Integrand size = 23, antiderivative size = 293 \[ \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {a x}{b^2}+\frac {2 a^{4/3} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^2 d}+\frac {2 a^{4/3} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} b^2 d}-\frac {2 a^{4/3} \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} b^2 d}-\frac {\cos (c+d x)}{b d}+\frac {\cos ^3(c+d x)}{3 b d} \]
-a*x/b^2-cos(d*x+c)/b/d+1/3*cos(d*x+c)^3/b/d+2/3*a^(4/3)*arctan((b^(1/3)+a ^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/b^2/d/(a^(2/3)-b^(2/3) )^(1/2)+2/3*a^(4/3)*arctan(((-1)^(2/3)*b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c)) /(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2))/b^2/d/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1 /2)-2/3*a^(4/3)*arctan((-1)^(1/3)*(b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+ 1/2*c))/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/b^2/d/(a^(2/3)-(-1)^(2/3)*b^(2 /3))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.38 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.56 \[ \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {12 a c+12 a d x+9 b \cos (c+d x)-b \cos (3 (c+d x))+8 i a^2 \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]}{12 b^2 d} \]
-1/12*(12*a*c + 12*a*d*x + 9*b*Cos[c + d*x] - b*Cos[3*(c + d*x)] + (8*I)*a ^2*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 - I*Log[1 - 2*Cos[c + d*x]* #1 + #1^2]*#1)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ])/(b^2*d)
Time = 0.62 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3699, 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.
| \(\displaystyle \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{a+b \sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {a^2}{b^2 \left (a+b \sin ^3(c+d x)\right )}-\frac {a}{b^2}+\frac {\sin ^3(c+d x)}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^{4/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a^{4/3} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 a^{4/3} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^2 d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {a x}{b^2}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x)}{b d}\) |
-((a*x)/b^2) + (2*a^(4/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt [a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/3)]*b^2*d) + (2*a^(4/3)*ArcTa n[((-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3 )*b^(2/3)]])/(3*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b^2*d) - (2*a^(4/3)*Arc Tan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2]))/Sqrt[a^(2 /3) - (-1)^(2/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^2*d) - Cos[c + d*x]/(b*d) + Cos[c + d*x]^3/(3*b*d)
3.2.89.3.1 Defintions of rubi rules used
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n) ^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4] || Gt Q[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.51 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.49
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 b}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b^{2}}}{d}\) | \(143\) |
| default | \(\frac {-\frac {2 \left (\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 b}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b^{2}}}{d}\) | \(143\) |
| risch | \(-\frac {a x}{b^{2}}-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}}{8 b d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b d}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{12} d^{6}-729 b^{14} d^{6}\right ) \textit {\_Z}^{6}+995328 a^{4} b^{8} d^{4} \textit {\_Z}^{4}+452984832 a^{6} b^{4} d^{2} \textit {\_Z}^{2}+68719476736 a^{8}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {243 b^{9} d^{5}}{536870912 a^{4}}-\frac {243 b^{11} d^{5}}{536870912 a^{6}}\right ) \textit {\_R}^{5}+\left (-\frac {81 i d^{4} b^{7}}{16777216 a^{3}}+\frac {81 i d^{4} b^{9}}{16777216 a^{5}}\right ) \textit {\_R}^{4}+\left (\frac {135 b^{5} d^{3}}{262144 a^{2}}+\frac {27 b^{7} d^{3}}{262144 a^{4}}\right ) \textit {\_R}^{3}-\frac {27 i d^{2} b^{3} \textit {\_R}^{2}}{4096 a}+\frac {9 b d \textit {\_R}}{64}-\frac {2 i a}{b}\right )\right )}{64}+\frac {\cos \left (3 d x +3 c \right )}{12 b d}\) | \(247\) |
1/d*(-2/b^2*((2*b*tan(1/2*d*x+1/2*c)^2+2/3*b)/(1+tan(1/2*d*x+1/2*c)^2)^3+a *arctan(tan(1/2*d*x+1/2*c)))+1/3*a^2/b^2*sum((_R^4+2*_R^2+1)/(_R^5*a+2*_R^ 3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_ Z^3*b+3*_Z^2*a+a)))
Result contains complex when optimal does not.
Time = 1.23 (sec) , antiderivative size = 29350, normalized size of antiderivative = 100.17 \[ \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\sin ^{6}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{6}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
-1/12*(96*a^2*b^2*d*integrate(-(8*a*cos(3*d*x + 3*c)^2 - b*cos(3*d*x + 3*c )*sin(6*d*x + 6*c) + 3*b*cos(3*d*x + 3*c)*sin(4*d*x + 4*c) + b*cos(6*d*x + 6*c)*sin(3*d*x + 3*c) - 3*b*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) + 8*a*sin(3 *d*x + 3*c)^2 - 3*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + (3*b*cos(2*d*x + 2 *c) - b)*sin(3*d*x + 3*c))/(b^4*cos(6*d*x + 6*c)^2 + 9*b^4*cos(4*d*x + 4*c )^2 + 64*a^2*b^2*cos(3*d*x + 3*c)^2 + 9*b^4*cos(2*d*x + 2*c)^2 + b^4*sin(6 *d*x + 6*c)^2 + 9*b^4*sin(4*d*x + 4*c)^2 + 64*a^2*b^2*sin(3*d*x + 3*c)^2 - 48*a*b^3*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 9*b^4*sin(2*d*x + 2*c)^2 - 6 *b^4*cos(2*d*x + 2*c) + b^4 - 2*(3*b^4*cos(4*d*x + 4*c) - 3*b^4*cos(2*d*x + 2*c) - 8*a*b^3*sin(3*d*x + 3*c) + b^4)*cos(6*d*x + 6*c) - 6*(3*b^4*cos(2 *d*x + 2*c) + 8*a*b^3*sin(3*d*x + 3*c) - b^4)*cos(4*d*x + 4*c) - 2*(8*a*b^ 3*cos(3*d*x + 3*c) + 3*b^4*sin(4*d*x + 4*c) - 3*b^4*sin(2*d*x + 2*c))*sin( 6*d*x + 6*c) + 6*(8*a*b^3*cos(3*d*x + 3*c) - 3*b^4*sin(2*d*x + 2*c))*sin(4 *d*x + 4*c) + 16*(3*a*b^3*cos(2*d*x + 2*c) - a*b^3)*sin(3*d*x + 3*c)), x) + 12*a*d*x - b*cos(3*d*x + 3*c) + 9*b*cos(d*x + c))/(b^2*d)
\[ \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{6}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
Time = 15.21 (sec) , antiderivative size = 1800, normalized size of antiderivative = 6.14 \[ \int \frac {\sin ^6(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
symsum(log((1073741824*a^13*tan(c/2 + (d*x)/2) + 2013265920*root(729*a^2*b ^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)*a^1 2*b^2 - 4831838208*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^2*a^10*b^5 + 268435456*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^2*a^12*b^3 + 33722204160*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27 *a^6*b^4*z^2 + a^8, z, k)^3*a^10*b^6 - 15703474176*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^4*a^8*b^9 + 4831838208*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6 *b^4*z^2 + a^8, z, k)^4*a^10*b^7 - 130459631616*root(729*a^2*b^12*z^6 - 72 9*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^5*a^6*b^12 + 15 4014842880*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6 *b^4*z^2 + a^8, z, k)^5*a^8*b^10 - 35332816896*root(729*a^2*b^12*z^6 - 729 *b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^6*a^6*b^13 + 217 43271936*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b ^4*z^2 + a^8, z, k)^6*a^8*b^11 - 130459631616*root(729*a^2*b^12*z^6 - 729* b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)^7*a^4*b^16 + 1223 05904640*root(729*a^2*b^12*z^6 - 729*b^14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b ^4*z^2 + a^8, z, k)^7*a^6*b^14 - 3221225472*root(729*a^2*b^12*z^6 - 729*b^ 14*z^6 + 243*a^4*b^8*z^4 + 27*a^6*b^4*z^2 + a^8, z, k)*a^11*b^3*tan(c/2...