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

Optimal. Leaf size=55 \[ \frac{\cot ^5(c+d x)}{5 a d}+\frac{\cot ^3(c+d x)}{3 a d}-\frac{\csc ^5(c+d x)}{5 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14281, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3872, 2839, 2606, 30, 2607, 14} \[ \frac{\cot ^5(c+d x)}{5 a d}+\frac{\cot ^3(c+d x)}{3 a d}-\frac{\csc ^5(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

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

Mathematica [B]  time = 0.500291, size = 116, normalized size = 2.11 \[ -\frac{\csc (c) (-54 \sin (c+d x)-18 \sin (2 (c+d x))+18 \sin (3 (c+d x))+9 \sin (4 (c+d x))-32 \sin (c+2 d x)+32 \sin (2 c+3 d x)+16 \sin (3 c+4 d x)+240 \sin (c)-96 \sin (d x)) \csc ^3(c+d x) \sec (c+d x)}{960 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + a*Sec[c + d*x]),x]

[Out]

-(Csc[c]*Csc[c + d*x]^3*Sec[c + d*x]*(240*Sin[c] - 96*Sin[d*x] - 54*Sin[c + d*x] - 18*Sin[2*(c + d*x)] + 18*Si
n[3*(c + d*x)] + 9*Sin[4*(c + d*x)] - 32*Sin[c + 2*d*x] + 32*Sin[2*c + 3*d*x] + 16*Sin[3*c + 4*d*x]))/(960*a*d
*(1 + Sec[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.057, size = 62, normalized size = 1.1 \begin{align*}{\frac{1}{16\,da} \left ( -{\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d/a*(-1/5*tan(1/2*d*x+1/2*c)^5-2/3*tan(1/2*d*x+1/2*c)^3-1/3/tan(1/2*d*x+1/2*c)^3-2/tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [A]  time = 0.977802, size = 130, normalized size = 2.36 \begin{align*} -\frac{\frac{\frac{10 \, \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} + \frac{5 \,{\left (\frac{6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a \sin \left (d x + c\right )^{3}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*((10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a + 5*(6*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + 1)*(cos(d*x + c) + 1)^3/(a*sin(d*x + c)^3))/d

________________________________________________________________________________________

Fricas [A]  time = 1.67212, size = 225, normalized size = 4.09 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 3}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a 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)^4/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\csc ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4/(a+a*sec(d*x+c)),x)

[Out]

Integral(csc(c + d*x)**4/(sec(c + d*x) + 1), x)/a

________________________________________________________________________________________

Giac [A]  time = 1.32857, size = 100, normalized size = 1.82 \begin{align*} -\frac{\frac{5 \,{\left (6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}} + \frac{3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}{a^{5}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(5*(6*tan(1/2*d*x + 1/2*c)^2 + 1)/(a*tan(1/2*d*x + 1/2*c)^3) + (3*a^4*tan(1/2*d*x + 1/2*c)^5 + 10*a^4*t
an(1/2*d*x + 1/2*c)^3)/a^5)/d