\(\int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx\) [269]

Optimal result

Integrand size = 19, antiderivative size = 84 \[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=-a x-\frac {\cot ^5(c+d x) (a+b \sec (c+d x))}{5 d}+\frac {\cot ^3(c+d x) (5 a+4 b \sec (c+d x))}{15 d}-\frac {\cot (c+d x) (15 a+8 b \sec (c+d x))}{15 d} \]

[Out]

-a*x-1/5*cot(d*x+c)^5*(a+b*sec(d*x+c))/d+1/15*cot(d*x+c)^3*(5*a+4*b*sec(d*x+c))/d-1/15*cot(d*x+c)*(15*a+8*b*se
c(d*x+c))/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3967, 8} \[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {\cot ^5(c+d x) (a+b \sec (c+d x))}{5 d}+\frac {\cot ^3(c+d x) (5 a+4 b \sec (c+d x))}{15 d}-\frac {\cot (c+d x) (15 a+8 b \sec (c+d x))}{15 d}-a x \]

[In]

Int[Cot[c + d*x]^6*(a + b*Sec[c + d*x]),x]

[Out]

-(a*x) - (Cot[c + d*x]^5*(a + b*Sec[c + d*x]))/(5*d) + (Cot[c + d*x]^3*(5*a + 4*b*Sec[c + d*x]))/(15*d) - (Cot
[c + d*x]*(15*a + 8*b*Sec[c + d*x]))/(15*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 ^5(c+d x) (a+b \sec (c+d x))}{5 d}+\frac {1}{5} \int \cot ^4(c+d x) (-5 a-4 b \sec (c+d x)) \, dx \\ & = -\frac {\cot ^5(c+d x) (a+b \sec (c+d x))}{5 d}+\frac {\cot ^3(c+d x) (5 a+4 b \sec (c+d x))}{15 d}+\frac {1}{15} \int \cot ^2(c+d x) (15 a+8 b \sec (c+d x)) \, dx \\ & = -\frac {\cot ^5(c+d x) (a+b \sec (c+d x))}{5 d}+\frac {\cot ^3(c+d x) (5 a+4 b \sec (c+d x))}{15 d}-\frac {\cot (c+d x) (15 a+8 b \sec (c+d x))}{15 d}+\frac {1}{15} \int -15 a \, dx \\ & = -a x-\frac {\cot ^5(c+d x) (a+b \sec (c+d x))}{5 d}+\frac {\cot ^3(c+d x) (5 a+4 b \sec (c+d x))}{15 d}-\frac {\cot (c+d x) (15 a+8 b \sec (c+d x))}{15 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 = 79, normalized size of antiderivative = 0.94 \[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {b \csc (c+d x)}{d}+\frac {2 b \csc ^3(c+d x)}{3 d}-\frac {b \csc ^5(c+d x)}{5 d}-\frac {a \cot ^5(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(c+d x)\right )}{5 d} \]

[In]

Integrate[Cot[c + d*x]^6*(a + b*Sec[c + d*x]),x]

[Out]

-((b*Csc[c + d*x])/d) + (2*b*Csc[c + d*x]^3)/(3*d) - (b*Csc[c + d*x]^5)/(5*d) - (a*Cot[c + d*x]^5*Hypergeometr
ic2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2])/(5*d)

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.54

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{5}}{5}+\frac {\cot \left (d x +c \right )^{3}}{3}-\cot \left (d x +c \right )-d x -c \right )+b \left (-\frac {\cos \left (d x +c \right )^{6}}{5 \sin \left (d x +c \right )^{5}}+\frac {\cos \left (d x +c \right )^{6}}{15 \sin \left (d x +c \right )^{3}}-\frac {\cos \left (d x +c \right )^{6}}{5 \sin \left (d x +c \right )}-\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}\right )}{d}\)	\(129\)
default	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{5}}{5}+\frac {\cot \left (d x +c \right )^{3}}{3}-\cot \left (d x +c \right )-d x -c \right )+b \left (-\frac {\cos \left (d x +c \right )^{6}}{5 \sin \left (d x +c \right )^{5}}+\frac {\cos \left (d x +c \right )^{6}}{15 \sin \left (d x +c \right )^{3}}-\frac {\cos \left (d x +c \right )^{6}}{5 \sin \left (d x +c \right )}-\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}\right )}{d}\)	\(129\)
risch	\(-a x -\frac {2 i \left (15 b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 a \,{\mathrm e}^{8 i \left (d x +c \right )}-20 b \,{\mathrm e}^{7 i \left (d x +c \right )}-90 a \,{\mathrm e}^{6 i \left (d x +c \right )}+58 b \,{\mathrm e}^{5 i \left (d x +c \right )}+140 a \,{\mathrm e}^{4 i \left (d x +c \right )}-20 b \,{\mathrm e}^{3 i \left (d x +c \right )}-70 a \,{\mathrm e}^{2 i \left (d x +c \right )}+15 b \,{\mathrm e}^{i \left (d x +c \right )}+23 a \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\)	\(137\)

[In]

int(cot(d*x+c)^6*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(-1/5*cot(d*x+c)^5+1/3*cot(d*x+c)^3-cot(d*x+c)-d*x-c)+b*(-1/5/sin(d*x+c)^5*cos(d*x+c)^6+1/15/sin(d*x+c)
^3*cos(d*x+c)^6-1/5/sin(d*x+c)*cos(d*x+c)^6-1/5*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.55 \[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {23 \, a \cos \left (d x + c\right )^{5} + 15 \, b \cos \left (d x + c\right )^{4} - 35 \, a \cos \left (d x + c\right )^{3} - 20 \, b \cos \left (d x + c\right )^{2} + 15 \, a \cos \left (d x + c\right ) + 15 \, {\left (a d x \cos \left (d x + c\right )^{4} - 2 \, a d x \cos \left (d x + c\right )^{2} + a d x\right )} \sin \left (d x + c\right ) + 8 \, b}{15 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cot(d*x+c)^6*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/15*(23*a*cos(d*x + c)^5 + 15*b*cos(d*x + c)^4 - 35*a*cos(d*x + c)^3 - 20*b*cos(d*x + c)^2 + 15*a*cos(d*x +
c) + 15*(a*d*x*cos(d*x + c)^4 - 2*a*d*x*cos(d*x + c)^2 + a*d*x)*sin(d*x + c) + 8*b)/((d*cos(d*x + c)^4 - 2*d*c
os(d*x + c)^2 + d)*sin(d*x + c))

Sympy [F]

\[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cot ^{6}{\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)**6*(a+b*sec(d*x+c)),x)

[Out]

Integral((a + b*sec(c + d*x))*cot(c + d*x)**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {{\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a + \frac {{\left (15 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{2} + 3\right )} b}{\sin \left (d x + c\right )^{5}}}{15 \, d} \]

[In]

integrate(cot(d*x+c)^6*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/15*((15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a + (15*sin(d*x + c)^4 - 10
*sin(d*x + c)^2 + 3)*b/sin(d*x + c)^5)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (78) = 156\).

Time = 0.33 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.02 \[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 25 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 480 \, {\left (d x + c\right )} a + 330 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 150 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {330 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 150 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 25 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a + 3 \, b}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

[In]

integrate(cot(d*x+c)^6*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/480*(3*a*tan(1/2*d*x + 1/2*c)^5 - 3*b*tan(1/2*d*x + 1/2*c)^5 - 35*a*tan(1/2*d*x + 1/2*c)^3 + 25*b*tan(1/2*d*
x + 1/2*c)^3 - 480*(d*x + c)*a + 330*a*tan(1/2*d*x + 1/2*c) - 150*b*tan(1/2*d*x + 1/2*c) - (330*a*tan(1/2*d*x
+ 1/2*c)^4 + 150*b*tan(1/2*d*x + 1/2*c)^4 - 35*a*tan(1/2*d*x + 1/2*c)^2 - 25*b*tan(1/2*d*x + 1/2*c)^2 + 3*a +
3*b)/tan(1/2*d*x + 1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 14.83 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.57 \[ \int \cot ^6(c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {a}{160}-\frac {b}{160}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\left (22\,a+10\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {7\,a}{3}-\frac {5\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{5}+\frac {b}{5}\right )}{32\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {7\,a}{96}-\frac {5\,b}{96}\right )}{d}-a\,x+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {11\,a}{16}-\frac {5\,b}{16}\right )}{d} \]

[In]

int(cot(c + d*x)^6*(a + b/cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)^5*(a/160 - b/160))/d - (cot(c/2 + (d*x)/2)^5*(a/5 + b/5 - tan(c/2 + (d*x)/2)^2*((7*a)/3 +
(5*b)/3) + tan(c/2 + (d*x)/2)^4*(22*a + 10*b)))/(32*d) - (tan(c/2 + (d*x)/2)^3*((7*a)/96 - (5*b)/96))/d - a*x
+ (tan(c/2 + (d*x)/2)*((11*a)/16 - (5*b)/16))/d