Integrand size = 13, antiderivative size = 61 \[ \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx=-\frac {b \left (a^2-b^2\right ) \log (b+a \sin (x))}{a^4}+\frac {\left (a^2-b^2\right ) \sin (x)}{a^3}+\frac {b \sin ^2(x)}{2 a^2}-\frac {\sin ^3(x)}{3 a} \]
Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx=\frac {6 b \left (-a^2+b^2\right ) \log (b+a \sin (x))+6 a \left (a^2-b^2\right ) \sin (x)+3 a^2 b \sin ^2(x)-2 a^3 \sin ^3(x)}{6 a^4} \]
(6*b*(-a^2 + b^2)*Log[b + a*Sin[x]] + 6*a*(a^2 - b^2)*Sin[x] + 3*a^2*b*Sin [x]^2 - 2*a^3*Sin[x]^3)/(6*a^4)
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3042, 4360, 3042, 3316, 27, 522, 2009}
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.
\(\displaystyle \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (x)^3}{a+b \csc (x)}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\sin (x) \cos ^3(x)}{a \sin (x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x) \cos (x)^3}{a \sin (x)+b}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \frac {\sin (x) \left (a^2-a^2 \sin ^2(x)\right )}{b+a \sin (x)}d(a \sin (x))}{a^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a \sin (x) \left (a^2-a^2 \sin ^2(x)\right )}{b+a \sin (x)}d(a \sin (x))}{a^4}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (-\sin ^2(x) a^2+\left (1-\frac {b^2}{a^2}\right ) a^2+b \sin (x) a+\frac {b^3-a^2 b}{b+a \sin (x)}\right )d(a \sin (x))}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} a^3 \sin ^3(x)+a \left (a^2-b^2\right ) \sin (x)-b \left (a^2-b^2\right ) \log (a \sin (x)+b)+\frac {1}{2} a^2 b \sin ^2(x)}{a^4}\) |
(-(b*(a^2 - b^2)*Log[b + a*Sin[x]]) + a*(a^2 - b^2)*Sin[x] + (a^2*b*Sin[x] ^2)/2 - (a^3*Sin[x]^3)/3)/a^4
3.1.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.65 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {\frac {\sin \left (x \right )^{3} a^{2}}{3}-\frac {b \sin \left (x \right )^{2} a}{2}-a^{2} \sin \left (x \right )+b^{2} \sin \left (x \right )}{a^{3}}-\frac {b \left (a^{2}-b^{2}\right ) \ln \left (a \sin \left (x \right )+b \right )}{a^{4}}\) | \(60\) |
parallelrisch | \(\frac {9 a^{3} \sin \left (x \right )-12 a \,b^{2} \sin \left (x \right )+\sin \left (3 x \right ) a^{3}-3 a^{2} b \cos \left (2 x \right )-12 a^{2} b \ln \left (\frac {a \sin \left (x \right )+b}{\cos \left (x \right )+1}\right )+12 b^{3} \ln \left (\frac {a \sin \left (x \right )+b}{\cos \left (x \right )+1}\right )+12 a^{2} b \ln \left (\frac {1}{\cos \left (x \right )+1}\right )-12 b^{3} \ln \left (\frac {1}{\cos \left (x \right )+1}\right )+3 a^{2} b}{12 a^{4}}\) | \(110\) |
norman | \(\frac {\frac {2 b \tan \left (\frac {x}{2}\right )^{2}}{a^{2}}+\frac {2 b \tan \left (\frac {x}{2}\right )^{4}}{a^{2}}+\frac {4 \left (a^{2}-3 b^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}}{3 a^{3}}+\frac {2 \left (a^{2}-b^{2}\right ) \tan \left (\frac {x}{2}\right )}{a^{3}}+\frac {2 \left (a^{2}-b^{2}\right ) \tan \left (\frac {x}{2}\right )^{5}}{a^{3}}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3}}+\frac {b \left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}{a^{4}}-\frac {b \left (a^{2}-b^{2}\right ) \ln \left (b \tan \left (\frac {x}{2}\right )^{2}+2 a \tan \left (\frac {x}{2}\right )+b \right )}{a^{4}}\) | \(152\) |
risch | \(\frac {i b x}{a^{2}}-\frac {i b^{3} x}{a^{4}}-\frac {b \,{\mathrm e}^{2 i x}}{8 a^{2}}-\frac {3 i {\mathrm e}^{i x}}{8 a}+\frac {i {\mathrm e}^{i x} b^{2}}{2 a^{3}}+\frac {3 i {\mathrm e}^{-i x}}{8 a}-\frac {i {\mathrm e}^{-i x} b^{2}}{2 a^{3}}-\frac {b \,{\mathrm e}^{-2 i x}}{8 a^{2}}-\frac {b \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a^{4}}+\frac {\sin \left (3 x \right )}{12 a}\) | \(154\) |
-1/a^3*(1/3*sin(x)^3*a^2-1/2*b*sin(x)^2*a-a^2*sin(x)+b^2*sin(x))-b*(a^2-b^ 2)*ln(a*sin(x)+b)/a^4
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx=-\frac {3 \, a^{2} b \cos \left (x\right )^{2} + 6 \, {\left (a^{2} b - b^{3}\right )} \log \left (a \sin \left (x\right ) + b\right ) - 2 \, {\left (a^{3} \cos \left (x\right )^{2} + 2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (x\right )}{6 \, a^{4}} \]
-1/6*(3*a^2*b*cos(x)^2 + 6*(a^2*b - b^3)*log(a*sin(x) + b) - 2*(a^3*cos(x) ^2 + 2*a^3 - 3*a*b^2)*sin(x))/a^4
Timed out. \[ \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx=-\frac {2 \, a^{2} \sin \left (x\right )^{3} - 3 \, a b \sin \left (x\right )^{2} - 6 \, {\left (a^{2} - b^{2}\right )} \sin \left (x\right )}{6 \, a^{3}} - \frac {{\left (a^{2} b - b^{3}\right )} \log \left (a \sin \left (x\right ) + b\right )}{a^{4}} \]
-1/6*(2*a^2*sin(x)^3 - 3*a*b*sin(x)^2 - 6*(a^2 - b^2)*sin(x))/a^3 - (a^2*b - b^3)*log(a*sin(x) + b)/a^4
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx=-\frac {2 \, a^{2} \sin \left (x\right )^{3} - 3 \, a b \sin \left (x\right )^{2} - 6 \, a^{2} \sin \left (x\right ) + 6 \, b^{2} \sin \left (x\right )}{6 \, a^{3}} - \frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a^{4}} \]
-1/6*(2*a^2*sin(x)^3 - 3*a*b*sin(x)^2 - 6*a^2*sin(x) + 6*b^2*sin(x))/a^3 - (a^2*b - b^3)*log(abs(a*sin(x) + b))/a^4
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^3(x)}{a+b \csc (x)} \, dx=\sin \left (x\right )\,\left (\frac {1}{a}-\frac {b^2}{a^3}\right )-\frac {{\sin \left (x\right )}^3}{3\,a}+\frac {b\,{\sin \left (x\right )}^2}{2\,a^2}-\frac {\ln \left (b+a\,\sin \left (x\right )\right )\,\left (a^2\,b-b^3\right )}{a^4} \]