\(\int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx\) [261]

Optimal result

Integrand size = 19, antiderivative size = 72 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {(2 a+b) \log (1-\cos (c+d x))}{4 d}-\frac {(2 a-b) \log (1+\cos (c+d x))}{4 d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3967, 3968, 2747, 647, 31} \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {(2 a+b) \log (1-\cos (c+d x))}{4 d}-\frac {(2 a-b) \log (\cos (c+d x)+1)}{4 d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d} \]

[In]

Int[Cot[c + d*x]^3*(a + b*Sec[c + d*x]),x]

[Out]

-1/4*((2*a + b)*Log[1 - Cos[c + d*x]])/d - ((2*a - b)*Log[1 + Cos[c + d*x]])/(4*d) - (Cot[c + d*x]^2*(a + b*Se
c[c + d*x]))/(2*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 ^2(c+d x) (a+b \sec (c+d x))}{2 d}+\frac {1}{2} \int \cot (c+d x) (-2 a-b \sec (c+d x)) \, dx \\ & = -\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}+\frac {1}{2} \int (-b-2 a \cos (c+d x)) \csc (c+d x) \, dx \\ & = -\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}+\frac {a \text {Subst}\left (\int \frac {-b+x}{4 a^2-x^2} \, dx,x,-2 a \cos (c+d x)\right )}{d} \\ & = -\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}+\frac {(2 a-b) \text {Subst}\left (\int \frac {1}{2 a-x} \, dx,x,-2 a \cos (c+d x)\right )}{4 d}+\frac {(2 a+b) \text {Subst}\left (\int \frac {1}{-2 a-x} \, dx,x,-2 a \cos (c+d x)\right )}{4 d} \\ & = -\frac {(2 a+b) \log (1-\cos (c+d x))}{4 d}-\frac {(2 a-b) \log (1+\cos (c+d x))}{4 d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.58 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {a \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{2 d}+\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d} \]

[In]

Integrate[Cot[c + d*x]^3*(a + b*Sec[c + d*x]),x]

[Out]

-1/8*(b*Csc[(c + d*x)/2]^2)/d + (b*Log[Cos[(c + d*x)/2]])/(2*d) - (b*Log[Sin[(c + d*x)/2]])/(2*d) - (a*(Cot[c
+ d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/(2*d) + (b*Sec[(c + d*x)/2]^2)/(8*d)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
default	\(\frac {a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
risch	\(i a x +\frac {2 i a c}{d}+\frac {b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{2 d}\)	\(139\)

[In]

int(cot(d*x+c)^3*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.38 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {2 \, b \cos \left (d x + c\right ) - {\left ({\left (2 \, a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, a + b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - 2 \, a - b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, a}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(cot(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cot ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)**3*(a+b*sec(d*x+c)),x)

[Out]

Integral((a + b*sec(c + d*x))*cot(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {{\left (2 \, a - b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) + {\left (2 \, a + b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (b \cos \left (d x + c\right ) + a\right )}}{\cos \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*((2*a - b)*log(cos(d*x + c) + 1) + (2*a + b)*log(cos(d*x + c) - 1) - 2*(b*cos(d*x + c) + a)/(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. 170 vs. \(2 (66) = 132\).

Time = 0.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.36 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {2 \, {\left (2 \, a + b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 8 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a + b + \frac {4 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(2*(2*a + b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 8*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) + 1)) - (a + b + 4*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) +
1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(c
os(d*x + c) + 1))/d

Mupad [B] (verification not implemented)

Time = 14.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a}{8}-\frac {b}{8}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a}{8}+\frac {b}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a+\frac {b}{2}\right )}{d} \]

[In]

int(cot(c + d*x)^3*(a + b/cos(c + d*x)),x)

[Out]

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