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]

-1/8*(7*I*A-B)*x/a^3+A*ln(sin(d*x+c))/a^3/d+1/6*(A+I*B)/d/(a+I*a*tan(d*x+c))^3+1/8*(3*A+I*B)/a/d/(a+I*a*tan(d*
x+c))^2+1/8*(7*A+I*B)/d/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.36, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 3475

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

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 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]

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*}

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Mathematica [A]  time = 1.33, 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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fricas [A]  time = 0.59, size = 103, normalized size = 0.79 \[ \frac {{\left ({\left (-180 i \, A + 12 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 \, A e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 6 \, {\left (11 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (5 \, A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, A + 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

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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giac [A]  time = 1.55, size = 145, normalized size = 1.11 \[ -\frac {\frac {6 \, {\left (15 \, A + i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {6 \, {\left (A - i \, B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} - \frac {96 \, A \log \left (\tan \left (d x + c\right )\right )}{a^{3}} - \frac {165 \, A \tan \left (d x + c\right )^{3} + 11 i \, B \tan \left (d x + c\right )^{3} - 579 i \, A \tan \left (d x + c\right )^{2} + 45 \, B \tan \left (d x + c\right )^{2} - 699 \, A \tan \left (d x + c\right ) - 69 i \, B \tan \left (d x + c\right ) + 301 i \, A - 51 \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

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(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

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maple [A]  time = 0.76, size = 218, normalized size = 1.66 \[ -\frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}+\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d}-\frac {3 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {15 \ln \left (\tan \left (d x +c \right )-i\right ) A}{16 d \,a^{3}}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right ) B}{16 d \,a^{3}} \]

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]

-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/a^3/d*A*ln(tan(d*x+c))-3/8/d/a^3/(tan(d*x+c)-
I)^2*A-1/8*I/a^3/d/(tan(d*x+c)-I)^2*B+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+1/8/d/a^3/(t
an(d*x+c)-I)*B-7/8*I/d/a^3/(tan(d*x+c)-I)*A-15/16/d/a^3*ln(tan(d*x+c)-I)*A-1/16*I/a^3/d*ln(tan(d*x+c)-I)*B

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

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mupad [B]  time = 6.58, size = 164, normalized size = 1.25 \[ \frac {\frac {17\,A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {7\,A}{8\,a^3}+\frac {B\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,5{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {3\,B}{8\,a^3}+\frac {A\,17{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {A\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (15\,A+B\,1{}\mathrm {i}\right )}{16\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(c + d*x)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

((17*A)/(12*a^3) - tan(c + d*x)^2*((7*A)/(8*a^3) + (B*1i)/(8*a^3)) + (B*5i)/(12*a^3) + tan(c + d*x)*((A*17i)/(
8*a^3) - (3*B)/(8*a^3)))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + (A*log(tan(c + d*x
)))/(a^3*d) + (log(tan(c + d*x) + 1i)*(A*1i + B)*1i)/(16*a^3*d) - (log(tan(c + d*x) - 1i)*(15*A + B*1i))/(16*a
^3*d)

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sympy [A]  time = 1.14, size = 303, normalized size = 2.31 \[ \frac {A \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} + \begin {cases} - \frac {\left (\left (- 512 A a^{6} d^{2} e^{6 i c} - 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 3840 A a^{6} d^{2} e^{8 i c} - 2304 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (- 16896 A a^{6} d^{2} e^{10 i c} - 4608 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- 15 i A + B}{8 a^{3}} - \frac {\left (15 i A e^{6 i c} + 11 i A e^{4 i c} + 5 i A e^{2 i c} + i A - B e^{6 i c} - 3 B e^{4 i c} - 3 B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (15 i A - B\right )}{8 a^{3}} \]

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) + (-1689
6*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(
24576*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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