Integrand size = 23, antiderivative size = 273 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {x}{2 b}-\frac {2 a \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{5/3} d}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} b^{5/3} d}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} b^{5/3} d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \]
1/2*x/b-1/2*cos(d*x+c)*sin(d*x+c)/b/d-2/3*a*arctan((b^(1/3)+a^(1/3)*tan(1/ 2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/b^(5/3)/d/(a^(2/3)-b^(2/3))^(1/2)+2 /3*a*arctanh((b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/((-1)^(1/3)*a ^(2/3)+b^(2/3))^(1/2))/b^(5/3)/d/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2)+2/3*a* arctanh((b^(1/3)-(-1)^(1/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(-(-1)^(2/3)*a^(2/ 3)+b^(2/3))^(1/2))/b^(5/3)/d/(-(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.33 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {6 (c+d x)-2 i a \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-4 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+2 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]-3 \sin (2 (c+d x))}{12 b d} \]
(6*(c + d*x) - (2*I)*a*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)] - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*# 1^2 + (2*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x] /(Cos[c + d*x] - #1)]*#1^4 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b* #1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1^5) & ] - 3*Sin[2*(c + d*x)])/(12*b*d)
Time = 0.79 (sec) , antiderivative size = 273, 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.
| \(\displaystyle \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^5}{a+b \sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3699 |
| \(\displaystyle \int \left (\frac {\sin ^2(c+d x)}{b}-\frac {a \sin ^2(c+d x)}{b \left (a+b \sin ^3(c+d x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{5/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {2 a \text {arctanh}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}+\frac {x}{2 b}\) |
x/(2*b) - (2*a*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^(5/3)*d) + (2*a*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3) ]])/(3*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*b^(5/3)*d) + (2*a*ArcTanh[(b^ (1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^( 2/3)]])/(3*Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]*b^(5/3)*d) - (Cos[c + d*x]*S in[c + d*x])/(2*b*d)
3.2.83.3.1 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 = 1.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.52
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {8 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {4 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\) | \(141\) |
| default | \(\frac {\frac {\frac {8 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {4 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\) | \(141\) |
| risch | \(\frac {x}{2 b}-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{10} d^{6}-729 b^{12} d^{6}\right ) \textit {\_Z}^{6}-248832 a^{2} b^{8} d^{4} \textit {\_Z}^{4}-28311552 a^{4} b^{4} d^{2} \textit {\_Z}^{2}-1073741824 a^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {243 i d^{5} b^{8}}{33554432 a^{3}}-\frac {243 i d^{5} b^{10}}{33554432 a^{5}}\right ) \textit {\_R}^{5}+\left (-\frac {81 i d^{4} b^{7}}{1048576 a^{3}}+\frac {81 i d^{4} b^{9}}{1048576 a^{5}}\right ) \textit {\_R}^{4}-\frac {81 i d^{3} b^{6} \textit {\_R}^{3}}{32768 a^{3}}+\left (\frac {9 i d^{2} b^{3}}{1024 a}+\frac {9 i d^{2} b^{5}}{512 a^{3}}\right ) \textit {\_R}^{2}-\frac {9 i d \,b^{2} \textit {\_R}}{32 a}+\frac {i b}{a}\right )\right )}{32}-\frac {\sin \left (2 d x +2 c \right )}{4 d b}\) | \(223\) |
1/d*(8/b*((1/8*tan(1/2*d*x+1/2*c)^3-1/8*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x +1/2*c)^2)^2+1/8*arctan(tan(1/2*d*x+1/2*c)))-4/3/b*a*sum(_R^2/(_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 = 1.28 (sec) , antiderivative size = 29175, normalized size of antiderivative = 106.87 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{5}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
-1/4*(4*b*d*integrate(2*(16*a^2*cos(3*d*x + 3*c)^2 + 16*a^2*sin(3*d*x + 3* c)^2 + 3*a*b*cos(d*x + c)*sin(2*d*x + 2*c) - 3*a*b*cos(2*d*x + 2*c)*sin(d* x + c) + a*b*sin(d*x + c) - (a*b*sin(5*d*x + 5*c) - 2*a*b*sin(3*d*x + 3*c) + a*b*sin(d*x + c))*cos(6*d*x + 6*c) - (8*a^2*cos(3*d*x + 3*c) + 3*a*b*si n(4*d*x + 4*c) - 3*a*b*sin(2*d*x + 2*c))*cos(5*d*x + 5*c) - 3*(2*a*b*sin(3 *d*x + 3*c) - a*b*sin(d*x + c))*cos(4*d*x + 4*c) - 2*(4*a^2*cos(d*x + c) + 3*a*b*sin(2*d*x + 2*c))*cos(3*d*x + 3*c) + (a*b*cos(5*d*x + 5*c) - 2*a*b* cos(3*d*x + 3*c) + a*b*cos(d*x + c))*sin(6*d*x + 6*c) + (3*a*b*cos(4*d*x + 4*c) - 3*a*b*cos(2*d*x + 2*c) - 8*a^2*sin(3*d*x + 3*c) + a*b)*sin(5*d*x + 5*c) + 3*(2*a*b*cos(3*d*x + 3*c) - a*b*cos(d*x + c))*sin(4*d*x + 4*c) + 2 *(3*a*b*cos(2*d*x + 2*c) - 4*a^2*sin(d*x + c) - a*b)*sin(3*d*x + 3*c))/(b^ 3*cos(6*d*x + 6*c)^2 + 9*b^3*cos(4*d*x + 4*c)^2 + 64*a^2*b*cos(3*d*x + 3*c )^2 + 9*b^3*cos(2*d*x + 2*c)^2 + b^3*sin(6*d*x + 6*c)^2 + 9*b^3*sin(4*d*x + 4*c)^2 + 64*a^2*b*sin(3*d*x + 3*c)^2 - 48*a*b^2*cos(3*d*x + 3*c)*sin(2*d *x + 2*c) + 9*b^3*sin(2*d*x + 2*c)^2 - 6*b^3*cos(2*d*x + 2*c) + b^3 - 2*(3 *b^3*cos(4*d*x + 4*c) - 3*b^3*cos(2*d*x + 2*c) - 8*a*b^2*sin(3*d*x + 3*c) + b^3)*cos(6*d*x + 6*c) - 6*(3*b^3*cos(2*d*x + 2*c) + 8*a*b^2*sin(3*d*x + 3*c) - b^3)*cos(4*d*x + 4*c) - 2*(8*a*b^2*cos(3*d*x + 3*c) + 3*b^3*sin(4*d *x + 4*c) - 3*b^3*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 6*(8*a*b^2*cos(3*d* x + 3*c) - 3*b^3*sin(2*d*x + 2*c))*sin(4*d*x + 4*c) + 16*(3*a*b^2*cos(2...
\[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{5}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
Time = 14.51 (sec) , antiderivative size = 1962, normalized size of antiderivative = 7.19 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
tan(c/2 + (d*x)/2)^3/(b*d + 2*b*d*tan(c/2 + (d*x)/2)^2 + b*d*tan(c/2 + (d* x)/2)^4) - tan(c/2 + (d*x)/2)/(b*d + 2*b*d*tan(c/2 + (d*x)/2)^2 + b*d*tan( c/2 + (d*x)/2)^4) + symsum(log((134217728*a^9*b^2 - 16777216*a^11 - 402653 184*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^ 2 + a^6, z, k)*a^8*b^4 + 50331648*a^10*b*tan(c/2 + (d*x)/2) - 2415919104*r oot(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a ^6, z, k)^2*a^7*b^6 + 914358272*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243 *a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^2*a^9*b^4 + 7247757312*root(729 *a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^3*a^6*b^8 - 478150656*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^ 8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^3*a^8*b^6 + 10871635968*root(729*a^2*b ^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^4*a ^5*b^10 - 21214789632*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z ^4 - 27*a^4*b^4*z^2 + a^6, z, k)^4*a^7*b^8 - 301989888*root(729*a^2*b^10*z ^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^4*a^9*b^ 6 - 32614907904*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 2 7*a^4*b^4*z^2 + a^6, z, k)^5*a^4*b^12 + 59567505408*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a^4*b^4*z^2 + a^6, z, k)^5*a^6*b^10 + 4529848320*root(729*a^2*b^10*z^6 - 729*b^12*z^6 + 243*a^2*b^8*z^4 - 27*a ^4*b^4*z^2 + a^6, z, k)^5*a^8*b^8 + 55717134336*root(729*a^2*b^10*z^6 -...