Integrand size = 23, antiderivative size = 43 \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{\sqrt {a+b} f} \] Output:
-arctanh((a+b)^(1/2)*sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b)^(1/2)/f
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.00 \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{\sqrt {2} \sqrt {a+b} f \sqrt {a+b \sec ^2(e+f x)}} \] Input:
Integrate[Csc[e + f*x]/Sqrt[a + b*Sec[e + f*x]^2],x]
Output:
-((ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]]*Sqrt[a + 2*b + a*Co s[2*e + 2*f*x]]*Sec[e + f*x])/(Sqrt[2]*Sqrt[a + b]*f*Sqrt[a + b*Sec[e + f* x]^2]))
Time = 0.25 (sec) , antiderivative size = 43, 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, 4622, 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.
\(\displaystyle \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (e+f x) \sqrt {a+b \sec (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int -\frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {\int \frac {1}{1-\frac {(a+b) \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f \sqrt {a+b}}\) |
Input:
Int[Csc[e + f*x]/Sqrt[a + b*Sec[e + f*x]^2],x]
Output:
-(ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]]/(Sqrt[a + b]*f))
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[((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])
Leaf count of result is larger than twice the leaf count of optimal. \(246\) vs. \(2(37)=74\).
Time = 1.33 (sec) , antiderivative size = 247, normalized size of antiderivative = 5.74
method | result | size |
default | \(-\frac {\left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-2 \cos \left (f x +e \right ) a +2 b}{\sqrt {a +b}\, \left (1+\cos \left (f x +e \right )\right )}\right )+\ln \left (-\frac {4 \left (\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sqrt {a +b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\cos \left (f x +e \right ) a +b \right )}{-1+\cos \left (f x +e \right )}\right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\sec \left (f x +e \right )+1\right )}{2 f \sqrt {a +b}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}\) | \(247\) |
Input:
int(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/f/(a+b)^(1/2)*(ln(2/(a+b)^(1/2)*((a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+c os(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ e))^2)^(1/2)-cos(f*x+e)*a+b)/(1+cos(f*x+e)))+ln(-4*((a+b)^(1/2)*((b+a*cos( f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+(a+b)^(1/2)*((b+a*cos(f*x+e)^ 2)/(1+cos(f*x+e))^2)^(1/2)+cos(f*x+e)*a+b)/(-1+cos(f*x+e))))*((b+a*cos(f*x +e)^2)/(1+cos(f*x+e))^2)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2)*(sec(f*x+e)+1)
Time = 0.14 (sec) , antiderivative size = 149, normalized size of antiderivative = 3.47 \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {\log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, \sqrt {a + b} f}, \frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right )}{{\left (a + b\right )} f}\right ] \] Input:
integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
[1/2*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/c os(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1))/(sqrt(a + b)* f), sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a*cos(f*x + e)^2 + b))/((a + b)*f)]
\[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \] Input:
integrate(csc(f*x+e)/(a+b*sec(f*x+e)**2)**(1/2),x)
Output:
Integral(csc(e + f*x)/sqrt(a + b*sec(e + f*x)**2), x)
\[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \] Input:
integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
integrate(csc(f*x + e)/sqrt(b*sec(f*x + e)^2 + a), x)
Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.14 \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a + b}}{\sqrt {-a - b}}\right )}{\sqrt {-a - b} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \] Input:
integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
arctan(sqrt(-a*sin(f*x + e)^2 + a + b)/sqrt(-a - b))/(sqrt(-a - b)*f*sgn(c os(f*x + e)))
Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \] Input:
int(1/(sin(e + f*x)*(a + b/cos(e + f*x)^2)^(1/2)),x)
Output:
int(1/(sin(e + f*x)*(a + b/cos(e + f*x)^2)^(1/2)), x)
\[ \int \frac {\csc (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \csc \left (f x +e \right )}{\sec \left (f x +e \right )^{2} b +a}d x \] Input:
int(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2),x)
Output:
int((sqrt(sec(e + f*x)**2*b + a)*csc(e + f*x))/(sec(e + f*x)**2*b + a),x)