\(\int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx\) [2]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5 \log (1-\sin (x))}{16 a}+\frac {11 \log (1+\sin (x))}{16 a}+\frac {1}{8 a (1-\sin (x))}-\frac {1}{8 a (1+\sin (x))^2}+\frac {3}{4 a (1+\sin (x))} \]

[Out]

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))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3964, 90} \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {1}{8 a (1-\sin (x))}+\frac {3}{4 a (\sin (x)+1)}-\frac {1}{8 a (\sin (x)+1)^2}+\frac {5 \log (1-\sin (x))}{16 a}+\frac {11 \log (\sin (x)+1)}{16 a} \]

[In]

Int[Tan[x]^3/(a + a*Csc[x]),x]

[Out]

(5*Log[1 - Sin[x]])/(16*a) + (11*Log[1 + Sin[x]])/(16*a) + 1/(8*a*(1 - Sin[x])) - 1/(8*a*(1 + Sin[x])^2) + 3/(
4*a*(1 + Sin[x]))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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]))

Rule 3964

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[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] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = a^4 \text {Subst}\left (\int \frac {x^4}{(a-a x)^2 (a+a x)^3} \, dx,x,\sin (x)\right ) \\ & = a^4 \text {Subst}\left (\int \left (\frac {1}{8 a^5 (-1+x)^2}+\frac {5}{16 a^5 (-1+x)}+\frac {1}{4 a^5 (1+x)^3}-\frac {3}{4 a^5 (1+x)^2}+\frac {11}{16 a^5 (1+x)}\right ) \, dx,x,\sin (x)\right ) \\ & = \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))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5 \log (1-\sin (x))+11 \log (1+\sin (x))+\frac {2 \left (-6-3 \sin (x)+5 \sin ^2(x)\right )}{(-1+\sin (x)) (1+\sin (x))^2}}{16 a} \]

[In]

Integrate[Tan[x]^3/(a + a*Csc[x]),x]

[Out]

(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)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {-\frac {1}{8 \left (\sin \left (x \right )-1\right )}+\frac {5 \ln \left (\sin \left (x \right )-1\right )}{16}-\frac {1}{8 \left (1+\sin \left (x \right )\right )^{2}}+\frac {3}{4 \left (1+\sin \left (x \right )\right )}+\frac {11 \ln \left (1+\sin \left (x \right )\right )}{16}}{a}\)	\(44\)
risch	\(-\frac {i x}{a}+\frac {i \left (5 \,{\mathrm e}^{i x}+6 i {\mathrm e}^{2 i x}+14 \,{\mathrm e}^{3 i x}-6 i {\mathrm e}^{4 i x}+5 \,{\mathrm e}^{5 i x}\right )}{4 \left ({\mathrm e}^{i x}-i\right )^{2} \left (i+{\mathrm e}^{i x}\right )^{4} a}+\frac {5 \ln \left ({\mathrm e}^{i x}-i\right )}{8 a}+\frac {11 \ln \left (i+{\mathrm e}^{i x}\right )}{8 a}\)	\(101\)

[In]

int(tan(x)^3/(a+a*csc(x)),x,method=_RETURNVERBOSE)

[Out]

1/a*(-1/8/(sin(x)-1)+5/16*ln(sin(x)-1)-1/8/(1+sin(x))^2+3/4/(1+sin(x))+11/16*ln(1+sin(x)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {10 \, \cos \left (x\right )^{2} + 11 \, {\left (\cos \left (x\right )^{2} \sin \left (x\right ) + \cos \left (x\right )^{2}\right )} \log \left (\sin \left (x\right ) + 1\right ) + 5 \, {\left (\cos \left (x\right )^{2} \sin \left (x\right ) + \cos \left (x\right )^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right ) + 6 \, \sin \left (x\right ) + 2}{16 \, {\left (a \cos \left (x\right )^{2} \sin \left (x\right ) + a \cos \left (x\right )^{2}\right )}} \]

[In]

integrate(tan(x)^3/(a+a*csc(x)),x, algorithm="fricas")

[Out]

1/16*(10*cos(x)^2 + 11*(cos(x)^2*sin(x) + cos(x)^2)*log(sin(x) + 1) + 5*(cos(x)^2*sin(x) + cos(x)^2)*log(-sin(
x) + 1) + 6*sin(x) + 2)/(a*cos(x)^2*sin(x) + a*cos(x)^2)

Sympy [F]

\[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {\int \frac {\tan ^{3}{\left (x \right )}}{\csc {\left (x \right )} + 1}\, dx}{a} \]

[In]

integrate(tan(x)**3/(a+a*csc(x)),x)

[Out]

Integral(tan(x)**3/(csc(x) + 1), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5 \, \sin \left (x\right )^{2} - 3 \, \sin \left (x\right ) - 6}{8 \, {\left (a \sin \left (x\right )^{3} + a \sin \left (x\right )^{2} - a \sin \left (x\right ) - a\right )}} + \frac {11 \, \log \left (\sin \left (x\right ) + 1\right )}{16 \, a} + \frac {5 \, \log \left (\sin \left (x\right ) - 1\right )}{16 \, a} \]

[In]

integrate(tan(x)^3/(a+a*csc(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {11 \, \log \left (\sin \left (x\right ) + 1\right )}{16 \, a} + \frac {5 \, \log \left (-\sin \left (x\right ) + 1\right )}{16 \, a} + \frac {5 \, \sin \left (x\right )^{2} - 3 \, \sin \left (x\right ) - 6}{8 \, a {\left (\sin \left (x\right ) + 1\right )}^{2} {\left (\sin \left (x\right ) - 1\right )}} \]

[In]

integrate(tan(x)^3/(a+a*csc(x)),x, algorithm="giac")

[Out]

11/16*log(sin(x) + 1)/a + 5/16*log(-sin(x) + 1)/a + 1/8*(5*sin(x)^2 - 3*sin(x) - 6)/(a*(sin(x) + 1)^2*(sin(x)
- 1))

Mupad [B] (verification not implemented)

Time = 19.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.97 \[ \int \frac {\tan ^3(x)}{a+a \csc (x)} \, dx=\frac {5\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}{8\,a}+\frac {11\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{8\,a}-\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}+\frac {-\frac {3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{4}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{2}+\frac {9\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{2}-\frac {3\,\mathrm {tan}\left (\frac {x}{2}\right )}{4}}{a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4-4\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+a} \]

[In]

int(tan(x)^3/(a + a/sin(x)),x)

[Out]

(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)