Integrand size = 13, antiderivative size = 46 \[ \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx=\frac {\text {arctanh}(\sin (x))}{8 a}+\frac {\sec ^4(x)}{4 a}+\frac {\sec (x) \tan (x)}{8 a}-\frac {\sec ^3(x) \tan (x)}{4 a} \]
Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx=\frac {\text {arctanh}(\sin (x))+\frac {1}{1-\sin (x)}+\frac {1}{(1+\sin (x))^2}}{8 a} \]
Time = 0.49 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 4360, 3042, 3314, 3042, 3086, 15, 3091, 3042, 4255, 3042, 4257}
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.
\(\displaystyle \int \frac {\sec ^3(x)}{a \csc (x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (x)^3 (a \csc (x)+a)}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\tan (x) \sec ^2(x)}{a \sin (x)+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{\cos (x)^3 (a \sin (x)+a)}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \sec ^4(x) \tan (x)dx}{a}-\frac {\int \sec ^3(x) \tan ^2(x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec (x)^4 \tan (x)dx}{a}-\frac {\int \sec (x)^3 \tan (x)^2dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \sec ^3(x)d\sec (x)}{a}-\frac {\int \sec (x)^3 \tan (x)^2dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\sec ^4(x)}{4 a}-\frac {\int \sec (x)^3 \tan (x)^2dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\sec ^4(x)}{4 a}-\frac {\frac {1}{4} \tan (x) \sec ^3(x)-\frac {1}{4} \int \sec ^3(x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^4(x)}{4 a}-\frac {\frac {1}{4} \tan (x) \sec ^3(x)-\frac {1}{4} \int \csc \left (x+\frac {\pi }{2}\right )^3dx}{a}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sec ^4(x)}{4 a}-\frac {\frac {1}{4} \left (-\frac {\int \sec (x)dx}{2}-\frac {1}{2} \tan (x) \sec (x)\right )+\frac {1}{4} \tan (x) \sec ^3(x)}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^4(x)}{4 a}-\frac {\frac {1}{4} \left (-\frac {1}{2} \int \csc \left (x+\frac {\pi }{2}\right )dx-\frac {1}{2} \tan (x) \sec (x)\right )+\frac {1}{4} \tan (x) \sec ^3(x)}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sec ^4(x)}{4 a}-\frac {\frac {1}{4} \left (-\frac {1}{2} \text {arctanh}(\sin (x))-\frac {1}{2} \tan (x) \sec (x)\right )+\frac {1}{4} \tan (x) \sec ^3(x)}{a}\) |
3.1.7.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
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 = 0.67 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {-\frac {1}{8 \left (\sin \left (x \right )-1\right )}-\frac {\ln \left (\sin \left (x \right )-1\right )}{16}+\frac {1}{8 \left (1+\sin \left (x \right )\right )^{2}}+\frac {\ln \left (1+\sin \left (x \right )\right )}{16}}{a}\) | \(36\) |
parallelrisch | \(\frac {\ln \left (\csc \left (x \right )-\cot \left (x \right )+1\right )-\ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )-2+\sec \left (x \right ) \tan \left (x \right )-2 \sec \left (x \right )^{3} \tan \left (x \right )+2 \sec \left (x \right )^{4}}{8 a}\) | \(47\) |
risch | \(-\frac {i \left (2 i {\mathrm e}^{4 i x}+{\mathrm e}^{5 i x}-2 i {\mathrm e}^{2 i x}-10 \,{\mathrm e}^{3 i x}+{\mathrm e}^{i x}\right )}{4 \left (i+{\mathrm e}^{i x}\right )^{4} \left ({\mathrm e}^{i x}-i\right )^{2} a}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{8 a}+\frac {\ln \left (i+{\mathrm e}^{i x}\right )}{8 a}\) | \(90\) |
norman | \(\frac {\frac {3 \tan \left (\frac {x}{2}\right )^{2}}{2 a}+\frac {3 \tan \left (\frac {x}{2}\right )^{4}}{2 a}+\frac {3 \tan \left (\frac {x}{2}\right )^{3}}{2 a}-\frac {\tan \left (\frac {x}{2}\right )}{4 a}-\frac {\tan \left (\frac {x}{2}\right )^{5}}{4 a}}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{4} \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{8 a}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{8 a}\) | \(97\) |
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx=-\frac {2 \, \cos \left (x\right )^{2} - {\left (\cos \left (x\right )^{2} \sin \left (x\right ) + \cos \left (x\right )^{2}\right )} \log \left (\sin \left (x\right ) + 1\right ) + {\left (\cos \left (x\right )^{2} \sin \left (x\right ) + \cos \left (x\right )^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, \sin \left (x\right ) - 6}{16 \, {\left (a \cos \left (x\right )^{2} \sin \left (x\right ) + a \cos \left (x\right )^{2}\right )}} \]
-1/16*(2*cos(x)^2 - (cos(x)^2*sin(x) + cos(x)^2)*log(sin(x) + 1) + (cos(x) ^2*sin(x) + cos(x)^2)*log(-sin(x) + 1) - 2*sin(x) - 6)/(a*cos(x)^2*sin(x) + a*cos(x)^2)
\[ \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx=\frac {\int \frac {\sec ^{3}{\left (x \right )}}{\csc {\left (x \right )} + 1}\, dx}{a} \]
Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx=-\frac {\sin \left (x\right )^{2} + \sin \left (x\right ) + 2}{8 \, {\left (a \sin \left (x\right )^{3} + a \sin \left (x\right )^{2} - a \sin \left (x\right ) - a\right )}} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{16 \, a} - \frac {\log \left (\sin \left (x\right ) - 1\right )}{16 \, a} \]
-1/8*(sin(x)^2 + sin(x) + 2)/(a*sin(x)^3 + a*sin(x)^2 - a*sin(x) - a) + 1/ 16*log(sin(x) + 1)/a - 1/16*log(sin(x) - 1)/a
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx=\frac {\log \left (\sin \left (x\right ) + 1\right )}{16 \, a} - \frac {\log \left (-\sin \left (x\right ) + 1\right )}{16 \, a} - \frac {\sin \left (x\right )^{2} + \sin \left (x\right ) + 2}{8 \, a {\left (\sin \left (x\right ) + 1\right )}^{2} {\left (\sin \left (x\right ) - 1\right )}} \]
1/16*log(sin(x) + 1)/a - 1/16*log(-sin(x) + 1)/a - 1/8*(sin(x)^2 + sin(x) + 2)/(a*(sin(x) + 1)^2*(sin(x) - 1))
Time = 18.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx=\frac {\mathrm {atanh}\left (\sin \left (x\right )\right )}{8\,a}+\frac {\frac {{\sin \left (x\right )}^2}{8}+\frac {\sin \left (x\right )}{8}+\frac {1}{4}}{-a\,{\sin \left (x\right )}^3-a\,{\sin \left (x\right )}^2+a\,\sin \left (x\right )+a} \]