Integrand size = 25, antiderivative size = 89 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f}-\frac {(a+2 b) \sqrt {a+b \sec ^2(e+f x)}}{b^2 f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f} \]
1/3*(a+b*sec(f*x+e)^2)^(3/2)/b^2/f-arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2 ))/f/a^(1/2)-(a+2*b)*(a+b*sec(f*x+e)^2)^(1/2)/b^2/f
Time = 1.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.20 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {-2 a (a+3 b)-b (a+6 b) \sec ^2(e+f x)+b^2 \sec ^4(e+f x)-3 b^2 \text {arctanh}\left (\sqrt {1+\frac {b \sec ^2(e+f x)}{a}}\right ) \sqrt {1+\frac {b \sec ^2(e+f x)}{a}}}{3 b^2 f \sqrt {a+b \sec ^2(e+f x)}} \]
(-2*a*(a + 3*b) - b*(a + 6*b)*Sec[e + f*x]^2 + b^2*Sec[e + f*x]^4 - 3*b^2* ArcTanh[Sqrt[1 + (b*Sec[e + f*x]^2)/a]]*Sqrt[1 + (b*Sec[e + f*x]^2)/a])/(3 *b^2*f*Sqrt[a + b*Sec[e + f*x]^2])
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4627, 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.
\(\displaystyle \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^5}{\sqrt {a+b \sec (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle \frac {\int \frac {\cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\sqrt {b \sec ^2(e+f x)+a}}d\sec ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {-a-2 b}{b \sqrt {b \sec ^2(e+f x)+a}}+\frac {\sqrt {b \sec ^2(e+f x)+a}}{b}+\frac {\cos (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}\right )d\sec ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {2 \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2}-\frac {2 (a+2 b) \sqrt {a+b \sec ^2(e+f x)}}{b^2}}{2 f}\) |
((-2*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/Sqrt[a] - (2*(a + 2*b)*S qrt[a + b*Sec[e + f*x]^2])/b^2 + (2*(a + b*Sec[e + f*x]^2)^(3/2))/(3*b^2)) /(2*f)
3.5.2.3.1 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_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si mp[1/f Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] , x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers Q[2*n, p])
Leaf count of result is larger than twice the leaf count of optimal. \(302\) vs. \(2(77)=154\).
Time = 8.11 (sec) , antiderivative size = 303, normalized size of antiderivative = 3.40
method | result | size |
default | \(-\frac {2 a^{\frac {5}{2}}+3 \ln \left (4 \cos \left (f x +e \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+4 \cos \left (f x +e \right ) a +4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}+6 a^{\frac {3}{2}} b +3 \ln \left (4 \cos \left (f x +e \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+4 \cos \left (f x +e \right ) a +4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \sec \left (f x +e \right )+a^{\frac {3}{2}} b \sec \left (f x +e \right )^{2}+6 \sqrt {a}\, b^{2} \sec \left (f x +e \right )^{2}-\sqrt {a}\, b^{2} \sec \left (f x +e \right )^{4}}{3 f \,b^{2} \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}\) | \(303\) |
-1/3/f/b^2/a^(1/2)/(a+b*sec(f*x+e)^2)^(1/2)*(2*a^(5/2)+3*ln(4*cos(f*x+e)*( (b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*cos(f*x+e)*a+4*a^(1/2 )*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos( f*x+e))^2)^(1/2)*b^2+6*a^(3/2)*b+3*ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*a^(1/2)+4*cos(f*x+e)*a+4*a^(1/2)*((b+a*cos(f*x+e)^2)/ (1+cos(f*x+e))^2)^(1/2))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*s ec(f*x+e)+a^(3/2)*b*sec(f*x+e)^2+6*a^(1/2)*b^2*sec(f*x+e)^2-a^(1/2)*b^2*se c(f*x+e)^4)
Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (77) = 154\).
Time = 0.62 (sec) , antiderivative size = 410, normalized size of antiderivative = 4.61 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \cos \left (f x + e\right )^{2} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) - 8 \, {\left (2 \, {\left (a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, a b^{2} f \cos \left (f x + e\right )^{2}}, \frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{2} - 4 \, {\left (2 \, {\left (a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, a b^{2} f \cos \left (f x + e\right )^{2}}\right ] \]
[1/24*(3*sqrt(a)*b^2*cos(f*x + e)^2*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b *cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b ^4 - 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) - 8*(2*(a^2 + 3*a*b)*cos(f*x + e)^2 - a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*b^2*f*cos(f*x + e)^2), 1/12*(3*sqrt(-a)*b^2*arct an(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt(( a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos( f*x + e)^2 + a*b^2))*cos(f*x + e)^2 - 4*(2*(a^2 + 3*a*b)*cos(f*x + e)^2 - a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*b^2*f*cos(f*x + e)^2) ]
\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^5}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]