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

Optimal. Leaf size=79 \[ \frac{a^2 (B+2 i A) \log (\sin (c+d x))}{d}-2 a^2 x (A-i B)-\frac{A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac{a^2 B \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.177857, antiderivative size = 79, 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 = {3593, 3589, 3475, 3531} \[ \frac{a^2 (B+2 i A) \log (\sin (c+d x))}{d}-2 a^2 x (A-i B)-\frac{A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac{a^2 B \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3593

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

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*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]

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]

Rubi steps

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

Mathematica [B]  time = 3.08541, size = 202, normalized size = 2.56 \[ \frac{a^2 (\cos (2 d x)+i \sin (2 d x)) (A \cot (c+d x)+B) \left (8 (A-i B) \sin (c+d x) \tan ^{-1}(\tan (3 c+d x))+\csc (c) \left (\cos (2 c+d x) \left ((-B-2 i A) \log \left (\sin ^2(c+d x)\right )+8 d x (A-i B)-B \log \left (\cos ^2(c+d x)\right )\right )+\cos (d x) \left ((B+2 i A) \log \left (\sin ^2(c+d x)\right )-8 d x (A-i B)+B \log \left (\cos ^2(c+d x)\right )\right )+4 A \sin (d x)\right )\right )}{4 d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(B + A*Cot[c + d*x])*(Cos[2*d*x] + I*Sin[2*d*x])*(Csc[c]*(Cos[2*c + d*x]*(8*(A - I*B)*d*x - B*Log[Cos[c +
d*x]^2] + ((-2*I)*A - B)*Log[Sin[c + d*x]^2]) + Cos[d*x]*(-8*(A - I*B)*d*x + B*Log[Cos[c + d*x]^2] + ((2*I)*A
+ B)*Log[Sin[c + d*x]^2]) + 4*A*Sin[d*x]) + 8*(A - I*B)*ArcTan[Tan[3*c + d*x]]*Sin[c + d*x]))/(4*d*(Cos[d*x]
+ I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [A]  time = 0.06, size = 100, normalized size = 1.3 \begin{align*} 2\,iB{a}^{2}x+{\frac{2\,iA{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,{a}^{2}Ax+{\frac{2\,iB{a}^{2}c}{d}}-{\frac{{a}^{2}A\cot \left ( dx+c \right ) }{d}}-2\,{\frac{A{a}^{2}c}{d}}+{\frac{{a}^{2}B\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}B\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))^2*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.54445, size = 101, normalized size = 1.28 \begin{align*} -\frac{{\left (d x + c\right )}{\left (2 \, A - 2 i \, B\right )} a^{2} -{\left (-i \, A - B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) -{\left (2 i \, A + B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac{A a^{2}}{\tan \left (d x + 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))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.55175, size = 266, normalized size = 3.37 \begin{align*} \frac{-2 i \, A a^{2} +{\left (B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) +{\left ({\left (2 i \, A + B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-2 i \, A - B\right )} a^{2}\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))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

(-2*I*A*a^2 + (B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2)*log(e^(2*I*d*x + 2*I*c) + 1) + ((2*I*A + B)*a^2*e^(2*I*d*x +
2*I*c) + (-2*I*A - B)*a^2)*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 = 5.21919, size = 121, normalized size = 1.53 \begin{align*} - \frac{2 i A a^{2} e^{- 2 i c}}{d \left (e^{2 i d x} - e^{- 2 i c}\right )} + \operatorname{RootSum}{\left (z^{2} d^{2} + z \left (- 2 i A a^{2} d - 2 B a^{2} d\right ) + 2 i A B a^{4} + B^{2} a^{4}, \left ( i \mapsto i \log{\left (\frac{i i d e^{- 2 i c}}{A a^{2}} + e^{2 i d x} + \frac{\left (A - i B\right ) e^{- 2 i c}}{A} \right )} \right )\right )} \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))**2*(A+B*tan(d*x+c)),x)

[Out]

-2*I*A*a**2*exp(-2*I*c)/(d*(exp(2*I*d*x) - exp(-2*I*c))) + RootSum(_z**2*d**2 + _z*(-2*I*A*a**2*d - 2*B*a**2*d
) + 2*I*A*B*a**4 + B**2*a**4, Lambda(_i, _i*log(_i*I*d*exp(-2*I*c)/(A*a**2) + exp(2*I*d*x) + (A - I*B)*exp(-2*
I*c)/A)))

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Giac [B]  time = 1.55165, size = 213, normalized size = 2.7 \begin{align*} \frac{2 \, B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, B a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \,{\left (i \, A a^{2} + B a^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 2 \,{\left (2 i \, A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{-4 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{2}}{\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))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + A*a^2*tan(1/2*d
*x + 1/2*c) - 8*(I*A*a^2 + B*a^2)*log(tan(1/2*d*x + 1/2*c) + I) + 2*(2*I*A*a^2 + B*a^2)*log(abs(tan(1/2*d*x +
1/2*c))) + (-4*I*A*a^2*tan(1/2*d*x + 1/2*c) - 2*B*a^2*tan(1/2*d*x + 1/2*c) - A*a^2)/tan(1/2*d*x + 1/2*c))/d