3.1.6 \(\int \frac {\sec ^2(x)}{a+a \csc (x)} \, dx\) [6]

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

[Out]

1/3*sec(x)^3/a-1/3*tan(x)^3/a

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Rubi [A]
time = 0.08, antiderivative size = 23, normalized size of antiderivative = 1.00, 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 = {3957, 2918, 2686, 30, 2687} \begin {gather*} \frac {\sec ^3(x)}{3 a}-\frac {\tan ^3(x)}{3 a} \end {gather*}

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 30

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

Rule 2686

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 2687

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])

Rule 2918

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

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 {\text {Subst}\left (\int x^2 \, dx,x,\sec (x)\right )}{a}-\frac {\text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(23)=46\).
time = 0.08, size = 56, normalized size = 2.43 \begin {gather*} -\frac {-3+\cos (2 x)-2 \sin (x)+\cos (x) (1+\sin (x))}{6 a \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2/(a + a*Csc[x]),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(46\) vs. \(2(19)=38\).
time = 0.08, size = 47, normalized size = 2.04

method	result	size
norman	\(\frac {-\frac {2 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a}-\frac {2}{3 a}-\frac {4 \tan \left (\frac {x}{2}\right )}{3 a}}{\left (\tan \left (\frac {x}{2}\right )-1\right ) \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}\)	\(44\)
risch	\(\frac {2 i \left (2 i {\mathrm e}^{i x}+3 \,{\mathrm e}^{2 i x}-1\right )}{3 \left ({\mathrm e}^{i x}-i\right ) \left (i+{\mathrm e}^{i x}\right )^{3} a}\)	\(44\)
default	\(\frac {-\frac {1}{2 \left (\tan \left (\frac {x}{2}\right )-1\right )}+\frac {2}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {4}{8 \tan \left (\frac {x}{2}\right )+8}}{a}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (19) = 38\).
time = 0.26, size = 67, normalized size = 2.91 \begin {gather*} \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 {gather*}

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 = 2.12, size = 25, normalized size = 1.09 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec ^{2}{\left (x \right )}}{\csc {\left (x \right )} + 1}\, dx}{a} \end {gather*}

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 = 0.43, size = 37, normalized size = 1.61 \begin {gather*} -\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 {gather*}

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)

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Mupad [B]
time = 0.27, size = 37, normalized size = 1.61 \begin {gather*} -\frac {2\,\left (3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{3\,a\,\left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^2*(a + a/sin(x))),x)

[Out]

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

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