3.6 \(\int \cot (c+d x) (a+a \sec (c+d x)) \, dx\)

Optimal. Leaf size=16 \[ \frac {a \log (1-\cos (c+d x))}{d} \]

[Out]

a*ln(1-cos(d*x+c))/d

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3879, 31} \[ \frac {a \log (1-\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[1 - Cos[c + d*x]])/d

Rule 31

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

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

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.81 \[ \frac {2 a \left (\log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(Log[Cos[(c + d*x)/2]] + Log[Tan[(c + d*x)/2]]))/d

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fricas [A]  time = 0.67, size = 16, normalized size = 1.00 \[ \frac {a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(-1/2*cos(d*x + c) + 1/2)/d

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giac [B]  time = 0.20, size = 58, normalized size = 3.62 \[ \frac {a \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)))
/d

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maple [A]  time = 0.44, size = 29, normalized size = 1.81 \[ -\frac {a \ln \left (\sec \left (d x +c \right )\right )}{d}+\frac {a \ln \left (-1+\sec \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d*ln(sec(d*x+c))+a/d*ln(-1+sec(d*x+c))

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maxima [A]  time = 0.52, size = 14, normalized size = 0.88 \[ \frac {a \log \left (\cos \left (d x + c\right ) - 1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(cos(d*x + c) - 1)/d

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mupad [B]  time = 1.24, size = 34, normalized size = 2.12 \[ \frac {a\,\left (2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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