Integrand size = 23, antiderivative size = 66 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f} \]
arctanh(sec(f*x+e)*b^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-cos(f*x+e)* (a+b*sec(f*x+e)^2)^(1/2)/f
Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.50 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \, dx=-\frac {\sqrt {2} \cos (e+f x) \left (-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b+a \cos ^2(e+f x)}}{\sqrt {b}}\right )+\sqrt {b+a \cos ^2(e+f x)}\right ) \sqrt {a+b \sec ^2(e+f x)}}{f \sqrt {a+2 b+a \cos (2 e+2 f x)}} \]
-((Sqrt[2]*Cos[e + f*x]*(-(Sqrt[b]*ArcTanh[Sqrt[b + a*Cos[e + f*x]^2]/Sqrt [b]]) + Sqrt[b + a*Cos[e + f*x]^2])*Sqrt[a + b*Sec[e + f*x]^2])/(f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]))
Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4622, 247, 224, 219}
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.
\(\displaystyle \int \sin (e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x) \sqrt {a+b \sec (e+f x)^2}dx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int \cos ^2(e+f x) \sqrt {b \sec ^2(e+f x)+a}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {b \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {b \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}\) |
(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]] - Cos[ e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/f
3.1.69.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si mp[1/(f*ff^m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p /x^(m + 1)), 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])
Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.41
method | result | size |
derivativedivides | \(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{f a \sec \left (f x +e \right )}+\frac {b \sec \left (f x +e \right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}{f a}+\frac {\sqrt {b}\, \ln \left (\sqrt {b}\, \sec \left (f x +e \right )+\sqrt {a +b \sec \left (f x +e \right )^{2}}\right )}{f}\) | \(93\) |
default | \(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{f a \sec \left (f x +e \right )}+\frac {b \sec \left (f x +e \right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}{f a}+\frac {\sqrt {b}\, \ln \left (\sqrt {b}\, \sec \left (f x +e \right )+\sqrt {a +b \sec \left (f x +e \right )^{2}}\right )}{f}\) | \(93\) |
-1/f/a/sec(f*x+e)*(a+b*sec(f*x+e)^2)^(3/2)+1/f*b/a*sec(f*x+e)*(a+b*sec(f*x +e)^2)^(1/2)+1/f*b^(1/2)*ln(b^(1/2)*sec(f*x+e)+(a+b*sec(f*x+e)^2)^(1/2))
Time = 0.39 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.76 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \, dx=\left [-\frac {2 \, \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b} \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) + \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{f}\right ] \]
[-1/2*(2*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b )*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2))/f, -(sqrt(-b)*arctan(sqrt(-b)*s qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + sqrt((a*cos(f *x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/f]
\[ \int \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.33 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \, dx=-\frac {2 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{2 \, f} \]
-1/2*(2*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b)*log((sqrt(a + b/ cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a + b/cos(f*x + e)^2)*cos(f* x + e) + sqrt(b))))/f
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \, dx=-\frac {{\left (\frac {b \arctan \left (\frac {\sqrt {a \cos \left (f x + e\right )^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + \sqrt {a \cos \left (f x + e\right )^{2} + b}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{f} \]
-(b*arctan(sqrt(a*cos(f*x + e)^2 + b)/sqrt(-b))/sqrt(-b) + sqrt(a*cos(f*x + e)^2 + b))*sgn(cos(f*x + e))/f
Time = 18.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.32 \[ \int \sqrt {a+b \sec ^2(e+f x)} \sin (e+f x) \, dx=-\frac {\cos \left (e+f\,x\right )\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{f}-\frac {\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,1{}\mathrm {i}}{\sqrt {a}\,\cos \left (e+f\,x\right )}\right )\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}\,1{}\mathrm {i}}{\sqrt {a}\,f\,\sqrt {\frac {b}{a\,{\cos \left (e+f\,x\right )}^2}+1}} \]