Integrand size = 23, antiderivative size = 42 \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{\sqrt {a} f} \]
Leaf count is larger than twice the leaf count of optimal. \(221\) vs. \(2(42)=84\).
Time = 3.14 (sec) , antiderivative size = 221, normalized size of antiderivative = 5.26 \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\cos (e+f x) \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{2 \sqrt {a} f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}} \]
(Cos[e + f*x]*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/ 2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2 ]])*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x ]^2])/(2*Sqrt[a]*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2 ]^4])
Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4147, 25, 291, 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.
\(\displaystyle \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (e+f x) \sqrt {a+b \tan (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4147 |
\(\displaystyle \frac {\int -\frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {\int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{\sqrt {a} f}\) |
3.2.19.3.1 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]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(306\) vs. \(2(36)=72\).
Time = 0.86 (sec) , antiderivative size = 307, normalized size of antiderivative = 7.31
method | result | size |
default | \(-\frac {\left (\ln \left (-\frac {4 \left (\cos \left (f x +e \right ) \sqrt {a}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\cos \left (f x +e \right ) a -b \cos \left (f x +e \right )+\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}+b \right )}{\cos \left (f x +e \right )-1}\right )+\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {a}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}-2 \cos \left (f x +e \right ) a +2 b \cos \left (f x +e \right )+2 b}{\sqrt {a}\, \left (\cos \left (f x +e \right )+1\right )}\right )\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\sec \left (f x +e \right )+1\right )}{2 f \sqrt {a}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\) | \(307\) |
-1/2/f/a^(1/2)*(ln(-4*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+ b)/(cos(f*x+e)+1)^2)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-b*co s(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))+ln(2/a^(1 /2)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^ 2)^(1/2)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2 )-cos(f*x+e)*a+b*cos(f*x+e)+b)/(cos(f*x+e)+1)))*((a*cos(f*x+e)^2-b*cos(f*x +e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*(sec(f*x+e)+1)
Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 3.19 \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {\log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, \sqrt {a} f}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right )}{a f}\right ] \]
[1/2*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e) ^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1))/(sqrt( a)*f), sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a)/(a*f)]
\[ \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]