\(\int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx\) [262]

Optimal result

Integrand size = 19, antiderivative size = 102 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {(8 a+3 b) \log (1-\cos (c+d x))}{16 d}+\frac {(8 a-3 b) \log (1+\cos (c+d x))}{16 d}-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d} \]

[Out]

1/16*(8*a+3*b)*ln(1-cos(d*x+c))/d+1/16*(8*a-3*b)*ln(1+cos(d*x+c))/d-1/4*cot(d*x+c)^4*(a+b*sec(d*x+c))/d+1/8*co
t(d*x+c)^2*(4*a+3*b*sec(d*x+c))/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3967, 3968, 2747, 647, 31} \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {(8 a+3 b) \log (1-\cos (c+d x))}{16 d}+\frac {(8 a-3 b) \log (\cos (c+d x)+1)}{16 d}-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d} \]

[In]

Int[Cot[c + d*x]^5*(a + b*Sec[c + d*x]),x]

[Out]

((8*a + 3*b)*Log[1 - Cos[c + d*x]])/(16*d) + ((8*a - 3*b)*Log[1 + Cos[c + d*x]])/(16*d) - (Cot[c + d*x]^4*(a +
b*Sec[c + d*x]))/(4*d) + (Cot[c + d*x]^2*(4*a + 3*b*Sec[c + d*x]))/(8*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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]

Rule 3968

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))/cot[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[(b + a*Sin[c + d*x])/Cos[
c + d*x], x] /; FreeQ[{a, b, c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {1}{4} \int \cot ^3(c+d x) (-4 a-3 b \sec (c+d x)) \, dx \\ & = -\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}+\frac {1}{8} \int \cot (c+d x) (8 a+3 b \sec (c+d x)) \, dx \\ & = -\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}+\frac {1}{8} \int (3 b+8 a \cos (c+d x)) \csc (c+d x) \, dx \\ & = -\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}-\frac {a \text {Subst}\left (\int \frac {3 b+x}{64 a^2-x^2} \, dx,x,8 a \cos (c+d x)\right )}{d} \\ & = -\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}-\frac {(8 a-3 b) \text {Subst}\left (\int \frac {1}{-8 a-x} \, dx,x,8 a \cos (c+d x)\right )}{16 d}-\frac {(8 a+3 b) \text {Subst}\left (\int \frac {1}{8 a-x} \, dx,x,8 a \cos (c+d x)\right )}{16 d} \\ & = \frac {(8 a+3 b) \log (1-\cos (c+d x))}{16 d}+\frac {(8 a-3 b) \log (1+\cos (c+d x))}{16 d}-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.72 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \cot ^4(c+d x)}{4 d}+\frac {5 b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {b \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {3 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {a \log (\tan (c+d x))}{d}-\frac {5 b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {b \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + b*Sec[c + d*x]),x]

[Out]

(a*Cot[c + d*x]^2)/(2*d) - (a*Cot[c + d*x]^4)/(4*d) + (5*b*Csc[(c + d*x)/2]^2)/(32*d) - (b*Csc[(c + d*x)/2]^4)
/(64*d) - (3*b*Log[Cos[(c + d*x)/2]])/(8*d) + (a*Log[Cos[c + d*x]])/d + (3*b*Log[Sin[(c + d*x)/2]])/(8*d) + (a
*Log[Tan[c + d*x]])/d - (5*b*Sec[(c + d*x)/2]^2)/(32*d) + (b*Sec[(c + d*x)/2]^4)/(64*d)

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )}{d}\)	\(111\)
default	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )}{d}\)	\(111\)
risch	\(-i a x -\frac {2 i a c}{d}-\frac {5 b \,{\mathrm e}^{7 i \left (d x +c \right )}+16 a \,{\mathrm e}^{6 i \left (d x +c \right )}+3 b \,{\mathrm e}^{5 i \left (d x +c \right )}-16 a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{3 i \left (d x +c \right )}+16 a \,{\mathrm e}^{2 i \left (d x +c \right )}+5 b \,{\mathrm e}^{i \left (d x +c \right )}}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{8 d}\)	\(188\)

[In]

int(cot(d*x+c)^5*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.65 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {10 \, b \cos \left (d x + c\right )^{3} + 16 \, a \cos \left (d x + c\right )^{2} - 6 \, b \cos \left (d x + c\right ) - {\left ({\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} + 8 \, a - 3 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{2} + 8 \, a + 3 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

[In]

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

[Out]

-1/16*(10*b*cos(d*x + c)^3 + 16*a*cos(d*x + c)^2 - 6*b*cos(d*x + c) - ((8*a - 3*b)*cos(d*x + c)^4 - 2*(8*a - 3
*b)*cos(d*x + c)^2 + 8*a - 3*b)*log(1/2*cos(d*x + c) + 1/2) - ((8*a + 3*b)*cos(d*x + c)^4 - 2*(8*a + 3*b)*cos(
d*x + c)^2 + 8*a + 3*b)*log(-1/2*cos(d*x + c) + 1/2) - 12*a)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

Sympy [F]

\[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cot ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)**5*(a+b*sec(d*x+c)),x)

[Out]

Integral((a + b*sec(c + d*x))*cot(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\left (8 \, a - 3 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 3 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, b \cos \left (d x + c\right )^{3} + 8 \, a \cos \left (d x + c\right )^{2} - 3 \, b \cos \left (d x + c\right ) - 6 \, a\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

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

[Out]

1/16*((8*a - 3*b)*log(cos(d*x + c) + 1) + (8*a + 3*b)*log(cos(d*x + c) - 1) - 2*(5*b*cos(d*x + c)^3 + 8*a*cos(
d*x + c)^2 - 3*b*cos(d*x + c) - 6*a)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (94) = 188\).

Time = 0.34 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.61 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {4 \, {\left (8 \, a + 3 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 64 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a + b + \frac {12 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {48 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {12 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{64 \, d} \]

[In]

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

[Out]

1/64*(4*(8*a + 3*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 64*a*log(abs(-(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) + 1)) - (a + b + 12*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b*(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 48*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 18*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(co
s(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 - 12*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)
/d

Mupad [B] (verification not implemented)

Time = 14.49 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.25 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a}{16}-\frac {b}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a}{64}-\frac {b}{64}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\left (-3\,a-2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{4}+\frac {b}{4}\right )}{16\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a+\frac {3\,b}{8}\right )}{d} \]

[In]

int(cot(c + d*x)^5*(a + b/cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)^2*((3*a)/16 - b/8))/d - (tan(c/2 + (d*x)/2)^4*(a/64 - b/64))/d - (a*log(tan(c/2 + (d*x)/2)
^2 + 1))/d - (cot(c/2 + (d*x)/2)^4*(a/4 + b/4 - tan(c/2 + (d*x)/2)^2*(3*a + 2*b)))/(16*d) + (log(tan(c/2 + (d*
x)/2))*(a + (3*b)/8))/d