Integrand size = 24, antiderivative size = 161 \[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {b \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} d}+\frac {b \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} d}+\frac {(a-3 b) \tan (c+d x)}{(a-b)^2 d}+\frac {\tan ^3(c+d x)}{3 (a-b) d} \] Output:
1/2*b*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(3/4)/(a^(1/2)- b^(1/2))^(5/2)/d+1/2*b*arctan((a^(1/2)+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/ a^(3/4)/(a^(1/2)+b^(1/2))^(5/2)/d+(a-3*b)*tan(d*x+c)/(a-b)^2/d+1/3*tan(d*x +c)^3/(a-b)/d
Time = 1.42 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {3 b \left (a-2 \sqrt {a} \sqrt {b}+b\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {3 \left (\sqrt {a}+\sqrt {b}\right )^2 b \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+4 (a-4 b) \tan (c+d x)+2 (a-b) \sec ^2(c+d x) \tan (c+d x)}{6 (a-b)^2 d} \] Input:
Integrate[Sec[c + d*x]^4/(a - b*Sin[c + d*x]^4),x]
Output:
((3*b*(a - 2*Sqrt[a]*Sqrt[b] + b)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x] )/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]]) - (3*(Sq rt[a] + Sqrt[b])^2*b*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + 4*(a - 4*b)*Tan[ c + d*x] + 2*(a - b)*Sec[c + d*x]^2*Tan[c + d*x])/(6*(a - b)^2*d)
Time = 0.49 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3703, 1484, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^4 \left (a-b \sin (c+d x)^4\right )}dx\) |
| \(\Big \downarrow \) 3703 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x)+1\right )^3}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1484 |
| \(\displaystyle \frac {\int \left (\frac {\tan ^2(c+d x)}{a-b}+\frac {b (a+3 b) \tan ^2(c+d x)+b (a+b)}{(a-b)^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}+\frac {a-3 b}{(a-b)^2}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2}}+\frac {b \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2}}+\frac {\tan ^3(c+d x)}{3 (a-b)}+\frac {(a-3 b) \tan (c+d x)}{(a-b)^2}}{d}\) |
Input:
Int[Sec[c + d*x]^4/(a - b*Sin[c + d*x]^4),x]
Output:
((b*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sq rt[a] - Sqrt[b])^(5/2)) + (b*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x]) /a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sqrt[b])^(5/2)) + ((a - 3*b)*Tan[c + d*x] )/(a - b)^2 + Tan[c + d*x]^3/(3*(a - b)))/d
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(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 = 2.65 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.43
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {a \tan \left (d x +c \right )^{3}}{3}-\frac {\tan \left (d x +c \right )^{3} b}{3}+\tan \left (d x +c \right ) a -3 \tan \left (d x +c \right ) b}{\left (a -b \right )^{2}}+\frac {b \left (\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b +3 a b +b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b -3 a b -b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{a -b}}{d}\) | \(230\) |
| default | \(\frac {\frac {\frac {a \tan \left (d x +c \right )^{3}}{3}-\frac {\tan \left (d x +c \right )^{3} b}{3}+\tan \left (d x +c \right ) a -3 \tan \left (d x +c \right ) b}{\left (a -b \right )^{2}}+\frac {b \left (\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b +3 a b +b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (a \sqrt {a b}+3 \sqrt {a b}\, b -3 a b -b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{a -b}}{d}\) | \(230\) |
| risch | \(-\frac {4 i \left (3 b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 a \,{\mathrm e}^{2 i \left (d x +c \right )}+9 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +4 b \right )}{3 d \left (a -b \right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+16 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16777216 a^{8} d^{4}-83886080 a^{7} b \,d^{4}+167772160 a^{6} b^{2} d^{4}-167772160 a^{5} b^{3} d^{4}+83886080 a^{4} b^{4} d^{4}-16777216 a^{3} b^{5} d^{4}\right ) \textit {\_Z}^{4}+\left (8192 a^{4} b^{2} d^{2}+81920 a^{3} b^{3} d^{2}+40960 a^{2} b^{4} d^{2}\right ) \textit {\_Z}^{2}+b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {524288 i d^{3} a^{9}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {1048576 i a^{8} b \,d^{3}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {2621440 i a^{7} b^{2} d^{3}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {10485760 i a^{6} b^{3} d^{3}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {13107200 i a^{5} b^{4} d^{3}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {7340032 i a^{4} b^{5} d^{3}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {1572864 i a^{3} b^{6} d^{3}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}\right ) \textit {\_R}^{3}+\left (-\frac {8192 d^{2} a^{7} b}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {40960 a^{6} b^{2} d^{2}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {81920 a^{5} b^{3} d^{2}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {81920 a^{4} b^{4} d^{2}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {40960 a^{3} b^{5} d^{2}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {8192 a^{2} b^{6} d^{2}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}\right ) \textit {\_R}^{2}+\left (\frac {128 i d \,a^{5} b^{2}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {3584 i a^{4} b^{3} d}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {8960 i a^{3} b^{4} d}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {3584 i a^{2} b^{5} d}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}+\frac {128 i d \,b^{6} a}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}\right ) \textit {\_R} -\frac {2 a^{3} b^{3}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {25 b^{4} a^{2}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {20 b^{5} a}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}-\frac {b^{6}}{5 b^{4} a^{2}+10 b^{5} a +b^{6}}\right )\right )\) | \(869\) |
Input:
int(sec(d*x+c)^4/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
1/d*(1/(a-b)^2*(1/3*a*tan(d*x+c)^3-1/3*tan(d*x+c)^3*b+tan(d*x+c)*a-3*tan(d *x+c)*b)+1/(a-b)*b*(1/2*(a*(a*b)^(1/2)+3*(a*b)^(1/2)*b+3*a*b+b^2)/(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))+1/2*(a*(a*b)^(1/2)+3*(a*b)^(1/2)*b-3*a*b-b^2)/(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))))
Leaf count of result is larger than twice the leaf count of optimal. 4113 vs. \(2 (123) = 246\).
Time = 0.63 (sec) , antiderivative size = 4113, normalized size of antiderivative = 25.55 \[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^4/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{a - b \sin ^{4}{\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**4/(a-b*sin(d*x+c)**4),x)
Output:
Integral(sec(c + d*x)**4/(a - b*sin(c + d*x)**4), x)
\[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sec \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sec(d*x+c)^4/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
Output:
-1/3*(36*(a - 2*b)*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) - 12*(b*sin(4*d*x + 4 *c) - (a - 3*b)*sin(2*d*x + 2*c))*cos(6*d*x + 6*c) - 3*((a^2 - 2*a*b + b^2 )*d*cos(6*d*x + 6*c)^2 + 9*(a^2 - 2*a*b + b^2)*d*cos(4*d*x + 4*c)^2 + 9*(a ^2 - 2*a*b + b^2)*d*cos(2*d*x + 2*c)^2 + (a^2 - 2*a*b + b^2)*d*sin(6*d*x + 6*c)^2 + 9*(a^2 - 2*a*b + b^2)*d*sin(4*d*x + 4*c)^2 + 18*(a^2 - 2*a*b + b ^2)*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*(a^2 - 2*a*b + b^2)*d*sin(2*d* x + 2*c)^2 + 6*(a^2 - 2*a*b + b^2)*d*cos(2*d*x + 2*c) + (a^2 - 2*a*b + b^2 )*d + 2*(3*(a^2 - 2*a*b + b^2)*d*cos(4*d*x + 4*c) + 3*(a^2 - 2*a*b + b^2)* d*cos(2*d*x + 2*c) + (a^2 - 2*a*b + b^2)*d)*cos(6*d*x + 6*c) + 6*(3*(a^2 - 2*a*b + b^2)*d*cos(2*d*x + 2*c) + (a^2 - 2*a*b + b^2)*d)*cos(4*d*x + 4*c) + 6*((a^2 - 2*a*b + b^2)*d*sin(4*d*x + 4*c) + (a^2 - 2*a*b + b^2)*d*sin(2 *d*x + 2*c))*sin(6*d*x + 6*c))*integrate(-8*(4*b^3*cos(6*d*x + 6*c)^2 + 4* b^3*cos(2*d*x + 2*c)^2 + 4*b^3*sin(6*d*x + 6*c)^2 + 4*b^3*sin(2*d*x + 2*c) ^2 - b^3*cos(2*d*x + 2*c) - 4*(8*a^2*b + 13*a*b^2 - 6*b^3)*cos(4*d*x + 4*c )^2 - 4*(8*a^2*b + 13*a*b^2 - 6*b^3)*sin(4*d*x + 4*c)^2 + 2*(4*a*b^2 - 11* b^3)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - (b^3*cos(6*d*x + 6*c) + b^3*cos(2 *d*x + 2*c) - 2*(a*b^2 + 2*b^3)*cos(4*d*x + 4*c))*cos(8*d*x + 8*c) + (8*b^ 3*cos(2*d*x + 2*c) - b^3 + 2*(4*a*b^2 - 11*b^3)*cos(4*d*x + 4*c))*cos(6*d* x + 6*c) + 2*(a*b^2 + 2*b^3 + (4*a*b^2 - 11*b^3)*cos(2*d*x + 2*c))*cos(4*d *x + 4*c) - (b^3*sin(6*d*x + 6*c) + b^3*sin(2*d*x + 2*c) - 2*(a*b^2 + 2...
Leaf count of result is larger than twice the leaf count of optimal. 2183 vs. \(2 (123) = 246\).
Time = 0.90 (sec) , antiderivative size = 2183, normalized size of antiderivative = 13.56 \[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^4/(a-b*sin(d*x+c)^4),x, algorithm="giac")
Output:
1/6*(2*(a^2*tan(d*x + c)^3 - 2*a*b*tan(d*x + c)^3 + b^2*tan(d*x + c)^3 + 3 *a^2*tan(d*x + c) - 12*a*b*tan(d*x + c) + 9*b^2*tan(d*x + c))/(a^3 - 3*a^2 *b + 3*a*b^2 - b^3) - 3*((3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)* a^3*b + 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 - 19*sqrt( a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^3 - 3*sqrt(a^2 - a*b + sqrt(a *b)*(a - b))*sqrt(a*b)*b^4)*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)^2*abs(-a + b) - (3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^7*b - 15*sqrt(a^2 - a*b + sqrt( a*b)*(a - b))*a^6*b^2 + 23*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^5*b^3 - 3 *sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^4*b^4 - 23*sqrt(a^2 - a*b + sqrt(a* b)*(a - b))*a^3*b^5 + 19*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^6 - 3*s qrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^7 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*b^8)*abs(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*abs(-a + b) - (9*sqrt(a^2 - a *b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^9*b - 69*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^8*b^2 + 216*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a* b)*a^7*b^3 - 352*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^6*b^4 + 3 06*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^5*b^5 - 114*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^4*b^6 - 16*sqrt(a^2 - a*b + sqrt(a*b) *(a - b))*sqrt(a*b)*a^3*b^7 + 24*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt( a*b)*a^2*b^8 - 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^9 - sqr t(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*b^10)*abs(-a + b))*(pi*floor...
Time = 37.97 (sec) , antiderivative size = 4664, normalized size of antiderivative = 28.97 \[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^4*(a - b*sin(c + d*x)^4)),x)
Output:
tan(c + d*x)^3/(3*d*(a - b)) - (tan(c + d*x)*((2*a)/(a - b)^2 - 3/(a - b)) )/d + (atan(((((16*a*b^6 - 32*a^2*b^5 + 32*a^4*b^3 - 16*a^5*b^2)/(3*a*b^2 - 3*a^2*b + a^3 - b^3) - (4*tan(c + d*x)*((5*a^2*(a^3*b^5)^(1/2) + b^2*(a^ 3*b^5)^(1/2) + 5*a^2*b^4 + 10*a^3*b^3 + a^4*b^2 + 10*a*b*(a^3*b^5)^(1/2))/ (16*(5*a^7*b - a^8 + a^3*b^5 - 5*a^4*b^4 + 10*a^5*b^3 - 10*a^6*b^2)))^(1/2 )*(16*a^7*b - 16*a^2*b^6 + 80*a^3*b^5 - 160*a^4*b^4 + 160*a^5*b^3 - 80*a^6 *b^2))/(3*a*b^2 - 3*a^2*b + a^3 - b^3))*((5*a^2*(a^3*b^5)^(1/2) + b^2*(a^3 *b^5)^(1/2) + 5*a^2*b^4 + 10*a^3*b^3 + a^4*b^2 + 10*a*b*(a^3*b^5)^(1/2))/( 16*(5*a^7*b - a^8 + a^3*b^5 - 5*a^4*b^4 + 10*a^5*b^3 - 10*a^6*b^2)))^(1/2) - (4*tan(c + d*x)*(15*a*b^5 + b^6 + 15*a^2*b^4 + a^3*b^3))/(3*a*b^2 - 3*a ^2*b + a^3 - b^3))*((5*a^2*(a^3*b^5)^(1/2) + b^2*(a^3*b^5)^(1/2) + 5*a^2*b ^4 + 10*a^3*b^3 + a^4*b^2 + 10*a*b*(a^3*b^5)^(1/2))/(16*(5*a^7*b - a^8 + a ^3*b^5 - 5*a^4*b^4 + 10*a^5*b^3 - 10*a^6*b^2)))^(1/2)*1i - (((16*a*b^6 - 3 2*a^2*b^5 + 32*a^4*b^3 - 16*a^5*b^2)/(3*a*b^2 - 3*a^2*b + a^3 - b^3) + (4* tan(c + d*x)*((5*a^2*(a^3*b^5)^(1/2) + b^2*(a^3*b^5)^(1/2) + 5*a^2*b^4 + 1 0*a^3*b^3 + a^4*b^2 + 10*a*b*(a^3*b^5)^(1/2))/(16*(5*a^7*b - a^8 + a^3*b^5 - 5*a^4*b^4 + 10*a^5*b^3 - 10*a^6*b^2)))^(1/2)*(16*a^7*b - 16*a^2*b^6 + 8 0*a^3*b^5 - 160*a^4*b^4 + 160*a^5*b^3 - 80*a^6*b^2))/(3*a*b^2 - 3*a^2*b + a^3 - b^3))*((5*a^2*(a^3*b^5)^(1/2) + b^2*(a^3*b^5)^(1/2) + 5*a^2*b^4 + 10 *a^3*b^3 + a^4*b^2 + 10*a*b*(a^3*b^5)^(1/2))/(16*(5*a^7*b - a^8 + a^3*b...
\[ \int \frac {\sec ^4(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {too large to display} \] Input:
int(sec(d*x+c)^4/(a-b*sin(d*x+c)^4),x)
Output:
( - 3072*cos(c + d*x)*int(tan((c + d*x)/2)**6/(tan((c + d*x)/2)**16*a - 4* tan((c + d*x)/2)**12*a - 16*tan((c + d*x)/2)**12*b + 64*tan((c + d*x)/2)** 10*b + 6*tan((c + d*x)/2)**8*a - 96*tan((c + d*x)/2)**8*b + 64*tan((c + d* x)/2)**6*b - 4*tan((c + d*x)/2)**4*a - 16*tan((c + d*x)/2)**4*b + a),x)*si n(c + d*x)**2*a*b**2*d + 3072*cos(c + d*x)*int(tan((c + d*x)/2)**6/(tan((c + d*x)/2)**16*a - 4*tan((c + d*x)/2)**12*a - 16*tan((c + d*x)/2)**12*b + 64*tan((c + d*x)/2)**10*b + 6*tan((c + d*x)/2)**8*a - 96*tan((c + d*x)/2)* *8*b + 64*tan((c + d*x)/2)**6*b - 4*tan((c + d*x)/2)**4*a - 16*tan((c + d* x)/2)**4*b + a),x)*a*b**2*d - 4608*cos(c + d*x)*int(tan((c + d*x)/2)**4/(t an((c + d*x)/2)**16*a - 4*tan((c + d*x)/2)**12*a - 16*tan((c + d*x)/2)**12 *b + 64*tan((c + d*x)/2)**10*b + 6*tan((c + d*x)/2)**8*a - 96*tan((c + d*x )/2)**8*b + 64*tan((c + d*x)/2)**6*b - 4*tan((c + d*x)/2)**4*a - 16*tan((c + d*x)/2)**4*b + a),x)*sin(c + d*x)**2*a*b**2*d + 12288*cos(c + d*x)*int( tan((c + d*x)/2)**4/(tan((c + d*x)/2)**16*a - 4*tan((c + d*x)/2)**12*a - 1 6*tan((c + d*x)/2)**12*b + 64*tan((c + d*x)/2)**10*b + 6*tan((c + d*x)/2)* *8*a - 96*tan((c + d*x)/2)**8*b + 64*tan((c + d*x)/2)**6*b - 4*tan((c + d* x)/2)**4*a - 16*tan((c + d*x)/2)**4*b + a),x)*sin(c + d*x)**2*b**3*d + 460 8*cos(c + d*x)*int(tan((c + d*x)/2)**4/(tan((c + d*x)/2)**16*a - 4*tan((c + d*x)/2)**12*a - 16*tan((c + d*x)/2)**12*b + 64*tan((c + d*x)/2)**10*b + 6*tan((c + d*x)/2)**8*a - 96*tan((c + d*x)/2)**8*b + 64*tan((c + d*x)/2...