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

Optimal. Leaf size=58 \[ -\frac{\csc ^2(c+d x)}{2 a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

-ArcTanh[Cos[c + d*x]]/(2*a*d) + (Cot[c + d*x]*Csc[c + d*x])/(2*a*d) - Csc[c + d*x]^2/(2*a*d)

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Rubi [A]  time = 0.0970141, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3872, 2706, 2606, 30, 2611, 3770} \[ -\frac{\csc ^2(c+d x)}{2 a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[c + d*x]]/(2*a*d) + (Cot[c + d*x]*Csc[c + d*x])/(2*a*d) - Csc[c + d*x]^2/(2*a*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0958943, size = 67, normalized size = 1.16 \[ -\frac{\sec (c+d x) \left (2 \cos ^2\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+1\right )}{2 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + 2*Cos[(c + d*x)/2]^2*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]))*Sec[c + d*x])/(2*a*d*(1 + Sec[c +
 d*x]))

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Maple [A]  time = 0.051, size = 54, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,da \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{4\,da}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{4\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)/(a+a*sec(d*x+c)),x)

[Out]

-1/2/a/d/(cos(d*x+c)+1)-1/4*ln(cos(d*x+c)+1)/a/d+1/4/a/d*ln(-1+cos(d*x+c))

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Maxima [A]  time = 1.00968, size = 63, normalized size = 1.09 \begin{align*} -\frac{\frac{\log \left (\cos \left (d x + c\right ) + 1\right )}{a} - \frac{\log \left (\cos \left (d x + c\right ) - 1\right )}{a} + \frac{2}{a \cos \left (d x + c\right ) + a}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.66203, size = 181, normalized size = 3.12 \begin{align*} -\frac{{\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2}{4 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\csc{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32623, size = 76, normalized size = 1.31 \begin{align*} \frac{\frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} + \frac{\cos \left (d x + c\right ) - 1}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + (cos(d*x + c) - 1)/(a*(cos(d*x + c) + 1)))/d