Integrand size = 10, antiderivative size = 384 \[ \int \frac {1}{a+b \sin ^5(x)} \, dx=\frac {2 \arctan \left (\frac {\sqrt [5]{b}+\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-b^{2/5}}}+\frac {2 \arctan \left (\frac {(-1)^{2/5} \sqrt [5]{b}+\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}+\frac {2 \arctan \left (\frac {(-1)^{4/5} \sqrt [5]{b}+\sqrt [5]{a} \tan \left (\frac {x}{2}\right )}{\sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}-\frac {2 \arctan \left (\frac {(-1)^{3/5} \left (\sqrt [5]{b}+(-1)^{2/5} \sqrt [5]{a} \tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}}-\frac {2 \arctan \left (\frac {\sqrt [5]{-1} \left (\sqrt [5]{b}+(-1)^{4/5} \sqrt [5]{a} \tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}} \]
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)+(-1)^(2/5)*a^(1/5)*t an(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)+(-1)^(4/5)*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*arcta n(((-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 = 5.13 (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 = 0.89 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 |
\(\displaystyle \int \left (-\frac {1}{5 a^{4/5} \left (-\sqrt [5]{a}-\sqrt [5]{b} \sin (x)\right )}-\frac {1}{5 a^{4/5} \left (\sqrt [5]{-1} \sqrt [5]{b} \sin (x)-\sqrt [5]{a}\right )}-\frac {1}{5 a^{4/5} \left (-\sqrt [5]{a}-(-1)^{2/5} \sqrt [5]{b} \sin (x)\right )}-\frac {1}{5 a^{4/5} \left ((-1)^{3/5} \sqrt [5]{b} \sin (x)-\sqrt [5]{a}\right )}-\frac {1}{5 a^{4/5} \left (-\sqrt [5]{a}-(-1)^{4/5} \sqrt [5]{b} \sin (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \arctan \left (\frac {\sqrt [5]{a} \tan \left (\frac {x}{2}\right )+\sqrt [5]{b}}{\sqrt {a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-b^{2/5}}}+\frac {2 \arctan \left (\frac {\sqrt [5]{a} \tan \left (\frac {x}{2}\right )+(-1)^{2/5} \sqrt [5]{b}}{\sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{4/5} b^{2/5}}}+\frac {2 \arctan \left (\frac {\sqrt [5]{a} \tan \left (\frac {x}{2}\right )+(-1)^{4/5} \sqrt [5]{b}}{\sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+(-1)^{3/5} b^{2/5}}}-\frac {2 \arctan \left (\frac {(-1)^{3/5} \left ((-1)^{2/5} \sqrt [5]{a} \tan \left (\frac {x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}+\sqrt [5]{-1} b^{2/5}}}-\frac {2 \arctan \left (\frac {\sqrt [5]{-1} \left ((-1)^{4/5} \sqrt [5]{a} \tan \left (\frac {x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt {a^{2/5}-(-1)^{2/5} b^{2/5}}}\) |
(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)^(2/5)*b^(1/5) + 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)]) + (2*ArcTan[((-1)^(4/5)*b^(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)^(3/5)*(b^(1/5) + (-1)^(2/5)*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)^(1/5)*(b^(1/5) + (-1)^(4/5)*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)])
3.3.49.3.1 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 = 1.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.28
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 = 19.69 (sec) , antiderivative size = 1515, normalized size of antiderivative = 3.95 \[ \int \frac {1}{a+b \sin ^5(x)} \, dx=\text {Too large to display} \]
symsum(log(-10995116277760*a*b^7*(16*tan(x/2) + 56*root(9765625*a^8*b^2*d^ 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)*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 - 19531 25*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* 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)^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^ 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)^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 - 976562...