Integrand size = 19, antiderivative size = 49 \[ \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx=-B \arctan \left (\frac {B \cos (x)}{\sqrt {A^2+B^2 \sin ^2(x)}}\right )-A \text {arctanh}\left (\frac {A \cos (x)}{\sqrt {A^2+B^2 \sin ^2(x)}}\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(49)=98\).
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.02 \[ \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx=-\sqrt {A^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {A^2} \cos (x)}{\sqrt {2 A^2+B^2-B^2 \cos (2 x)}}\right )+\sqrt {-B^2} \log \left (\sqrt {2} \sqrt {-B^2} \cos (x)+\sqrt {2 A^2+B^2-B^2 \cos (2 x)}\right ) \]
-(Sqrt[A^2]*ArcTanh[(Sqrt[2]*Sqrt[A^2]*Cos[x])/Sqrt[2*A^2 + B^2 - B^2*Cos[ 2*x]]]) + Sqrt[-B^2]*Log[Sqrt[2]*Sqrt[-B^2]*Cos[x] + Sqrt[2*A^2 + B^2 - B^ 2*Cos[2*x]]]
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3665, 301, 224, 216, 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.
\(\displaystyle \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {A^2+B^2 \sin (x)^2}}{\sin (x)}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\int \frac {\sqrt {A^2+B^2-B^2 \cos ^2(x)}}{1-\cos ^2(x)}d\cos (x)\) |
\(\Big \downarrow \) 301 |
\(\displaystyle A^2 \left (-\int \frac {1}{\left (1-\cos ^2(x)\right ) \sqrt {A^2+B^2-B^2 \cos ^2(x)}}d\cos (x)\right )-B^2 \int \frac {1}{\sqrt {A^2+B^2-B^2 \cos ^2(x)}}d\cos (x)\) |
\(\Big \downarrow \) 224 |
\(\displaystyle A^2 \left (-\int \frac {1}{\left (1-\cos ^2(x)\right ) \sqrt {A^2+B^2-B^2 \cos ^2(x)}}d\cos (x)\right )-B^2 \int \frac {1}{\frac {B^2 \cos ^2(x)}{A^2+B^2-B^2 \cos ^2(x)}+1}d\frac {\cos (x)}{\sqrt {A^2+B^2-B^2 \cos ^2(x)}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle A^2 \left (-\int \frac {1}{\left (1-\cos ^2(x)\right ) \sqrt {A^2+B^2-B^2 \cos ^2(x)}}d\cos (x)\right )-B \arctan \left (\frac {B \cos (x)}{\sqrt {A^2-B^2 \cos ^2(x)+B^2}}\right )\) |
\(\Big \downarrow \) 291 |
\(\displaystyle A^2 \left (-\int \frac {1}{1-\frac {A^2 \cos ^2(x)}{A^2+B^2-B^2 \cos ^2(x)}}d\frac {\cos (x)}{\sqrt {A^2+B^2-B^2 \cos ^2(x)}}\right )-B \arctan \left (\frac {B \cos (x)}{\sqrt {A^2-B^2 \cos ^2(x)+B^2}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -B \arctan \left (\frac {B \cos (x)}{\sqrt {A^2-B^2 \cos ^2(x)+B^2}}\right )-A \text {arctanh}\left (\frac {A \cos (x)}{\sqrt {A^2-B^2 \cos ^2(x)+B^2}}\right )\) |
-(B*ArcTan[(B*Cos[x])/Sqrt[A^2 + B^2 - B^2*Cos[x]^2]]) - A*ArcTanh[(A*Cos[ x])/Sqrt[A^2 + B^2 - B^2*Cos[x]^2]]
3.1.42.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 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), x_Symbol] :> Simp[b/ d Int[(a + b*x^2)^(p - 1), x], x] - Simp[(b*c - a*d)/d Int[(a + b*x^2)^ (p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4] || (EqQ[p, 2/3] && E qQ[b*c + 3*a*d, 0]))
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 3.04
method | result | size |
default | \(-\frac {\sqrt {\left (A^{2}+B^{2} \left (\sin ^{2}\left (x \right )\right )\right ) \left (\cos ^{2}\left (x \right )\right )}\, \left (A \,\operatorname {csgn}\left (A \right ) \ln \left (-\frac {A^{2} \left (\sin ^{2}\left (x \right )\right )-B^{2} \left (\sin ^{2}\left (x \right )\right )-2 \,\operatorname {csgn}\left (A \right ) A \sqrt {\left (A^{2}+B^{2} \left (\sin ^{2}\left (x \right )\right )\right ) \left (\cos ^{2}\left (x \right )\right )}-2 A^{2}}{\sin \left (x \right )^{2}}\right )-B \,\operatorname {csgn}\left (B \right ) \arctan \left (\frac {\operatorname {csgn}\left (B \right ) \left (2 B^{2} \left (\sin ^{2}\left (x \right )\right )+A^{2}-B^{2}\right )}{2 B \sqrt {\left (A^{2}+B^{2} \left (\sin ^{2}\left (x \right )\right )\right ) \left (\cos ^{2}\left (x \right )\right )}}\right )\right )}{2 \cos \left (x \right ) \sqrt {A^{2}+B^{2} \left (\sin ^{2}\left (x \right )\right )}}\) | \(149\) |
-1/2*((A^2+B^2*sin(x)^2)*cos(x)^2)^(1/2)*(A*csgn(A)*ln(-(A^2*sin(x)^2-B^2* sin(x)^2-2*csgn(A)*A*((A^2+B^2*sin(x)^2)*cos(x)^2)^(1/2)-2*A^2)/sin(x)^2)- B*csgn(B)*arctan(1/2*csgn(B)/B*(2*B^2*sin(x)^2+A^2-B^2)/((A^2+B^2*sin(x)^2 )*cos(x)^2)^(1/2)))/cos(x)/(A^2+B^2*sin(x)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (45) = 90\).
Time = 0.32 (sec) , antiderivative size = 244, normalized size of antiderivative = 4.98 \[ \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx=\frac {1}{2} \, B \arctan \left (-\frac {{\left (A^{4} + 2 \, A^{2} B^{2} + B^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - 2 \, {\left (2 \, B^{3} \cos \left (x\right )^{3} - {\left (A^{2} B + B^{3}\right )} \cos \left (x\right )\right )} \sqrt {-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}}}{4 \, B^{4} \cos \left (x\right )^{4} + A^{4} + 2 \, A^{2} B^{2} + B^{4} - {\left (A^{4} + 6 \, A^{2} B^{2} + 5 \, B^{4}\right )} \cos \left (x\right )^{2}}\right ) - \frac {1}{2} \, B \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right ) - \frac {1}{2} \, A \log \left (-B^{2} \cos \left (x\right )^{2} + A B \cos \left (x\right ) \sin \left (x\right ) + A^{2} + B^{2} + \sqrt {-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}} {\left (A \cos \left (x\right ) + B \sin \left (x\right )\right )}\right ) + \frac {1}{2} \, A \log \left (-B^{2} \cos \left (x\right )^{2} - A B \cos \left (x\right ) \sin \left (x\right ) + A^{2} + B^{2} - \sqrt {-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}} {\left (A \cos \left (x\right ) - B \sin \left (x\right )\right )}\right ) \]
1/2*B*arctan(-((A^4 + 2*A^2*B^2 + B^4)*cos(x)*sin(x) - 2*(2*B^3*cos(x)^3 - (A^2*B + B^3)*cos(x))*sqrt(-B^2*cos(x)^2 + A^2 + B^2))/(4*B^4*cos(x)^4 + A^4 + 2*A^2*B^2 + B^4 - (A^4 + 6*A^2*B^2 + 5*B^4)*cos(x)^2)) - 1/2*B*arcta n(sin(x)/cos(x)) - 1/2*A*log(-B^2*cos(x)^2 + A*B*cos(x)*sin(x) + A^2 + B^2 + sqrt(-B^2*cos(x)^2 + A^2 + B^2)*(A*cos(x) + B*sin(x))) + 1/2*A*log(-B^2 *cos(x)^2 - A*B*cos(x)*sin(x) + A^2 + B^2 - sqrt(-B^2*cos(x)^2 + A^2 + B^2 )*(A*cos(x) - B*sin(x)))
\[ \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx=\int \frac {\sqrt {A^{2} + B^{2} \sin ^{2}{\left (x \right )}}}{\sin {\left (x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (45) = 90\).
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.37 \[ \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx=-B \arcsin \left (\frac {B^{2} \cos \left (x\right )}{\sqrt {A^{2} B^{2} + B^{4}}}\right ) - \frac {1}{2} \, A \log \left (B^{2} - \frac {A^{2}}{\cos \left (x\right ) - 1} - \frac {\sqrt {-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}} A}{\cos \left (x\right ) - 1}\right ) + \frac {1}{2} \, A \log \left (-B^{2} + \frac {A^{2}}{\cos \left (x\right ) + 1} + \frac {\sqrt {-B^{2} \cos \left (x\right )^{2} + A^{2} + B^{2}} A}{\cos \left (x\right ) + 1}\right ) \]
-B*arcsin(B^2*cos(x)/sqrt(A^2*B^2 + B^4)) - 1/2*A*log(B^2 - A^2/(cos(x) - 1) - sqrt(-B^2*cos(x)^2 + A^2 + B^2)*A/(cos(x) - 1)) + 1/2*A*log(-B^2 + A^ 2/(cos(x) + 1) + sqrt(-B^2*cos(x)^2 + A^2 + B^2)*A/(cos(x) + 1))
\[ \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx=\int { \frac {\sqrt {B^{2} \sin \left (x\right )^{2} + A^{2}}}{\sin \left (x\right )} \,d x } \]
Timed out. \[ \int \csc (x) \sqrt {A^2+B^2 \sin ^2(x)} \, dx=\int \frac {\sqrt {A^2+B^2\,{\sin \left (x\right )}^2}}{\sin \left (x\right )} \,d x \]