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

3.2.4.1 Optimal result
3.2.4.2 Mathematica [A] (verified)
3.2.4.3 Rubi [A] (verified)
3.2.4.4 Maple [A] (verified)
3.2.4.5 Fricas [B] (verification not implemented)
3.2.4.6 Sympy [F]
3.2.4.7 Maxima [A] (verification not implemented)
3.2.4.8 Giac [A] (verification not implemented)
3.2.4.9 Mupad [B] (verification not implemented)

3.2.4.1 Optimal result

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

output
-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)
 
3.2.4.2 Mathematica [A] (verified)

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

input
Integrate[Csc[c + d*x]^2/(a + b*Sin[c + d*x]^2)^2,x]
 
output
-1/8*((2*a + b - b*Cos[2*(c + d*x)])*(b*(4*a + 3*b)*ArcTan[(Sqrt[a + b]*Ta 
n[c + d*x])/Sqrt[a]]*(2*a + b - b*Cos[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])*Csc 
[c + d*x]^4)/(a^(5/2)*(a + b)^(3/2)*d*(b + a*Csc[c + d*x]^2)^2)
 
3.2.4.3 Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3666, 365, 25, 298, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (c+d x)^2 \left (a+b \sin (c+d x)^2\right )^2}dx\)

\(\Big \downarrow \) 3666

\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right )^2}{\left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 365

\(\displaystyle \frac {\frac {\int -\frac {-a \tan ^2(c+d x)+a+3 b}{\left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{a}-\frac {\cot (c+d x)}{a \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \frac {-a \tan ^2(c+d x)+a+3 b}{\left ((a+b) \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{a}-\frac {\cot (c+d x)}{a \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {-\frac {\frac {b (4 a+3 b) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 a (a+b)}+\frac {\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{a}-\frac {\cot (c+d x)}{a \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {-\frac {\frac {b (4 a+3 b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{3/2}}+\frac {\left (2 a^2+4 a b+3 b^2\right ) \tan (c+d x)}{2 a (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}}{a}-\frac {\cot (c+d x)}{a \left ((a+b) \tan ^2(c+d x)+a\right )}}{d}\)

input
Int[Csc[c + d*x]^2/(a + b*Sin[c + d*x]^2)^2,x]
 
output
(-(Cot[c + d*x]/(a*(a + (a + b)*Tan[c + d*x]^2))) - ((b*(4*a + 3*b)*ArcTan 
[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(3/2)) + ((2*a^2 
+ 4*a*b + 3*b^2)*Tan[c + d*x])/(2*a*(a + b)*(a + (a + b)*Tan[c + d*x]^2))) 
/a)/d
 

3.2.4.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 298
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 365
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x 
_Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] 
- Simp[1/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p 
+ 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, 
 b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3666
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
 
3.2.4.4 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 \(-\frac {i \left (-4 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+8 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+14 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 a b -3 b^{2}\right )}{a^{2} \left (a +b \right ) d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{\sqrt {-a^{2}-a b}\, \left (a +b \right ) d a}+\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{\sqrt {-a^{2}-a b}\, \left (a +b \right ) d a}-\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d \,a^{2}}\) \(532\)

input
int(csc(d*x+c)^2/(a+b*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 
output
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))))
 
3.2.4.5 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 ] \]

input
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^2)^2,x, algorithm="fricas")
 
output
[-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))]
 
3.2.4.6 Sympy [F]

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

input
integrate(csc(d*x+c)**2/(a+b*sin(d*x+c)**2)**2,x)
 
output
Integral(csc(c + d*x)**2/(a + b*sin(c + d*x)**2)**2, x)
 
3.2.4.7 Maxima [A] (verification not implemented)

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

input
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^2)^2,x, algorithm="maxima")
 
output
-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
 
3.2.4.8 Giac [A] (verification not implemented)

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

input
integrate(csc(d*x+c)^2/(a+b*sin(d*x+c)^2)^2,x, algorithm="giac")
 
output
-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
 
3.2.4.9 Mupad [B] (verification not implemented)

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

input
int(1/(sin(c + d*x)^2*(a + b*sin(c + d*x)^2)^2),x)
 
output
- (1/a + (tan(c + d*x)^2*(4*a*b + 2*a^2 + 3*b^2))/(2*a^2*(a + b)))/(d*(a*t 
an(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))