3.62 \(\int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=103 \[ -\frac {1}{8 a d (1-\cos (c+d x))}-\frac {3}{4 a d (\cos (c+d x)+1)}+\frac {1}{8 a d (\cos (c+d x)+1)^2}-\frac {5 \log (1-\cos (c+d x))}{16 a d}-\frac {11 \log (\cos (c+d x)+1)}{16 a d} \]

[Out]

-1/8/a/d/(1-cos(d*x+c))+1/8/a/d/(1+cos(d*x+c))^2-3/4/a/d/(1+cos(d*x+c))-5/16*ln(1-cos(d*x+c))/a/d-11/16*ln(1+c
os(d*x+c))/a/d

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3879, 88} \[ -\frac {1}{8 a d (1-\cos (c+d x))}-\frac {3}{4 a d (\cos (c+d x)+1)}+\frac {1}{8 a d (\cos (c+d x)+1)^2}-\frac {5 \log (1-\cos (c+d x))}{16 a d}-\frac {11 \log (\cos (c+d x)+1)}{16 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(8*a*d*(1 - Cos[c + d*x])) + 1/(8*a*d*(1 + Cos[c + d*x])^2) - 3/(4*a*d*(1 + Cos[c + d*x])) - (5*Log[1 - Cos
[c + d*x]])/(16*a*d) - (11*Log[1 + Cos[c + d*x]])/(16*a*d)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3879

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[((a - b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x^(m + n), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx &=-\frac {a^4 \operatorname {Subst}\left (\int \frac {x^4}{(a-a x)^2 (a+a x)^3} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^4 \operatorname {Subst}\left (\int \left (\frac {1}{8 a^5 (-1+x)^2}+\frac {5}{16 a^5 (-1+x)}+\frac {1}{4 a^5 (1+x)^3}-\frac {3}{4 a^5 (1+x)^2}+\frac {11}{16 a^5 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {1}{8 a d (1-\cos (c+d x))}+\frac {1}{8 a d (1+\cos (c+d x))^2}-\frac {3}{4 a d (1+\cos (c+d x))}-\frac {5 \log (1-\cos (c+d x))}{16 a d}-\frac {11 \log (1+\cos (c+d x))}{16 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.62, size = 107, normalized size = 1.04 \[ -\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (2 \csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^4\left (\frac {1}{2} (c+d x)\right )+12 \sec ^2\left (\frac {1}{2} (c+d x)\right )+20 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+44 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{16 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(Cos[(c + d*x)/2]^2*(2*Csc[(c + d*x)/2]^2 + 44*Log[Cos[(c + d*x)/2]] + 20*Log[Sin[(c + d*x)/2]] + 12*Sec
[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^4)*Sec[c + d*x])/(a*d*(1 + Sec[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 139, normalized size = 1.35 \[ -\frac {10 \, \cos \left (d x + c\right )^{2} + 11 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 6 \, \cos \left (d x + c\right ) - 12}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(10*cos(d*x + c)^2 + 11*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2)
+ 5*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 6*cos(d*x + c) - 12)/
(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 157, normalized size = 1.52 \[ \frac {\frac {2 \, {\left (\frac {5 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {10 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} + \frac {32 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a} + \frac {\frac {10 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(2*(5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)*(cos(d*x + c) + 1)/(a*(cos(d*x + c) - 1)) - 10*log(abs(-
cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + 32*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a + (10*a
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/a^2)/d

________________________________________________________________________________________

maple [A]  time = 0.80, size = 90, normalized size = 0.87 \[ \frac {1}{8 a d \left (-1+\cos \left (d x +c \right )\right )}-\frac {5 \ln \left (-1+\cos \left (d x +c \right )\right )}{16 d a}+\frac {1}{8 a d \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {3}{4 a d \left (1+\cos \left (d x +c \right )\right )}-\frac {11 \ln \left (1+\cos \left (d x +c \right )\right )}{16 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/a/d/(-1+cos(d*x+c))-5/16/d/a*ln(-1+cos(d*x+c))+1/8/a/d/(1+cos(d*x+c))^2-3/4/a/d/(1+cos(d*x+c))-11/16*ln(1+
cos(d*x+c))/d/a

________________________________________________________________________________________

maxima [A]  time = 0.69, size = 91, normalized size = 0.88 \[ -\frac {\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 6\right )}}{a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a} + \frac {11 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {5 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.34, size = 76, normalized size = 0.74 \[ -\frac {\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32}}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((5*log(tan(c/2 + (d*x)/2)))/8 - log(tan(c/2 + (d*x)/2)^2 + 1) + cot(c/2 + (d*x)/2)^2/16 + (5*tan(c/2 + (d*x)
/2)^2)/16 - tan(c/2 + (d*x)/2)^4/32)/(a*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(c + d*x)**3/(sec(c + d*x) + 1), x)/a

________________________________________________________________________________________