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

Optimal. Leaf size=107 \[ -\frac{(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+4 a^3 x (B+i A)+\frac{a^3 A \log (\sin (c+d x))}{d}+\frac{i a B (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.280963, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3594, 3589, 3475, 3531} \[ -\frac{(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+4 a^3 x (B+i A)+\frac{a^3 A \log (\sin (c+d x))}{d}+\frac{i a B (a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\frac{i a B (a+i a \tan (c+d x))^2}{2 d}+\frac{1}{2} \int \cot (c+d x) (a+i a \tan (c+d x))^2 (2 a A+2 a (i A+2 B) \tan (c+d x)) \, dx\\ &=\frac{i a B (a+i a \tan (c+d x))^2}{2 d}-\frac{(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{1}{2} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (2 a^2 A+2 a^2 (3 i A+4 B) \tan (c+d x)\right ) \, dx\\ &=\frac{i a B (a+i a \tan (c+d x))^2}{2 d}-\frac{(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{1}{2} \int \cot (c+d x) \left (2 a^3 A+8 a^3 (i A+B) \tan (c+d x)\right ) \, dx-\left (a^3 (3 A-4 i B)\right ) \int \tan (c+d x) \, dx\\ &=4 a^3 (i A+B) x+\frac{a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+\frac{i a B (a+i a \tan (c+d x))^2}{2 d}-\frac{(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\left (a^3 A\right ) \int \cot (c+d x) \, dx\\ &=4 a^3 (i A+B) x+\frac{a^3 (3 A-4 i B) \log (\cos (c+d x))}{d}+\frac{a^3 A \log (\sin (c+d x))}{d}+\frac{i a B (a+i a \tan (c+d x))^2}{2 d}-\frac{(A-2 i B) \left (a^3+i a^3 \tan (c+d x)\right )}{d}\\ \end{align*}

Mathematica [B]  time = 7.65867, size = 281, normalized size = 2.63 \[ \frac{a^3 \sec (c) \sec ^2(c+d x) (\cos (3 d x)+i \sin (3 d x)) \left (2 \cos (c) \left ((3 A-4 i B) \log \left (\cos ^2(c+d x)\right )+A \log \left (\sin ^2(c+d x)\right )+8 i A d x+8 B d x-2 i B\right )+\cos (c+2 d x) \left ((3 A-4 i B) \log \left (\cos ^2(c+d x)\right )+8 d x (B+i A)+A \log \left (\sin ^2(c+d x)\right )\right )-4 i A \sin (c+2 d x)+8 i A d x \cos (3 c+2 d x)+3 A \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+A \cos (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )+4 i A \sin (c)-12 B \sin (c+2 d x)+8 B d x \cos (3 c+2 d x)-4 i B \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+12 B \sin (c)\right )}{8 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c]*Sec[c + d*x]^2*(Cos[3*d*x] + I*Sin[3*d*x])*((8*I)*A*d*x*Cos[3*c + 2*d*x] + 8*B*d*x*Cos[3*c + 2*d*x
] + 3*A*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - (4*I)*B*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] + A*Cos[3*c + 2*d*
x]*Log[Sin[c + d*x]^2] + 2*Cos[c]*((-2*I)*B + (8*I)*A*d*x + 8*B*d*x + (3*A - (4*I)*B)*Log[Cos[c + d*x]^2] + A*
Log[Sin[c + d*x]^2]) + Cos[c + 2*d*x]*(8*(I*A + B)*d*x + (3*A - (4*I)*B)*Log[Cos[c + d*x]^2] + A*Log[Sin[c + d
*x]^2]) + (4*I)*A*Sin[c] + 12*B*Sin[c] - (4*I)*A*Sin[c + 2*d*x] - 12*B*Sin[c + 2*d*x]))/(8*d*(Cos[d*x] + I*Sin
[d*x])^3)

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Maple [A]  time = 0.067, size = 135, normalized size = 1.3 \begin{align*} 4\,iAx{a}^{3}-{\frac{iA\tan \left ( dx+c \right ){a}^{3}}{d}}+{\frac{4\,iA{a}^{3}c}{d}}-{\frac{{\frac{i}{2}}B{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{4\,iB{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{A{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+4\,B{a}^{3}x-3\,{\frac{B{a}^{3}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{B{a}^{3}c}{d}}+{\frac{A{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)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.68772, size = 123, normalized size = 1.15 \begin{align*} -\frac{i \, B a^{3} \tan \left (d x + c\right )^{2} + 8 \,{\left (d x + c\right )}{\left (-i \, A - B\right )} a^{3} + 2 \,{\left (2 \, A - 2 i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, A a^{3} \log \left (\tan \left (d x + c\right )\right ) + 2 \,{\left (i \, A + 3 \, B\right )} a^{3} \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)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.44725, size = 466, normalized size = 4.36 \begin{align*} \frac{2 \,{\left (A - 4 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \,{\left (A - 3 i \, B\right )} a^{3} +{\left ({\left (3 \, A - 4 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \,{\left (3 \, A - 4 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (3 \, A - 4 i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) +{\left (A a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + A 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 )} + 2 \, 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)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 8.41282, size = 207, normalized size = 1.93 \begin{align*} \frac{\frac{\left (2 A a^{3} - 8 i B a^{3}\right ) e^{- 2 i c} e^{2 i d x}}{d} + \frac{\left (2 A a^{3} - 6 i B a^{3}\right ) e^{- 4 i c}}{d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \operatorname{RootSum}{\left (z^{2} d^{2} + z \left (- 4 A a^{3} d + 4 i B a^{3} d\right ) + 3 A^{2} a^{6} - 4 i A B a^{6}, \left ( i \mapsto i \log{\left (\frac{i i d}{i A a^{3} e^{2 i c} + 2 B a^{3} e^{2 i c}} - \frac{2 i A + 2 B}{i A e^{2 i c} + 2 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)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

((2*A*a**3 - 8*I*B*a**3)*exp(-2*I*c)*exp(2*I*d*x)/d + (2*A*a**3 - 6*I*B*a**3)*exp(-4*I*c)/d)/(exp(4*I*d*x) + 2
*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c)) + RootSum(_z**2*d**2 + _z*(-4*A*a**3*d + 4*I*B*a**3*d) + 3*A**2*a**6
- 4*I*A*B*a**6, Lambda(_i, _i*log(_i*I*d/(I*A*a**3*exp(2*I*c) + 2*B*a**3*exp(2*I*c)) - (2*I*A + 2*B)/(I*A*exp(
2*I*c) + 2*B*exp(2*I*c)) + exp(2*I*d*x))))

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Giac [B]  time = 1.5758, size = 362, normalized size = 3.38 \begin{align*} \frac{2 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 4 \,{\left (4 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 2 \,{\left (3 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 2 \,{\left (3 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{9 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 i \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 4 i \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 18 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 28 i \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 i \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, A a^{3} - 12 i \, B a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 4*(4*A*a^3 - 4*I*B*a^3)*log(tan(1/2*d*x + 1/2*c) + I) + 2*(3*A*a
^3 - 4*I*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*A*a^3 - 4*I*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)
) - (9*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 12*I*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 4*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 1
2*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 18*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 28*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 4*I*A*a
^3*tan(1/2*d*x + 1/2*c) + 12*B*a^3*tan(1/2*d*x + 1/2*c) + 9*A*a^3 - 12*I*B*a^3)/(tan(1/2*d*x + 1/2*c)^2 - 1)^2
)/d