Integrand size = 24, antiderivative size = 204 \[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{7/2} d}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{7/2} d}+\frac {\left (a^2-3 a b+6 b^2\right ) \tan (c+d x)}{(a-b)^3 d}+\frac {2 (a-2 b) \tan ^3(c+d x)}{3 (a-b)^2 d}+\frac {\tan ^5(c+d x)}{5 (a-b) d} \] Output:
-1/2*b^(3/2)*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))^(7/2)/d+1/2*b^(3/2)*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))^(7/2)/d+(a^2-3*a*b+6*b^2)*tan(d*x+c )/(a-b)^3/d+2/3*(a-2*b)*tan(d*x+c)^3/(a-b)^2/d+1/5*tan(d*x+c)^5/(a-b)/d
Time = 1.93 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.24 \[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {15 \left (\sqrt {a}-\sqrt {b}\right )^3 b^{3/2} \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {15 \left (\sqrt {a}+\sqrt {b}\right )^3 b^{3/2} \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+2 \left (8 a^2-21 a b+73 b^2\right ) \tan (c+d x)+4 (2 a-7 b) (a-b) \sec ^2(c+d x) \tan (c+d x)+6 (a-b)^2 \sec ^4(c+d x) \tan (c+d x)}{30 (a-b)^3 d} \] Input:
Integrate[Sec[c + d*x]^6/(a - b*Sin[c + d*x]^4),x]
Output:
((15*(Sqrt[a] - Sqrt[b])^3*b^(3/2)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x ])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]]) + (15*( Sqrt[a] + Sqrt[b])^3*b^(3/2)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sq rt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + 2*(8*a^2 - 21*a*b + 73*b^2)*Tan[c + d*x] + 4*(2*a - 7*b)*(a - b)*Sec[c + d*x]^2*Ta n[c + d*x] + 6*(a - b)^2*Sec[c + d*x]^4*Tan[c + d*x])/(30*(a - b)^3*d)
Time = 0.53 (sec) , antiderivative size = 193, 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.
| \(\displaystyle \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^6 \left (a-b \sin (c+d x)^4\right )}dx\) |
| \(\Big \downarrow \) 3703 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x)+1\right )^4}{(a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1484 |
| \(\displaystyle \frac {\int \left (\frac {\tan ^4(c+d x)}{a-b}+\frac {2 (a-2 b) \tan ^2(c+d x)}{(a-b)^2}+\frac {a^2-3 b a+6 b^2}{(a-b)^3}-\frac {4 (a+b) \tan ^2(c+d x) b^2+(3 a+b) b^2}{(a-b)^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{7/2}}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{7/2}}+\frac {\left (a^2-3 a b+6 b^2\right ) \tan (c+d x)}{(a-b)^3}+\frac {\tan ^5(c+d x)}{5 (a-b)}+\frac {2 (a-2 b) \tan ^3(c+d x)}{3 (a-b)^2}}{d}\) |
Input:
Int[Sec[c + d*x]^6/(a - b*Sin[c + d*x]^4),x]
Output:
(-1/2*(b^(3/2)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(a^ (3/4)*(Sqrt[a] - Sqrt[b])^(7/2)) + (b^(3/2)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b] ]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sqrt[b])^(7/2)) + ((a^2 - 3*a*b + 6*b^2)*Tan[c + d*x])/(a - b)^3 + (2*(a - 2*b)*Tan[c + d*x]^3)/(3*( a - b)^2) + Tan[c + d*x]^5/(5*(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 = 3.95 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.52
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {a^{2} \tan \left (d x +c \right )^{5}}{5}-\frac {2 a b \tan \left (d x +c \right )^{5}}{5}+\frac {b^{2} \tan \left (d x +c \right )^{5}}{5}+\frac {2 a^{2} \tan \left (d x +c \right )^{3}}{3}-2 a b \tan \left (d x +c \right )^{3}+\frac {4 b^{2} \tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right ) a^{2}-3 \tan \left (d x +c \right ) a b +6 \tan \left (d x +c \right ) b^{2}}{\left (a -b \right )^{3}}-\frac {b^{2} \left (\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b -a^{2}-6 a b -b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b +a^{2}+6 a b +b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2}}}{d}\) | \(311\) |
| default | \(\frac {\frac {\frac {a^{2} \tan \left (d x +c \right )^{5}}{5}-\frac {2 a b \tan \left (d x +c \right )^{5}}{5}+\frac {b^{2} \tan \left (d x +c \right )^{5}}{5}+\frac {2 a^{2} \tan \left (d x +c \right )^{3}}{3}-2 a b \tan \left (d x +c \right )^{3}+\frac {4 b^{2} \tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right ) a^{2}-3 \tan \left (d x +c \right ) a b +6 \tan \left (d x +c \right ) b^{2}}{\left (a -b \right )^{3}}-\frac {b^{2} \left (\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b -a^{2}-6 a b -b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b +a^{2}+6 a b +b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2}}}{d}\) | \(311\) |
| risch | \(\text {Expression too large to display}\) | \(1365\) |
Input:
int(sec(d*x+c)^6/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
1/d*(1/(a-b)^3*(1/5*a^2*tan(d*x+c)^5-2/5*a*b*tan(d*x+c)^5+1/5*b^2*tan(d*x+ c)^5+2/3*a^2*tan(d*x+c)^3-2*a*b*tan(d*x+c)^3+4/3*b^2*tan(d*x+c)^3+tan(d*x+ c)*a^2-3*tan(d*x+c)*a*b+6*tan(d*x+c)*b^2)-1/(a-b)^2*b^2*(1/2*(4*a*(a*b)^(1 /2)+4*(a*b)^(1/2)*b-a^2-6*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))+1/2*(4*a *(a*b)^(1/2)+4*(a*b)^(1/2)*b+a^2+6*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))))
Leaf count of result is larger than twice the leaf count of optimal. 5587 vs. \(2 (160) = 320\).
Time = 1.12 (sec) , antiderivative size = 5587, normalized size of antiderivative = 27.39 \[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^6/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**6/(a-b*sin(d*x+c)**4),x)
Output:
Timed out
\[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sec \left (d x + c\right )^{6}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sec(d*x+c)^6/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
Output:
1/15*(300*(a*b - 5*b^2)*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) - 10*(48*b^2*sin (6*d*x + 6*c) + 3*(a*b + 3*b^2)*sin(8*d*x + 8*c) + 2*(8*a^2 - 21*a*b + 49* b^2)*sin(4*d*x + 4*c) + 8*(a^2 - 3*a*b + 8*b^2)*sin(2*d*x + 2*c))*cos(10*d *x + 10*c) + 50*(6*(a*b - 5*b^2)*sin(6*d*x + 6*c) - 16*(a^2 - 3*a*b + 5*b^ 2)*sin(4*d*x + 4*c) - (8*a^2 - 27*a*b + 55*b^2)*sin(2*d*x + 2*c))*cos(8*d* x + 8*c) - 200*((8*a^2 - 21*a*b + 25*b^2)*sin(4*d*x + 4*c) + 4*(a^2 - 3*a* b + 5*b^2)*sin(2*d*x + 2*c))*cos(6*d*x + 6*c) + 15*((a^3 - 3*a^2*b + 3*a*b ^2 - b^3)*d*cos(10*d*x + 10*c)^2 + 25*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*co s(8*d*x + 8*c)^2 + 100*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*cos(6*d*x + 6*c)^ 2 + 100*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*cos(4*d*x + 4*c)^2 + 25*(a^3 - 3 *a^2*b + 3*a*b^2 - b^3)*d*cos(2*d*x + 2*c)^2 + (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*sin(10*d*x + 10*c)^2 + 25*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*sin(8*d *x + 8*c)^2 + 100*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*sin(6*d*x + 6*c)^2 + 1 00*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*sin(4*d*x + 4*c)^2 + 100*(a^3 - 3*a^2 *b + 3*a*b^2 - b^3)*d*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 25*(a^3 - 3*a^2* b + 3*a*b^2 - b^3)*d*sin(2*d*x + 2*c)^2 + 10*(a^3 - 3*a^2*b + 3*a*b^2 - b^ 3)*d*cos(2*d*x + 2*c) + (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d + 2*(5*(a^3 - 3* a^2*b + 3*a*b^2 - b^3)*d*cos(8*d*x + 8*c) + 10*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*cos(6*d*x + 6*c) + 10*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*cos(4*d*x + 4*c) + 5*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*cos(2*d*x + 2*c) + (a^3 - 3...
Leaf count of result is larger than twice the leaf count of optimal. 3106 vs. \(2 (160) = 320\).
Time = 0.90 (sec) , antiderivative size = 3106, normalized size of antiderivative = 15.23 \[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^6/(a-b*sin(d*x+c)^4),x, algorithm="giac")
Output:
1/30*(2*(3*a^4*tan(d*x + c)^5 - 12*a^3*b*tan(d*x + c)^5 + 18*a^2*b^2*tan(d *x + c)^5 - 12*a*b^3*tan(d*x + c)^5 + 3*b^4*tan(d*x + c)^5 + 10*a^4*tan(d* x + c)^3 - 50*a^3*b*tan(d*x + c)^3 + 90*a^2*b^2*tan(d*x + c)^3 - 70*a*b^3* tan(d*x + c)^3 + 20*b^4*tan(d*x + c)^3 + 15*a^4*tan(d*x + c) - 75*a^3*b*ta n(d*x + c) + 195*a^2*b^2*tan(d*x + c) - 225*a*b^3*tan(d*x + c) + 90*b^4*ta n(d*x + c))/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5) + 15 *(4*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b^2 - 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^3 - 7*sqrt(a^2 - a*b + sqrt(a*b )*(a - b))*sqrt(a*b)*a*b^4 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b) *b^5)*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)^2*abs(-a + b) - (9*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^9*b^2 - 69*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^8*b^3 + 216*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^7* b^4 - 352*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^6*b^5 + 306*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^5*b^6 - 114*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^ 4*b^7 - 16*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^3*b^8 + 24*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^9 - 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^ 10 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*b^11)*abs(a^5 - 5*a^4*b + 10*a^3* b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*abs(-a + b) - (3*sqrt(a^2 - a*b + sqrt(a *b)*(a - b))*sqrt(a*b)*a^14*b - 18*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqr t(a*b)*a^13*b^2 - 19*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^12...
Time = 38.93 (sec) , antiderivative size = 6534, normalized size of antiderivative = 32.03 \[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^6*(a - b*sin(c + d*x)^4)),x)
Output:
(atan(((((4*(4*a*b^8 - 4*a^2*b^7 - 24*a^3*b^6 + 56*a^4*b^5 - 44*a^5*b^4 + 12*a^6*b^3))/(5*a*b^4 - 5*a^4*b + a^5 - b^5 - 10*a^2*b^3 + 10*a^3*b^2) - ( 4*tan(c + d*x)*((7*a^3*(a^3*b^7)^(1/2) + b^3*(a^3*b^7)^(1/2) + 7*a^2*b^6 + 35*a^3*b^5 + 21*a^4*b^4 + a^5*b^3 + 21*a*b^2*(a^3*b^7)^(1/2) + 35*a^2*b*( a^3*b^7)^(1/2))/(16*(7*a^9*b - a^10 + a^3*b^7 - 7*a^4*b^6 + 21*a^5*b^5 - 3 5*a^6*b^4 + 35*a^7*b^3 - 21*a^8*b^2)))^(1/2)*(16*a^9*b - 16*a^2*b^8 + 112* a^3*b^7 - 336*a^4*b^6 + 560*a^5*b^5 - 560*a^6*b^4 + 336*a^7*b^3 - 112*a^8* b^2))/(5*a*b^4 - 5*a^4*b + a^5 - b^5 - 10*a^2*b^3 + 10*a^3*b^2))*((7*a^3*( a^3*b^7)^(1/2) + b^3*(a^3*b^7)^(1/2) + 7*a^2*b^6 + 35*a^3*b^5 + 21*a^4*b^4 + a^5*b^3 + 21*a*b^2*(a^3*b^7)^(1/2) + 35*a^2*b*(a^3*b^7)^(1/2))/(16*(7*a ^9*b - a^10 + a^3*b^7 - 7*a^4*b^6 + 21*a^5*b^5 - 35*a^6*b^4 + 35*a^7*b^3 - 21*a^8*b^2)))^(1/2) - (4*tan(c + d*x)*(28*a*b^7 + b^8 + 70*a^2*b^6 + 28*a ^3*b^5 + a^4*b^4))/(5*a*b^4 - 5*a^4*b + a^5 - b^5 - 10*a^2*b^3 + 10*a^3*b^ 2))*((7*a^3*(a^3*b^7)^(1/2) + b^3*(a^3*b^7)^(1/2) + 7*a^2*b^6 + 35*a^3*b^5 + 21*a^4*b^4 + a^5*b^3 + 21*a*b^2*(a^3*b^7)^(1/2) + 35*a^2*b*(a^3*b^7)^(1 /2))/(16*(7*a^9*b - a^10 + a^3*b^7 - 7*a^4*b^6 + 21*a^5*b^5 - 35*a^6*b^4 + 35*a^7*b^3 - 21*a^8*b^2)))^(1/2)*1i - (((4*(4*a*b^8 - 4*a^2*b^7 - 24*a^3* b^6 + 56*a^4*b^5 - 44*a^5*b^4 + 12*a^6*b^3))/(5*a*b^4 - 5*a^4*b + a^5 - b^ 5 - 10*a^2*b^3 + 10*a^3*b^2) + (4*tan(c + d*x)*((7*a^3*(a^3*b^7)^(1/2) + b ^3*(a^3*b^7)^(1/2) + 7*a^2*b^6 + 35*a^3*b^5 + 21*a^4*b^4 + a^5*b^3 + 21...
\[ \int \frac {\sec ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {too large to display} \] Input:
int(sec(d*x+c)^6/(a-b*sin(d*x+c)^4),x)
Output:
( - 15360*cos(c + d*x)*int(tan((c + d*x)/2)**6/(tan((c + d*x)/2)**20*a - 2 *tan((c + d*x)/2)**18*a - 3*tan((c + d*x)/2)**16*a - 16*tan((c + d*x)/2)** 16*b + 8*tan((c + d*x)/2)**14*a + 96*tan((c + d*x)/2)**14*b + 2*tan((c + d *x)/2)**12*a - 240*tan((c + d*x)/2)**12*b - 12*tan((c + d*x)/2)**10*a + 32 0*tan((c + d*x)/2)**10*b + 2*tan((c + d*x)/2)**8*a - 240*tan((c + d*x)/2)* *8*b + 8*tan((c + d*x)/2)**6*a + 96*tan((c + d*x)/2)**6*b - 3*tan((c + d*x )/2)**4*a - 16*tan((c + d*x)/2)**4*b - 2*tan((c + d*x)/2)**2*a + a),x)*sin (c + d*x)**4*a**2*b**2*d - 368640*cos(c + d*x)*int(tan((c + d*x)/2)**6/(ta n((c + d*x)/2)**20*a - 2*tan((c + d*x)/2)**18*a - 3*tan((c + d*x)/2)**16*a - 16*tan((c + d*x)/2)**16*b + 8*tan((c + d*x)/2)**14*a + 96*tan((c + d*x) /2)**14*b + 2*tan((c + d*x)/2)**12*a - 240*tan((c + d*x)/2)**12*b - 12*tan ((c + d*x)/2)**10*a + 320*tan((c + d*x)/2)**10*b + 2*tan((c + d*x)/2)**8*a - 240*tan((c + d*x)/2)**8*b + 8*tan((c + d*x)/2)**6*a + 96*tan((c + d*x)/ 2)**6*b - 3*tan((c + d*x)/2)**4*a - 16*tan((c + d*x)/2)**4*b - 2*tan((c + d*x)/2)**2*a + a),x)*sin(c + d*x)**4*a*b**3*d + 30720*cos(c + d*x)*int(tan ((c + d*x)/2)**6/(tan((c + d*x)/2)**20*a - 2*tan((c + d*x)/2)**18*a - 3*ta n((c + d*x)/2)**16*a - 16*tan((c + d*x)/2)**16*b + 8*tan((c + d*x)/2)**14* a + 96*tan((c + d*x)/2)**14*b + 2*tan((c + d*x)/2)**12*a - 240*tan((c + d* x)/2)**12*b - 12*tan((c + d*x)/2)**10*a + 320*tan((c + d*x)/2)**10*b + 2*t an((c + d*x)/2)**8*a - 240*tan((c + d*x)/2)**8*b + 8*tan((c + d*x)/2)**...