Integrand size = 39, antiderivative size = 120 \[ \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\cos ^5(a+b x)+\sin ^5(a+b x)} \, dx=\frac {\log (\cos (a+b x))}{b}+\frac {\log (1+\tan (a+b x))}{5 b}-\frac {4 \log \left (2-\left (1-\sqrt {5}\right ) \tan (a+b x)+2 \tan ^2(a+b x)\right )}{5 \left (1-\sqrt {5}\right ) b}-\frac {4 \log \left (2-\left (1+\sqrt {5}\right ) \tan (a+b x)+2 \tan ^2(a+b x)\right )}{5 \left (1+\sqrt {5}\right ) b} \]
ln(cos(b*x+a))/b+1/5*ln(1+tan(b*x+a))/b-4/5*ln(2-(-5^(1/2)+1)*tan(b*x+a)+2 *tan(b*x+a)^2)/b/(-5^(1/2)+1)-4/5*ln(2-(5^(1/2)+1)*tan(b*x+a)+2*tan(b*x+a) ^2)/b/(5^(1/2)+1)
Time = 8.95 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\cos ^5(a+b x)+\sin ^5(a+b x)} \, dx=\frac {\log (\cos (a+b x)+\sin (a+b x))-\left (-1+\sqrt {5}\right ) \log \left (1-\sqrt {5}+\sin (2 (a+b x))\right )+\left (1+\sqrt {5}\right ) \log \left (1+\sqrt {5}+\sin (2 (a+b x))\right )}{5 b} \]
(Log[Cos[a + b*x] + Sin[a + b*x]] - (-1 + Sqrt[5])*Log[1 - Sqrt[5] + Sin[2 *(a + b*x)]] + (1 + Sqrt[5])*Log[1 + Sqrt[5] + Sin[2*(a + b*x)]])/(5*b)
Time = 0.77 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 4889, 2462, 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.
| \(\displaystyle \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\sin ^5(a+b x)+\cos ^5(a+b x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (a+b x)^5-\sin (a+b x)^5}{\sin (a+b x)^5+\cos (a+b x)^5}dx\) |
| \(\Big \downarrow \) 4889 |
| \(\displaystyle \frac {\int \frac {1-\tan ^5(a+b x)}{\tan ^7(a+b x)+\tan ^5(a+b x)+\tan ^2(a+b x)+1}d\tan (a+b x)}{b}\) |
| \(\Big \downarrow \) 2462 |
| \(\displaystyle \frac {\int \left (-\frac {\tan (a+b x)}{\tan ^2(a+b x)+1}+\frac {1}{5 (\tan (a+b x)+1)}+\frac {2 \left (2 \tan ^3(a+b x)-4 \tan ^2(a+b x)+\tan (a+b x)+2\right )}{5 \left (\tan ^4(a+b x)-\tan ^3(a+b x)+\tan ^2(a+b x)-\tan (a+b x)+1\right )}\right )d\tan (a+b x)}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{2} \log \left (\tan ^2(a+b x)+1\right )-\frac {4 \log \left (2 \tan ^2(a+b x)-\left (1-\sqrt {5}\right ) \tan (a+b x)+2\right )}{5 \left (1-\sqrt {5}\right )}-\frac {4 \log \left (2 \tan ^2(a+b x)-\left (1+\sqrt {5}\right ) \tan (a+b x)+2\right )}{5 \left (1+\sqrt {5}\right )}+\frac {1}{5} \log (\tan (a+b x)+1)}{b}\) |
(Log[1 + Tan[a + b*x]]/5 - Log[1 + Tan[a + b*x]^2]/2 - (4*Log[2 - (1 - Sqr t[5])*Tan[a + b*x] + 2*Tan[a + b*x]^2])/(5*(1 - Sqrt[5])) - (4*Log[2 - (1 + Sqrt[5])*Tan[a + b*x] + 2*Tan[a + b*x]^2])/(5*(1 + Sqrt[5])))/b
3.10.42.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 1.42 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {-\frac {4 \left (\frac {\sqrt {5}}{4}-\frac {1}{4}\right ) \ln \left (-\sqrt {5}\, \tan \left (x b +a \right )+2 \tan \left (x b +a \right )^{2}-\tan \left (x b +a \right )+2\right )}{5}+\frac {4 \left (\frac {\sqrt {5}}{4}+\frac {1}{4}\right ) \ln \left (\sqrt {5}\, \tan \left (x b +a \right )+2 \tan \left (x b +a \right )^{2}-\tan \left (x b +a \right )+2\right )}{5}+\frac {\ln \left (1+\tan \left (x b +a \right )\right )}{5}-\frac {\ln \left (1+\tan \left (x b +a \right )^{2}\right )}{2}}{b}\) | \(111\) |
| default | \(\frac {-\frac {4 \left (\frac {\sqrt {5}}{4}-\frac {1}{4}\right ) \ln \left (-\sqrt {5}\, \tan \left (x b +a \right )+2 \tan \left (x b +a \right )^{2}-\tan \left (x b +a \right )+2\right )}{5}+\frac {4 \left (\frac {\sqrt {5}}{4}+\frac {1}{4}\right ) \ln \left (\sqrt {5}\, \tan \left (x b +a \right )+2 \tan \left (x b +a \right )^{2}-\tan \left (x b +a \right )+2\right )}{5}+\frac {\ln \left (1+\tan \left (x b +a \right )\right )}{5}-\frac {\ln \left (1+\tan \left (x b +a \right )^{2}\right )}{2}}{b}\) | \(111\) |
| risch | \(-i x -\frac {2 i a}{b}+\frac {\ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+i\right )}{5 b}+\frac {\ln \left ({\mathrm e}^{4 i \left (x b +a \right )}+2 i \left (\sqrt {5}+1\right ) {\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{5 b}+\frac {\ln \left ({\mathrm e}^{4 i \left (x b +a \right )}+2 i \left (\sqrt {5}+1\right ) {\mathrm e}^{2 i \left (x b +a \right )}-1\right ) \sqrt {5}}{5 b}+\frac {\ln \left ({\mathrm e}^{4 i \left (x b +a \right )}-2 i \left (\sqrt {5}-1\right ) {\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{5 b}-\frac {\ln \left ({\mathrm e}^{4 i \left (x b +a \right )}-2 i \left (\sqrt {5}-1\right ) {\mathrm e}^{2 i \left (x b +a \right )}-1\right ) \sqrt {5}}{5 b}\) | \(173\) |
1/b*(-4/5*(1/4*5^(1/2)-1/4)*ln(-5^(1/2)*tan(b*x+a)+2*tan(b*x+a)^2-tan(b*x+ a)+2)+4/5*(1/4*5^(1/2)+1/4)*ln(5^(1/2)*tan(b*x+a)+2*tan(b*x+a)^2-tan(b*x+a )+2)+1/5*ln(1+tan(b*x+a))-1/2*ln(1+tan(b*x+a)^2))
Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\cos ^5(a+b x)+\sin ^5(a+b x)} \, dx=\frac {2 \, \sqrt {5} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{4} - 2 \, {\left (\sqrt {5} + 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, \cos \left (b x + a\right )^{2} - \sqrt {5} - 3}{\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1}\right ) + 2 \, \log \left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) + \log \left (2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{10 \, b} \]
1/10*(2*sqrt(5)*log(-(2*cos(b*x + a)^4 - 2*(sqrt(5) + 1)*cos(b*x + a)*sin( b*x + a) - 2*cos(b*x + a)^2 - sqrt(5) - 3)/(cos(b*x + a)^4 - cos(b*x + a)^ 2 - cos(b*x + a)*sin(b*x + a) + 1)) + 2*log(cos(b*x + a)^4 - cos(b*x + a)^ 2 - cos(b*x + a)*sin(b*x + a) + 1) + log(2*cos(b*x + a)*sin(b*x + a) + 1)) /b
Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (100) = 200\).
Time = 84.52 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.16 \[ \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\cos ^5(a+b x)+\sin ^5(a+b x)} \, dx=\begin {cases} - \frac {47 \log {\left (\sin {\left (a + b x \right )} + \cos {\left (a + b x \right )} \right )}}{- 235 b + 105 \sqrt {5} b} + \frac {21 \sqrt {5} \log {\left (\sin {\left (a + b x \right )} + \cos {\left (a + b x \right )} \right )}}{- 235 b + 105 \sqrt {5} b} - \frac {26 \sqrt {5} \log {\left (16 \sin ^{2}{\left (a + b x \right )} - 8 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} + 8 \sqrt {5} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} + 16 \cos ^{2}{\left (a + b x \right )} \right )}}{- 235 b + 105 \sqrt {5} b} + \frac {58 \log {\left (16 \sin ^{2}{\left (a + b x \right )} - 8 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} + 8 \sqrt {5} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} + 16 \cos ^{2}{\left (a + b x \right )} \right )}}{- 235 b + 105 \sqrt {5} b} - \frac {152 \log {\left (16 \sin ^{2}{\left (a + b x \right )} - 8 \sqrt {5} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} - 8 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} + 16 \cos ^{2}{\left (a + b x \right )} \right )}}{- 235 b + 105 \sqrt {5} b} + \frac {68 \sqrt {5} \log {\left (16 \sin ^{2}{\left (a + b x \right )} - 8 \sqrt {5} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} - 8 \sin {\left (a + b x \right )} \cos {\left (a + b x \right )} + 16 \cos ^{2}{\left (a + b x \right )} \right )}}{- 235 b + 105 \sqrt {5} b} & \text {for}\: b \neq 0 \\\frac {x \left (- \sin ^{5}{\left (a \right )} + \cos ^{5}{\left (a \right )}\right )}{\sin ^{5}{\left (a \right )} + \cos ^{5}{\left (a \right )}} & \text {otherwise} \end {cases} \]
Piecewise((-47*log(sin(a + b*x) + cos(a + b*x))/(-235*b + 105*sqrt(5)*b) + 21*sqrt(5)*log(sin(a + b*x) + cos(a + b*x))/(-235*b + 105*sqrt(5)*b) - 26 *sqrt(5)*log(16*sin(a + b*x)**2 - 8*sin(a + b*x)*cos(a + b*x) + 8*sqrt(5)* sin(a + b*x)*cos(a + b*x) + 16*cos(a + b*x)**2)/(-235*b + 105*sqrt(5)*b) + 58*log(16*sin(a + b*x)**2 - 8*sin(a + b*x)*cos(a + b*x) + 8*sqrt(5)*sin(a + b*x)*cos(a + b*x) + 16*cos(a + b*x)**2)/(-235*b + 105*sqrt(5)*b) - 152* log(16*sin(a + b*x)**2 - 8*sqrt(5)*sin(a + b*x)*cos(a + b*x) - 8*sin(a + b *x)*cos(a + b*x) + 16*cos(a + b*x)**2)/(-235*b + 105*sqrt(5)*b) + 68*sqrt( 5)*log(16*sin(a + b*x)**2 - 8*sqrt(5)*sin(a + b*x)*cos(a + b*x) - 8*sin(a + b*x)*cos(a + b*x) + 16*cos(a + b*x)**2)/(-235*b + 105*sqrt(5)*b), Ne(b, 0)), (x*(-sin(a)**5 + cos(a)**5)/(sin(a)**5 + cos(a)**5), True))
\[ \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\cos ^5(a+b x)+\sin ^5(a+b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{5} - \sin \left (b x + a\right )^{5}}{\cos \left (b x + a\right )^{5} + \sin \left (b x + a\right )^{5}} \,d x } \]
Time = 0.48 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\cos ^5(a+b x)+\sin ^5(a+b x)} \, dx=-\frac {2 \, \sqrt {5} \log \left (-\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} \tan \left (b x + a\right ) + \tan \left (b x + a\right )^{2} + 1\right ) - 2 \, \sqrt {5} \log \left (\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} \tan \left (b x + a\right ) + \tan \left (b x + a\right )^{2} + 1\right ) - 2 \, \log \left (\tan \left (b x + a\right )^{4} - \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )^{2} - \tan \left (b x + a\right ) + 1\right ) + 5 \, \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - 2 \, \log \left ({\left | \tan \left (b x + a\right ) + 1 \right |}\right )}{10 \, b} \]
-1/10*(2*sqrt(5)*log(-1/2*(sqrt(5) + 1)*tan(b*x + a) + tan(b*x + a)^2 + 1) - 2*sqrt(5)*log(1/2*(sqrt(5) - 1)*tan(b*x + a) + tan(b*x + a)^2 + 1) - 2* log(tan(b*x + a)^4 - tan(b*x + a)^3 + tan(b*x + a)^2 - tan(b*x + a) + 1) + 5*log(tan(b*x + a)^2 + 1) - 2*log(abs(tan(b*x + a) + 1)))/b
Time = 28.34 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.88 \[ \int \frac {\cos ^5(a+b x)-\sin ^5(a+b x)}{\cos ^5(a+b x)+\sin ^5(a+b x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )-1\right )}{5\,b}-\frac {\ln \left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}{b}+\frac {\ln \left (2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+\sqrt {5}\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )-\sqrt {5}\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+1\right )\,\left (\sqrt {5}+1\right )}{5\,b}-\frac {\ln \left (2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-\sqrt {5}\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )+\sqrt {5}\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+1\right )\,\left (\sqrt {5}-1\right )}{5\,b} \]
log(tan(a/2 + (b*x)/2)^2 - 2*tan(a/2 + (b*x)/2) - 1)/(5*b) - log(tan(a/2 + (b*x)/2)^2 + 1)/b + (log(2*tan(a/2 + (b*x)/2)^2 - tan(a/2 + (b*x)/2) + ta n(a/2 + (b*x)/2)^3 + tan(a/2 + (b*x)/2)^4 + 5^(1/2)*tan(a/2 + (b*x)/2) - 5 ^(1/2)*tan(a/2 + (b*x)/2)^3 + 1)*(5^(1/2) + 1))/(5*b) - (log(2*tan(a/2 + ( b*x)/2)^2 - tan(a/2 + (b*x)/2) + tan(a/2 + (b*x)/2)^3 + tan(a/2 + (b*x)/2) ^4 - 5^(1/2)*tan(a/2 + (b*x)/2) + 5^(1/2)*tan(a/2 + (b*x)/2)^3 + 1)*(5^(1/ 2) - 1))/(5*b)