3.11 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=83 \[ -\frac{a \cot ^4(c+d x)}{4 d}-\frac{i a \cot ^3(c+d x)}{3 d}+\frac{a \cot ^2(c+d x)}{2 d}+\frac{i a \cot (c+d x)}{d}+\frac{a \log (\sin (c+d x))}{d}+i a x \]

[Out]

I*a*x + (I*a*Cot[c + d*x])/d + (a*Cot[c + d*x]^2)/(2*d) - ((I/3)*a*Cot[c + d*x]^3)/d - (a*Cot[c + d*x]^4)/(4*d
) + (a*Log[Sin[c + d*x]])/d

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Rubi [A]  time = 0.10865, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3529, 3531, 3475} \[ -\frac{a \cot ^4(c+d x)}{4 d}-\frac{i a \cot ^3(c+d x)}{3 d}+\frac{a \cot ^2(c+d x)}{2 d}+\frac{i a \cot (c+d x)}{d}+\frac{a \log (\sin (c+d x))}{d}+i a x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x]),x]

[Out]

I*a*x + (I*a*Cot[c + d*x])/d + (a*Cot[c + d*x]^2)/(2*d) - ((I/3)*a*Cot[c + d*x]^3)/d - (a*Cot[c + d*x]^4)/(4*d
) + (a*Log[Sin[c + d*x]])/d

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 ^5(c+d x) (a+i a \tan (c+d x)) \, dx &=-\frac{a \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) (i a-a \tan (c+d x)) \, dx\\ &=-\frac{i a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^4(c+d x)}{4 d}+\int \cot ^3(c+d x) (-a-i a \tan (c+d x)) \, dx\\ &=\frac{a \cot ^2(c+d x)}{2 d}-\frac{i a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^4(c+d x)}{4 d}+\int \cot ^2(c+d x) (-i a+a \tan (c+d x)) \, dx\\ &=\frac{i a \cot (c+d x)}{d}+\frac{a \cot ^2(c+d x)}{2 d}-\frac{i a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^4(c+d x)}{4 d}+\int \cot (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=i a x+\frac{i a \cot (c+d x)}{d}+\frac{a \cot ^2(c+d x)}{2 d}-\frac{i a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^4(c+d x)}{4 d}+a \int \cot (c+d x) \, dx\\ &=i a x+\frac{i a \cot (c+d x)}{d}+\frac{a \cot ^2(c+d x)}{2 d}-\frac{i a \cot ^3(c+d x)}{3 d}-\frac{a \cot ^4(c+d x)}{4 d}+\frac{a \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.368898, size = 84, normalized size = 1.01 \[ \frac{a \left (-\cot ^4(c+d x)+2 \cot ^2(c+d x)+4 \log (\tan (c+d x))+4 \log (\cos (c+d x))\right )}{4 d}-\frac{i a \cot ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\tan ^2(c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x]),x]

[Out]

((-I/3)*a*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/d + (a*(2*Cot[c + d*x]^2 - Cot[c +
d*x]^4 + 4*Log[Cos[c + d*x]] + 4*Log[Tan[c + d*x]]))/(4*d)

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Maple [A]  time = 0.041, size = 83, normalized size = 1. \begin{align*}{\frac{-{\frac{i}{3}}a \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{ia\cot \left ( dx+c \right ) }{d}}+iax+{\frac{iac}{d}}-{\frac{a \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{a \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{a\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)^5*(a+I*a*tan(d*x+c)),x)

[Out]

-1/3*I*a*cot(d*x+c)^3/d+I*a*cot(d*x+c)/d+I*a*x+I/d*a*c-1/4*a*cot(d*x+c)^4/d+1/2*a*cot(d*x+c)^2/d+a*ln(sin(d*x+
c))/d

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Maxima [A]  time = 2.00277, size = 112, normalized size = 1.35 \begin{align*} -\frac{-12 i \,{\left (d x + c\right )} a + 6 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \, a \log \left (\tan \left (d x + c\right )\right ) - \frac{12 i \, a \tan \left (d x + c\right )^{3} + 6 \, a \tan \left (d x + c\right )^{2} - 4 i \, a \tan \left (d x + c\right ) - 3 \, a}{\tan \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(-12*I*(d*x + c)*a + 6*a*log(tan(d*x + c)^2 + 1) - 12*a*log(tan(d*x + c)) - (12*I*a*tan(d*x + c)^3 + 6*a
*tan(d*x + c)^2 - 4*I*a*tan(d*x + c) - 3*a)/tan(d*x + c)^4)/d

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Fricas [B]  time = 2.22251, size = 456, normalized size = 5.49 \begin{align*} -\frac{24 \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 36 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 32 \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \,{\left (a e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 8 \, a}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/3*(24*a*e^(6*I*d*x + 6*I*c) - 36*a*e^(4*I*d*x + 4*I*c) + 32*a*e^(2*I*d*x + 2*I*c) - 3*(a*e^(8*I*d*x + 8*I*c
) - 4*a*e^(6*I*d*x + 6*I*c) + 6*a*e^(4*I*d*x + 4*I*c) - 4*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(2*I*d*x + 2*I*c) -
1) - 8*a)/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c
) + d)

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Sympy [B]  time = 8.0977, size = 165, normalized size = 1.99 \begin{align*} \frac{a \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{- \frac{8 a e^{- 2 i c} e^{6 i d x}}{d} + \frac{12 a e^{- 4 i c} e^{4 i d x}}{d} - \frac{32 a e^{- 6 i c} e^{2 i d x}}{3 d} + \frac{8 a e^{- 8 i c}}{3 d}}{e^{8 i d x} - 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} - 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c)),x)

[Out]

a*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-8*a*exp(-2*I*c)*exp(6*I*d*x)/d + 12*a*exp(-4*I*c)*exp(4*I*d*x)/d - 32*
a*exp(-6*I*c)*exp(2*I*d*x)/(3*d) + 8*a*exp(-8*I*c)/(3*d))/(exp(8*I*d*x) - 4*exp(-2*I*c)*exp(6*I*d*x) + 6*exp(-
4*I*c)*exp(4*I*d*x) - 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

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Giac [B]  time = 1.42836, size = 215, normalized size = 2.59 \begin{align*} -\frac{3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 384 \, a \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 192 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 120 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{400 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*a*tan(1/2*d*x + 1/2*c)^4 - 8*I*a*tan(1/2*d*x + 1/2*c)^3 - 36*a*tan(1/2*d*x + 1/2*c)^2 + 384*a*log(ta
n(1/2*d*x + 1/2*c) + I) - 192*a*log(abs(tan(1/2*d*x + 1/2*c))) + 120*I*a*tan(1/2*d*x + 1/2*c) + (400*a*tan(1/2
*d*x + 1/2*c)^4 - 120*I*a*tan(1/2*d*x + 1/2*c)^3 - 36*a*tan(1/2*d*x + 1/2*c)^2 + 8*I*a*tan(1/2*d*x + 1/2*c) +
3*a)/tan(1/2*d*x + 1/2*c)^4)/d