\(\int \frac {\csc ^2(c+d x)}{(a-b \sin ^4(c+d x))^2} \, dx\) [172]

Optimal result

Integrand size = 24, antiderivative size = 236 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\left (6 \sqrt {a}-5 \sqrt {b}\right ) \sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 a^{9/4} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} d}-\frac {\left (6 \sqrt {a}+5 \sqrt {b}\right ) \sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 a^{9/4} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} d}-\frac {\cot (c+d x)}{a^2 d}-\frac {b \tan (c+d x) \left (a+(a+b) \tan ^2(c+d x)\right )}{4 a^2 (a-b) d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \] Output:

1/8*(6*a^(1/2)-5*b^(1/2))*b^(1/2)*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c 
)/a^(1/4))/a^(9/4)/(a^(1/2)-b^(1/2))^(3/2)/d-1/8*(6*a^(1/2)+5*b^(1/2))*b^( 
1/2)*arctan((a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(9/4)/(a^(1/2)+b 
^(1/2))^(3/2)/d-cot(d*x+c)/a^2/d-1/4*b*tan(d*x+c)*(a+(a+b)*tan(d*x+c)^2)/a 
^2/(a-b)/d/(a+2*a*tan(d*x+c)^2+(a-b)*tan(d*x+c)^4)
 

Mathematica [A] (verified)

Time = 4.40 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.16 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {-\frac {\left (6 a \sqrt {b}+5 \sqrt {a} b\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\left (6 a \sqrt {b}-5 \sqrt {a} b\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {-a+\sqrt {a} \sqrt {b}}}-8 \sqrt {a} \cot (c+d x)-\frac {4 \sqrt {a} b (2 a+b-b \cos (2 (c+d x))) \sin (2 (c+d x))}{(a-b) (8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x)))}}{8 a^{5/2} d} \] Input:

Integrate[Csc[c + d*x]^2/(a - b*Sin[c + d*x]^4)^2,x]
 

Output:

(-(((6*a*Sqrt[b] + 5*Sqrt[a]*b)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/ 
Sqrt[a + Sqrt[a]*Sqrt[b]]])/((Sqrt[a] + Sqrt[b])*Sqrt[a + Sqrt[a]*Sqrt[b]] 
)) - ((6*a*Sqrt[b] - 5*Sqrt[a]*b)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x 
])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/((Sqrt[a] - Sqrt[b])*Sqrt[-a + Sqrt[a]*Sqr 
t[b]]) - 8*Sqrt[a]*Cot[c + d*x] - (4*Sqrt[a]*b*(2*a + b - b*Cos[2*(c + d*x 
)])*Sin[2*(c + d*x)])/((a - b)*(8*a - 3*b + 4*b*Cos[2*(c + d*x)] - b*Cos[4 
*(c + d*x)])))/(8*a^(5/2)*d)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3696, 1673, 27, 2195, 2009}

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.

\(\displaystyle \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3696

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

\(\Big \downarrow \) 1673

\(\displaystyle \frac {-\frac {\int -\frac {2 \cot ^2(c+d x) \left (\frac {b \left (4 a^2-b a-b^2\right ) \tan ^4(c+d x)}{a-b}+\frac {a (8 a-7 b) b \tan ^2(c+d x)}{a-b}+4 a b\right )}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{8 a^2 b}-\frac {b \tan (c+d x) \left ((a+b) \tan ^2(c+d x)+a\right )}{4 a^2 (a-b) \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\cot ^2(c+d x) \left (\frac {b \left (4 a^2-b a-b^2\right ) \tan ^4(c+d x)}{a-b}+\frac {a (8 a-7 b) b \tan ^2(c+d x)}{a-b}+4 a b\right )}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{4 a^2 b}-\frac {b \tan (c+d x) \left ((a+b) \tan ^2(c+d x)+a\right )}{4 a^2 (a-b) \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 2195

\(\displaystyle \frac {\frac {\int \left (\frac {\left ((7 a-5 b) \tan ^2(c+d x)+a\right ) b^2}{(a-b) \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}+4 \cot ^2(c+d x) b\right )d\tan (c+d x)}{4 a^2 b}-\frac {b \tan (c+d x) \left ((a+b) \tan ^2(c+d x)+a\right )}{4 a^2 (a-b) \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\frac {b^{3/2} \left (6 \sqrt {a}-5 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right )^{3/2}}-\frac {b^{3/2} \left (6 \sqrt {a}+5 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}+\sqrt {b}\right )^{3/2}}-4 b \cot (c+d x)}{4 a^2 b}-\frac {b \tan (c+d x) \left ((a+b) \tan ^2(c+d x)+a\right )}{4 a^2 (a-b) \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{d}\)

Input:

Int[Csc[c + d*x]^2/(a - b*Sin[c + d*x]^4)^2,x]
 

Output:

((((6*Sqrt[a] - 5*Sqrt[b])*b^(3/2)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + 
 d*x])/a^(1/4)])/(2*a^(1/4)*(Sqrt[a] - Sqrt[b])^(3/2)) - ((6*Sqrt[a] + 5*S 
qrt[b])*b^(3/2)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2 
*a^(1/4)*(Sqrt[a] + Sqrt[b])^(3/2)) - 4*b*Cot[c + d*x])/(4*a^2*b) - (b*Tan 
[c + d*x]*(a + (a + b)*Tan[c + d*x]^2))/(4*a^2*(a - b)*(a + 2*a*Tan[c + d* 
x]^2 + (a - b)*Tan[c + d*x]^4)))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1673

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 
4)^(p_), x_Symbol] :> With[{f = Coeff[PolynomialRemainder[x^m*(d + e*x^2)^q 
, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[x^m*(d + e*x^ 
2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a 
*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), 
 x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[x^m*(a + b*x^2 + c*x^4)^(p + 
 1)*Simp[ExpandToSum[(2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[x^m*(d + 
 e*x^2)^q, a + b*x^2 + c*x^4, x])/x^m + (b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5 
) - a*b*g)/x^m + c*(4*p + 7)*(b*f - 2*a*g)*x^(2 - m), x], x], x], x]] /; Fr 
eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[q, 1] 
&& ILtQ[m/2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2195

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d*x)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; 
FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]
 

rule 3042

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

rule 3696

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 
)/f   Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2) 
^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] & 
& IntegerQ[m/2] && IntegerQ[p]
 

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}+\frac {b \left (\frac {-\frac {\left (a +b \right ) \tan \left (d x +c \right )^{3}}{4 \left (a -b \right )}-\frac {a \tan \left (d x +c \right )}{4 \left (a -b \right )}}{\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b -6 a^{2}+4 a b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b +6 a^{2}-4 a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\)	\(269\)
default	\(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}+\frac {b \left (\frac {-\frac {\left (a +b \right ) \tan \left (d x +c \right )^{3}}{4 \left (a -b \right )}-\frac {a \tan \left (d x +c \right )}{4 \left (a -b \right )}}{\tan \left (d x +c \right )^{4} a -\tan \left (d x +c \right )^{4} b +2 a \tan \left (d x +c \right )^{2}+a}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b -6 a^{2}+4 a b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b +6 a^{2}-4 a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\)	\(269\)
risch	\(\text {Expression too large to display}\)	\(1107\)

Input:

int(csc(d*x+c)^2/(a-b*sin(d*x+c)^4)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/a^2/tan(d*x+c)+b/a^2*((-1/4*(a+b)/(a-b)*tan(d*x+c)^3-1/4*a/(a-b)*t 
an(d*x+c))/(tan(d*x+c)^4*a-tan(d*x+c)^4*b+2*a*tan(d*x+c)^2+a)+1/8*(7*a*(a* 
b)^(1/2)-5*(a*b)^(1/2)*b-6*a^2+4*a*b)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*( 
a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+1/8*( 
7*a*(a*b)^(1/2)-5*(a*b)^(1/2)*b+6*a^2-4*a*b)/(a*b)^(1/2)/(a-b)/(((a*b)^(1/ 
2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))) 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3648 vs. \(2 (186) = 372\).

Time = 0.87 (sec) , antiderivative size = 3648, normalized size of antiderivative = 15.46 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(csc(d*x+c)^2/(a-b*sin(d*x+c)^4)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:

integrate(csc(d*x+c)**2/(a-b*sin(d*x+c)**4)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int { \frac {\csc \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{2}} \,d x } \] Input:

integrate(csc(d*x+c)^2/(a-b*sin(d*x+c)^4)^2,x, algorithm="maxima")
 

Output:

1/2*(2*(48*a^2*b - 5*a*b^2 - 25*b^3)*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) + ( 
(6*a*b^2 - 5*b^3)*sin(8*d*x + 8*c) - 2*(13*a*b^2 - 10*b^3)*sin(6*d*x + 6*c 
) - 2*(32*a^2*b - 47*a*b^2 + 15*b^3)*sin(4*d*x + 4*c) - 2*(7*a*b^2 - 10*b^ 
3)*sin(2*d*x + 2*c))*cos(10*d*x + 10*c) + (2*(48*a^2*b - 5*a*b^2 - 25*b^3) 
*sin(6*d*x + 6*c) + 2*(112*a^2*b - 165*a*b^2 + 50*b^3)*sin(4*d*x + 4*c) + 
5*(8*a*b^2 - 15*b^3)*sin(2*d*x + 2*c))*cos(8*d*x + 8*c) + 2*(2*(256*a^3 - 
432*a^2*b + 210*a*b^2 - 25*b^3)*sin(4*d*x + 4*c) + (112*a^2*b - 165*a*b^2 
+ 50*b^3)*sin(2*d*x + 2*c))*cos(6*d*x + 6*c) + 2*((a^3*b^2 - a^2*b^3)*d*co 
s(10*d*x + 10*c)^2 + 25*(a^3*b^2 - a^2*b^3)*d*cos(8*d*x + 8*c)^2 + 4*(64*a 
^5 - 144*a^4*b + 105*a^3*b^2 - 25*a^2*b^3)*d*cos(6*d*x + 6*c)^2 + 4*(64*a^ 
5 - 144*a^4*b + 105*a^3*b^2 - 25*a^2*b^3)*d*cos(4*d*x + 4*c)^2 + 25*(a^3*b 
^2 - a^2*b^3)*d*cos(2*d*x + 2*c)^2 + (a^3*b^2 - a^2*b^3)*d*sin(10*d*x + 10 
*c)^2 + 25*(a^3*b^2 - a^2*b^3)*d*sin(8*d*x + 8*c)^2 + 4*(64*a^5 - 144*a^4* 
b + 105*a^3*b^2 - 25*a^2*b^3)*d*sin(6*d*x + 6*c)^2 + 4*(64*a^5 - 144*a^4*b 
 + 105*a^3*b^2 - 25*a^2*b^3)*d*sin(4*d*x + 4*c)^2 + 20*(8*a^4*b - 13*a^3*b 
^2 + 5*a^2*b^3)*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 25*(a^3*b^2 - a^2*b^ 
3)*d*sin(2*d*x + 2*c)^2 - 10*(a^3*b^2 - a^2*b^3)*d*cos(2*d*x + 2*c) + (a^3 
*b^2 - a^2*b^3)*d - 2*(5*(a^3*b^2 - a^2*b^3)*d*cos(8*d*x + 8*c) + 2*(8*a^4 
*b - 13*a^3*b^2 + 5*a^2*b^3)*d*cos(6*d*x + 6*c) - 2*(8*a^4*b - 13*a^3*b^2 
+ 5*a^2*b^3)*d*cos(4*d*x + 4*c) - 5*(a^3*b^2 - a^2*b^3)*d*cos(2*d*x + 2...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1545 vs. \(2 (186) = 372\).

Time = 0.75 (sec) , antiderivative size = 1545, normalized size of antiderivative = 6.55 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(csc(d*x+c)^2/(a-b*sin(d*x+c)^4)^2,x, algorithm="giac")
 

Output:

-1/8*(((21*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b - 57*sqrt(a 
^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 + 23*sqrt(a^2 - a*b - sqrt 
(a*b)*(a - b))*sqrt(a*b)*a*b^3 + 5*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqr 
t(a*b)*b^4)*(a^3 - a^2*b)^2*abs(-a + b) - (3*sqrt(a^2 - a*b - sqrt(a*b)*(a 
 - b))*a^7*b - 12*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^6*b^2 + 14*sqrt(a^ 
2 - a*b - sqrt(a*b)*(a - b))*a^5*b^3 - 4*sqrt(a^2 - a*b - sqrt(a*b)*(a - b 
))*a^4*b^4 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^3*b^5)*abs(-a^3 + a^2*b 
)*abs(-a + b) - 2*(9*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^10 - 
42*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^9*b + 66*sqrt(a^2 - a*b 
 - sqrt(a*b)*(a - b))*sqrt(a*b)*a^8*b^2 - 40*sqrt(a^2 - a*b - sqrt(a*b)*(a 
 - b))*sqrt(a*b)*a^7*b^3 + 5*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b) 
*a^6*b^4 + 2*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^5*b^5)*abs(-a 
 + b))*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sqrt((a^4 - a^3 
*b + sqrt((a^4 - a^3*b)^2 - (a^4 - a^3*b)*(a^4 - 2*a^3*b + a^2*b^2)))/(a^4 
 - 2*a^3*b + a^2*b^2))))/((3*a^12 - 21*a^11*b + 59*a^10*b^2 - 85*a^9*b^3 + 
 65*a^8*b^4 - 23*a^7*b^5 + a^6*b^6 + a^5*b^7)*abs(-a^3 + a^2*b)) + ((21*sq 
rt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b - 57*sqrt(a^2 - a*b + sq 
rt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 + 23*sqrt(a^2 - a*b + sqrt(a*b)*(a - b) 
)*sqrt(a*b)*a*b^3 + 5*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*b^4)*( 
a^3 - a^2*b)^2*abs(-a + b) - (3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^7...
 

Mupad [B] (verification not implemented)

Time = 40.45 (sec) , antiderivative size = 4411, normalized size of antiderivative = 18.69 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

int(1/(sin(c + d*x)^2*(a - b*sin(c + d*x)^4)^2),x)
 

Output:

(atan((((-(48*a^2*(a^9*b^3)^(1/2) + 25*b^2*(a^9*b^3)^(1/2) - 36*a^7*b - 15 
*a^5*b^3 + 47*a^6*b^2 - 69*a*b*(a^9*b^3)^(1/2))/(256*(3*a^11*b - a^12 + a^ 
9*b^3 - 3*a^10*b^2)))^(1/2)*(4096*a^10*b^8 - 24576*a^11*b^7 + 61440*a^12*b 
^6 - 81920*a^13*b^5 + 61440*a^14*b^4 - 24576*a^15*b^3 + 4096*a^16*b^2 + ta 
n(c + d*x)*(-(48*a^2*(a^9*b^3)^(1/2) + 25*b^2*(a^9*b^3)^(1/2) - 36*a^7*b - 
 15*a^5*b^3 + 47*a^6*b^2 - 69*a*b*(a^9*b^3)^(1/2))/(256*(3*a^11*b - a^12 + 
 a^9*b^3 - 3*a^10*b^2)))^(1/2)*(65536*a^19*b - 65536*a^12*b^8 + 458752*a^1 
3*b^7 - 1376256*a^14*b^6 + 2293760*a^15*b^5 - 2293760*a^16*b^4 + 1376256*a 
^17*b^3 - 458752*a^18*b^2)) + tan(c + d*x)*(6400*a^7*b^9 - 39424*a^8*b^8 + 
 93952*a^9*b^7 - 100352*a^10*b^6 + 26368*a^11*b^5 + 40448*a^12*b^4 - 36608 
*a^13*b^3 + 9216*a^14*b^2))*(-(48*a^2*(a^9*b^3)^(1/2) + 25*b^2*(a^9*b^3)^( 
1/2) - 36*a^7*b - 15*a^5*b^3 + 47*a^6*b^2 - 69*a*b*(a^9*b^3)^(1/2))/(256*( 
3*a^11*b - a^12 + a^9*b^3 - 3*a^10*b^2)))^(1/2)*1i - ((-(48*a^2*(a^9*b^3)^ 
(1/2) + 25*b^2*(a^9*b^3)^(1/2) - 36*a^7*b - 15*a^5*b^3 + 47*a^6*b^2 - 69*a 
*b*(a^9*b^3)^(1/2))/(256*(3*a^11*b - a^12 + a^9*b^3 - 3*a^10*b^2)))^(1/2)* 
(4096*a^10*b^8 - 24576*a^11*b^7 + 61440*a^12*b^6 - 81920*a^13*b^5 + 61440* 
a^14*b^4 - 24576*a^15*b^3 + 4096*a^16*b^2 - tan(c + d*x)*(-(48*a^2*(a^9*b^ 
3)^(1/2) + 25*b^2*(a^9*b^3)^(1/2) - 36*a^7*b - 15*a^5*b^3 + 47*a^6*b^2 - 6 
9*a*b*(a^9*b^3)^(1/2))/(256*(3*a^11*b - a^12 + a^9*b^3 - 3*a^10*b^2)))^(1/ 
2)*(65536*a^19*b - 65536*a^12*b^8 + 458752*a^13*b^7 - 1376256*a^14*b^6 ...
 

Reduce [F]

\[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {too large to display} \] Input:

int(csc(d*x+c)^2/(a-b*sin(d*x+c)^4)^2,x)
 

Output:

( - 8*cos(c + d*x)*sin(c + d*x)**4*a**3*b + 649*cos(c + d*x)*sin(c + d*x)* 
*4*a**2*b**2 + 32544*cos(c + d*x)*sin(c + d*x)**4*a*b**3 + 44160*cos(c + d 
*x)*sin(c + d*x)**4*b**4 + 8*cos(c + d*x)*sin(c + d*x)**2*a**3*b - 3720*co 
s(c + d*x)*sin(c + d*x)**2*a**2*b**2 - 58688*cos(c + d*x)*sin(c + d*x)**2* 
a*b**3 - 107520*cos(c + d*x)*sin(c + d*x)**2*b**4 + 8*cos(c + d*x)*a**4 - 
495*cos(c + d*x)*a**3*b - 9216*cos(c + d*x)*a**2*b**2 + 25984*cos(c + d*x) 
*a*b**3 + 92160*cos(c + d*x)*b**4 + 49152*int(tan((c + d*x)/2)**4/(8*tan(( 
c + d*x)/2)**16*a**3 - 495*tan((c + d*x)/2)**16*a**2*b + 64*tan((c + d*x)/ 
2)**14*a**3 - 3960*tan((c + d*x)/2)**14*a**2*b + 224*tan((c + d*x)/2)**12* 
a**3 - 14116*tan((c + d*x)/2)**12*a**2*b + 15840*tan((c + d*x)/2)**12*a*b* 
*2 + 448*tan((c + d*x)/2)**10*a**3 - 28744*tan((c + d*x)/2)**10*a**2*b + 6 
3360*tan((c + d*x)/2)**10*a*b**2 + 560*tan((c + d*x)/2)**8*a**3 - 36186*ta 
n((c + d*x)/2)**8*a**2*b + 97088*tan((c + d*x)/2)**8*a*b**2 - 126720*tan(( 
c + d*x)/2)**8*b**3 + 448*tan((c + d*x)/2)**6*a**3 - 28744*tan((c + d*x)/2 
)**6*a**2*b + 63360*tan((c + d*x)/2)**6*a*b**2 + 224*tan((c + d*x)/2)**4*a 
**3 - 14116*tan((c + d*x)/2)**4*a**2*b + 15840*tan((c + d*x)/2)**4*a*b**2 
+ 64*tan((c + d*x)/2)**2*a**3 - 3960*tan((c + d*x)/2)**2*a**2*b + 8*a**3 - 
 495*a**2*b),x)*sin(c + d*x)**5*a**4*b**3*d - 4173824*int(tan((c + d*x)/2) 
**4/(8*tan((c + d*x)/2)**16*a**3 - 495*tan((c + d*x)/2)**16*a**2*b + 64*ta 
n((c + d*x)/2)**14*a**3 - 3960*tan((c + d*x)/2)**14*a**2*b + 224*tan((c...