\(\int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [90]

Optimal result

Integrand size = 23, antiderivative size = 53 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {b \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2} \sqrt {a+b} d}-\frac {\cot (c+d x)}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3266, 464, 211} \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {b \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2} d \sqrt {a+b}}-\frac {\cot (c+d x)}{a d} \]

[In]

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

[Out]

-((b*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(3/2)*Sqrt[a + b]*d)) - Cot[c + d*x]/(a*d)

Rule 211

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

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 3266

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p +
1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1+x^2}{x^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {\cot (c+d x)}{a d}-\frac {b \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{a d} \\ & = -\frac {b \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2} \sqrt {a+b} d}-\frac {\cot (c+d x)}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {-\frac {b \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}-\sqrt {a} \cot (c+d x)}{a^{3/2} d} \]

[In]

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

[Out]

(-((b*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a + b]) - Sqrt[a]*Cot[c + d*x])/(a^(3/2)*d)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {-\frac {1}{a \tan \left (d x +c \right )}-\frac {b \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a \sqrt {a \left (a +b \right )}}}{d}\)	\(50\)
default	\(\frac {-\frac {1}{a \tan \left (d x +c \right )}-\frac {b \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a \sqrt {a \left (a +b \right )}}}{d}\)	\(50\)
risch	\(-\frac {2 i}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d a}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d a}\)	\(204\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (45) = 90\).

Time = 0.29 (sec) , antiderivative size = 313, normalized size of antiderivative = 5.91 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [-\frac {\sqrt {-a^{2} - a b} b \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) + 4 \, {\left (a^{2} + a b\right )} \cos \left (d x + c\right )}{4 \, {\left (a^{3} + a^{2} b\right )} d \sin \left (d x + c\right )}, \frac {\sqrt {a^{2} + a b} b \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{2} + a b\right )} \cos \left (d x + c\right )}{2 \, {\left (a^{3} + a^{2} b\right )} d \sin \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[-1/4*(sqrt(-a^2 - a*b)*b*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 -
4*((2*a + b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(-a^2 - a*b)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*c
os(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2))*sin(d*x + c) + 4*(a^2 + a*b)*cos(d*x + c))/
((a^3 + a^2*b)*d*sin(d*x + c)), 1/2*(sqrt(a^2 + a*b)*b*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2
+ a*b)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c) - 2*(a^2 + a*b)*cos(d*x + c))/((a^3 + a^2*b)*d*sin(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {b \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a} + \frac {1}{a \tan \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.57 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} b}{\sqrt {a^{2} + a b} a} + \frac {1}{a \tan \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.85 \[ \int \frac {\csc ^2(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\mathrm {cot}\left (c+d\,x\right )}{a\,d}-\frac {b\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )}{a^{3/2}\,d\,\sqrt {a+b}} \]

[In]

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

[Out]

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