\(\int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx\) [235]

Optimal result

Integrand size = 19, antiderivative size = 67 \[ \int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2826, 3855, 2739, 632, 210} \[ \int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 b \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}-\frac {\text {arctanh}(\cos (c+d x))}{a d} \]

[In]

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

[Out]

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

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*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 2826

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc (c+d x) \, dx}{a}-\frac {b \int \frac {1}{a+b \sin (c+d x)} \, dx}{a} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{a d}-\frac {(2 b) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \\ & = -\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {(4 b) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \\ & = -\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \]

[In]

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

[Out]

((-2*b*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - Log[Cos[(c + d*x)/2]] + Log[Sin[(c
+ d*x)/2]])/(a*d)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \sqrt {a^{2}-b^{2}}}}{d}\)	\(67\)
default	\(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \sqrt {a^{2}-b^{2}}}}{d}\)	\(67\)
risch	\(\frac {i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}\)	\(183\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 297, normalized size of antiderivative = 4.43 \[ \int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} b \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}, \frac {2 \, \sqrt {a^{2} - b^{2}} b \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(csc(d*x+c)/(a+b*sin(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*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24 \[ \int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{\sqrt {a^{2} - b^{2}} a} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.56 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.58 \[ \int \frac {\csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}+\frac {2\,b\,\mathrm {atanh}\left (\frac {\sqrt {b^2-a^2}\,\left (-1{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+2{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+4{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2\right )}{1{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3+3{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b-2{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^2-4{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )}{a\,d\,\sqrt {b^2-a^2}} \]

[In]

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

[Out]

log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) + (2*b*atanh(((b^2 - a^2)^(1/2)*(b^2*sin(c/2 + (d*x)/2)*4i -
a^2*sin(c/2 + (d*x)/2)*1i + a*b*cos(c/2 + (d*x)/2)*2i))/(a^3*cos(c/2 + (d*x)/2)*1i - b^3*sin(c/2 + (d*x)/2)*4i
- a*b^2*cos(c/2 + (d*x)/2)*2i + a^2*b*sin(c/2 + (d*x)/2)*3i)))/(a*d*(b^2 - a^2)^(1/2))