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

Optimal. Leaf size=102 \[ -\frac {(3 A+i B) \cot (c+d x)}{2 a d}-\frac {(-B+i A) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {x (3 A+i B)}{2 a} \]

[Out]

-1/2*(3*A+I*B)*x/a-1/2*(3*A+I*B)*cot(d*x+c)/a/d-(I*A-B)*ln(sin(d*x+c))/a/d+1/2*(A+I*B)*cot(d*x+c)/d/(a+I*a*tan
(d*x+c))

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Rubi [A]  time = 0.17, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3596, 3529, 3531, 3475} \[ -\frac {(3 A+i B) \cot (c+d x)}{2 a d}-\frac {(-B+i A) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot (c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {x (3 A+i B)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 3.17, size = 225, normalized size = 2.21 \[ \frac {(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left (\frac {1}{2} (B-i A) (\cos (c)-i \sin (c)) \cos (2 d x)-\frac {1}{2} (A+i B) (\cos (c)-i \sin (c)) \sin (2 d x)+2 d x (A+i B) (\cos (c)+i \sin (c))-d x (3 A+i B) (\cos (c)+i \sin (c))+(B-i A) (\cos (c)+i \sin (c)) \log \left (\sin ^2(c+d x)\right )-2 (A+i B) (\cos (c)+i \sin (c)) \tan ^{-1}(\tan (d x))+2 A (\cot (c)+i) \sin (d x) \csc (c+d x)\right )}{2 d (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[d*x] + I*Sin[d*x])*((((-I)*A + B)*Cos[2*d*x]*(Cos[c] - I*Sin[c]))/2 + 2*(A + I*B)*d*x*(Cos[c] + I*Sin[c]
) - (3*A + I*B)*d*x*(Cos[c] + I*Sin[c]) - 2*(A + I*B)*ArcTan[Tan[d*x]]*(Cos[c] + I*Sin[c]) + ((-I)*A + B)*Log[
Sin[c + d*x]^2]*(Cos[c] + I*Sin[c]) + 2*A*(I + Cot[c])*Csc[c + d*x]*Sin[d*x] - ((A + I*B)*(Cos[c] - I*Sin[c])*
Sin[2*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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fricas [A]  time = 0.58, size = 131, normalized size = 1.28 \[ -\frac {2 \, {\left (5 \, A + 3 i \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (2 \, {\left (5 \, A + 3 i \, B\right )} d x - 9 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left ({\left (-4 i \, A + 4 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (4 i \, A - 4 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - i \, A + B}{4 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.70, size = 135, normalized size = 1.32 \[ -\frac {\frac {2 \, {\left (-5 i \, A + 3 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {2 \, {\left (i \, A + B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac {8 \, {\left (i \, A - B\right )} \log \left (\tan \left (d x + c\right )\right )}{a} + \frac {A \tan \left (d x + c\right )^{2} - i \, B \tan \left (d x + c\right )^{2} - 13 i \, A \tan \left (d x + c\right ) + 3 \, B \tan \left (d x + c\right ) - 8 \, A}{{\left (-i \, \tan \left (d x + c\right )^{2} - \tan \left (d x + c\right )\right )} a}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*(-5*I*A + 3*B)*log(tan(d*x + c) - I)/a + 2*(I*A + B)*log(-I*tan(d*x + c) + 1)/a + 8*(I*A - B)*log(tan(
d*x + c))/a + (A*tan(d*x + c)^2 - I*B*tan(d*x + c)^2 - 13*I*A*tan(d*x + c) + 3*B*tan(d*x + c) - 8*A)/((-I*tan(
d*x + c)^2 - tan(d*x + c))*a))/d

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maple [A]  time = 0.58, size = 170, normalized size = 1.67 \[ -\frac {B \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {i A \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {i A \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {B \ln \left (\tan \left (d x +c \right )\right )}{a d}-\frac {A}{a d \tan \left (d x +c \right )}-\frac {A}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right ) A}{4 d a}-\frac {3 \ln \left (\tan \left (d x +c \right )-i\right ) B}{4 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/a*B*ln(tan(d*x+c)+I)-1/4*I/d/a*A*ln(tan(d*x+c)+I)-I/a/d*A*ln(tan(d*x+c))+1/a/d*B*ln(tan(d*x+c))-1/a/d*A
/tan(d*x+c)-1/2/d/a/(tan(d*x+c)-I)*A-1/2*I/d/a/(tan(d*x+c)-I)*B+5/4*I/a/d*ln(tan(d*x+c)-I)*A-3/4/d/a*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)^2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),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 = 6.46, size = 126, normalized size = 1.24 \[ -\frac {\frac {A}{a}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{2\,a}+\frac {A\,3{}\mathrm {i}}{2\,a}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{4\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-3\,B+A\,5{}\mathrm {i}\right )}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(c + d*x)^2*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i),x)

[Out]

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

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sympy [A]  time = 0.65, size = 160, normalized size = 1.57 \[ \frac {2 i A}{- a d e^{2 i c} e^{2 i d x} + a d} + \begin {cases} - \frac {\left (i A - B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 4 a d e^{2 i c} \neq 0 \\x \left (- \frac {- 5 A - 3 i B}{2 a} + \frac {i \left (5 i A e^{2 i c} + i A - 3 B e^{2 i c} - B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} - \frac {x \left (5 A + 3 i B\right )}{2 a} - \frac {i \left (A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*A/(-a*d*exp(2*I*c)*exp(2*I*d*x) + a*d) + Piecewise((-(I*A - B)*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(4*a*d
*exp(2*I*c), 0)), (x*(-(-5*A - 3*I*B)/(2*a) + I*(5*I*A*exp(2*I*c) + I*A - 3*B*exp(2*I*c) - B)*exp(-2*I*c)/(2*a
)), True)) - x*(5*A + 3*I*B)/(2*a) - I*(A + I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a*d)

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