Integrand size = 23, antiderivative size = 69 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b) \log (\cos (e+f x))}{b^2 f}-\frac {(a+b)^2 \log \left (b+a \cos ^2(e+f x)\right )}{2 a b^2 f}+\frac {\sec ^2(e+f x)}{2 b f} \]
Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.43 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (2 a (a+2 b) \log (\cos (e+f x))-(a+b)^2 \log \left (a+b-a \sin ^2(e+f x)\right )+a b \sec ^2(e+f x)\right )}{4 a b^2 f \left (a+b \sec ^2(e+f x)\right )} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(2*a*(a + 2*b)*Log[Cos[e + f*x]] - (a + b)^2*Log[a + b - a*Sin[e + f*x]^2] + a*b*Sec[e + f*x]^2))/(4* a*b^2*f*(a + b*Sec[e + f*x]^2))
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 354, 99, 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.
\(\displaystyle \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^5}{a+b \sec (e+f x)^2}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2 \sec ^3(e+f x)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2 \sec ^2(e+f x)}{a \cos ^2(e+f x)+b}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (\frac {(a+b)^2}{b^2 \left (a \cos ^2(e+f x)+b\right )}+\frac {\sec ^2(e+f x)}{b}+\frac {(-a-2 b) \sec (e+f x)}{b^2}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {(a+b)^2 \log \left (a \cos ^2(e+f x)+b\right )}{a b^2}-\frac {(a+2 b) \log \left (\cos ^2(e+f x)\right )}{b^2}-\frac {\sec (e+f x)}{b}}{2 f}\) |
-1/2*(-(((a + 2*b)*Log[Cos[e + f*x]^2])/b^2) + ((a + b)^2*Log[b + a*Cos[e + f*x]^2])/(a*b^2) - Sec[e + f*x]/b)/f
3.4.37.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 1.60 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {\frac {\left (a +2 b \right ) \ln \left (\cos \left (f x +e \right )\right )}{b^{2}}+\frac {1}{2 b \cos \left (f x +e \right )^{2}}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b^{2} a}}{f}\) | \(67\) |
default | \(\frac {\frac {\left (a +2 b \right ) \ln \left (\cos \left (f x +e \right )\right )}{b^{2}}+\frac {1}{2 b \cos \left (f x +e \right )^{2}}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 b^{2} a}}{f}\) | \(67\) |
risch | \(\frac {i x}{a}+\frac {2 i e}{a f}+\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )}}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{b^{2} f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}-\frac {a \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 b^{2} f}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{b f}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f}\) | \(207\) |
1/f*((a+2*b)/b^2*ln(cos(f*x+e))+1/2/b/cos(f*x+e)^2-1/2*(a^2+2*a*b+b^2)/b^2 /a*ln(b+a*cos(f*x+e)^2))
Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} \log \left (-\cos \left (f x + e\right )\right ) - a b}{2 \, a b^{2} f \cos \left (f x + e\right )^{2}} \]
-1/2*((a^2 + 2*a*b + b^2)*cos(f*x + e)^2*log(a*cos(f*x + e)^2 + b) - 2*(a^ 2 + 2*a*b)*cos(f*x + e)^2*log(-cos(f*x + e)) - a*b)/(a*b^2*f*cos(f*x + e)^ 2)
\[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}} - \frac {1}{b \sin \left (f x + e\right )^{2} - b} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a b^{2}}}{2 \, f} \]
1/2*((a + 2*b)*log(sin(f*x + e)^2 - 1)/b^2 - 1/(b*sin(f*x + e)^2 - b) - (a ^2 + 2*a*b + b^2)*log(a*sin(f*x + e)^2 - a - b)/(a*b^2))/f
Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (65) = 130\).
Time = 1.52 (sec) , antiderivative size = 373, normalized size of antiderivative = 5.41 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | -a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 2 \, a + 2 \, b \right |}\right )}{a^{2} b^{2} + a b^{3}} - \frac {\log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right )}{a} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right )}{b^{2}} + \frac {a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, a + 8 \, b}{b^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}}}{2 \, f} \]
-1/2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(abs(-a*((cos(f*x + e) + 1)/(cos( f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) - b*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) - 2*a + 2 *b))/(a^2*b^2 + a*b^3) - log(abs(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2))/a - (a + 2*b)*log(abs(-(cos(f* x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2 ))/b^2 + (a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(c os(f*x + e) + 1)) + 2*b*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) + 2*a + 8*b)/(b^2*((cos(f*x + e) + 1)/(cos(f *x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2)))/f
Time = 19.46 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{b\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,a\,f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,b\,f}-\frac {a\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,b^2\,f} \]