3.80 \(\int \frac{\csc (c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=60 \[ -\frac{3}{4 d \left (a^2 \cos (c+d x)+a^2\right )}-\frac{\tanh ^{-1}(\cos (c+d x))}{4 a^2 d}+\frac{1}{4 d (a \cos (c+d x)+a)^2} \]

[Out]

-ArcTanh[Cos[c + d*x]]/(4*a^2*d) + 1/(4*d*(a + a*Cos[c + d*x])^2) - 3/(4*d*(a^2 + a^2*Cos[c + d*x]))

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Rubi [A]  time = 0.126629, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3872, 2836, 12, 88, 206} \[ -\frac{3}{4 d \left (a^2 \cos (c+d x)+a^2\right )}-\frac{\tanh ^{-1}(\cos (c+d x))}{4 a^2 d}+\frac{1}{4 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]/(a + a*Sec[c + d*x])^2,x]

[Out]

-ArcTanh[Cos[c + d*x]]/(4*a^2*d) + 1/(4*d*(a + a*Cos[c + d*x])^2) - 3/(4*d*(a^2 + a^2*Cos[c + d*x]))

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 2836

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.174836, size = 83, normalized size = 1.38 \[ -\frac{\sec ^2(c+d x) \left (6 \cos ^2\left (\frac{1}{2} (c+d x)\right )+4 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )-1\right )}{4 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]/(a + a*Sec[c + d*x])^2,x]

[Out]

-((-1 + 6*Cos[(c + d*x)/2]^2 + 4*Cos[(c + d*x)/2]^4*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]))*Sec[c + d
*x]^2)/(4*a^2*d*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 0.06, size = 72, normalized size = 1.2 \begin{align*}{\frac{1}{4\,d{a}^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{3}{4\,d{a}^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{8\,d{a}^{2}}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{8\,d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)/(a+a*sec(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.00546, size = 100, normalized size = 1.67 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \cos \left (d x + c\right ) + 2\right )}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}} + \frac{\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} - \frac{\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, 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))^2,x, algorithm="maxima")

[Out]

-1/8*(2*(3*cos(d*x + c) + 2)/(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^2) + log(cos(d*x + c) + 1)/a^2 - log
(cos(d*x + c) - 1)/a^2)/d

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Fricas [A]  time = 1.70344, size = 294, normalized size = 4.9 \begin{align*} -\frac{{\left (\cos \left (d x + c\right )^{2} + 2 \, \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 )^{2} + 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 ) + 4}{8 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/8*((cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) - (cos(d*x + c)^2 + 2*cos(d*x + c) + 1
)*log(-1/2*cos(d*x + c) + 1/2) + 6*cos(d*x + c) + 4)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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 ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)/(a+a*sec(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.30401, size = 117, normalized size = 1.95 \begin{align*} \frac{\frac{2 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac{\frac{4 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{4}}}{16 \, 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))^2,x, algorithm="giac")

[Out]

1/16*(2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + (4*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) +
 a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/a^4)/d