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

Optimal. Leaf size=68 \[ \frac{1}{8 a (1-\sin (x))}+\frac{3}{4 a (\sin (x)+1)}-\frac{1}{8 a (\sin (x)+1)^2}+\frac{5 \log (1-\sin (x))}{16 a}+\frac{11 \log (\sin (x)+1)}{16 a} \]

[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.0703221, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3879, 88} \[ \frac{1}{8 a (1-\sin (x))}+\frac{3}{4 a (\sin (x)+1)}-\frac{1}{8 a (\sin (x)+1)^2}+\frac{5 \log (1-\sin (x))}{16 a}+\frac{11 \log (\sin (x)+1)}{16 a} \]

Antiderivative was successfully verified.

[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 3879

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]

Rule 88

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]))

Rubi steps

\begin{align*} \int \frac{\tan ^3(x)}{a+a \csc (x)} \, dx &=a^4 \operatorname{Subst}\left (\int \frac{x^4}{(a-a x)^2 (a+a x)^3} \, dx,x,\sin (x)\right )\\ &=a^4 \operatorname{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]  time = 0.131365, size = 50, normalized size = 0.74 \[ \frac{\frac{2 \left (5 \sin ^2(x)-3 \sin (x)-6\right )}{(\sin (x)-1) (\sin (x)+1)^2}+5 \log (1-\sin (x))+11 \log (\sin (x)+1)}{16 a} \]

Antiderivative was successfully verified.

[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]  time = 0.054, size = 55, normalized size = 0.8 \begin{align*} -{\frac{1}{8\,a \left ( \sin \left ( x \right ) +1 \right ) ^{2}}}+{\frac{3}{4\,a \left ( \sin \left ( x \right ) +1 \right ) }}+{\frac{11\,\ln \left ( \sin \left ( x \right ) +1 \right ) }{16\,a}}-{\frac{1}{8\,a \left ( \sin \left ( x \right ) -1 \right ) }}+{\frac{5\,\ln \left ( \sin \left ( x \right ) -1 \right ) }{16\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.962337, size = 78, normalized size = 1.15 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 0.522906, size = 227, normalized size = 3.34 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{3}{\left (x \right )}}{\csc{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)**3/(csc(x) + 1), x)/a

________________________________________________________________________________________

Giac [A]  time = 1.39496, size = 70, normalized size = 1.03 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))