Integrand size = 10, antiderivative size = 245 \[ \int \frac {1}{a+b \cos ^8(x)} \, dx=\frac {\arctan \left (\frac {\sqrt {\sqrt [4]{-a}-\sqrt [4]{b}} \cot (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-\sqrt [4]{b}}}+\frac {\arctan \left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \cot (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}+\frac {\arctan \left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \cot (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}+\frac {\arctan \left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \cot (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}} \] Output:
1/4*arctan(((-a)^(1/4)-b^(1/4))^(1/2)*cot(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^ (1/4)-b^(1/4))^(1/2)+1/4*arctan(((-a)^(1/4)-I*b^(1/4))^(1/2)*cot(x)/(-a)^( 1/8))/(-a)^(7/8)/((-a)^(1/4)-I*b^(1/4))^(1/2)+1/4*arctan(((-a)^(1/4)+I*b^( 1/4))^(1/2)*cot(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)+I*b^(1/4))^(1/2)+1/4 *arctan(((-a)^(1/4)+b^(1/4))^(1/2)*cot(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/ 4)+b^(1/4))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 5.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.70 \[ \int \frac {1}{a+b \cos ^8(x)} \, dx=8 \text {RootSum}\left [b+8 b \text {$\#$1}+28 b \text {$\#$1}^2+56 b \text {$\#$1}^3+256 a \text {$\#$1}^4+70 b \text {$\#$1}^4+56 b \text {$\#$1}^5+28 b \text {$\#$1}^6+8 b \text {$\#$1}^7+b \text {$\#$1}^8\&,\frac {2 \arctan \left (\frac {\sin (2 x)}{\cos (2 x)-\text {$\#$1}}\right ) \text {$\#$1}^3-i \log \left (1-2 \cos (2 x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3}{b+7 b \text {$\#$1}+21 b \text {$\#$1}^2+128 a \text {$\#$1}^3+35 b \text {$\#$1}^3+35 b \text {$\#$1}^4+21 b \text {$\#$1}^5+7 b \text {$\#$1}^6+b \text {$\#$1}^7}\&\right ] \] Input:
Integrate[(a + b*Cos[x]^8)^(-1),x]
Output:
8*RootSum[b + 8*b*#1 + 28*b*#1^2 + 56*b*#1^3 + 256*a*#1^4 + 70*b*#1^4 + 56 *b*#1^5 + 28*b*#1^6 + 8*b*#1^7 + b*#1^8 & , (2*ArcTan[Sin[2*x]/(Cos[2*x] - #1)]*#1^3 - I*Log[1 - 2*Cos[2*x]*#1 + #1^2]*#1^3)/(b + 7*b*#1 + 21*b*#1^2 + 128*a*#1^3 + 35*b*#1^3 + 35*b*#1^4 + 21*b*#1^5 + 7*b*#1^6 + b*#1^7) & ]
Time = 0.72 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3690, 3042, 3660, 216}
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.
\(\displaystyle \int \frac {1}{a+b \cos ^8(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a+b \sin \left (x+\frac {\pi }{2}\right )^8}dx\) |
\(\Big \downarrow \) 3690 |
\(\displaystyle \frac {\int \frac {1}{1-\frac {\sqrt [4]{b} \cos ^2(x)}{\sqrt [4]{-a}}}dx}{4 a}+\frac {\int \frac {1}{1-\frac {i \sqrt [4]{b} \cos ^2(x)}{\sqrt [4]{-a}}}dx}{4 a}+\frac {\int \frac {1}{\frac {i \sqrt [4]{b} \cos ^2(x)}{\sqrt [4]{-a}}+1}dx}{4 a}+\frac {\int \frac {1}{\frac {\sqrt [4]{b} \cos ^2(x)}{\sqrt [4]{-a}}+1}dx}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{1-\frac {\sqrt [4]{b} \sin \left (x+\frac {\pi }{2}\right )^2}{\sqrt [4]{-a}}}dx}{4 a}+\frac {\int \frac {1}{1-\frac {i \sqrt [4]{b} \sin \left (x+\frac {\pi }{2}\right )^2}{\sqrt [4]{-a}}}dx}{4 a}+\frac {\int \frac {1}{\frac {i \sqrt [4]{b} \sin \left (x+\frac {\pi }{2}\right )^2}{\sqrt [4]{-a}}+1}dx}{4 a}+\frac {\int \frac {1}{\frac {\sqrt [4]{b} \sin \left (x+\frac {\pi }{2}\right )^2}{\sqrt [4]{-a}}+1}dx}{4 a}\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle -\frac {\int \frac {1}{\left (1-\frac {i \sqrt [4]{b}}{\sqrt [4]{-a}}\right ) \cot ^2(x)+1}d\cot (x)}{4 a}-\frac {\int \frac {1}{\left (\frac {i \sqrt [4]{b}}{\sqrt [4]{-a}}+1\right ) \cot ^2(x)+1}d\cot (x)}{4 a}-\frac {\int \frac {1}{\left (\frac {\sqrt [4]{b}}{\sqrt [4]{-a}}+1\right ) \cot ^2(x)+1}d\cot (x)}{4 a}-\frac {\int \frac {1}{\left (\frac {\sqrt [4]{b} a}{(-a)^{5/4}}+1\right ) \cot ^2(x)+1}d\cot (x)}{4 a}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\sqrt [8]{-a} \arctan \left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \cot (x)}{\sqrt [8]{-a}}\right )}{4 a \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}-\frac {\sqrt [8]{-a} \arctan \left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \cot (x)}{\sqrt [8]{-a}}\right )}{4 a \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}-\frac {\sqrt [8]{-a} \arctan \left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \cot (x)}{\sqrt [8]{-a}}\right )}{4 a \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}}-\frac {(-a)^{5/8} \arctan \left (\frac {\sqrt {a \sqrt [4]{b}+(-a)^{5/4}} \cot (x)}{(-a)^{5/8}}\right )}{4 a \sqrt {a \sqrt [4]{b}+(-a)^{5/4}}}\) |
Input:
Int[(a + b*Cos[x]^8)^(-1),x]
Output:
-1/4*((-a)^(1/8)*ArcTan[(Sqrt[(-a)^(1/4) - I*b^(1/4)]*Cot[x])/(-a)^(1/8)]) /(a*Sqrt[(-a)^(1/4) - I*b^(1/4)]) - ((-a)^(1/8)*ArcTan[(Sqrt[(-a)^(1/4) + I*b^(1/4)]*Cot[x])/(-a)^(1/8)])/(4*a*Sqrt[(-a)^(1/4) + I*b^(1/4)]) - ((-a) ^(1/8)*ArcTan[(Sqrt[(-a)^(1/4) + b^(1/4)]*Cot[x])/(-a)^(1/8)])/(4*a*Sqrt[( -a)^(1/4) + b^(1/4)]) - ((-a)^(5/8)*ArcTan[(Sqrt[(-a)^(5/4) + a*b^(1/4)]*C ot[x])/(-a)^(5/8)])/(4*a*Sqrt[(-a)^(5/4) + a*b^(1/4)])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[1/(a + (a + b)*ff^2*x^ 2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(-1), x_Symbol] :> Module[{ k}, Simp[2/(a*n) Sum[Int[1/(1 - Sin[e + f*x]^2/((-1)^(4*(k/n))*Rt[-a/b, n /2])), x], {k, 1, n/2}], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[n/2]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.31
method | result | size |
default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8} a +4 \textit {\_Z}^{6} a +6 \textit {\_Z}^{4} a +4 \textit {\_Z}^{2} a +a +b \right )}{\sum }\frac {\left (\textit {\_R}^{6}+3 \textit {\_R}^{4}+3 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (x \right )-\textit {\_R} \right )}{\textit {\_R}^{7}+3 \textit {\_R}^{5}+3 \textit {\_R}^{3}+\textit {\_R}}}{8 a}\) | \(76\) |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (16777216 a^{8}+16777216 a^{7} b \right ) \textit {\_Z}^{8}+1048576 a^{6} \textit {\_Z}^{6}+24576 a^{4} \textit {\_Z}^{4}+256 a^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (\frac {4194304 i a^{8}}{b}+4194304 i a^{7}\right ) \textit {\_R}^{7}+\left (-\frac {524288 a^{7}}{b}-524288 a^{6}\right ) \textit {\_R}^{6}+\left (\frac {196608 i a^{6}}{b}-65536 i a^{5}\right ) \textit {\_R}^{5}+\left (-\frac {24576 a^{5}}{b}+8192 a^{4}\right ) \textit {\_R}^{4}+\left (\frac {3072 i a^{4}}{b}+1024 i a^{3}\right ) \textit {\_R}^{3}+\left (-\frac {384 a^{3}}{b}-128 a^{2}\right ) \textit {\_R}^{2}+\left (\frac {16 i a^{2}}{b}-16 i a \right ) \textit {\_R} -\frac {2 a}{b}+1\right )\) | \(193\) |
Input:
int(1/(a+b*cos(x)^8),x,method=_RETURNVERBOSE)
Output:
1/8/a*sum((_R^6+3*_R^4+3*_R^2+1)/(_R^7+3*_R^5+3*_R^3+_R)*ln(tan(x)-_R),_R= RootOf(_Z^8*a+4*_Z^6*a+6*_Z^4*a+4*_Z^2*a+a+b))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 665467 vs. \(2 (165) = 330\).
Time = 6.80 (sec) , antiderivative size = 665467, normalized size of antiderivative = 2716.19 \[ \int \frac {1}{a+b \cos ^8(x)} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*cos(x)^8),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {1}{a+b \cos ^8(x)} \, dx=\int \frac {1}{a + b \cos ^{8}{\left (x \right )}}\, dx \] Input:
integrate(1/(a+b*cos(x)**8),x)
Output:
Integral(1/(a + b*cos(x)**8), x)
\[ \int \frac {1}{a+b \cos ^8(x)} \, dx=\int { \frac {1}{b \cos \left (x\right )^{8} + a} \,d x } \] Input:
integrate(1/(a+b*cos(x)^8),x, algorithm="maxima")
Output:
integrate(1/(b*cos(x)^8 + a), x)
\[ \int \frac {1}{a+b \cos ^8(x)} \, dx=\int { \frac {1}{b \cos \left (x\right )^{8} + a} \,d x } \] Input:
integrate(1/(a+b*cos(x)^8),x, algorithm="giac")
Output:
integrate(1/(b*cos(x)^8 + a), x)
Time = 1.67 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.88 \[ \int \frac {1}{a+b \cos ^8(x)} \, dx=\sum _{k=1}^8\ln \left ({\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^4\,a^5\,b^5\,\left ({\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^2\,a^2\,64+1\right )\,\left (\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )\,a\,\mathrm {tan}\left (x\right )\,8-1\right )\,4096\right )\,\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right ) \] Input:
int(1/(a + b*cos(x)^8),x)
Output:
symsum(log(4096*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d ^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^4*a^5*b^5*(64*root(16777216*a^ 7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^2*a^2 + 1)*(8*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 104 8576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)*a*tan(x) - 1))*root( 16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k), k, 1, 8)
\[ \int \frac {1}{a+b \cos ^8(x)} \, dx=\int \frac {1}{\cos \left (x \right )^{8} b +a}d x \] Input:
int(1/(a+b*cos(x)^8),x)
Output:
int(1/(cos(x)**8*b + a),x)