3.8 \(\int \frac{\sec ^4(x)}{a+a \csc (x)} \, dx\)

Optimal. Leaf size=34 \[ -\frac{\tan ^5(x)}{5 a}-\frac{\tan ^3(x)}{3 a}+\frac{\sec ^5(x)}{5 a} \]

[Out]

Sec[x]^5/(5*a) - Tan[x]^3/(3*a) - Tan[x]^5/(5*a)

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Rubi [A]  time = 0.12078, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3872, 2839, 2606, 30, 2607, 14} \[ -\frac{\tan ^5(x)}{5 a}-\frac{\tan ^3(x)}{3 a}+\frac{\sec ^5(x)}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4/(a + a*Csc[x]),x]

[Out]

Sec[x]^5/(5*a) - Tan[x]^3/(3*a) - Tan[x]^5/(5*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])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\sec ^4(x)}{a+a \csc (x)} \, dx &=\int \frac{\sec ^3(x) \tan (x)}{a+a \sin (x)} \, dx\\ &=\frac{\int \sec ^5(x) \tan (x) \, dx}{a}-\frac{\int \sec ^4(x) \tan ^2(x) \, dx}{a}\\ &=\frac{\operatorname{Subst}\left (\int x^4 \, dx,x,\sec (x)\right )}{a}-\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (x)\right )}{a}\\ &=\frac{\sec ^5(x)}{5 a}-\frac{\operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (x)\right )}{a}\\ &=\frac{\sec ^5(x)}{5 a}-\frac{\tan ^3(x)}{3 a}-\frac{\tan ^5(x)}{5 a}\\ \end{align*}

Mathematica [B]  time = 0.143172, size = 85, normalized size = 2.5 \[ -\frac{-96 \sin (x)+18 \sin (2 x)-32 \sin (3 x)+9 \sin (4 x)+54 \cos (x)+32 \cos (2 x)+18 \cos (3 x)+16 \cos (4 x)-240}{960 a \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )^3 \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^4/(a + a*Csc[x]),x]

[Out]

-(-240 + 54*Cos[x] + 32*Cos[2*x] + 18*Cos[3*x] + 16*Cos[4*x] - 96*Sin[x] + 18*Sin[2*x] - 32*Sin[3*x] + 9*Sin[4
*x])/(960*a*(Cos[x/2] - Sin[x/2])^3*(Cos[x/2] + Sin[x/2])^5)

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Maple [B]  time = 0.053, size = 87, normalized size = 2.6 \begin{align*} 4\,{\frac{1}{a} \left ( -1/4\, \left ( \tan \left ( x/2 \right ) +1 \right ) ^{-4}+1/10\, \left ( \tan \left ( x/2 \right ) +1 \right ) ^{-5}+1/3\, \left ( \tan \left ( x/2 \right ) +1 \right ) ^{-3}-1/4\, \left ( \tan \left ( x/2 \right ) +1 \right ) ^{-2}+{\frac{3}{32\,\tan \left ( x/2 \right ) +32}}-1/24\, \left ( \tan \left ( x/2 \right ) -1 \right ) ^{-3}-1/16\, \left ( \tan \left ( x/2 \right ) -1 \right ) ^{-2}-{\frac{3}{32\,\tan \left ( x/2 \right ) -32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^4/(a+a*csc(x)),x)

[Out]

4/a*(-1/4/(tan(1/2*x)+1)^4+1/10/(tan(1/2*x)+1)^5+1/3/(tan(1/2*x)+1)^3-1/4/(tan(1/2*x)+1)^2+3/32/(tan(1/2*x)+1)
-1/24/(tan(1/2*x)-1)^3-1/16/(tan(1/2*x)-1)^2-3/32/(tan(1/2*x)-1))

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Maxima [B]  time = 0.983455, size = 225, normalized size = 6.62 \begin{align*} \frac{2 \,{\left (\frac{6 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{9 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac{8 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{5 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac{10 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + 3\right )}}{15 \,{\left (a + \frac{2 \, a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac{2 \, a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac{6 \, a \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{6 \, a \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac{2 \, a \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} - \frac{2 \, a \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}} - \frac{a \sin \left (x\right )^{8}}{{\left (\cos \left (x\right ) + 1\right )}^{8}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+a*csc(x)),x, algorithm="maxima")

[Out]

2/15*(6*sin(x)/(cos(x) + 1) + 9*sin(x)^2/(cos(x) + 1)^2 - 8*sin(x)^3/(cos(x) + 1)^3 + 5*sin(x)^4/(cos(x) + 1)^
4 + 10*sin(x)^5/(cos(x) + 1)^5 + 15*sin(x)^6/(cos(x) + 1)^6 + 3)/(a + 2*a*sin(x)/(cos(x) + 1) - 2*a*sin(x)^2/(
cos(x) + 1)^2 - 6*a*sin(x)^3/(cos(x) + 1)^3 + 6*a*sin(x)^5/(cos(x) + 1)^5 + 2*a*sin(x)^6/(cos(x) + 1)^6 - 2*a*
sin(x)^7/(cos(x) + 1)^7 - a*sin(x)^8/(cos(x) + 1)^8)

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Fricas [A]  time = 0.461498, size = 127, normalized size = 3.74 \begin{align*} -\frac{2 \, \cos \left (x\right )^{4} - \cos \left (x\right )^{2} -{\left (2 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right ) - 4}{15 \,{\left (a \cos \left (x\right )^{3} \sin \left (x\right ) + a \cos \left (x\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+a*csc(x)),x, algorithm="fricas")

[Out]

-1/15*(2*cos(x)^4 - cos(x)^2 - (2*cos(x)^2 + 1)*sin(x) - 4)/(a*cos(x)^3*sin(x) + a*cos(x)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{4}{\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)**4/(a+a*csc(x)),x)

[Out]

Integral(sec(x)**4/(csc(x) + 1), x)/a

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Giac [B]  time = 1.43957, size = 101, normalized size = 2.97 \begin{align*} -\frac{9 \, \tan \left (\frac{1}{2} \, x\right )^{2} - 12 \, \tan \left (\frac{1}{2} \, x\right ) + 7}{24 \, a{\left (\tan \left (\frac{1}{2} \, x\right ) - 1\right )}^{3}} + \frac{45 \, \tan \left (\frac{1}{2} \, x\right )^{4} + 60 \, \tan \left (\frac{1}{2} \, x\right )^{3} + 70 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 20 \, \tan \left (\frac{1}{2} \, x\right ) + 13}{120 \, a{\left (\tan \left (\frac{1}{2} \, x\right ) + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(a+a*csc(x)),x, algorithm="giac")

[Out]

-1/24*(9*tan(1/2*x)^2 - 12*tan(1/2*x) + 7)/(a*(tan(1/2*x) - 1)^3) + 1/120*(45*tan(1/2*x)^4 + 60*tan(1/2*x)^3 +
70*tan(1/2*x)^2 + 20*tan(1/2*x) + 13)/(a*(tan(1/2*x) + 1)^5)