Integrand size = 13, antiderivative size = 85 \[ \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx=-\frac {a \log (1-\sin (x))}{4 (a+b)^2}+\frac {a \log (1+\sin (x))}{4 (a-b)^2}-\frac {a^2 b \log (b+a \sin (x))}{\left (a^2-b^2\right )^2}-\frac {\sec ^2(x) (b-a \sin (x))}{2 \left (a^2-b^2\right )} \]
-1/4*a*ln(1-sin(x))/(a+b)^2+1/4*a*ln(1+sin(x))/(a-b)^2-a^2*b*ln(b+a*sin(x) )/(a^2-b^2)^2-1/2*sec(x)^2*(b-a*sin(x))/(a^2-b^2)
Time = 0.60 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx=\frac {\csc (x) \left (-\frac {2 a \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )}{(a+b)^2}+\frac {2 a \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )}{(a-b)^2}-\frac {4 a^2 b \log (b+a \sin (x))}{\left (a^2-b^2\right )^2}+\frac {1}{(a+b) \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^2}+\frac {1}{(-a+b) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}\right ) (b+a \sin (x))}{4 (a+b \csc (x))} \]
(Csc[x]*((-2*a*Log[Cos[x/2] - Sin[x/2]])/(a + b)^2 + (2*a*Log[Cos[x/2] + S in[x/2]])/(a - b)^2 - (4*a^2*b*Log[b + a*Sin[x]])/(a^2 - b^2)^2 + 1/((a + b)*(Cos[x/2] - Sin[x/2])^2) + 1/((-a + b)*(Cos[x/2] + Sin[x/2])^2))*(b + a *Sin[x]))/(4*(a + b*Csc[x]))
Time = 0.43 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.51, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3042, 4360, 3042, 3316, 27, 593, 25, 657, 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.
| \(\displaystyle \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (x)^3 (a+b \csc (x))}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\tan (x) \sec ^2(x)}{a \sin (x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (x)}{\cos (x)^3 (a \sin (x)+b)}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle a^3 \int \frac {\sin (x)}{(b+a \sin (x)) \left (a^2-a^2 \sin ^2(x)\right )^2}d(a \sin (x))\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle a^2 \int \frac {a \sin (x)}{(b+a \sin (x)) \left (a^2-a^2 \sin ^2(x)\right )^2}d(a \sin (x))\) |
| \(\Big \downarrow \) 593 |
| \(\displaystyle a^2 \left (\frac {\int -\frac {b-a \sin (x)}{(b+a \sin (x)) \left (a^2-a^2 \sin ^2(x)\right )}d(a \sin (x))}{2 \left (a^2-b^2\right )}-\frac {b-a \sin (x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \sin ^2(x)\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle a^2 \left (-\frac {\int \frac {b-a \sin (x)}{(b+a \sin (x)) \left (a^2-a^2 \sin ^2(x)\right )}d(a \sin (x))}{2 \left (a^2-b^2\right )}-\frac {b-a \sin (x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \sin ^2(x)\right )}\right )\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle a^2 \left (-\frac {\int \left (\frac {-a-b}{2 a (a-b) (\sin (x) a+a)}+\frac {b-a}{2 a (a+b) (a-a \sin (x))}+\frac {2 b}{(a-b) (a+b) (b+a \sin (x))}\right )d(a \sin (x))}{2 \left (a^2-b^2\right )}-\frac {b-a \sin (x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \sin ^2(x)\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a^2 \left (-\frac {b-a \sin (x)}{2 \left (a^2-b^2\right ) \left (a^2-a^2 \sin ^2(x)\right )}-\frac {\frac {2 b \log (a \sin (x)+b)}{a^2-b^2}+\frac {(a-b) \log (a-a \sin (x))}{2 a (a+b)}-\frac {(a+b) \log (a \sin (x)+a)}{2 a (a-b)}}{2 \left (a^2-b^2\right )}\right )\) |
a^2*(-1/2*(((a - b)*Log[a - a*Sin[x]])/(2*a*(a + b)) - ((a + b)*Log[a + a* Sin[x]])/(2*a*(a - b)) + (2*b*Log[b + a*Sin[x]])/(a^2 - b^2))/(a^2 - b^2) - (b - a*Sin[x])/(2*(a^2 - b^2)*(a^2 - a^2*Sin[x]^2)))
3.1.15.3.1 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[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b* x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
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 = 0.96 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (x \right )-1\right )}-\frac {a \ln \left (\sin \left (x \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {b \,a^{2} \ln \left (a \sin \left (x \right )+b \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (x \right )\right )}+\frac {a \ln \left (1+\sin \left (x \right )\right )}{4 \left (a -b \right )^{2}}\) | \(89\) |
| parallelrisch | \(\frac {-2 a^{2} b \ln \left (\frac {a \sin \left (x \right )+b}{\cos \left (x \right )+1}\right )+2 a^{2} b \ln \left (\frac {\csc \left (x \right )}{2}-\frac {\cot \left (x \right )}{2}+\frac {1}{2}\right )-a \left (a -b \right )^{2} \ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )+\left (a^{3}+a \,b^{2}\right ) \ln \left (\csc \left (x \right )-\cot \left (x \right )+1\right )+\tan \left (x \right ) \left (a -b \right ) \left (a +b \right ) \left (a \sec \left (x \right )-b \tan \left (x \right )\right )}{2 \left (a -b \right )^{2} \left (a +b \right )^{2}}\) | \(112\) |
| norman | \(\frac {\frac {a \tan \left (\frac {x}{2}\right )}{a^{2}-b^{2}}+\frac {a \tan \left (\frac {x}{2}\right )^{3}}{a^{2}-b^{2}}-\frac {2 b \tan \left (\frac {x}{2}\right )^{2}}{a^{2}-b^{2}}}{\left (\tan \left (\frac {x}{2}\right )^{2}-1\right )^{2}}+\frac {a \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a^{2}-4 a b +2 b^{2}}-\frac {a \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {a^{2} b \ln \left (b \tan \left (\frac {x}{2}\right )^{2}+2 a \tan \left (\frac {x}{2}\right )+b \right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) | \(157\) |
| risch | \(\frac {i x a}{2 a^{2}+4 a b +2 b^{2}}-\frac {i x a}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i x \,a^{2} b}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {i a \,{\mathrm e}^{3 i x}-i {\mathrm e}^{i x} a +2 b \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2} \left (-a^{2}+b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{i x}-i\right )}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {a \ln \left (i+{\mathrm e}^{i x}\right )}{2 a^{2}-4 a b +2 b^{2}}-\frac {a^{2} b \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) | \(204\) |
-1/(4*a+4*b)/(sin(x)-1)-1/4*a/(a+b)^2*ln(sin(x)-1)-b*a^2/(a+b)^2/(a-b)^2*l n(a*sin(x)+b)-1/(4*a-4*b)/(1+sin(x))+1/4*a*ln(1+sin(x))/(a-b)^2
Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx=-\frac {4 \, a^{2} b \cos \left (x\right )^{2} \log \left (a \sin \left (x\right ) + b\right ) - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2}} \]
-1/4*(4*a^2*b*cos(x)^2*log(a*sin(x) + b) - (a^3 + 2*a^2*b + a*b^2)*cos(x)^ 2*log(sin(x) + 1) + (a^3 - 2*a^2*b + a*b^2)*cos(x)^2*log(-sin(x) + 1) + 2* a^2*b - 2*b^3 - 2*(a^3 - a*b^2)*sin(x))/((a^4 - 2*a^2*b^2 + b^4)*cos(x)^2)
\[ \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx=\int \frac {\sec ^{3}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx=-\frac {a^{2} b \log \left (a \sin \left (x\right ) + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {a \log \left (\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {a \log \left (\sin \left (x\right ) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a \sin \left (x\right ) - b}{2 \, {\left ({\left (a^{2} - b^{2}\right )} \sin \left (x\right )^{2} - a^{2} + b^{2}\right )}} \]
-a^2*b*log(a*sin(x) + b)/(a^4 - 2*a^2*b^2 + b^4) + 1/4*a*log(sin(x) + 1)/( a^2 - 2*a*b + b^2) - 1/4*a*log(sin(x) - 1)/(a^2 + 2*a*b + b^2) - 1/2*(a*si n(x) - b)/((a^2 - b^2)*sin(x)^2 - a^2 + b^2)
Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx=-\frac {a^{3} b \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {a \log \left (\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {a \log \left (-\sin \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {a^{2} b - b^{3} - {\left (a^{3} - a b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} {\left (\sin \left (x\right ) + 1\right )} {\left (\sin \left (x\right ) - 1\right )}} \]
-a^3*b*log(abs(a*sin(x) + b))/(a^5 - 2*a^3*b^2 + a*b^4) + 1/4*a*log(sin(x) + 1)/(a^2 - 2*a*b + b^2) - 1/4*a*log(-sin(x) + 1)/(a^2 + 2*a*b + b^2) + 1 /2*(a^2*b - b^3 - (a^3 - a*b^2)*sin(x))/((a + b)^2*(a - b)^2*(sin(x) + 1)* (sin(x) - 1))
Time = 19.52 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^3(x)}{a+b \csc (x)} \, dx=\ln \left (b+a\,\sin \left (x\right )\right )\,\left (\frac {a}{4\,{\left (a+b\right )}^2}-\frac {a}{4\,{\left (a-b\right )}^2}\right )-\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\sin \left (x\right )}{2\,\left (a^2-b^2\right )}}{{\cos \left (x\right )}^2}+\frac {a\,\ln \left (\sin \left (x\right )+1\right )}{4\,{\left (a-b\right )}^2}-\frac {a\,\ln \left (\sin \left (x\right )-1\right )}{4\,{\left (a+b\right )}^2} \]