3.69 \(\int \frac{\cot ^4(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=88 \[ \frac{\cot ^5(c+d x) (1-\sec (c+d x))}{5 a d}-\frac{\cot ^3(c+d x) (5-4 \sec (c+d x))}{15 a d}+\frac{\cot (c+d x) (15-8 \sec (c+d x))}{15 a d}+\frac{x}{a} \]

[Out]

x/a + (Cot[c + d*x]*(15 - 8*Sec[c + d*x]))/(15*a*d) - (Cot[c + d*x]^3*(5 - 4*Sec[c + d*x]))/(15*a*d) + (Cot[c
+ d*x]^5*(1 - Sec[c + d*x]))/(5*a*d)

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Rubi [A]  time = 0.127382, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3888, 3882, 8} \[ \frac{\cot ^5(c+d x) (1-\sec (c+d x))}{5 a d}-\frac{\cot ^3(c+d x) (5-4 \sec (c+d x))}{15 a d}+\frac{\cot (c+d x) (15-8 \sec (c+d x))}{15 a d}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

x/a + (Cot[c + d*x]*(15 - 8*Sec[c + d*x]))/(15*a*d) - (Cot[c + d*x]^3*(5 - 4*Sec[c + d*x]))/(15*a*d) + (Cot[c
+ d*x]^5*(1 - Sec[c + d*x]))/(5*a*d)

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\cot ^4(c+d x)}{a+a \sec (c+d x)} \, dx &=\frac{\int \cot ^6(c+d x) (-a+a \sec (c+d x)) \, dx}{a^2}\\ &=\frac{\cot ^5(c+d x) (1-\sec (c+d x))}{5 a d}+\frac{\int \cot ^4(c+d x) (5 a-4 a \sec (c+d x)) \, dx}{5 a^2}\\ &=-\frac{\cot ^3(c+d x) (5-4 \sec (c+d x))}{15 a d}+\frac{\cot ^5(c+d x) (1-\sec (c+d x))}{5 a d}+\frac{\int \cot ^2(c+d x) (-15 a+8 a \sec (c+d x)) \, dx}{15 a^2}\\ &=\frac{\cot (c+d x) (15-8 \sec (c+d x))}{15 a d}-\frac{\cot ^3(c+d x) (5-4 \sec (c+d x))}{15 a d}+\frac{\cot ^5(c+d x) (1-\sec (c+d x))}{5 a d}+\frac{\int 15 a \, dx}{15 a^2}\\ &=\frac{x}{a}+\frac{\cot (c+d x) (15-8 \sec (c+d x))}{15 a d}-\frac{\cot ^3(c+d x) (5-4 \sec (c+d x))}{15 a d}+\frac{\cot ^5(c+d x) (1-\sec (c+d x))}{5 a d}\\ \end{align*}

Mathematica [B]  time = 0.810169, size = 254, normalized size = 2.89 \[ \frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \csc ^3(c+d x) \sec (c+d x) (534 \sin (c+d x)+178 \sin (2 (c+d x))-178 \sin (3 (c+d x))-89 \sin (4 (c+d x))-520 \sin (2 c+d x)-248 \sin (c+2 d x)-120 \sin (3 c+2 d x)+248 \sin (2 c+3 d x)+120 \sin (4 c+3 d x)+184 \sin (3 c+4 d x)-360 d x \cos (2 c+d x)+120 d x \cos (c+2 d x)-120 d x \cos (3 c+2 d x)-120 d x \cos (2 c+3 d x)+120 d x \cos (4 c+3 d x)-60 d x \cos (3 c+4 d x)+60 d x \cos (5 c+4 d x)-200 \sin (c)-584 \sin (d x)+360 d x \cos (d x))}{1920 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

(Csc[c/2]*Csc[c + d*x]^3*Sec[c/2]*Sec[c + d*x]*(360*d*x*Cos[d*x] - 360*d*x*Cos[2*c + d*x] + 120*d*x*Cos[c + 2*
d*x] - 120*d*x*Cos[3*c + 2*d*x] - 120*d*x*Cos[2*c + 3*d*x] + 120*d*x*Cos[4*c + 3*d*x] - 60*d*x*Cos[3*c + 4*d*x
] + 60*d*x*Cos[5*c + 4*d*x] - 200*Sin[c] - 584*Sin[d*x] + 534*Sin[c + d*x] + 178*Sin[2*(c + d*x)] - 178*Sin[3*
(c + d*x)] - 89*Sin[4*(c + d*x)] - 520*Sin[2*c + d*x] - 248*Sin[c + 2*d*x] - 120*Sin[3*c + 2*d*x] + 248*Sin[2*
c + 3*d*x] + 120*Sin[4*c + 3*d*x] + 184*Sin[3*c + 4*d*x]))/(1920*a*d*(1 + Sec[c + d*x]))

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Maple [A]  time = 0.068, size = 113, normalized size = 1.3 \begin{align*} -{\frac{1}{80\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{1}{48\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{3}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4/(a+a*sec(d*x+c)),x)

[Out]

-1/80/a/d*tan(1/2*d*x+1/2*c)^5+1/8/a/d*tan(1/2*d*x+1/2*c)^3-1/a/d*tan(1/2*d*x+1/2*c)+2/d/a*arctan(tan(1/2*d*x+
1/2*c))-1/48/a/d/tan(1/2*d*x+1/2*c)^3+3/8/a/d/tan(1/2*d*x+1/2*c)

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Maxima [A]  time = 1.71367, size = 185, normalized size = 2.1 \begin{align*} -\frac{\frac{3 \,{\left (\frac{80 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a} - \frac{480 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{5 \,{\left (\frac{18 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a \sin \left (d x + c\right )^{3}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/240*(3*(80*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a - 480*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 5*(18*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 1)*(cos(d*x + c) + 1)^3/(a*sin(d*x + c)^3))/d

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Fricas [A]  time = 1.13097, size = 342, normalized size = 3.89 \begin{align*} \frac{23 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} - 27 \, \cos \left (d x + c\right )^{2} + 15 \,{\left (d x \cos \left (d x + c\right )^{3} + d x \cos \left (d x + c\right )^{2} - d x \cos \left (d x + c\right ) - d x\right )} \sin \left (d x + c\right ) - 7 \, \cos \left (d x + c\right ) + 8}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/15*(23*cos(d*x + c)^4 + 8*cos(d*x + c)^3 - 27*cos(d*x + c)^2 + 15*(d*x*cos(d*x + c)^3 + d*x*cos(d*x + c)^2 -
d*x*cos(d*x + c) - d*x)*sin(d*x + c) - 7*cos(d*x + c) + 8)/((a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*co
s(d*x + c) - a*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cot ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4/(a+a*sec(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**4/(sec(c + d*x) + 1), x)/a

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Giac [A]  time = 1.36358, size = 132, normalized size = 1.5 \begin{align*} \frac{\frac{240 \,{\left (d x + c\right )}}{a} + \frac{5 \,{\left (18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}} - \frac{3 \,{\left (a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 80 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{5}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

1/240*(240*(d*x + c)/a + 5*(18*tan(1/2*d*x + 1/2*c)^2 - 1)/(a*tan(1/2*d*x + 1/2*c)^3) - 3*(a^4*tan(1/2*d*x + 1
/2*c)^5 - 10*a^4*tan(1/2*d*x + 1/2*c)^3 + 80*a^4*tan(1/2*d*x + 1/2*c))/a^5)/d