Integrand size = 23, antiderivative size = 71 \[ \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\log (\cos (e+f x))}{(a-b) f}-\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{2 (a-b) b^2 f}+\frac {\tan ^2(e+f x)}{2 b f} \]
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {-\frac {2 \log (\cos (e+f x))}{a-b}-\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{(a-b) b^2}+\frac {\tan ^2(e+f x)}{b}}{2 f} \]
((-2*Log[Cos[e + f*x]])/(a - b) - (a^2*Log[a + b*Tan[e + f*x]^2])/((a - b) *b^2) + Tan[e + f*x]^2/b)/(2*f)
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4153, 354, 93, 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.
\(\displaystyle \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^5}{a+b \tan (e+f x)^2}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan ^5(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 93 |
\(\displaystyle \frac {\int \left (-\frac {a^2}{(a-b) b \left (b \tan ^2(e+f x)+a\right )}+\frac {1}{b}+\frac {1}{(a-b) \left (\tan ^2(e+f x)+1\right )}\right )d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{b^2 (a-b)}+\frac {\log \left (\tan ^2(e+f x)+1\right )}{a-b}+\frac {\tan ^2(e+f x)}{b}}{2 f}\) |
(Log[1 + Tan[e + f*x]^2]/(a - b) - (a^2*Log[a + b*Tan[e + f*x]^2])/((a - b )*b^2) + Tan[e + f*x]^2/b)/(2*f)
3.3.11.3.1 Defintions of rubi rules used
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
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[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )^{2}}{2 b}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 b^{2} \left (a -b \right )}}{f}\) | \(67\) |
default | \(\frac {\frac {\tan \left (f x +e \right )^{2}}{2 b}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 b^{2} \left (a -b \right )}}{f}\) | \(67\) |
norman | \(\frac {\tan \left (f x +e \right )^{2}}{2 b f}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \left (a -b \right )}-\frac {a^{2} \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right ) b^{2} f}\) | \(72\) |
parallelrisch | \(\frac {\tan \left (f x +e \right )^{2} a b -b^{2} \tan \left (f x +e \right )^{2}+\ln \left (1+\tan \left (f x +e \right )^{2}\right ) b^{2}-a^{2} \ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right ) b^{2} f}\) | \(74\) |
risch | \(-\frac {i x}{a -b}-\frac {2 i a x}{b^{2}}-\frac {2 i a e}{b^{2} f}-\frac {2 i x}{b}-\frac {2 i e}{b f}+\frac {2 i a^{2} x}{\left (a -b \right ) b^{2}}+\frac {2 i a^{2} e}{\left (a -b \right ) b^{2} f}+\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )}}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{b^{2} f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}-\frac {a^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 \left (a -b \right ) b^{2} f}\) | \(206\) |
1/f*(1/2*tan(f*x+e)^2/b+1/2/(a-b)*ln(1+tan(f*x+e)^2)-1/2*a^2/b^2/(a-b)*ln( a+b*tan(f*x+e)^2))
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30 \[ \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {a^{2} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a b^{2} - b^{3}\right )} f} \]
-1/2*(a^2*log((b*tan(f*x + e)^2 + a)/(tan(f*x + e)^2 + 1)) - (a*b - b^2)*t an(f*x + e)^2 - (a^2 - b^2)*log(1/(tan(f*x + e)^2 + 1)))/((a*b^2 - b^3)*f)
Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (53) = 106\).
Time = 8.68 (sec) , antiderivative size = 338, normalized size of antiderivative = 4.76 \[ \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\begin {cases} \tilde {\infty } x \tan ^{3}{\left (e \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {\frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {\tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {\tan ^{2}{\left (e + f x \right )}}{2 f}}{a} & \text {for}\: b = 0 \\- \frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {\tan ^{4}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {2}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text {for}\: a = b \\\frac {x \tan ^{5}{\left (e \right )}}{a + b \tan ^{2}{\left (e \right )}} & \text {for}\: f = 0 \\- \frac {a^{2} \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f - 2 b^{3} f} - \frac {a^{2} \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f - 2 b^{3} f} + \frac {a b \tan ^{2}{\left (e + f x \right )}}{2 a b^{2} f - 2 b^{3} f} + \frac {b^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a b^{2} f - 2 b^{3} f} - \frac {b^{2} \tan ^{2}{\left (e + f x \right )}}{2 a b^{2} f - 2 b^{3} f} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x*tan(e)**3, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((log(tan(e + f*x)**2 + 1)/(2*f) + tan(e + f*x)**4/(4*f) - tan(e + f*x)**2/(2*f))/a, Eq (b, 0)), (-2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(2*b*f*tan(e + f*x)* *2 + 2*b*f) - 2*log(tan(e + f*x)**2 + 1)/(2*b*f*tan(e + f*x)**2 + 2*b*f) + tan(e + f*x)**4/(2*b*f*tan(e + f*x)**2 + 2*b*f) - 2/(2*b*f*tan(e + f*x)** 2 + 2*b*f), Eq(a, b)), (x*tan(e)**5/(a + b*tan(e)**2), Eq(f, 0)), (-a**2*l og(-sqrt(-a/b) + tan(e + f*x))/(2*a*b**2*f - 2*b**3*f) - a**2*log(sqrt(-a/ b) + tan(e + f*x))/(2*a*b**2*f - 2*b**3*f) + a*b*tan(e + f*x)**2/(2*a*b**2 *f - 2*b**3*f) + b**2*log(tan(e + f*x)**2 + 1)/(2*a*b**2*f - 2*b**3*f) - b **2*tan(e + f*x)**2/(2*a*b**2*f - 2*b**3*f), True))
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07 \[ \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {a^{2} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a b^{2} - b^{3}} - \frac {{\left (a + b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}} + \frac {1}{b \sin \left (f x + e\right )^{2} - b}}{2 \, f} \]
-1/2*(a^2*log(-(a - b)*sin(f*x + e)^2 + a)/(a*b^2 - b^3) - (a + b)*log(sin (f*x + e)^2 - 1)/b^2 + 1/(b*sin(f*x + e)^2 - b))/f
Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (67) = 134\).
Time = 1.58 (sec) , antiderivative size = 317, normalized size of antiderivative = 4.46 \[ \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {a^{3} \log \left ({\left | -a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 2 \, a + 4 \, b \right |}\right )}{a^{2} b^{2} - a b^{3}} - \frac {\log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right )}{a - b} - \frac {{\left (a + b\right )} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right )}{b^{2}} + \frac {a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, a + 6 \, b}{b^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}}}{2 \, f} \]
-1/2*(a^3*log(abs(-a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e ) - 1)/(cos(f*x + e) + 1)) - 2*a + 4*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 - b) - (a + 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)/(c os(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 + 6*b)/(b^2*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(c os(f*x + e) + 1) + 2)))/f
Time = 11.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,\left (a-b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,b\,f}-\frac {a^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}{2\,f\,\left (a\,b^2-b^3\right )} \]