\(\int \frac {\csc ^2(c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\) [104]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {b (4 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{5/2} (a+b)^{3/2} d}-\frac {\cot (c+d x)}{a d \left (a+b \sin ^2(c+d x)\right )}-\frac {\left (2 a b+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 (a+b) d \left (a+b \sin ^2(c+d x)\right )} \]

[Out]

-1/2*b*(4*a+3*b)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(5/2)/(a+b)^(3/2)/d-cot(d*x+c)/a/d/(a+b*sin(d*x+c)^2
)-1/2*(2*a*b+3*b^2)*cos(d*x+c)*sin(d*x+c)/a^2/(a+b)/d/(a+b*sin(d*x+c)^2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3266, 473, 393, 211} \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {b (4 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{5/2} d (a+b)^{3/2}}-\frac {\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a^2 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\cot (c+d x)}{a d \left ((a+b) \tan ^2(c+d x)+a\right )} \]

[In]

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

[Out]

-1/2*(b*(4*a + 3*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(5/2)*(a + b)^(3/2)*d) - Cot[c + d*x]/(a*d*
(a + (a + b)*Tan[c + d*x]^2)) - ((2*a^2 + 4*a*b + 3*b^2)*Tan[c + d*x])/(2*a^2*(a + b)*d*(a + (a + b)*Tan[c + d
*x]^2))

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 393

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

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

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 {\left (1+x^2\right )^2}{x^2 \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {\cot (c+d x)}{a d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {-a-3 b+a x^2}{\left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{a d} \\ & = -\frac {\cot (c+d x)}{a d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {(b (4 a+3 b)) \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{2 a^2 (a+b) d} \\ & = -\frac {b (4 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{5/2} (a+b)^{3/2} d}-\frac {\cot (c+d x)}{a d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.22 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {(2 a+b-b \cos (2 (c+d x))) \left (b (4 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right ) (2 a+b-b \cos (2 (c+d x)))+\sqrt {a} \sqrt {a+b} \left (4 a^2+6 a b+3 b^2-b (2 a+3 b) \cos (2 (c+d x))\right ) \cot (c+d x)\right ) \csc ^4(c+d x)}{8 a^{5/2} (a+b)^{3/2} d \left (b+a \csc ^2(c+d x)\right )^2} \]

[In]

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

[Out]

-1/8*((2*a + b - b*Cos[2*(c + d*x)])*(b*(4*a + 3*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]*(2*a + b - b*Co
s[2*(c + d*x)]) + Sqrt[a]*Sqrt[a + b]*(4*a^2 + 6*a*b + 3*b^2 - b*(2*a + 3*b)*Cos[2*(c + d*x)])*Cot[c + d*x])*C
sc[c + d*x]^4)/(a^(5/2)*(a + b)^(3/2)*d*(b + a*Csc[c + d*x]^2)^2)

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}-\frac {b \left (\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (4 a +3 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{a^{2}}}{d}\)	\(103\)
default	\(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}-\frac {b \left (\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\left (4 a +3 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{a^{2}}}{d}\)	\(103\)
risch	\(-\frac {i \left (-4 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+8 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+14 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 a b -3 b^{2}\right )}{a^{2} \left (a +b \right ) d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{\sqrt {-a^{2}-a b}\, \left (a +b \right ) d a}+\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{\sqrt {-a^{2}-a b}\, \left (a +b \right ) d a}-\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}\)	\(532\)

[In]

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

[Out]

1/d*(-1/a^2/tan(d*x+c)-1/a^2*b*(1/2*b/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+1/2*(4*a+3*b)/(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. 243 vs. \(2 (115) = 230\).

Time = 0.30 (sec) , antiderivative size = 588, normalized size of antiderivative = 4.63 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [-\frac {4 \, {\left (2 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3} - {\left (4 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} - a 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 (2 \, a^{4} + 6 \, a^{3} b + 7 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )}{8 \, {\left ({\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (2 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3} - {\left (4 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} + a 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 (2 \, a^{4} + 6 \, a^{3} b + 7 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[-1/8*(4*(2*a^3*b + 5*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^3 - (4*a^2*b + 7*a*b^2 + 3*b^3 - (4*a*b^2 + 3*b^3)*cos(d
*x + c)^2)*sqrt(-a^2 - a*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
*cos(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*(2*a^4 + 6*a^3*b + 7*a^2
*b^2 + 3*a*b^3)*cos(d*x + c))/(((a^5*b + 2*a^4*b^2 + a^3*b^3)*d*cos(d*x + c)^2 - (a^6 + 3*a^5*b + 3*a^4*b^2 +
a^3*b^3)*d)*sin(d*x + c)), -1/4*(2*(2*a^3*b + 5*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^3 + (4*a^2*b + 7*a*b^2 + 3*b^3
- (4*a*b^2 + 3*b^3)*cos(d*x + c)^2)*sqrt(a^2 + a*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*(2*a^4 + 6*a^3*b + 7*a^2*b^2 + 3*a*b^3)*cos(d*x + c))/(((a^
5*b + 2*a^4*b^2 + a^3*b^3)*d*cos(d*x + c)^2 - (a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d)*sin(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.05 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (4 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} a}} + \frac {{\left (2 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )} \tan \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, a b}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{4} + a^{3} b\right )} \tan \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

-1/2*((4*a*b + 3*b^2)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^3 + a^2*b)*sqrt((a + b)*a)) + ((2*a^2 +
4*a*b + 3*b^2)*tan(d*x + c)^2 + 2*a^2 + 2*a*b)/((a^4 + 2*a^3*b + a^2*b^2)*tan(d*x + c)^3 + (a^4 + a^3*b)*tan(
d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.41 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (4 \, a b + 3 \, b^{2}\right )}}{{\left (a^{3} + a^{2} b\right )} \sqrt {a^{2} + a b}} + \frac {2 \, a^{2} \tan \left (d x + c\right )^{2} + 4 \, a b \tan \left (d x + c\right )^{2} + 3 \, b^{2} \tan \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, a b}{{\left (a \tan \left (d x + c\right )^{3} + b \tan \left (d x + c\right )^{3} + a \tan \left (d x + c\right )\right )} {\left (a^{3} + a^{2} b\right )}}}{2 \, d} \]

[In]

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

[Out]

-1/2*((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))
)*(4*a*b + 3*b^2)/((a^3 + a^2*b)*sqrt(a^2 + a*b)) + (2*a^2*tan(d*x + c)^2 + 4*a*b*tan(d*x + c)^2 + 3*b^2*tan(d
*x + c)^2 + 2*a^2 + 2*a*b)/((a*tan(d*x + c)^3 + b*tan(d*x + c)^3 + a*tan(d*x + c))*(a^3 + a^2*b)))/d

Mupad [B] (verification not implemented)

Time = 13.73 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.04 \[ \int \frac {\csc ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {1}{a}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+4\,a\,b+3\,b^2\right )}{2\,a^2\,\left (a+b\right )}}{d\,\left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+a\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^3+b\,a^2\right )\,\left (4\,a+3\,b\right )}{a^{5/2}\,\sqrt {a+b}\,\left (3\,b^2+4\,a\,b\right )}\right )\,\left (4\,a+3\,b\right )}{2\,a^{5/2}\,d\,{\left (a+b\right )}^{3/2}} \]

[In]

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

[Out]

- (1/a + (tan(c + d*x)^2*(4*a*b + 2*a^2 + 3*b^2))/(2*a^2*(a + b)))/(d*(a*tan(c + d*x) + tan(c + d*x)^3*(a + b)
)) - (b*atan((b*tan(c + d*x)*(a^2*b + a^3)*(4*a + 3*b))/(a^(5/2)*(a + b)^(1/2)*(4*a*b + 3*b^2)))*(4*a + 3*b))/
(2*a^(5/2)*d*(a + b)^(3/2))