Integrand size = 24, antiderivative size = 236 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {\left (6 \sqrt {a}-5 \sqrt {b}\right ) \sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 a^{9/4} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} d}-\frac {\left (6 \sqrt {a}+5 \sqrt {b}\right ) \sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 a^{9/4} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} d}-\frac {\cot (c+d x)}{a^2 d}-\frac {b \tan (c+d x) \left (a+(a+b) \tan ^2(c+d x)\right )}{4 a^2 (a-b) d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )} \]
-cot(d*x+c)/a^2/d+1/8*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*( 6*a^(1/2)-5*b^(1/2))*b^(1/2)/a^(9/4)/d/(a^(1/2)-b^(1/2))^(3/2)-1/8*arctan( (a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*b^(1/2)*(6*a^(1/2)+5*b^(1/2))/ a^(9/4)/d/(a^(1/2)+b^(1/2))^(3/2)-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)
Time = 2.32 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.16 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\frac {-\frac {\left (6 a \sqrt {b}+5 \sqrt {a} b\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\left (6 a \sqrt {b}-5 \sqrt {a} b\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {-a+\sqrt {a} \sqrt {b}}}-8 \sqrt {a} \cot (c+d x)-\frac {4 \sqrt {a} b (2 a+b-b \cos (2 (c+d x))) \sin (2 (c+d x))}{(a-b) (8 a-3 b+4 b \cos (2 (c+d x))-b \cos (4 (c+d x)))}}{8 a^{5/2} d} \]
(-(((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)
Time = 0.71 (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 |
\(\displaystyle \frac {\frac {\frac {b^{3/2} \left (6 \sqrt {a}-5 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right )^{3/2}}-\frac {b^{3/2} \left (6 \sqrt {a}+5 \sqrt {b}\right ) \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a} \left (\sqrt {a}+\sqrt {b}\right )^{3/2}}-4 b \cot (c+d x)}{4 a^2 b}-\frac {b \tan (c+d x) \left ((a+b) \tan ^2(c+d x)+a\right )}{4 a^2 (a-b) \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}}{d}\) |
((((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
3.3.23.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
Time = 2.62 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}+\frac {b \left (\frac {-\frac {\left (a +b \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{4 \left (a -b \right )}-\frac {a \tan \left (d x +c \right )}{4 \left (a -b \right )}}{\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b -6 a^{2}+4 a b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b +6 a^{2}-4 a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\) | \(269\) |
default | \(\frac {-\frac {1}{a^{2} \tan \left (d x +c \right )}+\frac {b \left (\frac {-\frac {\left (a +b \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{4 \left (a -b \right )}-\frac {a \tan \left (d x +c \right )}{4 \left (a -b \right )}}{\left (\tan ^{4}\left (d x +c \right )\right ) a -b \left (\tan ^{4}\left (d x +c \right )\right )+2 a \left (\tan ^{2}\left (d x +c \right )\right )+a}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b -6 a^{2}+4 a b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (7 a \sqrt {a b}-5 \sqrt {a b}\, b +6 a^{2}-4 a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{8 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\) | \(269\) |
risch | \(\text {Expression too large to display}\) | \(1107\) |
1/d*(-1/a^2/tan(d*x+c)+1/a^2*b*((-1/4*(a+b)/(a-b)*tan(d*x+c)^3-1/4*a/(a-b) *tan(d*x+c))/(tan(d*x+c)^4*a-b*tan(d*x+c)^4+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) )))
Leaf count of result is larger than twice the leaf count of optimal. 3648 vs. \(2 (186) = 372\).
Time = 1.10 (sec) , antiderivative size = 3648, normalized size of antiderivative = 15.46 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \]
-1/32*(8*(4*a*b - 5*b^2)*cos(d*x + c)^5 - 8*(7*a*b - 10*b^2)*cos(d*x + c)^ 3 - ((a^3*b - a^2*b^2)*d*cos(d*x + c)^4 - 2*(a^3*b - a^2*b^2)*d*cos(d*x + c)^2 - (a^4 - 2*a^3*b + a^2*b^2)*d)*sqrt(-((a^7 - 3*a^6*b + 3*a^5*b^2 - a^ 4*b^3)*d^2*sqrt((2304*a^4*b^3 - 6624*a^3*b^4 + 7161*a^2*b^5 - 3450*a*b^6 + 625*b^7)/((a^15 - 6*a^14*b + 15*a^13*b^2 - 20*a^12*b^3 + 15*a^11*b^4 - 6* a^10*b^5 + a^9*b^6)*d^4)) + 36*a^2*b - 47*a*b^2 + 15*b^3)/((a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*d^2))*log(432*a^3*b^2 - 921*a^2*b^3 + 2625/4*a*b^4 - 625/4*b^5 - 1/4*(1728*a^3*b^2 - 3684*a^2*b^3 + 2625*a*b^4 - 625*b^5)*cos (d*x + c)^2 + 1/2*((7*a^11 - 26*a^10*b + 36*a^9*b^2 - 22*a^8*b^3 + 5*a^7*b ^4)*d^3*sqrt((2304*a^4*b^3 - 6624*a^3*b^4 + 7161*a^2*b^5 - 3450*a*b^6 + 62 5*b^7)/((a^15 - 6*a^14*b + 15*a^13*b^2 - 20*a^12*b^3 + 15*a^11*b^4 - 6*a^1 0*b^5 + a^9*b^6)*d^4))*cos(d*x + c)*sin(d*x + c) - 2*(144*a^6*b - 303*a^5* b^2 + 213*a^4*b^3 - 50*a^3*b^4)*d*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*d^2*sqrt((2304*a^4*b^3 - 6624*a^3*b^4 + 71 61*a^2*b^5 - 3450*a*b^6 + 625*b^7)/((a^15 - 6*a^14*b + 15*a^13*b^2 - 20*a^ 12*b^3 + 15*a^11*b^4 - 6*a^10*b^5 + a^9*b^6)*d^4)) + 36*a^2*b - 47*a*b^2 + 15*b^3)/((a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*d^2)) + 1/4*(2*(36*a^9 - 1 33*a^8*b + 183*a^7*b^2 - 111*a^6*b^3 + 25*a^5*b^4)*d^2*cos(d*x + c)^2 - (3 6*a^9 - 133*a^8*b + 183*a^7*b^2 - 111*a^6*b^3 + 25*a^5*b^4)*d^2)*sqrt((230 4*a^4*b^3 - 6624*a^3*b^4 + 7161*a^2*b^5 - 3450*a*b^6 + 625*b^7)/((a^15 ...
Timed out. \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\int { \frac {\csc \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right )^{4} - a\right )}^{2}} \,d x } \]
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...
Leaf count of result is larger than twice the leaf count of optimal. 1545 vs. \(2 (186) = 372\).
Time = 0.89 (sec) , antiderivative size = 1545, normalized size of antiderivative = 6.55 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \]
-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...
Time = 18.23 (sec) , antiderivative size = 4411, normalized size of antiderivative = 18.69 \[ \int \frac {\csc ^2(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx=\text {Too large to display} \]
(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 ...