Integrand size = 23, antiderivative size = 80 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{(a+b)^{3/2} f}-\frac {b \sec (e+f x)}{a (a+b) f \sqrt {a+b \sec ^2(e+f x)}} \]
-arctanh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))/(a+b)^(3/2)/f-b* sec(f*x+e)/a/(a+b)/f/(a+b*sec(f*x+e)^2)^(1/2)
Time = 0.57 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.41 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^3(e+f x) \left (b \sqrt {a+b}+a \text {arctanh}\left (\frac {\sqrt {a+b-a \sin ^2(e+f x)}}{\sqrt {a+b}}\right ) \sqrt {a+b-a \sin ^2(e+f x)}\right )}{2 a (a+b)^{3/2} f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]
-1/2*((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*(b*Sqrt[a + b] + a*Arc Tanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]]*Sqrt[a + b - a*Sin[e + f* x]^2]))/(a*(a + b)^(3/2)*f*(a + b*Sec[e + f*x]^2)^(3/2))
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4622, 25, 296, 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.
| \(\displaystyle \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x) \left (a+b \sec (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4622 |
| \(\displaystyle \frac {\int -\frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 296 |
| \(\displaystyle \frac {-\frac {\int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a}}d\sec (e+f x)}{a+b}-\frac {b \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {-\frac {\int \frac {1}{1-\frac {(a+b) \sec ^2(e+f x)}{b \sec ^2(e+f x)+a}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a}}}{a+b}-\frac {b \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{(a+b)^{3/2}}-\frac {b \sec (e+f x)}{a (a+b) \sqrt {a+b \sec ^2(e+f x)}}}{f}\) |
(-(ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]]/(a + b)^ (3/2)) - (b*Sec[e + f*x])/(a*(a + b)*Sqrt[a + b*Sec[e + f*x]^2]))/f
3.2.9.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
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. \(1060\) vs. \(2(72)=144\).
Time = 1.18 (sec) , antiderivative size = 1061, normalized size of antiderivative = 13.26
-1/2/f/(a+b)^(5/2)/a*(b+a*cos(f*x+e)^2)*(cos(f*x+e)*ln(2/(a+b)^(1/2)*(((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+((b+a*cos(f *x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(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^2+cos(f*x+e)*ln(2/(a+b)^( 1/2)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+( (b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(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+cos(f*x+e)*ln (-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(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^2+cos(f*x+e)*ln (-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x+e)+(( b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(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+2*(a+b)^(3/2) *b+ln(2/(a+b)^(1/2)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/ 2)*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(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^ 2+ln(2/(a+b)^(1/2)*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2 )*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(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 +ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)*cos(f*x...
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (72) = 144\).
Time = 0.33 (sec) , antiderivative size = 344, normalized size of antiderivative = 4.30 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {2 \, {\left (a b + b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - {\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, \frac {{\left (a^{2} \cos \left (f x + e\right )^{2} + a b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) - {\left (a b + b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f}\right ] \]
[-1/2*(2*(a*b + b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - (a^2*cos(f*x + e)^2 + a*b)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*s qrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)))/((a^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x + e)^2 + (a^3*b + 2*a^2*b^2 + a*b^3)*f), ((a^2*cos(f*x + e)^2 + a*b)*sqrt(-a - b)*a rctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e )/(a + b)) - (a*b + b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f *x + e))/((a^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x + e)^2 + (a^3*b + 2*a^2*b^2 + a*b^3)*f)]
\[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 503 vs. \(2 (72) = 144\).
Time = 0.86 (sec) , antiderivative size = 503, normalized size of antiderivative = 6.29 \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {2 \, {\left (\frac {{\left (a b^{2} + b^{3}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{3} b \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 2 \, a^{2} b^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + a b^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} + \frac {a b^{2} + b^{3}}{a^{3} b \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 2 \, a^{2} b^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + a b^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}\right )}}{\sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}} - \frac {\log \left ({\left | -\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b} - \sqrt {a + b} \right |}\right )}{{\left (a + b\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} - \frac {\log \left ({\left | {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} \sqrt {a + b} - a + b \right |}\right )}{{\left (a + b\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} + \frac {\log \left ({\left | {\left (\sqrt {a + b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b}\right )} \sqrt {a + b} - a - b \right |}\right )}{{\left (a + b\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}}{2 \, f} \]
-1/2*(2*((a*b^2 + b^3)*tan(1/2*f*x + 1/2*e)^2/(a^3*b*sgn(cos(f*x + e)) + 2 *a^2*b^2*sgn(cos(f*x + e)) + a*b^3*sgn(cos(f*x + e))) + (a*b^2 + b^3)/(a^3 *b*sgn(cos(f*x + e)) + 2*a^2*b^2*sgn(cos(f*x + e)) + a*b^3*sgn(cos(f*x + e ))))/sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/ 2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - log(abs(-sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 + sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b) - sqrt(a + b)))/((a + b)^(3/2)*sgn(cos(f*x + e))) - log(abs((sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))*sqrt(a + b) - a + b))/((a + b)^(3/2)*sgn(cos(f*x + e))) + log(abs((s qrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan( 1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) ^2 + a + b))*sqrt(a + b) - a - b))/((a + b)^(3/2)*sgn(cos(f*x + e))))/f
Timed out. \[ \int \frac {\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]