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

Optimal. Leaf size=19 \[ \frac {\sec ^2(x)}{2 a}-\frac {\sec (x)}{a} \]

[Out]

-sec(x)/a+1/2*sec(x)^2/a

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Rubi [A]  time = 0.06, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2706, 2606, 30, 8} \[ \frac {\sec ^2(x)}{2 a}-\frac {\sec (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3/(a + a*Cos[x]),x]

[Out]

-(Sec[x]/a) + Sec[x]^2/(2*a)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^3(x)}{a+a \cos (x)} \, dx &=-\frac {\int \sec (x) \tan (x) \, dx}{a}+\frac {\int \sec ^2(x) \tan (x) \, dx}{a}\\ &=-\frac {\operatorname {Subst}(\int 1 \, dx,x,\sec (x))}{a}+\frac {\operatorname {Subst}(\int x \, dx,x,\sec (x))}{a}\\ &=-\frac {\sec (x)}{a}+\frac {\sec ^2(x)}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.89 \[ \frac {2 \sin ^4\left (\frac {x}{2}\right ) \sec ^2(x)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3/(a + a*Cos[x]),x]

[Out]

(2*Sec[x]^2*Sin[x/2]^4)/a

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fricas [A]  time = 0.74, size = 15, normalized size = 0.79 \[ -\frac {2 \, \cos \relax (x) - 1}{2 \, a \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*cos(x) - 1)/(a*cos(x)^2)

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giac [A]  time = 0.36, size = 15, normalized size = 0.79 \[ -\frac {2 \, \cos \relax (x) - 1}{2 \, a \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*cos(x) - 1)/(a*cos(x)^2)

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maple [A]  time = 0.05, size = 18, normalized size = 0.95 \[ \frac {\frac {1}{2 \cos \relax (x )^{2}}-\frac {1}{\cos \relax (x )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(1/2/cos(x)^2-1/cos(x))

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maxima [A]  time = 0.66, size = 15, normalized size = 0.79 \[ -\frac {2 \, \cos \relax (x) - 1}{2 \, a \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*cos(x) - 1)/(a*cos(x)^2)

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mupad [B]  time = 0.34, size = 13, normalized size = 0.68 \[ -\frac {\cos \relax (x)-\frac {1}{2}}{a\,{\cos \relax (x)}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3/(a + a*cos(x)),x)

[Out]

-(cos(x) - 1/2)/(a*cos(x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)**3/(cos(x) + 1), x)/a

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