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

Optimal. Leaf size=68 \[ -\frac{a (B+i A) \cot (c+d x)}{d}-\frac{a (A-i B) \log (\sin (c+d x))}{d}-a x (B+i A)-\frac{a A \cot ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.121235, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3591, 3529, 3531, 3475} \[ -\frac{a (B+i A) \cot (c+d x)}{d}-\frac{a (A-i B) \log (\sin (c+d x))}{d}-a x (B+i A)-\frac{a A \cot ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(I*A + B)*x) - (a*(I*A + B)*Cot[c + d*x])/d - (a*A*Cot[c + d*x]^2)/(2*d) - (a*(A - I*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 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(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*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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

Mathematica [C]  time = 0.361013, size = 76, normalized size = 1.12 \[ -\frac{a \left (2 (B+i A) \cot (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-\tan ^2(c+d x)\right )+2 (A-i B) (\log (\tan (c+d x))+\log (\cos (c+d x)))+A \cot ^2(c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.064, size = 101, normalized size = 1.5 \begin{align*} -iAax-{\frac{iA\cot \left ( dx+c \right ) a}{d}}-{\frac{iAac}{d}}+{\frac{iBa\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{Aa \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{Aa\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-aBx-{\frac{\cot \left ( dx+c \right ) Ba}{d}}-{\frac{Bac}{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))*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.70647, size = 113, normalized size = 1.66 \begin{align*} \frac{2 \,{\left (d x + c\right )}{\left (-i \, A - B\right )} a +{\left (A - i \, B\right )} a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \,{\left (A - i \, B\right )} a \log \left (\tan \left (d x + c\right )\right ) + \frac{2 \,{\left (-i \, A - B\right )} a \tan \left (d x + c\right ) - A a}{\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))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.40163, size = 302, normalized size = 4.44 \begin{align*} \frac{2 \,{\left (2 \, A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \,{\left (A - i \, B\right )} a -{\left ({\left (A - i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (A - i \, B\right )} a\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)^3*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

(2*(2*A - I*B)*a*e^(2*I*d*x + 2*I*c) - 2*(A - I*B)*a - ((A - I*B)*a*e^(4*I*d*x + 4*I*c) - 2*(A - I*B)*a*e^(2*I
*d*x + 2*I*c) + (A - I*B)*a)*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 = 5.83582, size = 109, normalized size = 1.6 \begin{align*} \frac{a \left (- A + i B\right ) \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{- \frac{\left (2 A a - 2 i B a\right ) e^{- 4 i c}}{d} + \frac{\left (4 A a - 2 i B a\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))*(A+B*tan(d*x+c)),x)

[Out]

a*(-A + I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-(2*A*a - 2*I*B*a)*exp(-4*I*c)/d + (4*A*a - 2*I*B*a)*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 = 1.45869, size = 220, normalized size = 3.24 \begin{align*} -\frac{A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 i \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 16 \,{\left (A a - i \, B a\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 8 \,{\left (A a - i \, B a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{12 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 i \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 i \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, 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}}}{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))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a*tan(1/2*d*x + 1/2*c)^2 - 4*I*A*a*tan(1/2*d*x + 1/2*c) - 4*B*a*tan(1/2*d*x + 1/2*c) - 16*(A*a - I*B*a
)*log(tan(1/2*d*x + 1/2*c) + I) + 8*(A*a - I*B*a)*log(abs(tan(1/2*d*x + 1/2*c))) - (12*A*a*tan(1/2*d*x + 1/2*c
)^2 - 12*I*B*a*tan(1/2*d*x + 1/2*c)^2 - 4*I*A*a*tan(1/2*d*x + 1/2*c) - 4*B*a*tan(1/2*d*x + 1/2*c) - A*a)/tan(1
/2*d*x + 1/2*c)^2)/d