3.6 \(\int \frac{\sec ^2(x)}{a+a \csc (x)} \, dx\)

Optimal. Leaf size=23 \[ \frac{\sec ^3(x)}{3 a}-\frac{\tan ^3(x)}{3 a} \]

[Out]

Sec[x]^3/(3*a) - Tan[x]^3/(3*a)

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Rubi [A]  time = 0.107117, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3872, 2839, 2606, 30, 2607} \[ \frac{\sec ^3(x)}{3 a}-\frac{\tan ^3(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2/(a + a*Csc[x]),x]

[Out]

Sec[x]^3/(3*a) - Tan[x]^3/(3*a)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

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 30

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

Rule 2607

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

Rubi steps

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

Mathematica [B]  time = 0.0769092, size = 56, normalized size = 2.43 \[ -\frac{-2 \sin (x)+\cos (2 x)+(\sin (x)+1) \cos (x)-3}{6 a \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right ) \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2/(a + a*Csc[x]),x]

[Out]

-(-3 + Cos[2*x] - 2*Sin[x] + Cos[x]*(1 + Sin[x]))/(6*a*(Cos[x/2] - Sin[x/2])*(Cos[x/2] + Sin[x/2])^3)

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Maple [B]  time = 0.047, size = 47, normalized size = 2. \begin{align*} 4\,{\frac{1}{a} \left ( 1/6\, \left ( \tan \left ( x/2 \right ) +1 \right ) ^{-3}-1/4\, \left ( \tan \left ( x/2 \right ) +1 \right ) ^{-2}+1/8\, \left ( \tan \left ( x/2 \right ) +1 \right ) ^{-1}-1/8\, \left ( \tan \left ( x/2 \right ) -1 \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2/(a+a*csc(x)),x)

[Out]

4/a*(1/6/(tan(1/2*x)+1)^3-1/4/(tan(1/2*x)+1)^2+1/8/(tan(1/2*x)+1)-1/8/(tan(1/2*x)-1))

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Maxima [B]  time = 0.972871, size = 90, normalized size = 3.91 \begin{align*} \frac{2 \,{\left (\frac{2 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{3 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}}{3 \,{\left (a + \frac{2 \, a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac{2 \, a \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac{a \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+a*csc(x)),x, algorithm="maxima")

[Out]

2/3*(2*sin(x)/(cos(x) + 1) + 3*sin(x)^2/(cos(x) + 1)^2 + 1)/(a + 2*a*sin(x)/(cos(x) + 1) - 2*a*sin(x)^3/(cos(x
) + 1)^3 - a*sin(x)^4/(cos(x) + 1)^4)

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Fricas [A]  time = 0.453665, size = 80, normalized size = 3.48 \begin{align*} -\frac{\cos \left (x\right )^{2} - \sin \left (x\right ) - 2}{3 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+a*csc(x)),x, algorithm="fricas")

[Out]

-1/3*(cos(x)^2 - sin(x) - 2)/(a*cos(x)*sin(x) + a*cos(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{2}{\left (x \right )}}{\csc{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2/(a+a*csc(x)),x)

[Out]

Integral(sec(x)**2/(csc(x) + 1), x)/a

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Giac [A]  time = 1.38948, size = 50, normalized size = 2.17 \begin{align*} -\frac{1}{2 \, a{\left (\tan \left (\frac{1}{2} \, x\right ) - 1\right )}} + \frac{3 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}{6 \, a{\left (\tan \left (\frac{1}{2} \, x\right ) + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2/(a+a*csc(x)),x, algorithm="giac")

[Out]

-1/2/(a*(tan(1/2*x) - 1)) + 1/6*(3*tan(1/2*x)^2 + 1)/(a*(tan(1/2*x) + 1)^3)