Integrand size = 23, antiderivative size = 79 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{a^{3/2} f}+\frac {b \cos (e+f x)}{a (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}} \]
-arctanh(cos(f*x+e)*a^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/a^(3/2)/f+b*cos(f* x+e)/a/(a+b)/f/(a+b-b*cos(f*x+e)^2)^(1/2)
Time = 0.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.18 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-\text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \cos (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\frac {\sqrt {2} \sqrt {a} b \cos (e+f x)}{(a+b) \sqrt {2 a+b-b \cos (2 (e+f x))}}}{a^{3/2} f} \]
(-ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)] ]] + (Sqrt[2]*Sqrt[a]*b*Cos[e + f*x])/((a + b)*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]))/(a^(3/2)*f)
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, 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.
\(\displaystyle \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (e+f x) \left (a+b \sin (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \left (-b \cos ^2(e+f x)+a+b\right )^{3/2}}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 296 |
\(\displaystyle -\frac {\frac {\int \frac {1}{\left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{a}-\frac {b \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {\frac {\int \frac {1}{1-\frac {a \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}}{a}-\frac {b \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b \cos (e+f x)}{a (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\) |
-((ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/a^(3/2) - (b*Cos[e + f*x])/(a*(a + b)*Sqrt[a + b - b*Cos[e + f*x]^2]))/f)
3.2.55.3.1 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. \(2(71)=142\).
Time = 1.02 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.09
method | result | size |
default | \(\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\ln \left (\frac {2 a +\left (-a +b \right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )^{2}}\right )}{2 a^{\frac {3}{2}}}+\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a \left (a +b \right ) \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(165\) |
(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)*(-1/2/a^(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)+1/a*b*cos(f*x+e)^2/(a+b)/(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2))/cos (f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (71) = 142\).
Time = 0.38 (sec) , antiderivative size = 422, normalized size of antiderivative = 5.34 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) - {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}, -\frac {2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) - {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right )}{2 \, {\left ({\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}\right ] \]
[-1/4*(4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*b*cos(f*x + e) - ((a*b + b^2)*c os(f*x + e)^2 - a^2 - 2*a*b - b^2)*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)/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)))/((a^3*b + a^2*b ^2)*f*cos(f*x + e)^2 - (a^4 + 2*a^3*b + a^2*b^2)*f), -1/2*(2*sqrt(-b*cos(f *x + e)^2 + a + b)*a*b*cos(f*x + e) - ((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*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))))/((a^3*b + a^2*b^2)*f*cos(f*x + e)^2 - (a^4 + 2*a^3*b + a^2*b^2)* f)]
\[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (71) = 142\).
Time = 0.34 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.09 \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {2 \, b^{2} \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a^{2} b + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b^{2}} - \frac {\log \left (b - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) - 1} - \frac {a}{\cos \left (f x + e\right ) - 1}\right )}{a^{\frac {3}{2}}} + \frac {\log \left (-b + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) + 1} + \frac {a}{\cos \left (f x + e\right ) + 1}\right )}{a^{\frac {3}{2}}}}{2 \, f} \]
1/2*(2*b^2*cos(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*a^2*b + sqrt(-b*c os(f*x + e)^2 + a + b)*a*b^2) - log(b - sqrt(-b*cos(f*x + e)^2 + a + b)*sq rt(a)/(cos(f*x + e) - 1) - a/(cos(f*x + e) - 1))/a^(3/2) + log(-b + sqrt(- b*cos(f*x + e)^2 + a + b)*sqrt(a)/(cos(f*x + e) + 1) + a/(cos(f*x + e) + 1 ))/a^(3/2))/f
\[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\csc \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]