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

Optimal. Leaf size=101 \[ \frac {17}{8 a^3 d (\cos (c+d x)+1)}-\frac {7}{8 a^3 d (\cos (c+d x)+1)^2}+\frac {1}{6 a^3 d (\cos (c+d x)+1)^3}+\frac {\log (1-\cos (c+d x))}{16 a^3 d}+\frac {15 \log (\cos (c+d x)+1)}{16 a^3 d} \]

[Out]

1/6/a^3/d/(1+cos(d*x+c))^3-7/8/a^3/d/(1+cos(d*x+c))^2+17/8/a^3/d/(1+cos(d*x+c))+1/16*ln(1-cos(d*x+c))/a^3/d+15
/16*ln(1+cos(d*x+c))/a^3/d

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Rubi [A]  time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3879, 88} \[ \frac {17}{8 a^3 d (\cos (c+d x)+1)}-\frac {7}{8 a^3 d (\cos (c+d x)+1)^2}+\frac {1}{6 a^3 d (\cos (c+d x)+1)^3}+\frac {\log (1-\cos (c+d x))}{16 a^3 d}+\frac {15 \log (\cos (c+d x)+1)}{16 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(6*a^3*d*(1 + Cos[c + d*x])^3) - 7/(8*a^3*d*(1 + Cos[c + d*x])^2) + 17/(8*a^3*d*(1 + Cos[c + d*x])) + Log[1
- Cos[c + d*x]]/(16*a^3*d) + (15*Log[1 + Cos[c + d*x]])/(16*a^3*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 (c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {a^2 \operatorname {Subst}\left (\int \frac {x^4}{(a-a x) (a+a x)^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^2 \operatorname {Subst}\left (\int \left (-\frac {1}{16 a^5 (-1+x)}+\frac {1}{2 a^5 (1+x)^4}-\frac {7}{4 a^5 (1+x)^3}+\frac {17}{8 a^5 (1+x)^2}-\frac {15}{16 a^5 (1+x)}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {1}{6 a^3 d (1+\cos (c+d x))^3}-\frac {7}{8 a^3 d (1+\cos (c+d x))^2}+\frac {17}{8 a^3 d (1+\cos (c+d x))}+\frac {\log (1-\cos (c+d x))}{16 a^3 d}+\frac {15 \log (1+\cos (c+d x))}{16 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 97, normalized size = 0.96 \[ \frac {\sec ^3(c+d x) \left (102 \cos ^4\left (\frac {1}{2} (c+d x)\right )-21 \cos ^2\left (\frac {1}{2} (c+d x)\right )+12 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+2\right )}{12 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 - 21*Cos[(c + d*x)/2]^2 + 102*Cos[(c + d*x)/2]^4 + 12*Cos[(c + d*x)/2]^6*(15*Log[Cos[(c + d*x)/2]] + Log[S
in[(c + d*x)/2]]))*Sec[c + d*x]^3)/(12*a^3*d*(1 + Sec[c + d*x])^3)

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fricas [A]  time = 0.66, size = 151, normalized size = 1.50 \[ \frac {102 \, \cos \left (d x + c\right )^{2} + 45 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 162 \, \cos \left (d x + c\right ) + 68}{48 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(102*cos(d*x + c)^2 + 45*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) +
1/2) + 3*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 162*cos(d*x +
c) + 68)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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giac [A]  time = 0.34, size = 143, normalized size = 1.42 \[ \frac {\frac {6 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} - \frac {96 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} - \frac {\frac {66 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {15 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{6} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{9}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(6*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 - 96*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) +
1) + 1))/a^3 - (66*a^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 15*a^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)
^2 + 2*a^6*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/a^9)/d

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maple [A]  time = 0.82, size = 90, normalized size = 0.89 \[ \frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{16 d \,a^{3}}+\frac {1}{6 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {7}{8 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {17}{8 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\cos \left (d x +c \right )\right )}{16 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d/a^3*ln(-1+cos(d*x+c))+1/6/d/a^3/(1+cos(d*x+c))^3-7/8/d/a^3/(1+cos(d*x+c))^2+17/8/d/a^3/(1+cos(d*x+c))+1
5/16*ln(1+cos(d*x+c))/a^3/d

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maxima [A]  time = 0.37, size = 98, normalized size = 0.97 \[ \frac {\frac {2 \, {\left (51 \, \cos \left (d x + c\right )^{2} + 81 \, \cos \left (d x + c\right ) + 34\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}} + \frac {45 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(51*cos(d*x + c)^2 + 81*cos(d*x + c) + 34)/(a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x
+ c) + a^3) + 45*log(cos(d*x + c) + 1)/a^3 + 3*log(cos(d*x + c) - 1)/a^3)/d

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mupad [B]  time = 1.24, size = 75, normalized size = 0.74 \[ \frac {\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c/2 + (d*x)/2))/8 - log(tan(c/2 + (d*x)/2)^2 + 1) + (11*tan(c/2 + (d*x)/2)^2)/16 - (5*tan(c/2 + (d*x)
/2)^4)/32 + tan(c/2 + (d*x)/2)^6/48)/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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