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

Optimal. Leaf size=73 \[ -\frac{2 \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^3(c+d x)}{3 a^2 d}+\frac{2 \csc ^5(c+d x)}{5 a^2 d}-\frac{2 \csc ^3(c+d x)}{3 a^2 d} \]

[Out]

-Cot[c + d*x]^3/(3*a^2*d) - (2*Cot[c + d*x]^5)/(5*a^2*d) - (2*Csc[c + d*x]^3)/(3*a^2*d) + (2*Csc[c + d*x]^5)/(
5*a^2*d)

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Rubi [A]  time = 0.199552, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3872, 2711, 2607, 30, 2606, 14} \[ -\frac{2 \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^3(c+d x)}{3 a^2 d}+\frac{2 \csc ^5(c+d x)}{5 a^2 d}-\frac{2 \csc ^3(c+d x)}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2/(a + a*Sec[c + d*x])^2,x]

[Out]

-Cot[c + d*x]^3/(3*a^2*d) - (2*Cot[c + d*x]^5)/(5*a^2*d) - (2*Csc[c + d*x]^3)/(3*a^2*d) + (2*Csc[c + d*x]^5)/(
5*a^2*d)

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 2711

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

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

Mathematica [A]  time = 0.4256, size = 105, normalized size = 1.44 \[ \frac{\csc (c) (55 \sin (c+d x)+44 \sin (2 (c+d x))+11 \sin (3 (c+d x))-60 \sin (2 c+d x)+16 \sin (c+2 d x)+4 \sin (2 c+3 d x)-80 \sin (c)+80 \sin (d x)) \csc (c+d x) \sec ^2(c+d x)}{240 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2/(a + a*Sec[c + d*x])^2,x]

[Out]

(Csc[c]*Csc[c + d*x]*Sec[c + d*x]^2*(-80*Sin[c] + 80*Sin[d*x] + 55*Sin[c + d*x] + 44*Sin[2*(c + d*x)] + 11*Sin
[3*(c + d*x)] - 60*Sin[2*c + d*x] + 16*Sin[c + 2*d*x] + 4*Sin[2*c + 3*d*x]))/(240*a^2*d*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 0.058, size = 60, normalized size = 0.8 \begin{align*}{\frac{1}{8\,d{a}^{2}} \left ({\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) - \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a^2*(1/5*tan(1/2*d*x+1/2*c)^5-1/3*tan(1/2*d*x+1/2*c)^3-tan(1/2*d*x+1/2*c)-1/tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.01255, size = 122, normalized size = 1.67 \begin{align*} -\frac{\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} + \frac{15 \,{\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} \sin \left (d x + c\right )}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*((15*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 3*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5)/a^2 + 15*(cos(d*x + c) + 1)/(a^2*sin(d*x + c)))/d

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Fricas [A]  time = 1.60875, size = 180, normalized size = 2.47 \begin{align*} -\frac{\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 4}{15 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + 8*cos(d*x + c) + 4)/((a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) +
a^2*d)*sin(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.29461, size = 100, normalized size = 1.37 \begin{align*} -\frac{\frac{15}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{3 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{10}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15/(a^2*tan(1/2*d*x + 1/2*c)) - (3*a^8*tan(1/2*d*x + 1/2*c)^5 - 5*a^8*tan(1/2*d*x + 1/2*c)^3 - 15*a^8*
tan(1/2*d*x + 1/2*c))/a^10)/d