Integrand size = 11, antiderivative size = 379 \[ \int \frac {1}{a-b \sin ^5(x)} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [5]{b}-\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-b^{2/5}}}-\frac {2 \arctan \left (\frac {(-1)^{2/5} \sqrt [5]{b}-\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}-\frac {2 \arctan \left (\frac {(-1)^{4/5} \sqrt [5]{b}-\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}+\frac {2 \arctan \left (\frac {\sqrt [5]{-1} \sqrt [5]{b}+\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}}+\frac {2 \arctan \left (\frac {(-1)^{3/5} \sqrt [5]{b}+\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}} \]
-2/5*arctan((b^(1/5)-a^(1/5)*tan(1/2*x))/(a^(2/5)-b^(2/5))^(1/2))/a^(4/5)/ (a^(2/5)-b^(2/5))^(1/2)+2/5*arctan(((-1)^(3/5)*b^(1/5)+a^(1/5)*tan(1/2*x)) /(a^(2/5)+(-1)^(1/5)*b^(2/5))^(1/2))/a^(4/5)/(a^(2/5)+(-1)^(1/5)*b^(2/5))^ (1/2)+2/5*arctan(((-1)^(1/5)*b^(1/5)+a^(1/5)*tan(1/2*x))/(a^(2/5)-(-1)^(2/ 5)*b^(2/5))^(1/2))/a^(4/5)/(a^(2/5)-(-1)^(2/5)*b^(2/5))^(1/2)-2/5*arctan(( (-1)^(4/5)*b^(1/5)-a^(1/5)*tan(1/2*x))/(a^(2/5)+(-1)^(3/5)*b^(2/5))^(1/2)) /a^(4/5)/(a^(2/5)+(-1)^(3/5)*b^(2/5))^(1/2)-2/5*arctan(((-1)^(2/5)*b^(1/5) -a^(1/5)*tan(1/2*x))/(a^(2/5)-(-1)^(4/5)*b^(2/5))^(1/2))/a^(4/5)/(a^(2/5)- (-1)^(4/5)*b^(2/5))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 5.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.39 \[ \int \frac {1}{a-b \sin ^5(x)} \, dx=-\frac {8}{5} i \text {RootSum}\left [-i b+5 i b \text {$\#$1}^2-10 i b \text {$\#$1}^4+32 a \text {$\#$1}^5+10 i b \text {$\#$1}^6-5 i b \text {$\#$1}^8+i b \text {$\#$1}^{10}\&,\frac {2 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^3-i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3}{b-4 b \text {$\#$1}^2-16 i a \text {$\#$1}^3+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8}\&\right ] \]
((-8*I)/5)*RootSum[(-I)*b + (5*I)*b*#1^2 - (10*I)*b*#1^4 + 32*a*#1^5 + (10 *I)*b*#1^6 - (5*I)*b*#1^8 + I*b*#1^10 & , (2*ArcTan[Sin[x]/(Cos[x] - #1)]* #1^3 - I*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^3)/(b - 4*b*#1^2 - (16*I)*a*#1^3 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8) & ]
Time = 0.73 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3692, 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.
\(\displaystyle \int \frac {1}{a-b \sin ^5(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a-b \sin (x)^5}dx\) |
\(\Big \downarrow \) 3692 |
\(\displaystyle \int \left (\frac {1}{5 a^{4/5} \left (\sqrt [5]{a}-\sqrt [5]{b} \sin (x)\right )}+\frac {1}{5 a^{4/5} \left (\sqrt [5]{a}+\sqrt [5]{-1} \sqrt [5]{b} \sin (x)\right )}+\frac {1}{5 a^{4/5} \left (\sqrt [5]{a}-(-1)^{2/5} \sqrt [5]{b} \sin (x)\right )}+\frac {1}{5 a^{4/5} \left (\sqrt [5]{a}+(-1)^{3/5} \sqrt [5]{b} \sin (x)\right )}+\frac {1}{5 a^{4/5} \left (\sqrt [5]{a}-(-1)^{4/5} \sqrt [5]{b} \sin (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \arctan \left (\frac {\sqrt [5]{b}-\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-b^{2/5}}}+\frac {2 \arctan \left (\frac {\sqrt [5]{a} \tan \left (\frac {x}{2}\right )+\sqrt [5]{-1} \sqrt [5]{b}}{\sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}}+\frac {2 \arctan \left (\frac {\sqrt [5]{a} \tan \left (\frac {x}{2}\right )+(-1)^{3/5} \sqrt [5]{b}}{\sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}}-\frac {2 \arctan \left (\frac {(-1)^{4/5} \left (\sqrt [5]{-1} \sqrt [5]{a} \tan \left (\frac {x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}-\frac {2 \arctan \left (\frac {(-1)^{2/5} \left ((-1)^{3/5} \sqrt [5]{a} \tan \left (\frac {x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}\) |
(-2*ArcTan[(b^(1/5) - a^(1/5)*Tan[x/2])/Sqrt[a^(2/5) - b^(2/5)]])/(5*a^(4/ 5)*Sqrt[a^(2/5) - b^(2/5)]) + (2*ArcTan[((-1)^(1/5)*b^(1/5) + a^(1/5)*Tan[ x/2])/Sqrt[a^(2/5) - (-1)^(2/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - (-1)^ (2/5)*b^(2/5)]) + (2*ArcTan[((-1)^(3/5)*b^(1/5) + a^(1/5)*Tan[x/2])/Sqrt[a ^(2/5) + (-1)^(1/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) + (-1)^(1/5)*b^(2/5 )]) - (2*ArcTan[((-1)^(4/5)*(b^(1/5) + (-1)^(1/5)*a^(1/5)*Tan[x/2]))/Sqrt[ a^(2/5) + (-1)^(3/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) + (-1)^(3/5)*b^(2/ 5)]) - (2*ArcTan[((-1)^(2/5)*(b^(1/5) + (-1)^(3/5)*a^(1/5)*Tan[x/2]))/Sqrt [a^(2/5) - (-1)^(4/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - (-1)^(4/5)*b^(2 /5)])
3.3.52.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f , n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.01 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.29
method | result | size |
default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{10}+5 a \,\textit {\_Z}^{8}+10 a \,\textit {\_Z}^{6}-32 b \,\textit {\_Z}^{5}+10 a \,\textit {\_Z}^{4}+5 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{8}+4 \textit {\_R}^{6}+6 \textit {\_R}^{4}+4 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{9} a +4 \textit {\_R}^{7} a +6 \textit {\_R}^{5} a -16 \textit {\_R}^{4} b +4 \textit {\_R}^{3} a +\textit {\_R} a}\right )}{5}\) | \(109\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (9765625 a^{10}-9765625 a^{8} b^{2}\right ) \textit {\_Z}^{10}+1953125 a^{8} \textit {\_Z}^{8}+156250 a^{6} \textit {\_Z}^{6}+6250 a^{4} \textit {\_Z}^{4}+125 a^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i x}+\left (-\frac {11718750 a^{10}}{b}+11718750 a^{8} b \right ) \textit {\_R}^{9}+\left (-\frac {1171875 i a^{9}}{b}+1171875 i a^{7} b \right ) \textit {\_R}^{8}+\left (-\frac {2109375 a^{8}}{b}-234375 a^{6} b \right ) \textit {\_R}^{7}+\left (-\frac {218750 i a^{7}}{b}-15625 i a^{5} b \right ) \textit {\_R}^{6}+\left (-\frac {143750 a^{6}}{b}+3125 a^{4} b \right ) \textit {\_R}^{5}-\frac {15625 i a^{5} \textit {\_R}^{4}}{b}-\frac {4375 a^{4} \textit {\_R}^{3}}{b}-\frac {500 i a^{3} \textit {\_R}^{2}}{b}-\frac {50 a^{2} \textit {\_R}}{b}-\frac {6 i a}{b}\right )\) | \(216\) |
1/5*sum((_R^8+4*_R^6+6*_R^4+4*_R^2+1)/(_R^9*a+4*_R^7*a+6*_R^5*a-16*_R^4*b+ 4*_R^3*a+_R*a)*ln(tan(1/2*x)-_R),_R=RootOf(_Z^10*a+5*_Z^8*a+10*_Z^6*a-32*_ Z^5*b+10*_Z^4*a+5*_Z^2*a+a))
Exception generated. \[ \int \frac {1}{a-b \sin ^5(x)} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {1}{a-b \sin ^5(x)} \, dx=\int \frac {1}{a - b \sin ^{5}{\left (x \right )}}\, dx \]
\[ \int \frac {1}{a-b \sin ^5(x)} \, dx=\int { -\frac {1}{b \sin \left (x\right )^{5} - a} \,d x } \]
\[ \int \frac {1}{a-b \sin ^5(x)} \, dx=\int { -\frac {1}{b \sin \left (x\right )^{5} - a} \,d x } \]
Time = 18.73 (sec) , antiderivative size = 1515, normalized size of antiderivative = 4.00 \[ \int \frac {1}{a-b \sin ^5(x)} \, dx=\text {Too large to display} \]
symsum(log(10995116277760*a*b^7*(16*tan(x/2) + 56*root(9765625*a^8*b^2*d^1 0 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d^4 - 125*a^2*d^2 - 1, d, k)*a + 5425*root(9765625*a^8*b^2*d^10 - 9765625*a^10*d ^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d^4 - 125*a^2*d^2 - 1, d , k)^3*a^3 + 196875*root(9765625*a^8*b^2*d^10 - 9765625*a^10*d^10 - 195312 5*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d^4 - 125*a^2*d^2 - 1, d, k)^5*a^5 + 3171875*root(9765625*a^8*b^2*d^10 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d^4 - 125*a^2*d^2 - 1, d, k)^7*a^7 + 19140625*r oot(9765625*a^8*b^2*d^10 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^ 6*d^6 - 6250*a^4*d^4 - 125*a^2*d^2 - 1, d, k)^9*a^9 + 1560*root(9765625*a^ 8*b^2*d^10 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a ^4*d^4 - 125*a^2*d^2 - 1, d, k)^2*a^2*tan(x/2) + 57000*root(9765625*a^8*b^ 2*d^10 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d ^4 - 125*a^2*d^2 - 1, d, k)^4*a^4*tan(x/2) + 925000*root(9765625*a^8*b^2*d ^10 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d^4 - 125*a^2*d^2 - 1, d, k)^6*a^6*tan(x/2) + 5625000*root(9765625*a^8*b^2*d^1 0 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d^4 - 125*a^2*d^2 - 1, d, k)^8*a^8*tan(x/2) - 14000*root(9765625*a^8*b^2*d^10 - 9765625*a^10*d^10 - 1953125*a^8*d^8 - 156250*a^6*d^6 - 6250*a^4*d^4 - 125* a^2*d^2 - 1, d, k)^4*a^3*b - 175000*root(9765625*a^8*b^2*d^10 - 9765625...