3.21 \(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=116 \[ \frac{(-B+i A) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{a^3 (B+3 i A) \log (\sin (c+d x))}{d}+\frac{a^3 (3 B+i A) \log (\cos (c+d x))}{d}-4 a^3 x (A-i B)-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d} \]

[Out]

-4*a^3*(A - I*B)*x + (a^3*(I*A + 3*B)*Log[Cos[c + d*x]])/d + (a^3*((3*I)*A + B)*Log[Sin[c + d*x]])/d - (a*A*Co
t[c + d*x]*(a + I*a*Tan[c + d*x])^2)/d + ((I*A - B)*(a^3 + I*a^3*Tan[c + d*x]))/d

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Rubi [A]  time = 0.29589, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3593, 3594, 3589, 3475, 3531} \[ \frac{(-B+i A) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{a^3 (B+3 i A) \log (\sin (c+d x))}{d}+\frac{a^3 (3 B+i A) \log (\cos (c+d x))}{d}-4 a^3 x (A-i B)-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-4*a^3*(A - I*B)*x + (a^3*(I*A + 3*B)*Log[Cos[c + d*x]])/d + (a^3*((3*I)*A + B)*Log[Sin[c + d*x]])/d - (a*A*Co
t[c + d*x]*(a + I*a*Tan[c + d*x])^2)/d + ((I*A - B)*(a^3 + I*a^3*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 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && 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))^3 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d}+\int \cot (c+d x) (a+i a \tan (c+d x))^2 (a (3 i A+B)+a (A+i B) \tan (c+d x)) \, dx\\ &=-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d}+\frac{(i A-B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\int \cot (c+d x) (a+i a \tan (c+d x)) \left (a^2 (3 i A+B)-a^2 (A-3 i B) \tan (c+d x)\right ) \, dx\\ &=-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d}+\frac{(i A-B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-\left (a^3 (i A+3 B)\right ) \int \tan (c+d x) \, dx+\int \cot (c+d x) \left (a^3 (3 i A+B)-4 a^3 (A-i B) \tan (c+d x)\right ) \, dx\\ &=-4 a^3 (A-i B) x+\frac{a^3 (i A+3 B) \log (\cos (c+d x))}{d}-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d}+\frac{(i A-B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\left (a^3 (3 i A+B)\right ) \int \cot (c+d x) \, dx\\ &=-4 a^3 (A-i B) x+\frac{a^3 (i A+3 B) \log (\cos (c+d x))}{d}+\frac{a^3 (3 i A+B) \log (\sin (c+d x))}{d}-\frac{a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d}+\frac{(i A-B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}\\ \end{align*}

Mathematica [B]  time = 4.20123, size = 291, normalized size = 2.51 \[ \frac{a^3 \csc (c) \sec (c) \csc (c+d x) \sec (c+d x) \left (4 (3 A-i B) \sin (2 c) \sin (2 (c+d x)) \tan ^{-1}(\tan (4 c+d x))+\cos (2 d x) \left ((B+3 i A) \log \left (\sin ^2(c+d x)\right )+(3 B+i A) \log \left (\cos ^2(c+d x)\right )+2 d x (-7 A+5 i B)\right )+4 A \sin (2 (c+d x))+14 A d x \cos (4 c+2 d x)-i A \cos (4 c+2 d x) \log \left (\cos ^2(c+d x)\right )-3 i A \cos (4 c+2 d x) \log \left (\sin ^2(c+d x)\right )-4 A \sin (2 c)+4 A \sin (2 d x)+4 i B \sin (2 (c+d x))-10 i B d x \cos (4 c+2 d x)-3 B \cos (4 c+2 d x) \log \left (\cos ^2(c+d x)\right )-B \cos (4 c+2 d x) \log \left (\sin ^2(c+d x)\right )-4 i B \sin (2 c)-4 i B \sin (2 d x)\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(a^3*Csc[c]*Csc[c + d*x]*Sec[c]*Sec[c + d*x]*(14*A*d*x*Cos[4*c + 2*d*x] - (10*I)*B*d*x*Cos[4*c + 2*d*x] - I*A*
Cos[4*c + 2*d*x]*Log[Cos[c + d*x]^2] - 3*B*Cos[4*c + 2*d*x]*Log[Cos[c + d*x]^2] - (3*I)*A*Cos[4*c + 2*d*x]*Log
[Sin[c + d*x]^2] - B*Cos[4*c + 2*d*x]*Log[Sin[c + d*x]^2] + Cos[2*d*x]*(2*(-7*A + (5*I)*B)*d*x + (I*A + 3*B)*L
og[Cos[c + d*x]^2] + ((3*I)*A + B)*Log[Sin[c + d*x]^2]) - 4*A*Sin[2*c] - (4*I)*B*Sin[2*c] + 4*A*Sin[2*d*x] - (
4*I)*B*Sin[2*d*x] + 4*A*Sin[2*(c + d*x)] + (4*I)*B*Sin[2*(c + d*x)] + 4*(3*A - I*B)*ArcTan[Tan[4*c + d*x]]*Sin
[2*c]*Sin[2*(c + d*x)]))/(16*d)

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Maple [A]  time = 0.061, size = 134, normalized size = 1.2 \begin{align*} 4\,iBx{a}^{3}+{\frac{iA{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,iA{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-4\,A{a}^{3}x-{\frac{iB\tan \left ( dx+c \right ){a}^{3}}{d}}+{\frac{4\,iB{a}^{3}c}{d}}-{\frac{A\cot \left ( dx+c \right ){a}^{3}}{d}}-4\,{\frac{A{a}^{3}c}{d}}+3\,{\frac{B{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{B{a}^{3}\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))^3*(A+B*tan(d*x+c)),x)

[Out]

4*I*B*x*a^3+I/d*A*a^3*ln(cos(d*x+c))+3*I/d*A*a^3*ln(sin(d*x+c))-4*A*a^3*x-I/d*B*tan(d*x+c)*a^3+4*I/d*B*a^3*c-1
/d*A*cot(d*x+c)*a^3-4/d*A*a^3*c+3/d*B*a^3*ln(cos(d*x+c))+1/d*B*a^3*ln(sin(d*x+c))

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Maxima [A]  time = 1.56119, size = 115, normalized size = 0.99 \begin{align*} -\frac{{\left (d x + c\right )}{\left (4 \, A - 4 i \, B\right )} a^{3} + 2 \,{\left (i \, A + B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) -{\left (3 i \, A + B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) + i \, B a^{3} \tan \left (d x + c\right ) + \frac{A a^{3}}{\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))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.40567, size = 360, normalized size = 3.1 \begin{align*} \frac{{\left (-2 i \, A + 2 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-2 i \, A - 2 \, B\right )} a^{3} +{\left ({\left (i \, A + 3 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-i \, A - 3 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) +{\left ({\left (3 i \, A + B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-3 i \, A - B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 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))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 3.94357, size = 199, normalized size = 1.72 \begin{align*} \frac{- \frac{\left (2 i A a^{3} - 2 B a^{3}\right ) e^{- 2 i c} e^{2 i d x}}{d} - \frac{\left (2 i A a^{3} + 2 B a^{3}\right ) e^{- 4 i c}}{d}}{e^{4 i d x} - e^{- 4 i c}} + \operatorname{RootSum}{\left (z^{2} d^{2} + z \left (- 4 i A a^{3} d - 4 B a^{3} d\right ) - 3 A^{2} a^{6} + 10 i A B a^{6} + 3 B^{2} a^{6}, \left ( i \mapsto i \log{\left (\frac{i i d}{A a^{3} e^{2 i c} + i B a^{3} e^{2 i c}} + \frac{2 A - 2 i B}{A e^{2 i c} + i B e^{2 i c}} + e^{2 i d x} \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))**3*(A+B*tan(d*x+c)),x)

[Out]

(-(2*I*A*a**3 - 2*B*a**3)*exp(-2*I*c)*exp(2*I*d*x)/d - (2*I*A*a**3 + 2*B*a**3)*exp(-4*I*c)/d)/(exp(4*I*d*x) -
exp(-4*I*c)) + RootSum(_z**2*d**2 + _z*(-4*I*A*a**3*d - 4*B*a**3*d) - 3*A**2*a**6 + 10*I*A*B*a**6 + 3*B**2*a**
6, Lambda(_i, _i*log(_i*I*d/(A*a**3*exp(2*I*c) + I*B*a**3*exp(2*I*c)) + (2*A - 2*I*B)/(A*exp(2*I*c) + I*B*exp(
2*I*c)) + exp(2*I*d*x))))

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Giac [B]  time = 1.59254, size = 351, normalized size = 3.03 \begin{align*} \frac{3 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 48 \,{\left (i \, A a^{3} + B a^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 6 \,{\left (i \, A a^{3} + 3 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 6 \,{\left (i \, A a^{3} + 3 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 6 \,{\left (-3 i \, A a^{3} - B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{-10 i \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 14 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 i \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 10 i \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 14 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{6 \, 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))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*A*a^3*tan(1/2*d*x + 1/2*c) - 48*(I*A*a^3 + B*a^3)*log(tan(1/2*d*x + 1/2*c) + I) + 6*(I*A*a^3 + 3*B*a^3)
*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(I*A*a^3 + 3*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(-3*I*A*a^3
- B*a^3)*log(abs(tan(1/2*d*x + 1/2*c))) + (-10*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 14*B*a^3*tan(1/2*d*x + 1/2*c)
^3 - 3*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 10*I*A*a^3*tan(1/2*d*x + 1/2*c) + 14
*B*a^3*tan(1/2*d*x + 1/2*c) + 3*A*a^3)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)))/d