\(\int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [39]

Optimal result

Integrand size = 22, antiderivative size = 29 \[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {\sec (c+d x)}{a d} \]

[Out]

-arctanh(cos(d*x+c))/a/d+sec(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3254, 2702, 327, 213} \[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\sec (c+d x)}{a d}-\frac {\text {arctanh}(\cos (c+d x))}{a d} \]

[In]

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

[Out]

-(ArcTanh[Cos[c + d*x]]/(a*d)) + Sec[c + d*x]/(a*d)

Rule 213

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

Rule 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 3254

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {-\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {\sec (c+d x)}{d}}{a} \]

[In]

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

[Out]

(-(Log[Cos[(c + d*x)/2]]/d) + Log[Sin[(c + d*x)/2]]/d + Sec[c + d*x]/d)/a

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2}+\frac {1}{\cos \left (d x +c \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2}}{d a}\)	\(39\)
default	\(\frac {-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2}+\frac {1}{\cos \left (d x +c \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2}}{d a}\)	\(39\)
norman	\(-\frac {2}{d a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\)	\(42\)
parallelrisch	\(\frac {-2+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(59\)
risch	\(\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}\)	\(71\)

[In]

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

[Out]

1/d/a*(-1/2*ln(1+cos(d*x+c))+1/cos(d*x+c)+1/2*ln(cos(d*x+c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {\cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2}{2 \, a d \cos \left (d x + c\right )} \]

[In]

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

[Out]

-1/2*(cos(d*x + c)*log(1/2*cos(d*x + c) + 1/2) - cos(d*x + c)*log(-1/2*cos(d*x + c) + 1/2) - 2)/(a*d*cos(d*x +
c))

Sympy [F]

\[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=- \frac {\int \frac {\csc {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \]

[In]

integrate(csc(d*x+c)/(a-a*sin(d*x+c)**2),x)

[Out]

-Integral(csc(c + d*x)/(sin(c + d*x)**2 - 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {\frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a} - \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a} - \frac {2}{a \cos \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

-1/2*(log(cos(d*x + c) + 1)/a - log(cos(d*x + c) - 1)/a - 2/(a*cos(d*x + c)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (29) = 58\).

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14 \[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} + \frac {4}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{2 \, d} \]

[In]

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

[Out]

1/2*(log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + 4/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)))/
d

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\csc (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {1}{a\,d\,\cos \left (c+d\,x\right )}-\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{a\,d} \]

[In]

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

[Out]

1/(a*d*cos(c + d*x)) - atanh(cos(c + d*x))/(a*d)