\(\int \frac {\tan (x)}{a+b \csc (x)} \, dx\) [14]

Optimal result

Integrand size = 11, antiderivative size = 66 \[ \int \frac {\tan (x)}{a+b \csc (x)} \, dx=-\frac {\log (1-\csc (x))}{2 (a+b)}-\frac {\log (1+\csc (x))}{2 (a-b)}+\frac {b^2 \log (a+b \csc (x))}{a \left (a^2-b^2\right )}-\frac {\log (\sin (x))}{a} \]

[Out]

-1/2*ln(1-csc(x))/(a+b)-1/2*ln(1+csc(x))/(a-b)+b^2*ln(a+b*csc(x))/a/(a^2-b^2)-ln(sin(x))/a

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3970, 908} \[ \int \frac {\tan (x)}{a+b \csc (x)} \, dx=\frac {b^2 \log (a+b \csc (x))}{a \left (a^2-b^2\right )}-\frac {\log (1-\csc (x))}{2 (a+b)}-\frac {\log (\csc (x)+1)}{2 (a-b)}-\frac {\log (\sin (x))}{a} \]

[In]

Int[Tan[x]/(a + b*Csc[x]),x]

[Out]

-1/2*Log[1 - Csc[x]]/(a + b) - Log[1 + Csc[x]]/(2*(a - b)) + (b^2*Log[a + b*Csc[x]])/(a*(a^2 - b^2)) - Log[Sin
[x]]/a

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3970

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[(b^2 - x^2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = b^2 \text {Subst}\left (\int \frac {1}{x (a+x) \left (b^2-x^2\right )} \, dx,x,b \csc (x)\right ) \\ & = b^2 \text {Subst}\left (\int \left (\frac {1}{2 b^2 (a+b) (b-x)}+\frac {1}{a b^2 x}+\frac {1}{a (a-b) (a+b) (a+x)}-\frac {1}{2 (a-b) b^2 (b+x)}\right ) \, dx,x,b \csc (x)\right ) \\ & = -\frac {\log (1-\csc (x))}{2 (a+b)}-\frac {\log (1+\csc (x))}{2 (a-b)}+\frac {b^2 \log (a+b \csc (x))}{a \left (a^2-b^2\right )}-\frac {\log (\sin (x))}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (x)}{a+b \csc (x)} \, dx=-\frac {a (a-b) \log (1-\sin (x))+a (a+b) \log (1+\sin (x))-2 b^2 \log (b+a \sin (x))}{2 a (a-b) (a+b)} \]

[In]

Integrate[Tan[x]/(a + b*Csc[x]),x]

[Out]

-1/2*(a*(a - b)*Log[1 - Sin[x]] + a*(a + b)*Log[1 + Sin[x]] - 2*b^2*Log[b + a*Sin[x]])/(a*(a - b)*(a + b))

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {\ln \left (1+\sin \left (x \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (x \right )-1\right )}{2 a +2 b}+\frac {b^{2} \ln \left (a \sin \left (x \right )+b \right )}{\left (a +b \right ) \left (a -b \right ) a}\)	\(60\)
risch	\(-\frac {i x}{a}+\frac {i x}{a -b}+\frac {i x}{a +b}-\frac {2 i x \,b^{2}}{a \left (a^{2}-b^{2}\right )}-\frac {\ln \left (i+{\mathrm e}^{i x}\right )}{a -b}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{a +b}+\frac {b^{2} \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a \left (a^{2}-b^{2}\right )}\)	\(122\)

[In]

int(tan(x)/(a+b*csc(x)),x,method=_RETURNVERBOSE)

[Out]

-1/(2*a-2*b)*ln(1+sin(x))-1/(2*a+2*b)*ln(sin(x)-1)+b^2/(a+b)/(a-b)/a*ln(a*sin(x)+b)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (x)}{a+b \csc (x)} \, dx=\frac {2 \, b^{2} \log \left (a \sin \left (x\right ) + b\right ) - {\left (a^{2} + a b\right )} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a^{2} - a b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{3} - a b^{2}\right )}} \]

[In]

integrate(tan(x)/(a+b*csc(x)),x, algorithm="fricas")

[Out]

1/2*(2*b^2*log(a*sin(x) + b) - (a^2 + a*b)*log(sin(x) + 1) - (a^2 - a*b)*log(-sin(x) + 1))/(a^3 - a*b^2)

Sympy [F]

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

[In]

integrate(tan(x)/(a+b*csc(x)),x)

[Out]

Integral(tan(x)/(a + b*csc(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int \frac {\tan (x)}{a+b \csc (x)} \, dx=\frac {b^{2} \log \left (a \sin \left (x\right ) + b\right )}{a^{3} - a b^{2}} - \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a - b\right )}} - \frac {\log \left (\sin \left (x\right ) - 1\right )}{2 \, {\left (a + b\right )}} \]

[In]

integrate(tan(x)/(a+b*csc(x)),x, algorithm="maxima")

[Out]

b^2*log(a*sin(x) + b)/(a^3 - a*b^2) - 1/2*log(sin(x) + 1)/(a - b) - 1/2*log(sin(x) - 1)/(a + b)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int \frac {\tan (x)}{a+b \csc (x)} \, dx=\frac {b^{2} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a^{3} - a b^{2}} - \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a - b\right )}} - \frac {\log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}} \]

[In]

integrate(tan(x)/(a+b*csc(x)),x, algorithm="giac")

[Out]

b^2*log(abs(a*sin(x) + b))/(a^3 - a*b^2) - 1/2*log(sin(x) + 1)/(a - b) - 1/2*log(-sin(x) + 1)/(a + b)

Mupad [B] (verification not implemented)

Time = 19.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.21 \[ \int \frac {\tan (x)}{a+b \csc (x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{a-b}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}{a+b}+\frac {b^2\,\ln \left (b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+b\right )}{a\,\left (a^2-b^2\right )} \]

[In]

int(tan(x)/(a + b/sin(x)),x)

[Out]

log(tan(x/2)^2 + 1)/a - log(tan(x/2) + 1)/(a - b) - log(tan(x/2) - 1)/(a + b) + (b^2*log(b + 2*a*tan(x/2) + b*
tan(x/2)^2))/(a*(a^2 - b^2))