Integrand size = 23, antiderivative size = 62 \[ \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{f} \]
Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.95 \[ \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {-\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\sqrt {a+b \tan ^2(e+f x)}}{f} \]
(-(Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]]) + Sqrt[a + b*Tan[e + f*x]^2])/f
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4153, 353, 60, 73, 221}
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.
\(\displaystyle \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x) \sqrt {a+b \tan (e+f x)^2}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\tan (e+f x) \sqrt {b \tan ^2(e+f x)+a}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {\int \frac {\sqrt {b \tan ^2(e+f x)+a}}{\tan ^2(e+f x)+1}d\tan ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a-b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan ^2(e+f x)+2 \sqrt {a+b \tan ^2(e+f x)}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 (a-b) \int \frac {1}{\frac {\tan ^4(e+f x)}{b}-\frac {a}{b}+1}d\sqrt {b \tan ^2(e+f x)+a}}{b}+2 \sqrt {a+b \tan ^2(e+f x)}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {a+b \tan ^2(e+f x)}-2 \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{2 f}\) |
(-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*f)
3.3.95.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
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 = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.47
method | result | size |
derivativedivides | \(\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{f}-\frac {b \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}\) | \(91\) |
default | \(\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{f}-\frac {b \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}\) | \(91\) |
(a+b*tan(f*x+e)^2)^(1/2)/f-1/f*b/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1 /2)/(-a+b)^(1/2))+1/f*a/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b )^(1/2))
Time = 0.31 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.45 \[ \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {\sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{4 \, f}, \frac {\sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{2 \, f}\right ] \]
[1/4*(sqrt(a - b)*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)*sqrt(a - b ) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 4*sqrt (b*tan(f*x + e)^2 + a))/f, 1/2*(sqrt(-a + b)*arctan(2*sqrt(b*tan(f*x + e)^ 2 + a)*sqrt(-a + b)/(b*tan(f*x + e)^2 + 2*a - b)) + 2*sqrt(b*tan(f*x + e)^ 2 + a))/f]
\[ \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx \]
\[ \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right ) \,d x } \]
Timed out. \[ \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 11.55 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{f}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sqrt {a-b}}\right )\,\sqrt {a-b}}{f} \]