\(\int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx\) [225]

Optimal result

Integrand size = 24, antiderivative size = 34 \[ \int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

-1/2*arctanh(sin(d*x+c))/d+1/2*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {294, 212} \[ \int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx=\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d} \]

[In]

Int[Tan[c + d*x]/(Csc[c + d*x] - Sin[c + d*x]),x]

[Out]

-1/2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 212

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

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

Integrate[Tan[c + d*x]/(Csc[c + d*x] - Sin[c + d*x]),x]

[Out]

-1/2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53

method	result	size
derivativedivides	\(\frac {-\frac {1}{4 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}-\frac {1}{4 \left (\sin \left (d x +c \right )+1\right )}-\frac {\ln \left (\sin \left (d x +c \right )+1\right )}{4}}{d}\)	\(52\)
default	\(\frac {-\frac {1}{4 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}-\frac {1}{4 \left (\sin \left (d x +c \right )+1\right )}-\frac {\ln \left (\sin \left (d x +c \right )+1\right )}{4}}{d}\)	\(52\)
risch	\(-\frac {i \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(78\)

[In]

int(tan(d*x+c)/(csc(d*x+c)-sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/4/(sin(d*x+c)-1)+1/4*ln(sin(d*x+c)-1)-1/4/(sin(d*x+c)+1)-1/4*ln(sin(d*x+c)+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (30) = 60\).

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx=-\frac {\cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-1/4*(cos(d*x + c)^2*log(sin(d*x + c) + 1) - cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*sin(d*x + c))/(d*cos(d*
x + c)^2)

Sympy [F]

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

[In]

integrate(tan(d*x+c)/(csc(d*x+c)-sin(d*x+c)),x)

[Out]

Integral(tan(c + d*x)/(-sin(c + d*x) + csc(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx=-\frac {\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )}{4 \, d} \]

[In]

integrate(tan(d*x+c)/(csc(d*x+c)-sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx=-\frac {\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, d} \]

[In]

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

[Out]

-1/4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(abs(sin(d*x + c) + 1)) - log(abs(sin(d*x + c) - 1)))/d

Mupad [B] (verification not implemented)

Time = 23.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.03 \[ \int \frac {\tan (c+d x)}{\csc (c+d x)-\sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

int(-tan(c + d*x)/(sin(c + d*x) - 1/sin(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^3)/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) - atanh(ta
n(c/2 + (d*x)/2))/d