Integrand size = 12, antiderivative size = 41 \[ \int \arctan \left (\sqrt {-1+\sec (x)}\right ) \sin (x) \, dx=\frac {1}{2} \arctan \left (\sqrt {-1+\sec (x)}\right )-\arctan \left (\sqrt {-1+\sec (x)}\right ) \cos (x)+\frac {1}{2} \cos (x) \sqrt {-1+\sec (x)} \]
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.71 \[ \int \arctan \left (\sqrt {-1+\sec (x)}\right ) \sin (x) \, dx=-\arctan \left (\sqrt {-1+\sec (x)}\right ) \cos (x)+\frac {1}{2} \left (\cos (x)+\arctan \left (\frac {\tan \left (\frac {x}{2}\right )}{\sqrt {\frac {\cos (x)}{1+\cos (x)}}}\right ) \sqrt {\frac {\cos (x)}{1+\cos (x)}} \cot \left (\frac {x}{2}\right )\right ) \sqrt {-1+\sec (x)} \]
-(ArcTan[Sqrt[-1 + Sec[x]]]*Cos[x]) + ((Cos[x] + ArcTan[Tan[x/2]/Sqrt[Cos[ x]/(1 + Cos[x])]]*Sqrt[Cos[x]/(1 + Cos[x])]*Cot[x/2])*Sqrt[-1 + Sec[x]])/2
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4835, 5726, 27, 773, 52, 73, 216}
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.
\(\displaystyle \int \sin (x) \arctan \left (\sqrt {\sec (x)-1}\right ) \, dx\) |
\(\Big \downarrow \) 4835 |
\(\displaystyle -\int \arctan \left (\sqrt {\sec (x)-1}\right )d\cos (x)\) |
\(\Big \downarrow \) 5726 |
\(\displaystyle \int -\frac {1}{2 \sqrt {\sec (x)-1}}d\cos (x)-\cos (x) \arctan \left (\sqrt {\sec (x)-1}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \cos (x) \left (-\arctan \left (\sqrt {\sec (x)-1}\right )\right )-\frac {1}{2} \int \frac {1}{\sqrt {\sec (x)-1}}d\cos (x)\) |
\(\Big \downarrow \) 773 |
\(\displaystyle \frac {1}{2} \int \frac {\sec ^2(x)}{\sqrt {\sec (x)-1}}d\sec (x)-\cos (x) \arctan \left (\sqrt {\sec (x)-1}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {\sec (x)}{\sqrt {\sec (x)-1}}d\sec (x)+\sqrt {\sec (x)-1} \sec (x)\right )-\cos (x) \arctan \left (\sqrt {\sec (x)-1}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\int \cos (x)d\sqrt {\sec (x)-1}+\sqrt {\sec (x)-1} \sec (x)\right )-\cos (x) \arctan \left (\sqrt {\sec (x)-1}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\arctan \left (\sqrt {\sec (x)-1}\right )+\sqrt {\sec (x)-1} \sec (x)\right )-\cos (x) \arctan \left (\sqrt {\sec (x)-1}\right )\) |
-(ArcTan[Sqrt[-1 + Sec[x]]]*Cos[x]) + (ArcTan[Sqrt[-1 + Sec[x]]] + Sqrt[-1 + Sec[x]]*Sec[x])/2
3.1.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^ 2, x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && !IntegerQ[p]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Cos[c*(a + b*x)], x]}, Simp[-d/(b*c) Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b* x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])
Int[ArcTan[u_], x_Symbol] :> Simp[x*ArcTan[u], x] - Int[SimplifyIntegrand[x *(D[u, x]/(1 + u^2)), x], x] /; InverseFunctionFreeQ[u, x]
Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(-\frac {\arctan \left (\sqrt {-1+\sec \left (x \right )}\right )}{\sec \left (x \right )}+\frac {\sqrt {-1+\sec \left (x \right )}}{2 \sec \left (x \right )}+\frac {\arctan \left (\sqrt {-1+\sec \left (x \right )}\right )}{2}\) | \(36\) |
default | \(-\frac {\arctan \left (\sqrt {-1+\sec \left (x \right )}\right )}{\sec \left (x \right )}+\frac {\sqrt {-1+\sec \left (x \right )}}{2 \sec \left (x \right )}+\frac {\arctan \left (\sqrt {-1+\sec \left (x \right )}\right )}{2}\) | \(36\) |
-1/sec(x)*arctan((-1+sec(x))^(1/2))+1/2*(-1+sec(x))^(1/2)/sec(x)+1/2*arcta n((-1+sec(x))^(1/2))
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \arctan \left (\sqrt {-1+\sec (x)}\right ) \sin (x) \, dx=-\frac {1}{2} \, {\left (2 \, \cos \left (x\right ) - 1\right )} \arctan \left (\sqrt {\sec \left (x\right ) - 1}\right ) + \frac {1}{2} \, \sqrt {-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right )}} \cos \left (x\right ) \]
\[ \int \arctan \left (\sqrt {-1+\sec (x)}\right ) \sin (x) \, dx=\int \sin {\left (x \right )} \operatorname {atan}{\left (\sqrt {\sec {\left (x \right )} - 1} \right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.46 \[ \int \arctan \left (\sqrt {-1+\sec (x)}\right ) \sin (x) \, dx=-\arctan \left (\sqrt {-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right )}}\right ) \cos \left (x\right ) - \frac {\sqrt {-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right )}}}{2 \, {\left (\frac {\cos \left (x\right ) - 1}{\cos \left (x\right )} - 1\right )}} + \frac {1}{2} \, \arctan \left (\sqrt {-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right )}}\right ) \]
-arctan(sqrt(-(cos(x) - 1)/cos(x)))*cos(x) - 1/2*sqrt(-(cos(x) - 1)/cos(x) )/((cos(x) - 1)/cos(x) - 1) + 1/2*arctan(sqrt(-(cos(x) - 1)/cos(x)))
\[ \int \arctan \left (\sqrt {-1+\sec (x)}\right ) \sin (x) \, dx=\int { \arctan \left (\sqrt {\sec \left (x\right ) - 1}\right ) \sin \left (x\right ) \,d x } \]
Time = 0.42 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.46 \[ \int \arctan \left (\sqrt {-1+\sec (x)}\right ) \sin (x) \, dx=-\mathrm {atan}\left (\sqrt {\frac {1}{\cos \left (x\right )}-1}\right )\,\cos \left (x\right )-\frac {\cos \left (x\right )\,\left (\frac {3\,\mathrm {asin}\left (\sqrt {\cos \left (x\right )}\right )}{2\,{\cos \left (x\right )}^{3/2}}-\frac {3\,\sqrt {1-\cos \left (x\right )}}{2\,\cos \left (x\right )}\right )\,\sqrt {1-\cos \left (x\right )}}{3\,\sqrt {\frac {1}{\cos \left (x\right )}-1}} \]