Integrand size = 24, antiderivative size = 182 \[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {(8 a+3 b) x}{8 b^2}+\frac {a^{5/4} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^2 d}+\frac {a^{5/4} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^2 d}+\frac {5 \cos (c+d x) \sin (c+d x)}{8 b d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{4 b d} \] Output:
-1/8*(8*a+3*b)*x/b^2+1/2*a^(5/4)*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c) /a^(1/4))/(a^(1/2)-b^(1/2))^(1/2)/b^2/d+1/2*a^(5/4)*arctan((a^(1/2)+b^(1/2 ))^(1/2)*tan(d*x+c)/a^(1/4))/(a^(1/2)+b^(1/2))^(1/2)/b^2/d+5/8*cos(d*x+c)* sin(d*x+c)/b/d-1/4*cos(d*x+c)^3*sin(d*x+c)/b/d
Time = 3.01 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {4 (8 a+3 b) (c+d x)-\frac {16 a^{3/2} \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {16 a^{3/2} \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}-8 b \sin (2 (c+d x))+b \sin (4 (c+d x))}{32 b^2 d} \] Input:
Integrate[Sin[c + d*x]^8/(a - b*Sin[c + d*x]^4),x]
Output:
-1/32*(4*(8*a + 3*b)*(c + d*x) - (16*a^(3/2)*ArcTan[((Sqrt[a] + Sqrt[b])*T an[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]*Sqrt[b]] + (16*a ^(3/2)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b ]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]] - 8*b*Sin[2*(c + d*x)] + b*Sin[4*(c + d*x) ])/(b^2*d)
Time = 0.45 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3696, 1610, 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.
| \(\displaystyle \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^8}{a-b \sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\tan ^8(c+d x)}{\left (\tan ^2(c+d x)+1\right )^3 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 1610 |
| \(\displaystyle \frac {\int \left (\frac {\left (\tan ^2(c+d x)+1\right ) a^2}{b^2 \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}+\frac {-a-b}{b^2 \left (\tan ^2(c+d x)+1\right )}+\frac {2}{b \left (\tan ^2(c+d x)+1\right )^2}-\frac {1}{b \left (\tan ^2(c+d x)+1\right )^3}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^{5/4} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {a^{5/4} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {(a+b) \arctan (\tan (c+d x))}{b^2}+\frac {5 \arctan (\tan (c+d x))}{8 b}+\frac {5 \tan (c+d x)}{8 b \left (\tan ^2(c+d x)+1\right )}-\frac {\tan (c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}}{d}\) |
Input:
Int[Sin[c + d*x]^8/(a - b*Sin[c + d*x]^4),x]
Output:
((5*ArcTan[Tan[c + d*x]])/(8*b) - ((a + b)*ArcTan[Tan[c + d*x]])/b^2 + (a^ (5/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt [a] - Sqrt[b]]*b^2) + (a^(5/4)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x ])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^2) - Tan[c + d*x]/(4*b*(1 + Tan[ c + d*x]^2)^2) + (5*Tan[c + d*x])/(8*b*(1 + Tan[c + d*x]^2)))/d
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a + b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 *a*c, 0] && IntegerQ[q] && IntegerQ[m]
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 = 3.38 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.15
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{2} \left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{b^{2}}-\frac {\frac {-\frac {5 \tan \left (d x +c \right )^{3} b}{8}-\frac {3 \tan \left (d x +c \right ) b}{8}}{\left (\tan \left (d x +c \right )^{2}+1\right )^{2}}+\frac {\left (8 a +3 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{b^{2}}}{d}\) | \(209\) |
| default | \(\frac {\frac {a^{2} \left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{b^{2}}-\frac {\frac {-\frac {5 \tan \left (d x +c \right )^{3} b}{8}-\frac {3 \tan \left (d x +c \right ) b}{8}}{\left (\tan \left (d x +c \right )^{2}+1\right )^{2}}+\frac {\left (8 a +3 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{b^{2}}}{d}\) | \(209\) |
| risch | \(-\frac {a x}{b^{2}}-\frac {3 x}{8 b}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a \,b^{8} d^{4}-b^{9} d^{4}\right ) \textit {\_Z}^{4}+8192 a^{3} b^{4} d^{2} \textit {\_Z}^{2}+16777216 a^{5}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {i b^{5} d^{3}}{131072 a^{2}}-\frac {i b^{6} d^{3}}{131072 a^{3}}\right ) \textit {\_R}^{3}+\left (-\frac {b^{3} d^{2}}{2048 a}+\frac {b^{4} d^{2}}{2048 a^{2}}\right ) \textit {\_R}^{2}+\left (\frac {i b d}{32}+\frac {i b^{2} d}{32 a}\right ) \textit {\_R} -\frac {2 a}{b}-1\right )\right )}{256}-\frac {\sin \left (4 d x +4 c \right )}{32 b d}\) | \(209\) |
Input:
int(sin(d*x+c)^8/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
Output:
1/d*(a^2/b^2*(a-b)*(1/2*((a*b)^(1/2)+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))+1/2*( (a*b)^(1/2)-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/b^2*((-5/8*tan(d*x+c)^3*b -3/8*tan(d*x+c)*b)/(tan(d*x+c)^2+1)^2+1/8*(8*a+3*b)*arctan(tan(d*x+c))))
Leaf count of result is larger than twice the leaf count of optimal. 1311 vs. \(2 (140) = 280\).
Time = 0.26 (sec) , antiderivative size = 1311, normalized size of antiderivative = 7.20 \[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
Output:
-1/8*(b^2*d*sqrt(-((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d ^4)) + a^3)/((a*b^4 - b^5)*d^2))*log(1/4*a^3*cos(d*x + c)^2 - 1/4*a^3 - 1/ 4*(2*(a^2*b^3 - a*b^4)*d^2*cos(d*x + c)^2 - (a^2*b^3 - a*b^4)*d^2)*sqrt(a^ 5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + 1/2*(a^2*b^2*d*cos(d*x + c)*sin(d*x + c) - (a*b^5 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4))*cos(d*x + c)*sin(d*x + c))*sqrt(-((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + a^3)/((a*b^4 - b^5)*d^2))) - b^2*d*sqrt(-((a*b^4 - b^5)*d^2* sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + a^3)/((a*b^4 - b^5)*d^2))*log( 1/4*a^3*cos(d*x + c)^2 - 1/4*a^3 - 1/4*(2*(a^2*b^3 - a*b^4)*d^2*cos(d*x + c)^2 - (a^2*b^3 - a*b^4)*d^2)*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - 1/2*(a^2*b^2*d*cos(d*x + c)*sin(d*x + c) - (a*b^5 - b^6)*d^3*sqrt(a^5/((a^ 2*b^7 - 2*a*b^8 + b^9)*d^4))*cos(d*x + c)*sin(d*x + c))*sqrt(-((a*b^4 - b^ 5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + a^3)/((a*b^4 - b^5)*d^2 ))) - b^2*d*sqrt(((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^ 4)) - a^3)/((a*b^4 - b^5)*d^2))*log(-1/4*a^3*cos(d*x + c)^2 + 1/4*a^3 - 1/ 4*(2*(a^2*b^3 - a*b^4)*d^2*cos(d*x + c)^2 - (a^2*b^3 - a*b^4)*d^2)*sqrt(a^ 5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + 1/2*(a^2*b^2*d*cos(d*x + c)*sin(d*x + c) + (a*b^5 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4))*cos(d*x + c)*sin(d*x + c))*sqrt(((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - a^3)/((a*b^4 - b^5)*d^2))) + b^2*d*sqrt(((a*b^4 - b^5)*d^2...
Timed out. \[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**8/(a-b*sin(d*x+c)**4),x)
Output:
Timed out
\[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{8}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:
integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
Output:
-1/32*(512*a^2*b^2*d*integrate((b*cos(8*d*x + 8*c)*cos(4*d*x + 4*c) - 4*b* cos(6*d*x + 6*c)*cos(4*d*x + 4*c) - 2*(8*a - 3*b)*cos(4*d*x + 4*c)^2 + b*s in(8*d*x + 8*c)*sin(4*d*x + 4*c) - 4*b*sin(6*d*x + 6*c)*sin(4*d*x + 4*c) - 2*(8*a - 3*b)*sin(4*d*x + 4*c)^2 - 4*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - (4*b*cos(2*d*x + 2*c) - b)*cos(4*d*x + 4*c))/(b^4*cos(8*d*x + 8*c)^2 + 1 6*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)*sin(4*d*x + 4*c)^2 + 16*(8*a*b^3 - 3*b^4 )*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - 2*(4*b^4*cos(6*d*x + 6*c) + 4*b^4*co s(2*d*x + 2*c) - b^4 + 2*(8*a*b^3 - 3*b^4)*cos(4*d*x + 4*c))*cos(8*d*x + 8 *c) + 8*(4*b^4*cos(2*d*x + 2*c) - b^4 + 2*(8*a*b^3 - 3*b^4)*cos(4*d*x + 4* c))*cos(6*d*x + 6*c) - 4*(8*a*b^3 - 3*b^4 - 4*(8*a*b^3 - 3*b^4)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 4*(2*b^4*sin(6*d*x + 6*c) + 2*b^4*sin(2*d*x + 2 *c) + (8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c))*sin(8*d*x + 8*c) + 16*(2*b^4*sin (2*d*x + 2*c) + (8*a*b^3 - 3*b^4)*sin(4*d*x + 4*c))*sin(6*d*x + 6*c)), x) + 4*(8*a + 3*b)*d*x + b*sin(4*d*x + 4*c) - 8*b*sin(2*d*x + 2*c))/(b^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (140) = 280\).
Time = 0.80 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.53 \[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {4 \, {\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{3} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} b - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a b^{2} + \sqrt {a^{2} b^{4} - {\left (a b^{2} - b^{3}\right )} a b^{2}}}{a b^{2} - b^{3}}}}\right )\right )} {\left | -a + b \right |}}{3 \, a^{4} b^{2} - 12 \, a^{3} b^{3} + 14 \, a^{2} b^{4} - 4 \, a b^{5} - b^{6}} + \frac {4 \, {\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{3} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} b - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (d x + c\right )}{\sqrt {\frac {a b^{2} - \sqrt {a^{2} b^{4} - {\left (a b^{2} - b^{3}\right )} a b^{2}}}{a b^{2} - b^{3}}}}\right )\right )} {\left | -a + b \right |}}{3 \, a^{4} b^{2} - 12 \, a^{3} b^{3} + 14 \, a^{2} b^{4} - 4 \, a b^{5} - b^{6}} - \frac {{\left (d x + c\right )} {\left (8 \, a + 3 \, b\right )}}{b^{2}} + \frac {5 \, \tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2} b}}{8 \, d} \] Input:
integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4),x, algorithm="giac")
Output:
1/8*(4*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^3 - 6*sqrt(a^2 - a*b + sqr t(a*b)*(a - b))*a^2*b - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^2)*(pi*flo or((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^4*b^2 - 12*a^3*b^3 + 14*a^2*b^4 - 4*a*b^5 - b^6) + 4*(3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))* a^3 - 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b - sqrt(a^2 - a*b - sqrt( a*b)*(a - b))*a*b^2)*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + c)/s qrt((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^4*b^2 - 12*a^3*b^3 + 14*a^2*b^4 - 4*a*b^5 - b^6) - (d*x + c)*(8* a + 3*b)/b^2 + (5*tan(d*x + c)^3 + 3*tan(d*x + c))/((tan(d*x + c)^2 + 1)^2 *b))/d
Time = 40.02 (sec) , antiderivative size = 5022, normalized size of antiderivative = 27.59 \[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:
int(sin(c + d*x)^8/(a - b*sin(c + d*x)^4),x)
Output:
(atan(((((((2048*a^3*b^10 + 8192*a^4*b^9 - 22528*a^5*b^8 + 12288*a^6*b^7)/ (64*b^5) - (tan(c + d*x)*(-((a^5*b^9)^(1/2) + a^3*b^4)/(16*(a*b^8 - b^9))) ^(1/2)*(12288*a^2*b^11 - 12288*a^3*b^10 - 12288*a^4*b^9 + 12288*a^5*b^8))/ (16*b^4))*(-((a^5*b^9)^(1/2) + a^3*b^4)/(16*(a*b^8 - b^9)))^(1/2) - (tan(c + d*x)*(432*a^2*b^9 + 1584*a^3*b^8 - 880*a^4*b^7 - 5488*a^5*b^6 + 2048*a^ 6*b^5 + 2304*a^7*b^4))/(16*b^4))*(-((a^5*b^9)^(1/2) + a^3*b^4)/(16*(a*b^8 - b^9)))^(1/2) - (144*a^3*b^8 + 624*a^4*b^7 + 112*a^5*b^6 - 1648*a^6*b^5 + 1536*a^7*b^4 - 768*a^8*b^3)/(64*b^5))*(-((a^5*b^9)^(1/2) + a^3*b^4)/(16*( a*b^8 - b^9)))^(1/2) + (tan(c + d*x)*(9*a^4*b^5 - 96*a^9 - 336*a^8*b + 93* a^5*b^4 + 259*a^6*b^3 + 71*a^7*b^2))/(16*b^4))*(-((a^5*b^9)^(1/2) + a^3*b^ 4)/(16*(a*b^8 - b^9)))^(1/2)*1i - (((((2048*a^3*b^10 + 8192*a^4*b^9 - 2252 8*a^5*b^8 + 12288*a^6*b^7)/(64*b^5) + (tan(c + d*x)*(-((a^5*b^9)^(1/2) + a ^3*b^4)/(16*(a*b^8 - b^9)))^(1/2)*(12288*a^2*b^11 - 12288*a^3*b^10 - 12288 *a^4*b^9 + 12288*a^5*b^8))/(16*b^4))*(-((a^5*b^9)^(1/2) + a^3*b^4)/(16*(a* b^8 - b^9)))^(1/2) + (tan(c + d*x)*(432*a^2*b^9 + 1584*a^3*b^8 - 880*a^4*b ^7 - 5488*a^5*b^6 + 2048*a^6*b^5 + 2304*a^7*b^4))/(16*b^4))*(-((a^5*b^9)^( 1/2) + a^3*b^4)/(16*(a*b^8 - b^9)))^(1/2) - (144*a^3*b^8 + 624*a^4*b^7 + 1 12*a^5*b^6 - 1648*a^6*b^5 + 1536*a^7*b^4 - 768*a^8*b^3)/(64*b^5))*(-((a^5* b^9)^(1/2) + a^3*b^4)/(16*(a*b^8 - b^9)))^(1/2) - (tan(c + d*x)*(9*a^4*b^5 - 96*a^9 - 336*a^8*b + 93*a^5*b^4 + 259*a^6*b^3 + 71*a^7*b^2))/(16*b^4...
\[ \int \frac {\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\sin \left (d x +c \right )^{8}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:
int(sin(d*x+c)^8/(a-b*sin(d*x+c)^4),x)
Output:
- int(sin(c + d*x)**8/(sin(c + d*x)**4*b - a),x)