Integrand size = 25, antiderivative size = 117 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f} \]
-arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))*(a-b)^(1/2)/f+(a+b*tan(f*x+ e)^2)^(1/2)/f-1/3*(a+b)*(a+b*tan(f*x+e)^2)^(3/2)/b^2/f+1/5*(a+b*tan(f*x+e) ^2)^(5/2)/b^2/f
Time = 1.47 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {-15 \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\frac {\sqrt {a+b \tan ^2(e+f x)} \left (-2 a^2-5 a b+15 b^2+(a-5 b) b \tan ^2(e+f x)+3 b^2 \tan ^4(e+f x)\right )}{b^2}}{15 f} \]
(-15*Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + (Sqrt[a + b*Tan[e + f*x]^2]*(-2*a^2 - 5*a*b + 15*b^2 + (a - 5*b)*b*Tan[e + f*x]^2 + 3*b^2*Tan[e + f*x]^4))/b^2)/(15*f)
Time = 0.33 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4153, 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.
\(\displaystyle \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^5 \sqrt {a+b \tan (e+f x)^2}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan ^5(e+f x) \sqrt {b \tan ^2(e+f x)+a}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\int \frac {\tan ^4(e+f x) \sqrt {b \tan ^2(e+f x)+a}}{\tan ^2(e+f x)+1}d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {\left (b \tan ^2(e+f x)+a\right )^{3/2}}{b}+\frac {(-a-b) \sqrt {b \tan ^2(e+f x)+a}}{b}+\frac {\sqrt {b \tan ^2(e+f x)+a}}{\tan ^2(e+f x)+1}\right )d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-2 \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\frac {2 \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2}-\frac {2 (a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2}+2 \sqrt {a+b \tan ^2(e+f x)}}{2 f}\) |
(-2*Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + 2*Sqrt[a + b*Tan[e + f*x]^2] - (2*(a + b)*(a + b*Tan[e + f*x]^2)^(3/2))/(3*b^2) + (2*(a + b*Tan[e + f*x]^2)^(5/2))/(5*b^2))/(2*f)
3.3.93.3.1 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[((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 = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{15 b^{2}}-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b}+b \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\right )+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\) | \(157\) |
default | \(\frac {\frac {\tan \left (f x +e \right )^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{15 b^{2}}-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b}+b \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\right )+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\) | \(157\) |
1/f*(1/5*tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^(3/2)/b-2/15*a/b^2*(a+b*tan(f*x+e )^2)^(3/2)-1/3*(a+b*tan(f*x+e)^2)^(3/2)/b+b*(1/b*(a+b*tan(f*x+e)^2)^(1/2)- 1/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2)))+a/(-a+b)^(1/ 2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2)))
Time = 0.32 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.74 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {15 \, \sqrt {a - b} b^{2} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (3 \, b^{2} \tan \left (f x + e\right )^{4} + {\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, b^{2} f}, \frac {15 \, \sqrt {-a + b} b^{2} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (3 \, b^{2} \tan \left (f x + e\right )^{4} + {\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, b^{2} f}\right ] \]
[1/60*(15*sqrt(a - b)*b^2*log(-(b^2*tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan (f*x + e)^2 - 4*(b*tan(f*x + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sq rt(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 4*(3*b^2*tan(f*x + e)^4 + (a*b - 5*b^2)*tan(f*x + e)^2 - 2*a^2 - 5*a*b + 15*b^2)*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/30*(15*sqrt(-a + b)*b^2*ar ctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + e)^2 + 2*a - b )) + 2*(3*b^2*tan(f*x + e)^4 + (a*b - 5*b^2)*tan(f*x + e)^2 - 2*a^2 - 5*a* b + 15*b^2)*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f)]
\[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \tan ^{5}{\left (e + f x \right )}\, dx \]
\[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 19.81 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}}{5\,b^2\,f}-\left (\frac {2\,a}{3\,b^2\,f}-\frac {a-b}{3\,b^2\,f}\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}-\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\left (\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\left (a-b\right )-\frac {a^2}{b^2\,f}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}}{\sqrt {a-b}}\right )\,\sqrt {a-b}\,1{}\mathrm {i}}{f} \]