\(\int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx\) [206]

Optimal result

Integrand size = 21, antiderivative size = 84 \[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {2 a b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}+\frac {(b-a \cos (c+d x)) \csc (c+d x)}{\left (a^2-b^2\right ) d} \]

[Out]

-2*a*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b)^(3/2)/d+(b-a*cos(d*x+c))*csc(d*x+
c)/(a^2-b^2)/d

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2945, 12, 2738, 214} \[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\csc (c+d x) (b-a \cos (c+d x))}{d \left (a^2-b^2\right )}-\frac {2 a b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}} \]

[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 12

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

Rule 214

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

Rule 2738

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 2945

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

Rubi steps \begin{align*} \text {integral}& = -\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) \text {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 \text {arctanh}\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] (verified)

Time = 0.48 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.40 \[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {a^2-b^2} (b-a \cos (c+d x))+2 a b \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) \sin (c+d x)\right )}{2 (a-b) (a+b) \sqrt {a^2-b^2} d} \]

[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)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a -2 b}-\frac {2 a b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{2 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(96\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a -2 b}-\frac {2 a b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{2 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(96\)
risch	\(-\frac {2 i \left (-b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{d \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(209\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.57 \[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\left [-\frac {\sqrt {a^{2} - b^{2}} a b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \sin \left (d x + c\right )}, -\frac {\sqrt {-a^{2} + b^{2}} a b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - a^{2} b + b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \sin \left (d x + c\right )}\right ] \]

[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))]

Sympy [F]

\[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.54 \[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a b}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a - b} + \frac {1}{{\left (a + b\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 13.81 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.30 \[ \int \frac {\csc ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (2\,a-2\,b\right )}-\frac {a-b}{d\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )\,\left (2\,a-2\,b\right )}-\frac {2\,a\,b\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{{\left (a+b\right )}^{3/2}\,\sqrt {a-b}}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

[In]

int(1/(sin(c + d*x)^2*(a + b/cos(c + d*x))),x)

[Out]

tan(c/2 + (d*x)/2)/(d*(2*a - 2*b)) - (a - b)/(d*tan(c/2 + (d*x)/2)*(a + b)*(2*a - 2*b)) - (2*a*b*atanh((tan(c/
2 + (d*x)/2)*(a^2 - b^2))/((a + b)^(3/2)*(a - b)^(1/2))))/(d*(a + b)^(3/2)*(a - b)^(3/2))