Integrand size = 24, antiderivative size = 252 \[ \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {17 x}{16 b}-\frac {4 (a+b) x}{b^2}-\frac {(a+3 b) x}{2 b^2}-\frac {\left (\sqrt {a}-\sqrt {b}\right )^{9/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/2} d}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^{9/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/2} d}-\frac {17 \cos (c+d x) \sin (c+d x)}{16 b d}-\frac {(a+3 b) \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac {17 \cos ^3(c+d x) \sin (c+d x)}{24 b d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 b d} \]
-17/16*x/b-4*(a+b)*x/b^2-1/2*(a+3*b)*x/b^2-17/16*cos(d*x+c)*sin(d*x+c)/b/d -1/2*(a+3*b)*cos(d*x+c)*sin(d*x+c)/b^2/d-17/24*cos(d*x+c)^3*sin(d*x+c)/b/d -1/6*cos(d*x+c)^5*sin(d*x+c)/b/d-1/2*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d* x+c)/a^(1/4))*(a^(1/2)-b^(1/2))^(9/2)/a^(3/4)/b^(5/2)/d+1/2*arctan((a^(1/2 )+b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))*(a^(1/2)+b^(1/2))^(9/2)/a^(3/4)/b^(5/ 2)/d
Time = 1.61 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {36 b (24 a+35 b) (c+d x)-\frac {96 \left (\sqrt {a}+\sqrt {b}\right )^5 \sqrt {b} \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {96 \left (\sqrt {a}-\sqrt {b}\right )^5 \sqrt {b} \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+3 b (16 a+95 b) \sin (2 (c+d x))+21 b^2 \sin (4 (c+d x))+b^2 \sin (6 (c+d x))}{192 b^3 d} \]
-1/192*(36*b*(24*a + 35*b)*(c + d*x) - (96*(Sqrt[a] + Sqrt[b])^5*Sqrt[b]*A rcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt [a]*Sqrt[a + Sqrt[a]*Sqrt[b]]) - (96*(Sqrt[a] - Sqrt[b])^5*Sqrt[b]*ArcTanh [((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]* Sqrt[-a + Sqrt[a]*Sqrt[b]]) + 3*b*(16*a + 95*b)*Sin[2*(c + d*x)] + 21*b^2* Sin[4*(c + d*x)] + b^2*Sin[6*(c + d*x)])/(b^3*d)
Time = 0.58 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.10, 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.
| \(\displaystyle \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^{10}}{a-b \sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3703 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right )^4 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1484 |
| \(\displaystyle \frac {\int \left (\frac {-a-3 b}{b^2 \left (\tan ^2(c+d x)+1\right )^2}-\frac {4 (a+b)}{b^2 \left (\tan ^2(c+d x)+1\right )}+\frac {5 a^2+10 b a+b^2+4 \left (a^2-b^2\right ) \tan ^2(c+d x)}{b^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac {2}{b \left (\tan ^2(c+d x)+1\right )^3}-\frac {1}{b \left (\tan ^2(c+d x)+1\right )^4}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\left (\sqrt {a}-\sqrt {b}\right )^{9/2} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/2}}+\frac {\left (\sqrt {a}+\sqrt {b}\right )^{9/2} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/2}}-\frac {4 (a+b) \arctan (\tan (c+d x))}{b^2}-\frac {(a+3 b) \arctan (\tan (c+d x))}{2 b^2}-\frac {(a+3 b) \tan (c+d x)}{2 b^2 \left (\tan ^2(c+d x)+1\right )}-\frac {17 \arctan (\tan (c+d x))}{16 b}-\frac {17 \tan (c+d x)}{16 b \left (\tan ^2(c+d x)+1\right )}-\frac {17 \tan (c+d x)}{24 b \left (\tan ^2(c+d x)+1\right )^2}-\frac {\tan (c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
((-17*ArcTan[Tan[c + d*x]])/(16*b) - (4*(a + b)*ArcTan[Tan[c + d*x]])/b^2 - ((a + 3*b)*ArcTan[Tan[c + d*x]])/(2*b^2) - ((Sqrt[a] - Sqrt[b])^(9/2)*Ar cTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(5/2)) + ((Sqrt[a] + Sqrt[b])^(9/2)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x]) /a^(1/4)])/(2*a^(3/4)*b^(5/2)) - Tan[c + d*x]/(6*b*(1 + Tan[c + d*x]^2)^3) - (17*Tan[c + d*x])/(24*b*(1 + Tan[c + d*x]^2)^2) - (17*Tan[c + d*x])/(16 *b*(1 + Tan[c + d*x]^2)) - ((a + 3*b)*Tan[c + d*x])/(2*b^2*(1 + Tan[c + d* x]^2)))/d
3.5.11.3.1 Defintions of rubi rules used
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 = 4.14 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (a -b \right ) \left (\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b +a^{2}+6 a b +b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b -a^{2}-6 a b -b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b^{2}}-\frac {\frac {\left (\frac {a}{2}+\frac {41 b}{16}\right ) \left (\tan ^{5}\left (d x +c \right )\right )+\left (a +\frac {35 b}{6}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {a}{2}+\frac {55 b}{16}\right ) \tan \left (d x +c \right )}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}+\frac {3 \left (24 a +35 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{16}}{b^{2}}}{d}\) | \(258\) |
| default | \(\frac {\frac {\left (a -b \right ) \left (\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b +a^{2}+6 a b +b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\left (4 a \sqrt {a b}+4 \sqrt {a b}\, b -a^{2}-6 a b -b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{b^{2}}-\frac {\frac {\left (\frac {a}{2}+\frac {41 b}{16}\right ) \left (\tan ^{5}\left (d x +c \right )\right )+\left (a +\frac {35 b}{6}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {a}{2}+\frac {55 b}{16}\right ) \tan \left (d x +c \right )}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}+\frac {3 \left (24 a +35 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{16}}{b^{2}}}{d}\) | \(258\) |
| risch | \(\text {Expression too large to display}\) | \(1905\) |
1/d*(1/b^2*(a-b)*(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)^(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)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^( 1/2)+a)*(a-b))^(1/2)))-1/b^2*(((1/2*a+41/16*b)*tan(d*x+c)^5+(a+35/6*b)*tan (d*x+c)^3+(1/2*a+55/16*b)*tan(d*x+c))/(1+tan(d*x+c)^2)^3+3/16*(24*a+35*b)* arctan(tan(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 2948 vs. \(2 (200) = 400\).
Time = 1.63 (sec) , antiderivative size = 2948, normalized size of antiderivative = 11.70 \[ \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
1/48*(6*b^2*d*sqrt((a*b^5*d^2*sqrt((81*a^8 + 1512*a^7*b + 9324*a^6*b^2 + 2 1816*a^5*b^3 + 21942*a^4*b^4 + 9240*a^3*b^5 + 1548*a^2*b^6 + 72*a*b^7 + b^ 8)/(a^3*b^9*d^4)) - a^4 - 36*a^3*b - 126*a^2*b^2 - 84*a*b^3 - 9*b^4)/(a*b^ 5*d^2))*log(9/4*a^8 + 12*a^7*b - 39*a^6*b^2 + 143/2*a^4*b^4 - 52*a^3*b^5 - 3*a^2*b^6 + 8*a*b^7 + 1/4*b^8 - 1/4*(9*a^8 + 48*a^7*b - 156*a^6*b^2 + 286 *a^4*b^4 - 208*a^3*b^5 - 12*a^2*b^6 + 32*a*b^7 + b^8)*cos(d*x + c)^2 + 1/2 *(4*(a^4*b^7 + a^3*b^8)*d^3*sqrt((81*a^8 + 1512*a^7*b + 9324*a^6*b^2 + 218 16*a^5*b^3 + 21942*a^4*b^4 + 9240*a^3*b^5 + 1548*a^2*b^6 + 72*a*b^7 + b^8) /(a^3*b^9*d^4))*cos(d*x + c)*sin(d*x + c) + (9*a^7*b^2 + 138*a^6*b^3 + 639 *a^5*b^4 + 876*a^4*b^5 + 343*a^3*b^6 + 42*a^2*b^7 + a*b^8)*d*cos(d*x + c)* sin(d*x + c))*sqrt((a*b^5*d^2*sqrt((81*a^8 + 1512*a^7*b + 9324*a^6*b^2 + 2 1816*a^5*b^3 + 21942*a^4*b^4 + 9240*a^3*b^5 + 1548*a^2*b^6 + 72*a*b^7 + b^ 8)/(a^3*b^9*d^4)) - a^4 - 36*a^3*b - 126*a^2*b^2 - 84*a*b^3 - 9*b^4)/(a*b^ 5*d^2)) + 1/4*(2*(a^6*b^4 - 4*a^5*b^5 + 6*a^4*b^6 - 4*a^3*b^7 + a^2*b^8)*d ^2*cos(d*x + c)^2 - (a^6*b^4 - 4*a^5*b^5 + 6*a^4*b^6 - 4*a^3*b^7 + a^2*b^8 )*d^2)*sqrt((81*a^8 + 1512*a^7*b + 9324*a^6*b^2 + 21816*a^5*b^3 + 21942*a^ 4*b^4 + 9240*a^3*b^5 + 1548*a^2*b^6 + 72*a*b^7 + b^8)/(a^3*b^9*d^4))) - 6* b^2*d*sqrt((a*b^5*d^2*sqrt((81*a^8 + 1512*a^7*b + 9324*a^6*b^2 + 21816*a^5 *b^3 + 21942*a^4*b^4 + 9240*a^3*b^5 + 1548*a^2*b^6 + 72*a*b^7 + b^8)/(a^3* b^9*d^4)) - a^4 - 36*a^3*b - 126*a^2*b^2 - 84*a*b^3 - 9*b^4)/(a*b^5*d^2...
Timed out. \[ \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\cos \left (d x + c\right )^{10}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \]
-1/192*(192*b^2*d*integrate(-4*(4*(a^2*b + 10*a*b^2 + 5*b^3)*cos(6*d*x + 6 *c)^2 + 4*(72*a^3 + 53*a^2*b - 54*a*b^2 + 9*b^3)*cos(4*d*x + 4*c)^2 + 4*(a ^2*b + 10*a*b^2 + 5*b^3)*cos(2*d*x + 2*c)^2 + 4*(a^2*b + 10*a*b^2 + 5*b^3) *sin(6*d*x + 6*c)^2 + 4*(72*a^3 + 53*a^2*b - 54*a*b^2 + 9*b^3)*sin(4*d*x + 4*c)^2 + 2*(8*a^3 + 113*a^2*b + 50*a*b^2 - 27*b^3)*sin(4*d*x + 4*c)*sin(2 *d*x + 2*c) + 4*(a^2*b + 10*a*b^2 + 5*b^3)*sin(2*d*x + 2*c)^2 - ((a^2*b + 10*a*b^2 + 5*b^3)*cos(6*d*x + 6*c) + 2*(9*a^2*b + 10*a*b^2 - 3*b^3)*cos(4* d*x + 4*c) + (a^2*b + 10*a*b^2 + 5*b^3)*cos(2*d*x + 2*c))*cos(8*d*x + 8*c) - (a^2*b + 10*a*b^2 + 5*b^3 - 2*(8*a^3 + 113*a^2*b + 50*a*b^2 - 27*b^3)*c os(4*d*x + 4*c) - 8*(a^2*b + 10*a*b^2 + 5*b^3)*cos(2*d*x + 2*c))*cos(6*d*x + 6*c) - 2*(9*a^2*b + 10*a*b^2 - 3*b^3 - (8*a^3 + 113*a^2*b + 50*a*b^2 - 27*b^3)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - (a^2*b + 10*a*b^2 + 5*b^3)*co s(2*d*x + 2*c) - ((a^2*b + 10*a*b^2 + 5*b^3)*sin(6*d*x + 6*c) + 2*(9*a^2*b + 10*a*b^2 - 3*b^3)*sin(4*d*x + 4*c) + (a^2*b + 10*a*b^2 + 5*b^3)*sin(2*d *x + 2*c))*sin(8*d*x + 8*c) + 2*((8*a^3 + 113*a^2*b + 50*a*b^2 - 27*b^3)*s in(4*d*x + 4*c) + 4*(a^2*b + 10*a*b^2 + 5*b^3)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))/(b^4*cos(8*d*x + 8*c)^2 + 16*b^4*cos(6*d*x + 6*c)^2 + 16*b^4*cos( 2*d*x + 2*c)^2 + b^4*sin(8*d*x + 8*c)^2 + 16*b^4*sin(6*d*x + 6*c)^2 + 16*b ^4*sin(2*d*x + 2*c)^2 - 8*b^4*cos(2*d*x + 2*c) + b^4 + 4*(64*a^2*b^2 - 48* a*b^3 + 9*b^4)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b^2 - 48*a*b^3 + 9*b^4)*s...
Leaf count of result is larger than twice the leaf count of optimal. 896 vs. \(2 (200) = 400\).
Time = 1.01 (sec) , antiderivative size = 896, normalized size of antiderivative = 3.56 \[ \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {24 \, {\left (15 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{4} b - 62 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} b^{3} - 16 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b^{4} - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} b^{5} - 3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{4} - 24 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{3} b + 46 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} b^{2} + 40 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b^{3} + 5 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a b^{2} + \sqrt {a^{2} b^{4} - {\left (a b^{2} - b^{3}\right )} a b^{2}}}{a b^{2} - b^{3}}}}\right )\right )} {\left | -a + b \right |}}{3 \, a^{5} b^{3} - 12 \, a^{4} b^{4} + 14 \, a^{3} b^{5} - 4 \, a^{2} b^{6} - a b^{7}} + \frac {24 \, {\left (15 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{4} b - 62 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} b^{3} - 16 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b^{4} - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} b^{5} + 3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{4} + 24 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{3} b - 46 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} b^{2} - 40 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b^{3} - 5 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a b^{2} - \sqrt {a^{2} b^{4} - {\left (a b^{2} - b^{3}\right )} a b^{2}}}{a b^{2} - b^{3}}}}\right )\right )} {\left | -a + b \right |}}{3 \, a^{5} b^{3} - 12 \, a^{4} b^{4} + 14 \, a^{3} b^{5} - 4 \, a^{2} b^{6} - a b^{7}} - \frac {9 \, {\left (d x + c\right )} {\left (24 \, a + 35 \, b\right )}}{b^{2}} - \frac {24 \, a \tan \left (d x + c\right )^{5} + 123 \, b \tan \left (d x + c\right )^{5} + 48 \, a \tan \left (d x + c\right )^{3} + 280 \, b \tan \left (d x + c\right )^{3} + 24 \, a \tan \left (d x + c\right ) + 165 \, b \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} b^{2}}}{48 \, d} \]
1/48*(24*(15*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^4*b - 62*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b^3 - 16*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b ^4 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*b^5 - 3*sqrt(a^2 - a*b + sqrt(a*b )*(a - b))*sqrt(a*b)*a^4 - 24*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b )*a^3*b + 46*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 + 40*sq rt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^3 + 5*sqrt(a^2 - a*b + sqr t(a*b)*(a - b))*sqrt(a*b)*b^4)*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan( d*x + c)/sqrt((a*b^2 + sqrt(a^2*b^4 - (a*b^2 - b^3)*a*b^2))/(a*b^2 - b^3)) ))*abs(-a + b)/(3*a^5*b^3 - 12*a^4*b^4 + 14*a^3*b^5 - 4*a^2*b^6 - a*b^7) + 24*(15*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^4*b - 62*sqrt(a^2 - a*b - sq rt(a*b)*(a - b))*a^2*b^3 - 16*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b^4 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*b^5 + 3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^4 + 24*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^3 *b - 46*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^2 - 40*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)*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/sqrt((a*b^2 - sqrt(a^2*b^4 - (a*b^2 - b^3)*a*b^2))/(a*b^2 - b^3))))*ab s(-a + b)/(3*a^5*b^3 - 12*a^4*b^4 + 14*a^3*b^5 - 4*a^2*b^6 - a*b^7) - 9*(d *x + c)*(24*a + 35*b)/b^2 - (24*a*tan(d*x + c)^5 + 123*b*tan(d*x + c)^5 + 48*a*tan(d*x + c)^3 + 280*b*tan(d*x + c)^3 + 24*a*tan(d*x + c) + 165*b*...
Time = 18.63 (sec) , antiderivative size = 10319, normalized size of antiderivative = 40.95 \[ \int \frac {\cos ^{10}(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
(atan(((((tan(c + d*x)*(123962*a*b^10 - 3776*a^10*b - 128*a^11 + 11153*b^1 1 - 387826*a^2*b^9 + 2370*a^3*b^8 + 780960*a^4*b^7 - 444642*a^5*b^6 - 3875 34*a^6*b^5 + 261366*a^7*b^4 + 118095*a^8*b^3 - 74000*a^9*b^2))/(64*b^6) + (3*(((9873*a*b^13)/16 + 8*b^14 + (198963*a^2*b^12)/16 - (13467*a^3*b^11)/8 - (240165*a^4*b^10)/8 + (68805*a^5*b^9)/16 + (307047*a^6*b^8)/16 - 1929*a ^7*b^7 - 2766*a^8*b^6 - 152*a^9*b^5)/b^8 + (3*((3*((64*a*b^15 + 1216*a^2*b ^14 - 2064*a^3*b^13 - 480*a^4*b^12 + 1968*a^5*b^11 - 704*a^6*b^10)/b^8 - ( 3*tan(c + d*x)*(a*24i + b*35i)*(49152*a^2*b^13 - 49152*a^3*b^12 - 49152*a^ 4*b^11 + 49152*a^5*b^10))/(2048*b^8))*(a*24i + b*35i))/(32*b^2) - (tan(c + d*x)*(617264*a^2*b^11 - 1024*b^13 - 10240*a*b^12 + 46512*a^3*b^10 - 91953 6*a^4*b^9 - 469488*a^5*b^8 + 498944*a^6*b^7 + 232448*a^7*b^6 + 5120*a^8*b^ 5))/(64*b^6))*(a*24i + b*35i))/(32*b^2))*(a*24i + b*35i))/(32*b^2))*(a*24i + b*35i)*3i)/(32*b^2) + (((tan(c + d*x)*(123962*a*b^10 - 3776*a^10*b - 12 8*a^11 + 11153*b^11 - 387826*a^2*b^9 + 2370*a^3*b^8 + 780960*a^4*b^7 - 444 642*a^5*b^6 - 387534*a^6*b^5 + 261366*a^7*b^4 + 118095*a^8*b^3 - 74000*a^9 *b^2))/(64*b^6) - (3*(((9873*a*b^13)/16 + 8*b^14 + (198963*a^2*b^12)/16 - (13467*a^3*b^11)/8 - (240165*a^4*b^10)/8 + (68805*a^5*b^9)/16 + (307047*a^ 6*b^8)/16 - 1929*a^7*b^7 - 2766*a^8*b^6 - 152*a^9*b^5)/b^8 + (3*((3*((64*a *b^15 + 1216*a^2*b^14 - 2064*a^3*b^13 - 480*a^4*b^12 + 1968*a^5*b^11 - 704 *a^6*b^10)/b^8 + (3*tan(c + d*x)*(a*24i + b*35i)*(49152*a^2*b^13 - 4915...