3.347 \(\int \frac{1}{(\sin (x)+\tan (x))^3} \, dx\)

Optimal. Leaf size=60 \[ -\frac{1}{32 (1-\cos (x))}-\frac{1}{16 (\cos (x)+1)}-\frac{3}{32 (\cos (x)+1)^2}+\frac{1}{6 (\cos (x)+1)^3}-\frac{1}{16 (\cos (x)+1)^4}+\frac{1}{32} \tanh ^{-1}(\cos (x)) \]

[Out]

ArcTanh[Cos[x]]/32 - 1/(32*(1 - Cos[x])) - 1/(16*(1 + Cos[x])^4) + 1/(6*(1 + Cos[x])^3) - 3/(32*(1 + Cos[x])^2
) - 1/(16*(1 + Cos[x]))

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Rubi [A]  time = 0.0716931, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {4397, 2707, 88, 207} \[ -\frac{1}{32 (1-\cos (x))}-\frac{1}{16 (\cos (x)+1)}-\frac{3}{32 (\cos (x)+1)^2}+\frac{1}{6 (\cos (x)+1)^3}-\frac{1}{16 (\cos (x)+1)^4}+\frac{1}{32} \tanh ^{-1}(\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[(Sin[x] + Tan[x])^(-3),x]

[Out]

ArcTanh[Cos[x]]/32 - 1/(32*(1 - Cos[x])) - 1/(16*(1 + Cos[x])^4) + 1/(6*(1 + Cos[x])^3) - 3/(32*(1 + Cos[x])^2
) - 1/(16*(1 + Cos[x]))

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

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]))

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(\sin (x)+\tan (x))^3} \, dx &=\int \frac{\cot ^3(x)}{(1+\cos (x))^3} \, dx\\ &=-\operatorname{Subst}\left (\int \frac{x^3}{(1-x)^2 (1+x)^5} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{1}{32 (-1+x)^2}-\frac{1}{4 (1+x)^5}+\frac{1}{2 (1+x)^4}-\frac{3}{16 (1+x)^3}-\frac{1}{16 (1+x)^2}+\frac{1}{32 \left (-1+x^2\right )}\right ) \, dx,x,\cos (x)\right )\\ &=-\frac{1}{32 (1-\cos (x))}-\frac{1}{16 (1+\cos (x))^4}+\frac{1}{6 (1+\cos (x))^3}-\frac{3}{32 (1+\cos (x))^2}-\frac{1}{16 (1+\cos (x))}-\frac{1}{32} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\cos (x)\right )\\ &=\frac{1}{32} \tanh ^{-1}(\cos (x))-\frac{1}{32 (1-\cos (x))}-\frac{1}{16 (1+\cos (x))^4}+\frac{1}{6 (1+\cos (x))^3}-\frac{3}{32 (1+\cos (x))^2}-\frac{1}{16 (1+\cos (x))}\\ \end{align*}

Mathematica [A]  time = 0.0178364, size = 83, normalized size = 1.38 \[ -\frac{1}{64} \csc ^2\left (\frac{x}{2}\right )-\frac{1}{256} \sec ^8\left (\frac{x}{2}\right )+\frac{1}{48} \sec ^6\left (\frac{x}{2}\right )-\frac{3}{128} \sec ^4\left (\frac{x}{2}\right )-\frac{1}{32} \sec ^2\left (\frac{x}{2}\right )-\frac{1}{32} \log \left (\sin \left (\frac{x}{2}\right )\right )+\frac{1}{32} \log \left (\cos \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[x] + Tan[x])^(-3),x]

[Out]

-Csc[x/2]^2/64 + Log[Cos[x/2]]/32 - Log[Sin[x/2]]/32 - Sec[x/2]^2/32 - (3*Sec[x/2]^4)/128 + Sec[x/2]^6/48 - Se
c[x/2]^8/256

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Maple [A]  time = 0.052, size = 56, normalized size = 0.9 \begin{align*} -{\frac{1}{16\, \left ( 1+\cos \left ( x \right ) \right ) ^{4}}}+{\frac{1}{6\, \left ( 1+\cos \left ( x \right ) \right ) ^{3}}}-{\frac{3}{32\, \left ( 1+\cos \left ( x \right ) \right ) ^{2}}}-{\frac{1}{16+16\,\cos \left ( x \right ) }}+{\frac{\ln \left ( 1+\cos \left ( x \right ) \right ) }{64}}+{\frac{1}{-32+32\,\cos \left ( x \right ) }}-{\frac{\ln \left ( -1+\cos \left ( x \right ) \right ) }{64}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)+tan(x))^3,x)

[Out]

-1/16/(1+cos(x))^4+1/6/(1+cos(x))^3-3/32/(1+cos(x))^2-1/16/(1+cos(x))+1/64*ln(1+cos(x))+1/32/(-1+cos(x))-1/64*
ln(-1+cos(x))

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Maxima [A]  time = 1.00228, size = 99, normalized size = 1.65 \begin{align*} -\frac{{\left (\cos \left (x\right ) + 1\right )}^{2}}{64 \, \sin \left (x\right )^{2}} - \frac{\sin \left (x\right )^{2}}{32 \,{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{\sin \left (x\right )^{4}}{64 \,{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac{\sin \left (x\right )^{6}}{192 \,{\left (\cos \left (x\right ) + 1\right )}^{6}} - \frac{\sin \left (x\right )^{8}}{256 \,{\left (\cos \left (x\right ) + 1\right )}^{8}} - \frac{1}{32} \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sin(x)+tan(x))^3,x, algorithm="maxima")

[Out]

-1/64*(cos(x) + 1)^2/sin(x)^2 - 1/32*sin(x)^2/(cos(x) + 1)^2 + 1/64*sin(x)^4/(cos(x) + 1)^4 + 1/192*sin(x)^6/(
cos(x) + 1)^6 - 1/256*sin(x)^8/(cos(x) + 1)^8 - 1/32*log(sin(x)/(cos(x) + 1))

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Fricas [B]  time = 2.24772, size = 424, normalized size = 7.07 \begin{align*} -\frac{6 \, \cos \left (x\right )^{4} + 18 \, \cos \left (x\right )^{3} - 50 \, \cos \left (x\right )^{2} - 3 \,{\left (\cos \left (x\right )^{5} + 3 \, \cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{2} - 3 \, \cos \left (x\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) + 3 \,{\left (\cos \left (x\right )^{5} + 3 \, \cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{2} - 3 \, \cos \left (x\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - 54 \, \cos \left (x\right ) - 16}{192 \,{\left (\cos \left (x\right )^{5} + 3 \, \cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{2} - 3 \, \cos \left (x\right ) - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sin(x)+tan(x))^3,x, algorithm="fricas")

[Out]

-1/192*(6*cos(x)^4 + 18*cos(x)^3 - 50*cos(x)^2 - 3*(cos(x)^5 + 3*cos(x)^4 + 2*cos(x)^3 - 2*cos(x)^2 - 3*cos(x)
- 1)*log(1/2*cos(x) + 1/2) + 3*(cos(x)^5 + 3*cos(x)^4 + 2*cos(x)^3 - 2*cos(x)^2 - 3*cos(x) - 1)*log(-1/2*cos(
x) + 1/2) - 54*cos(x) - 16)/(cos(x)^5 + 3*cos(x)^4 + 2*cos(x)^3 - 2*cos(x)^2 - 3*cos(x) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sin(x)+tan(x))**3,x)

[Out]

Integral((sin(x) + tan(x))**(-3), x)

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Giac [B]  time = 1.13901, size = 128, normalized size = 2.13 \begin{align*} \frac{{\left (\frac{\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1} + 1\right )}{\left (\cos \left (x\right ) + 1\right )}}{64 \,{\left (\cos \left (x\right ) - 1\right )}} + \frac{\cos \left (x\right ) - 1}{32 \,{\left (\cos \left (x\right ) + 1\right )}} + \frac{{\left (\cos \left (x\right ) - 1\right )}^{2}}{64 \,{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac{{\left (\cos \left (x\right ) - 1\right )}^{3}}{192 \,{\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac{{\left (\cos \left (x\right ) - 1\right )}^{4}}{256 \,{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac{1}{64} \, \log \left (-\frac{\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sin(x)+tan(x))^3,x, algorithm="giac")

[Out]

1/64*((cos(x) - 1)/(cos(x) + 1) + 1)*(cos(x) + 1)/(cos(x) - 1) + 1/32*(cos(x) - 1)/(cos(x) + 1) + 1/64*(cos(x)
- 1)^2/(cos(x) + 1)^2 - 1/192*(cos(x) - 1)^3/(cos(x) + 1)^3 - 1/256*(cos(x) - 1)^4/(cos(x) + 1)^4 - 1/64*log(
-(cos(x) - 1)/(cos(x) + 1))