Integrand size = 25, antiderivative size = 89 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 a^{3/2} f}-\frac {\sqrt {a+b-b \cos ^2(e+f x)} \cot (e+f x) \csc (e+f x)}{2 a f} \]
-1/2*(a-b)*arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/a^(3/2)/ f-1/2*cot(f*x+e)*csc(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/a/f
Time = 0.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {-2 (a-b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )-\sqrt {2} \sqrt {a} \sqrt {2 a+b-b \cos (2 (e+f x))} \cot (e+f x) \csc (e+f x)}{4 a^{3/2} f} \]
(-2*(a - b)*ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2* (e + f*x)]]] - Sqrt[2]*Sqrt[a]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*Cot[e + f*x]*Csc[e + f*x])/(4*a^(3/2)*f)
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3665, 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.
\(\displaystyle \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (e+f x)^3 \sqrt {a+b \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1}{\left (1-\cos ^2(e+f x)\right )^2 \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 296 |
\(\displaystyle -\frac {\frac {(a-b) \int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{2 a}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 a \left (1-\cos ^2(e+f x)\right )}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {\frac {(a-b) \int \frac {1}{1-\frac {a \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}}{2 a}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 a \left (1-\cos ^2(e+f x)\right )}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 a^{3/2}}+\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 a \left (1-\cos ^2(e+f x)\right )}}{f}\) |
-((((a - b)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]] )/(2*a^(3/2)) + (Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/(2*a*(1 - Co s[e + f*x]^2)))/f)
3.2.47.3.1 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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[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. \(164\) vs. \(2(77)=154\).
Time = 1.08 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.85
method | result | size |
default | \(\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{2 a \sin \left (f x +e \right )^{2}}+\frac {\left (-a +b \right ) \ln \left (\frac {2 a +\left (-a +b \right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )^{2}}\right )}{4 a^{\frac {3}{2}}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(165\) |
(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)*(-1/2/a/sin(f*x+e)^2*(-(-b*sin(f *x+e)^2-a)*cos(f*x+e)^2)^(1/2)+1/4*(-a+b)/a^(3/2)*ln((2*a+(-a+b)*sin(f*x+e )^2+2*a^(1/2)*(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2))/sin(f*x+e)^2))/co s(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.35 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.90 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a \cos \left (f x + e\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right )}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}, \frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a \cos \left (f x + e\right )}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}\right ] \]
[1/8*(4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos(f*x + e) - ((a - b)*cos(f*x + e)^2 - a + b)*sqrt(a)*log(2*((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 2*(3*a ^2 + 2*a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + (a + b)*cos (f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) + a^2 + 2*a*b + b^2)/(c os(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)))/(a^2*f*cos(f*x + e)^2 - a^2*f), 1/ 4*(((a - b)*cos(f*x + e)^2 - a + b)*sqrt(-a)*arctan(-1/2*((a - b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(a*b*cos(f*x + e) ^3 - (a^2 + a*b)*cos(f*x + e))) + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*a*cos( f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f)]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]