\(\int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx\) [268]

Optimal result

Integrand size = 19, antiderivative size = 55 \[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=a x-\frac {\cot ^3(c+d x) (a+b \sec (c+d x))}{3 d}+\frac {\cot (c+d x) (3 a+2 b \sec (c+d x))}{3 d} \]

[Out]

a*x-1/3*cot(d*x+c)^3*(a+b*sec(d*x+c))/d+1/3*cot(d*x+c)*(3*a+2*b*sec(d*x+c))/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3967, 8} \[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {\cot ^3(c+d x) (a+b \sec (c+d x))}{3 d}+\frac {\cot (c+d x) (3 a+2 b \sec (c+d x))}{3 d}+a x \]

[In]

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

[Out]

a*x - (Cot[c + d*x]^3*(a + b*Sec[c + d*x]))/(3*d) + (Cot[c + d*x]*(3*a + 2*b*Sec[c + d*x]))/(3*d)

Rule 8

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

Rule 3967

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cot ^3(c+d x) (a+b \sec (c+d x))}{3 d}+\frac {1}{3} \int \cot ^2(c+d x) (-3 a-2 b \sec (c+d x)) \, dx \\ & = -\frac {\cot ^3(c+d x) (a+b \sec (c+d x))}{3 d}+\frac {\cot (c+d x) (3 a+2 b \sec (c+d x))}{3 d}+\frac {1}{3} \int 3 a \, dx \\ & = a x-\frac {\cot ^3(c+d x) (a+b \sec (c+d x))}{3 d}+\frac {\cot (c+d x) (3 a+2 b \sec (c+d x))}{3 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.13 \[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {b \csc (c+d x)}{d}-\frac {b \csc ^3(c+d x)}{3 d}-\frac {a \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d} \]

[In]

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

[Out]

(b*Csc[c + d*x])/d - (b*Csc[c + d*x]^3)/(3*d) - (a*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*
x]^2])/(3*d)

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.56

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+b \left (-\frac {\cos \left (d x +c \right )^{4}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{4}}{3 \sin \left (d x +c \right )}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}\right )}{d}\)	\(86\)
default	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+b \left (-\frac {\cos \left (d x +c \right )^{4}}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos \left (d x +c \right )^{4}}{3 \sin \left (d x +c \right )}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}\right )}{d}\)	\(86\)
risch	\(a x +\frac {2 i \left (3 b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 a \,{\mathrm e}^{4 i \left (d x +c \right )}-2 b \,{\mathrm e}^{3 i \left (d x +c \right )}-6 a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 b \,{\mathrm e}^{i \left (d x +c \right )}+4 a \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}\)	\(88\)

[In]

int(cot(d*x+c)^4*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.58 \[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {4 \, a \cos \left (d x + c\right )^{3} + 3 \, b \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) + 3 \, {\left (a d x \cos \left (d x + c\right )^{2} - a d x\right )} \sin \left (d x + c\right ) - 2 \, b}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

1/3*(4*a*cos(d*x + c)^3 + 3*b*cos(d*x + c)^2 - 3*a*cos(d*x + c) + 3*(a*d*x*cos(d*x + c)^2 - a*d*x)*sin(d*x + c
) - 2*b)/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

Sympy [F]

\[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cot ^{4}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.07 \[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a + \frac {{\left (3 \, \sin \left (d x + c\right )^{2} - 1\right )} b}{\sin \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

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

[Out]

1/3*((3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a + (3*sin(d*x + c)^2 - 1)*b/sin(d*x + c)^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (51) = 102\).

Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.04 \[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, {\left (d x + c\right )} a - 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

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

[Out]

1/24*(a*tan(1/2*d*x + 1/2*c)^3 - b*tan(1/2*d*x + 1/2*c)^3 + 24*(d*x + c)*a - 15*a*tan(1/2*d*x + 1/2*c) + 9*b*t
an(1/2*d*x + 1/2*c) + (15*a*tan(1/2*d*x + 1/2*c)^2 + 9*b*tan(1/2*d*x + 1/2*c)^2 - a - b)/tan(1/2*d*x + 1/2*c)^
3)/d

Mupad [B] (verification not implemented)

Time = 14.61 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.64 \[ \int \cot ^4(c+d x) (a+b \sec (c+d x)) \, dx=a\,x+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a}{24}-\frac {b}{24}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\left (-5\,a-3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{3}+\frac {b}{3}\right )}{8\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a}{8}-\frac {3\,b}{8}\right )}{d} \]

[In]

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

[Out]

a*x + (tan(c/2 + (d*x)/2)^3*(a/24 - b/24))/d - (cot(c/2 + (d*x)/2)^3*(a/3 + b/3 - tan(c/2 + (d*x)/2)^2*(5*a +
3*b)))/(8*d) - (tan(c/2 + (d*x)/2)*((5*a)/8 - (3*b)/8))/d