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

Optimal. Leaf size=94 \[ -\frac{a^2 (2 B+3 i A) \cot (c+d x)}{2 d}-\frac{2 a^2 (A-i B) \log (\sin (c+d x))}{d}-2 a^2 x (B+i A)-\frac{A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.207893, antiderivative size = 94, normalized size of antiderivative = 1., 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 = {3593, 3591, 3531, 3475} \[ -\frac{a^2 (2 B+3 i A) \cot (c+d x)}{2 d}-\frac{2 a^2 (A-i B) \log (\sin (c+d x))}{d}-2 a^2 x (B+i A)-\frac{A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*a^2*(I*A + B)*x - (a^2*((3*I)*A + 2*B)*Cot[c + d*x])/(2*d) - (2*a^2*(A - I*B)*Log[Sin[c + d*x]])/d - (A*Cot
[c + d*x]^2*(a^2 + I*a^2*Tan[c + d*x]))/(2*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 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 ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=-\frac{A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}+\frac{1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x)) (a (3 i A+2 B)-a (A-2 i B) \tan (c+d x)) \, dx\\ &=-\frac{a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac{A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}+\frac{1}{2} \int \cot (c+d x) \left (-4 a^2 (A-i B)-4 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=-2 a^2 (i A+B) x-\frac{a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac{A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}-\left (2 a^2 (A-i B)\right ) \int \cot (c+d x) \, dx\\ &=-2 a^2 (i A+B) x-\frac{a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac{2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac{A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\\ \end{align*}

Mathematica [B]  time = 2.3557, size = 302, normalized size = 3.21 \[ \frac{a^2 \csc (c) \csc ^2(c+d x) (\cos (2 d x)+i \sin (2 d x)) \left (8 (B+i A) \sin (c) \sin ^2(c+d x) \tan ^{-1}(\tan (3 c+d x))+2 (B+2 i A) \cos (c)-8 i A d x \sin (c)-4 i A d x \sin (c+2 d x)+4 i A d x \sin (3 c+2 d x)-4 i A \cos (c+2 d x)-2 A \sin (c) \log \left (\sin ^2(c+d x)\right )-A \sin (c+2 d x) \log \left (\sin ^2(c+d x)\right )+A \sin (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )-2 A \sin (c)-8 B d x \sin (c)-4 B d x \sin (c+2 d x)+4 B d x \sin (3 c+2 d x)-2 B \cos (c+2 d x)+2 i B \sin (c) \log \left (\sin ^2(c+d x)\right )+i B \sin (c+2 d x) \log \left (\sin ^2(c+d x)\right )-i B \sin (3 c+2 d x) \log \left (\sin ^2(c+d x)\right )\right )}{4 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.071, size = 119, normalized size = 1.3 \begin{align*} -2\,{\frac{{a}^{2}A\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,{a}^{2}Bx-2\,{\frac{B{a}^{2}c}{d}}-2\,iA{a}^{2}x-{\frac{2\,iA\cot \left ( dx+c \right ){a}^{2}}{d}}-{\frac{2\,iA{a}^{2}c}{d}}+{\frac{2\,iB{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}A \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{\cot \left ( dx+c \right ) B{a}^{2}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.66555, size = 130, normalized size = 1.38 \begin{align*} -\frac{4 \,{\left (d x + c\right )}{\left (i \, A + B\right )} a^{2} - 2 \,{\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (2 \, A - 2 i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac{2 \,{\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - A a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.40485, size = 316, normalized size = 3.36 \begin{align*} \frac{2 \,{\left ({\left (3 \, A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (2 \, A - i \, B\right )} a^{2} -{\left ({\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (A - i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\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)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

2*((3*A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - (2*A - I*B)*a^2 - ((A - I*B)*a^2*e^(4*I*d*x + 4*I*c) - 2*(A - I*B)*a^
2*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^2)*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 [A]  time = 5.36212, size = 119, normalized size = 1.27 \begin{align*} \frac{2 a^{2} \left (- A + i B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{- \frac{\left (4 A a^{2} - 2 i B a^{2}\right ) e^{- 4 i c}}{d} + \frac{\left (6 A a^{2} - 2 i B a^{2}\right ) e^{- 2 i c} e^{2 i d x}}{d}}{e^{4 i d x} - 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

2*a**2*(-A + I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-(4*A*a**2 - 2*I*B*a**2)*exp(-4*I*c)/d + (6*A*a**2 - 2*
I*B*a**2)*exp(-2*I*c)*exp(2*I*d*x)/d)/(exp(4*I*d*x) - 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c))

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Giac [B]  time = 1.61919, size = 254, normalized size = 2.7 \begin{align*} -\frac{A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 16 \,{\left (2 \, A a^{2} - 2 i \, B a^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 16 \,{\left (A a^{2} - i \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{24 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 i \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 i \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, 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}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*I*A*a^2*tan(1/2*d*x + 1/2*c) - 4*B*a^2*tan(1/2*d*x + 1/2*c) - 16*(2*A*a
^2 - 2*I*B*a^2)*log(tan(1/2*d*x + 1/2*c) + I) + 16*(A*a^2 - I*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (24*A*a^
2*tan(1/2*d*x + 1/2*c)^2 - 24*I*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*I*A*a^2*tan(1/2*d*x + 1/2*c) - 4*B*a^2*tan(1/
2*d*x + 1/2*c) - A*a^2)/tan(1/2*d*x + 1/2*c)^2)/d