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]

1/32*arctanh(cos(x))-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 = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 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])

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 4397

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

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*}

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Mathematica [A]  time = 0.02, 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]

-1/64*Csc[x/2]^2 + 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 -
Sec[x/2]^8/256

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fricas [B]  time = 1.82, size = 130, normalized size = 2.17 \[ -\frac {6 \, \cos \relax (x)^{4} + 18 \, \cos \relax (x)^{3} - 50 \, \cos \relax (x)^{2} - 3 \, {\left (\cos \relax (x)^{5} + 3 \, \cos \relax (x)^{4} + 2 \, \cos \relax (x)^{3} - 2 \, \cos \relax (x)^{2} - 3 \, \cos \relax (x) - 1\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 3 \, {\left (\cos \relax (x)^{5} + 3 \, \cos \relax (x)^{4} + 2 \, \cos \relax (x)^{3} - 2 \, \cos \relax (x)^{2} - 3 \, \cos \relax (x) - 1\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 54 \, \cos \relax (x) - 16}{192 \, {\left (\cos \relax (x)^{5} + 3 \, \cos \relax (x)^{4} + 2 \, \cos \relax (x)^{3} - 2 \, \cos \relax (x)^{2} - 3 \, \cos \relax (x) - 1\right )}} \]

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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giac [B]  time = 0.16, size = 95, normalized size = 1.58 \[ \frac {{\left (\frac {\cos \relax (x) - 1}{\cos \relax (x) + 1} + 1\right )} {\left (\cos \relax (x) + 1\right )}}{64 \, {\left (\cos \relax (x) - 1\right )}} + \frac {\cos \relax (x) - 1}{32 \, {\left (\cos \relax (x) + 1\right )}} + \frac {{\left (\cos \relax (x) - 1\right )}^{2}}{64 \, {\left (\cos \relax (x) + 1\right )}^{2}} - \frac {{\left (\cos \relax (x) - 1\right )}^{3}}{192 \, {\left (\cos \relax (x) + 1\right )}^{3}} - \frac {{\left (\cos \relax (x) - 1\right )}^{4}}{256 \, {\left (\cos \relax (x) + 1\right )}^{4}} - \frac {1}{64} \, \log \left (-\frac {\cos \relax (x) - 1}{\cos \relax (x) + 1}\right ) \]

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))

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maple [A]  time = 0.13, size = 56, normalized size = 0.93 \[ \frac {1}{-32+32 \cos \relax (x )}-\frac {\ln \left (-1+\cos \relax (x )\right )}{64}-\frac {1}{16 \left (1+\cos \relax (x )\right )^{4}}+\frac {1}{6 \left (1+\cos \relax (x )\right )^{3}}-\frac {3}{32 \left (1+\cos \relax (x )\right )^{2}}-\frac {1}{16 \left (1+\cos \relax (x )\right )}+\frac {\ln \left (1+\cos \relax (x )\right )}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/(-1+cos(x))-1/64*ln(-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))+1/64*
ln(1+cos(x))

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maxima [A]  time = 0.31, size = 73, normalized size = 1.22 \[ -\frac {{\left (\cos \relax (x) + 1\right )}^{2}}{64 \, \sin \relax (x)^{2}} - \frac {\sin \relax (x)^{2}}{32 \, {\left (\cos \relax (x) + 1\right )}^{2}} + \frac {\sin \relax (x)^{4}}{64 \, {\left (\cos \relax (x) + 1\right )}^{4}} + \frac {\sin \relax (x)^{6}}{192 \, {\left (\cos \relax (x) + 1\right )}^{6}} - \frac {\sin \relax (x)^{8}}{256 \, {\left (\cos \relax (x) + 1\right )}^{8}} - \frac {1}{32} \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) \]

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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mupad [B]  time = 2.39, size = 48, normalized size = 0.80 \[ \frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{64}-\frac {1}{64\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{32}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{32}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^6}{192}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^8}{256} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x) + tan(x))^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\sin {\relax (x )} + \tan {\relax (x )}\right )^{3}}\, dx \]

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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