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

Optimal. Leaf size=84 \[ \frac{\csc (c+d x) (b-a \cos (c+d x))}{d \left (a^2-b^2\right )}-\frac{2 a b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

(-2*a*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*d) + ((b - a*Cos[c +
d*x])*Csc[c + d*x])/((a^2 - b^2)*d)

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Rubi [A]  time = 0.148712, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2866, 12, 2659, 208} \[ \frac{\csc (c+d x) (b-a \cos (c+d x))}{d \left (a^2-b^2\right )}-\frac{2 a b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)*d) + ((b - a*Cos[c +
d*x])*Csc[c + d*x])/((a^2 - b^2)*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 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 12

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.195303, size = 118, normalized size = 1.4 \[ \frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{a^2-b^2} (b-a \cos (c+d x))+2 a b \sin (c+d x) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )\right )}{2 d (a-b) (a+b) \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Sqrt[a^2 - b^2]*(b - a*Cos[c + d*x]) + 2*a*b*ArcTanh[((-a + b)*Tan[(c + d*
x)/2])/Sqrt[a^2 - b^2]]*Sin[c + d*x]))/(2*(a - b)*(a + b)*Sqrt[a^2 - b^2]*d)

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Maple [A]  time = 0.058, size = 96, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{1}{2\,a-2\,b}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{ab}{ \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{2\,a+2\,b} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86099, size = 680, normalized size = 8.1 \begin{align*} \left [-\frac{\sqrt{a^{2} - b^{2}} a b \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \sin \left (d x + c\right )}, -\frac{\sqrt{-a^{2} + b^{2}} a b \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - a^{2} b + b^{3} +{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(a^2 - b^2)*a*b*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d
*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 2*a^2
*b + 2*b^3 + 2*(a^3 - a*b^2)*cos(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*sin(d*x + c)), -(sqrt(-a^2 + b^2)*a*b*ar
ctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - a^2*b + b^3 + (a^3 - a*
b^2)*cos(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.37457, size = 174, normalized size = 2.07 \begin{align*} -\frac{\frac{4 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )} a b}{{\left (a^{2} - b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a - b} + \frac{1}{{\left (a + b\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(4*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x +
1/2*c))/sqrt(-a^2 + b^2)))*a*b/((a^2 - b^2)*sqrt(-a^2 + b^2)) - tan(1/2*d*x + 1/2*c)/(a - b) + 1/((a + b)*tan(
1/2*d*x + 1/2*c)))/d