Integrand size = 13, antiderivative size = 68 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5 \log (1-\sin (x))}{16 a}+\frac {11 \log (1+\sin (x))}{16 a}+\frac {1}{8 a (1-\sin (x))}-\frac {1}{8 a (1+\sin (x))^2}+\frac {3}{4 a (1+\sin (x))} \] Output:
5/16*ln(1-sin(x))/a+11/16*ln(1+sin(x))/a+1/8/a/(1-sin(x))-1/8/a/(1+sin(x)) ^2+3/4/a/(1+sin(x))
Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5 \log (1-\sin (x))+11 \log (1+\sin (x))+\frac {2 \left (-6-3 \sin (x)+5 \sin ^2(x)\right )}{(-1+\sin (x)) (1+\sin (x))^2}}{16 a} \] Input:
Integrate[Tan[x]^3/(a + a*Csc[x]),x]
Output:
(5*Log[1 - Sin[x]] + 11*Log[1 + Sin[x]] + (2*(-6 - 3*Sin[x] + 5*Sin[x]^2)) /((-1 + Sin[x])*(1 + Sin[x])^2))/(16*a)
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4367, 27, 99, 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.
\(\displaystyle \int \frac {\tan ^3(x)}{a \csc (x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cot (x)^3 (a \csc (x)+a)}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle a^4 \int \frac {\sin ^4(x)}{a^5 (1-\sin (x))^2 (\sin (x)+1)^3}d\sin (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sin ^4(x)}{(1-\sin (x))^2 (\sin (x)+1)^3}d\sin (x)}{a}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {11}{16 (\sin (x)+1)}-\frac {3}{4 (\sin (x)+1)^2}+\frac {1}{4 (\sin (x)+1)^3}+\frac {5}{16 (\sin (x)-1)}+\frac {1}{8 (\sin (x)-1)^2}\right )d\sin (x)}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{8 (1-\sin (x))}+\frac {3}{4 (\sin (x)+1)}-\frac {1}{8 (\sin (x)+1)^2}+\frac {5}{16} \log (1-\sin (x))+\frac {11}{16} \log (\sin (x)+1)}{a}\) |
Input:
Int[Tan[x]^3/(a + a*Csc[x]),x]
Output:
((5*Log[1 - Sin[x]])/16 + (11*Log[1 + Sin[x]])/16 + 1/(8*(1 - Sin[x])) - 1 /(8*(1 + Sin[x])^2) + 3/(4*(1 + Sin[x])))/a
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[1/(a^(m - n - 1)*b^n*d) Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /; Fr eeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && Integer Q[n]
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.65
method | result | size |
default | \(\frac {-\frac {1}{8 \left (1+\sin \left (x \right )\right )^{2}}+\frac {3}{4 \left (1+\sin \left (x \right )\right )}+\frac {11 \ln \left (1+\sin \left (x \right )\right )}{16}-\frac {1}{8 \left (\sin \left (x \right )-1\right )}+\frac {5 \ln \left (\sin \left (x \right )-1\right )}{16}}{a}\) | \(44\) |
risch | \(-\frac {i x}{a}+\frac {i \left (6 i {\mathrm e}^{2 i x}+5 \,{\mathrm e}^{i x}-6 i {\mathrm e}^{4 i x}+14 \,{\mathrm e}^{3 i x}+5 \,{\mathrm e}^{5 i x}\right )}{4 \left ({\mathrm e}^{i x}-i\right )^{2} \left ({\mathrm e}^{i x}+i\right )^{4} a}+\frac {5 \ln \left ({\mathrm e}^{i x}-i\right )}{8 a}+\frac {11 \ln \left ({\mathrm e}^{i x}+i\right )}{8 a}\) | \(101\) |
Input:
int(tan(x)^3/(a+a*csc(x)),x,method=_RETURNVERBOSE)
Output:
1/a*(-1/8/(1+sin(x))^2+3/4/(1+sin(x))+11/16*ln(1+sin(x))-1/8/(sin(x)-1)+5/ 16*ln(sin(x)-1))
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {10 \, \cos \left (x\right )^{2} + 11 \, {\left (\cos \left (x\right )^{2} \sin \left (x\right ) + \cos \left (x\right )^{2}\right )} \log \left (\sin \left (x\right ) + 1\right ) + 5 \, {\left (\cos \left (x\right )^{2} \sin \left (x\right ) + \cos \left (x\right )^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 6 \, \sin \left (x\right ) + 2}{16 \, {\left (a \cos \left (x\right )^{2} \sin \left (x\right ) + a \cos \left (x\right )^{2}\right )}} \] Input:
integrate(tan(x)^3/(a+a*csc(x)),x, algorithm="fricas")
Output:
1/16*(10*cos(x)^2 + 11*(cos(x)^2*sin(x) + cos(x)^2)*log(sin(x) + 1) + 5*(c os(x)^2*sin(x) + cos(x)^2)*log(-sin(x) + 1) + 6*sin(x) + 2)/(a*cos(x)^2*si n(x) + a*cos(x)^2)
\[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {\int \frac {\tan ^{3}{\left (x \right )}}{\csc {\left (x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(x)**3/(a+a*csc(x)),x)
Output:
Integral(tan(x)**3/(csc(x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5 \, \sin \left (x\right )^{2} - 3 \, \sin \left (x\right ) - 6}{8 \, {\left (a \sin \left (x\right )^{3} + a \sin \left (x\right )^{2} - a \sin \left (x\right ) - a\right )}} + \frac {11 \, \log \left (\sin \left (x\right ) + 1\right )}{16 \, a} + \frac {5 \, \log \left (\sin \left (x\right ) - 1\right )}{16 \, a} \] Input:
integrate(tan(x)^3/(a+a*csc(x)),x, algorithm="maxima")
Output:
1/8*(5*sin(x)^2 - 3*sin(x) - 6)/(a*sin(x)^3 + a*sin(x)^2 - a*sin(x) - a) + 11/16*log(sin(x) + 1)/a + 5/16*log(sin(x) - 1)/a
Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {11 \, \log \left (\sin \left (x\right ) + 1\right )}{16 \, a} + \frac {5 \, \log \left (-\sin \left (x\right ) + 1\right )}{16 \, a} + \frac {5 \, \sin \left (x\right )^{2} - 3 \, \sin \left (x\right ) - 6}{8 \, a {\left (\sin \left (x\right ) + 1\right )}^{2} {\left (\sin \left (x\right ) - 1\right )}} \] Input:
integrate(tan(x)^3/(a+a*csc(x)),x, algorithm="giac")
Output:
11/16*log(sin(x) + 1)/a + 5/16*log(-sin(x) + 1)/a + 1/8*(5*sin(x)^2 - 3*si n(x) - 6)/(a*(sin(x) + 1)^2*(sin(x) - 1))
Time = 15.42 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.97 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}{8\,a}+\frac {11\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{8\,a}-\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}+\frac {-\frac {3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{4}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{2}+\frac {9\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{2}-\frac {3\,\mathrm {tan}\left (\frac {x}{2}\right )}{4}}{a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4-4\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+a} \] Input:
int(tan(x)^3/(a + a/sin(x)),x)
Output:
(5*log(tan(x/2) - 1))/(8*a) + (11*log(tan(x/2) + 1))/(8*a) - log(tan(x/2)^ 2 + 1)/a + (tan(x/2)^2/2 - (3*tan(x/2))/4 + (9*tan(x/2)^3)/2 + tan(x/2)^4/ 2 - (3*tan(x/2)^5)/4)/(a + 2*a*tan(x/2) - a*tan(x/2)^2 - 4*a*tan(x/2)^3 - a*tan(x/2)^4 + 2*a*tan(x/2)^5 + a*tan(x/2)^6)
Time = 0.17 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.66 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {-8 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) \sin \left (x \right )^{3}-8 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) \sin \left (x \right )^{2}+8 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right ) \sin \left (x \right )+8 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2}+1\right )+5 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{3}+5 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2}-5 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )-5 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right )+11 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{3}+11 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2}-11 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )-11 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right )-3 \sin \left (x \right )^{3}+2 \sin \left (x \right )^{2}-3}{8 a \left (\sin \left (x \right )^{3}+\sin \left (x \right )^{2}-\sin \left (x \right )-1\right )} \] Input:
int(tan(x)^3/(a+a*csc(x)),x)
Output:
( - 8*log(tan(x/2)**2 + 1)*sin(x)**3 - 8*log(tan(x/2)**2 + 1)*sin(x)**2 + 8*log(tan(x/2)**2 + 1)*sin(x) + 8*log(tan(x/2)**2 + 1) + 5*log(tan(x/2) - 1)*sin(x)**3 + 5*log(tan(x/2) - 1)*sin(x)**2 - 5*log(tan(x/2) - 1)*sin(x) - 5*log(tan(x/2) - 1) + 11*log(tan(x/2) + 1)*sin(x)**3 + 11*log(tan(x/2) + 1)*sin(x)**2 - 11*log(tan(x/2) + 1)*sin(x) - 11*log(tan(x/2) + 1) - 3*sin (x)**3 + 2*sin(x)**2 - 3)/(8*a*(sin(x)**3 + sin(x)**2 - sin(x) - 1))