3.5 \(\int \cot ^2(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=44 \[ \frac{a (B+i A) \log (\sin (c+d x))}{d}-a x (A-i B)-\frac{a A \cot (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0840953, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3591, 3531, 3475} \[ \frac{a (B+i A) \log (\sin (c+d x))}{d}-a x (A-i B)-\frac{a A \cot (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*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*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 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 \cot ^2(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot (c+d x)}{d}+\int \cot (c+d x) (a (i A+B)-a (A-i B) \tan (c+d x)) \, dx\\ &=-a (A-i B) x-\frac{a A \cot (c+d x)}{d}+(a (i A+B)) \int \cot (c+d x) \, dx\\ &=-a (A-i B) x-\frac{a A \cot (c+d x)}{d}+\frac{a (i A+B) \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.197743, size = 84, normalized size = 1.91 \[ -\frac{a A \cot (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-\tan ^2(c+d x)\right )}{d}+\frac{i a A (\log (\tan (c+d x))+\log (\cos (c+d x)))}{d}+\frac{a B (\log (\tan (c+d x))+\log (\cos (c+d x)))}{d}+i a B x \]

Antiderivative was successfully verified.

[In]

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

[Out]

I*a*B*x - (a*A*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d + (I*a*A*(Log[Cos[c + d*x]] +
Log[Tan[c + d*x]]))/d + (a*B*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]]))/d

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Maple [A]  time = 0.046, size = 71, normalized size = 1.6 \begin{align*} iBax+{\frac{iAa\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-Axa+{\frac{iBac}{d}}-{\frac{Aa\cot \left ( dx+c \right ) }{d}}-{\frac{Aac}{d}}+{\frac{aB\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.66823, size = 86, normalized size = 1.95 \begin{align*} -\frac{2 \,{\left (d x + c\right )}{\left (A - i \, B\right )} a +{\left (i \, A + B\right )} a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \,{\left (i \, A + B\right )} a \log \left (\tan \left (d x + c\right )\right ) + \frac{2 \, A a}{\tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.38598, size = 162, normalized size = 3.68 \begin{align*} \frac{-2 i \, A a +{\left ({\left (i \, A + B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-i \, A - B\right )} a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.39197, size = 58, normalized size = 1.32 \begin{align*} - \frac{2 i A a e^{- 2 i c}}{d \left (e^{2 i d x} - e^{- 2 i c}\right )} + \frac{a \left (i A + B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*A*a*exp(-2*I*c)/(d*(exp(2*I*d*x) - exp(-2*I*c))) + a*(I*A + B)*log(exp(2*I*d*x) - exp(-2*I*c))/d

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Giac [B]  time = 1.4037, size = 142, normalized size = 3.23 \begin{align*} \frac{A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \,{\left (-i \, A a - B a\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 2 \,{\left (i \, A a + B a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{-2 i \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(A*a*tan(1/2*d*x + 1/2*c) + 4*(-I*A*a - B*a)*log(tan(1/2*d*x + 1/2*c) + I) + 2*(I*A*a + B*a)*log(abs(tan(1
/2*d*x + 1/2*c))) + (-2*I*A*a*tan(1/2*d*x + 1/2*c) - 2*B*a*tan(1/2*d*x + 1/2*c) - A*a)/tan(1/2*d*x + 1/2*c))/d