + I*B)/(8*d*(a^3 + I*a^3*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 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))^3} \, dx &=\frac{A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\cot (c+d x) (6 a A-3 a (i A-B) \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac{3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\cot (c+d x) \left (24 a^2 A-6 a^2 (3 i A-B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac{A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac{3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac{7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\int \cot (c+d x) \left (48 a^3 A-6 a^3 (7 i A-B) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=-\frac{(7 i A-B) x}{8 a^3}+\frac{A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac{3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac{7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{A \int \cot (c+d x) \, dx}{a^3}\\ &=-\frac{(7 i A-B) x}{8 a^3}+\frac{A \log (\sin (c+d x))}{a^3 d}+\frac{A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac{3 A+i B}{8 a d (a+i a \tan (c+d x))^2}+\frac{7 A+i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.06443, size = 180, normalized size = 1.37 \[ \frac{\sec ^3(c+d x) ((-27 B+81 i A) \cos (c+d x)+2 \cos (3 (c+d x)) (48 i A \log (\sin (c+d x))+42 A d x+i A+6 i B d x-B)-51 A \sin (c+d x)+2 A \sin (3 (c+d x))+84 i A d x \sin (3 (c+d x))-96 A \sin (3 (c+d x)) \log (\sin (c+d x))-9 i B \sin (c+d x)+2 i B \sin (3 (c+d x))-12 B d x \sin (3 (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(((81*I)*A - 27*B)*Cos[c + d*x] + 2*Cos[3*(c + d*x)]*(I*A - B + 42*A*d*x + (6*I)*B*d*x + (48*I
)*A*Log[Sin[c + d*x]]) - 51*A*Sin[c + d*x] - (9*I)*B*Sin[c + d*x] + 2*A*Sin[3*(c + d*x)] + (2*I)*B*Sin[3*(c +
d*x)] + (84*I)*A*d*x*Sin[3*(c + d*x)] - 12*B*d*x*Sin[3*(c + d*x)] - 96*A*Log[Sin[c + d*x]]*Sin[3*(c + d*x)]))/
(96*a^3*d*(-I + Tan[c + d*x])^3)

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Maple [A]  time = 0.125, size = 218, normalized size = 1.7 \begin{align*} -{\frac{3\,A}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{8}}B}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{{\frac{7\,i}{8}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{B}{8\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{{a}^{3}d}}-{\frac{15\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{16\,{a}^{3}d}}+{\frac{{\frac{i}{6}}A}{{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{B}{6\,{a}^{3}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{16\,{a}^{3}d}}+{\frac{{\frac{i}{16}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{3}d}}+{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{3}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))^3,x)

[Out]

-3/8/d/a^3/(tan(d*x+c)-I)^2*A-1/8*I/d/a^3/(tan(d*x+c)-I)^2*B-7/8*I/d/a^3/(tan(d*x+c)-I)*A+1/8/d/a^3/(tan(d*x+c
)-I)*B-1/16*I/d/a^3*ln(tan(d*x+c)-I)*B-15/16/d/a^3*ln(tan(d*x+c)-I)*A+1/6*I/d/a^3/(tan(d*x+c)-I)^3*A-1/6/d/a^3
/(tan(d*x+c)-I)^3*B-1/16/d/a^3*A*ln(tan(d*x+c)+I)+1/16*I/d/a^3*B*ln(tan(d*x+c)+I)+1/d/a^3*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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.48293, size = 305, normalized size = 2.33 \begin{align*} \frac{{\left ({\left (-180 i \, A + 12 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 \, A e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 6 \,{\left (11 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \,{\left (5 \, A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, A + 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/96*((-180*I*A + 12*B)*d*x*e^(6*I*d*x + 6*I*c) + 96*A*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) - 1) + 6*(1
1*A + 3*I*B)*e^(4*I*d*x + 4*I*c) + 3*(5*A + 3*I*B)*e^(2*I*d*x + 2*I*c) + 2*A + 2*I*B)*e^(-6*I*d*x - 6*I*c)/(a^
3*d)

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Sympy [A]  time = 11.2294, size = 294, normalized size = 2.24 \begin{align*} \frac{A \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} + \begin{cases} \frac{\left (\left (512 A a^{6} d^{2} e^{6 i c} + 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (3840 A a^{6} d^{2} e^{8 i c} + 2304 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (16896 A a^{6} d^{2} e^{10 i c} + 4608 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text{for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac{15 i A - B}{8 a^{3}} - \frac{\left (15 i A e^{6 i c} + 11 i A e^{4 i c} + 5 i A e^{2 i c} + i A - B e^{6 i c} - 3 B e^{4 i c} - 3 B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (- 15 i A + B\right )}{8 a^{3}} \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))**3,x)

[Out]

A*log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d) + Piecewise((((512*A*a**6*d**2*exp(6*I*c) + 512*I*B*a**6*d**2*exp(6
*I*c))*exp(-6*I*d*x) + (3840*A*a**6*d**2*exp(8*I*c) + 2304*I*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (16896*A*
a**6*d**2*exp(10*I*c) + 4608*I*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(2457
6*a**9*d**3*exp(12*I*c), 0)), (x*((15*I*A - B)/(8*a**3) - (15*I*A*exp(6*I*c) + 11*I*A*exp(4*I*c) + 5*I*A*exp(2
*I*c) + I*A - B*exp(6*I*c) - 3*B*exp(4*I*c) - 3*B*exp(2*I*c) - B)*exp(-6*I*c)/(8*a**3)), True)) + x*(-15*I*A +
B)/(8*a**3)

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Giac [A]  time = 1.39853, size = 197, normalized size = 1.5 \begin{align*} -\frac{\frac{6 \,{\left (15 \, A + i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac{6 \,{\left (A - i \, B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} - \frac{96 \, A \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac{165 \, A \tan \left (d x + c\right )^{3} + 11 i \, B \tan \left (d x + c\right )^{3} - 579 i \, A \tan \left (d x + c\right )^{2} + 45 \, B \tan \left (d x + c\right )^{2} - 699 \, A \tan \left (d x + c\right ) - 69 i \, B \tan \left (d x + c\right ) + 301 i \, A - 51 \, B}{a^{3}{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, 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))^3,x, algorithm="giac")

[Out]

-1/96*(6*(15*A + I*B)*log(tan(d*x + c) - I)/a^3 + 6*(A - I*B)*log(I*tan(d*x + c) - 1)/a^3 - 96*A*log(abs(tan(d
*x + c)))/a^3 - (165*A*tan(d*x + c)^3 + 11*I*B*tan(d*x + c)^3 - 579*I*A*tan(d*x + c)^2 + 45*B*tan(d*x + c)^2 -
699*A*tan(d*x + c) - 69*I*B*tan(d*x + c) + 301*I*A - 51*B)/(a^3*(tan(d*x + c) - I)^3))/d