3.105 \(\int \frac{\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=103 \[ \frac{4 \cot ^9(c+d x)}{9 a^3 d}+\frac{\cot ^7(c+d x)}{a^3 d}+\frac{3 \cot ^5(c+d x)}{5 a^3 d}-\frac{4 \csc ^9(c+d x)}{9 a^3 d}+\frac{\csc ^7(c+d x)}{a^3 d}-\frac{3 \csc ^5(c+d x)}{5 a^3 d} \]

[Out]

(3*Cot[c + d*x]^5)/(5*a^3*d) + Cot[c + d*x]^7/(a^3*d) + (4*Cot[c + d*x]^9)/(9*a^3*d) - (3*Csc[c + d*x]^5)/(5*a
^3*d) + Csc[c + d*x]^7/(a^3*d) - (4*Csc[c + d*x]^9)/(9*a^3*d)

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Rubi [A]  time = 0.378706, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2875, 2873, 2607, 14, 2606, 270} \[ \frac{4 \cot ^9(c+d x)}{9 a^3 d}+\frac{\cot ^7(c+d x)}{a^3 d}+\frac{3 \cot ^5(c+d x)}{5 a^3 d}-\frac{4 \csc ^9(c+d x)}{9 a^3 d}+\frac{\csc ^7(c+d x)}{a^3 d}-\frac{3 \csc ^5(c+d x)}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + a*Sec[c + d*x])^3,x]

[Out]

(3*Cot[c + d*x]^5)/(5*a^3*d) + Cot[c + d*x]^7/(a^3*d) + (4*Cot[c + d*x]^9)/(9*a^3*d) - (3*Csc[c + d*x]^5)/(5*a
^3*d) + Csc[c + d*x]^7/(a^3*d) - (4*Csc[c + d*x]^9)/(9*a^3*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 2875

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

Rule 2873

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

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\int \frac{\cot ^3(c+d x) \csc (c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=-\frac{\int (-a+a \cos (c+d x))^3 \cot ^3(c+d x) \csc ^7(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (-a^3 \cot ^6(c+d x) \csc ^4(c+d x)+3 a^3 \cot ^5(c+d x) \csc ^5(c+d x)-3 a^3 \cot ^4(c+d x) \csc ^6(c+d x)+a^3 \cot ^3(c+d x) \csc ^7(c+d x)\right ) \, dx}{a^6}\\ &=-\frac{\int \cot ^6(c+d x) \csc ^4(c+d x) \, dx}{a^3}+\frac{\int \cot ^3(c+d x) \csc ^7(c+d x) \, dx}{a^3}+\frac{3 \int \cot ^5(c+d x) \csc ^5(c+d x) \, dx}{a^3}-\frac{3 \int \cot ^4(c+d x) \csc ^6(c+d x) \, dx}{a^3}\\ &=-\frac{\operatorname{Subst}\left (\int x^6 \left (-1+x^2\right ) \, dx,x,\csc (c+d x)\right )}{a^3 d}-\frac{\operatorname{Subst}\left (\int x^6 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\csc (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,-\cot (c+d x)\right )}{a^3 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-x^6+x^8\right ) \, dx,x,\csc (c+d x)\right )}{a^3 d}-\frac{\operatorname{Subst}\left (\int \left (x^6+x^8\right ) \, dx,x,-\cot (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\csc (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,-\cot (c+d x)\right )}{a^3 d}\\ &=\frac{3 \cot ^5(c+d x)}{5 a^3 d}+\frac{\cot ^7(c+d x)}{a^3 d}+\frac{4 \cot ^9(c+d x)}{9 a^3 d}-\frac{3 \csc ^5(c+d x)}{5 a^3 d}+\frac{\csc ^7(c+d x)}{a^3 d}-\frac{4 \csc ^9(c+d x)}{9 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.713802, size = 175, normalized size = 1.7 \[ -\frac{\csc (c) (-1764 \sin (c+d x)-1323 \sin (2 (c+d x))+98 \sin (3 (c+d x))+588 \sin (4 (c+d x))+294 \sin (5 (c+d x))+49 \sin (6 (c+d x))+3456 \sin (2 c+d x)-1152 \sin (c+2 d x)+2880 \sin (3 c+2 d x)-128 \sin (2 c+3 d x)-768 \sin (3 c+4 d x)-384 \sin (4 c+5 d x)-64 \sin (5 c+6 d x)+5376 \sin (c)-1152 \sin (d x)) \csc ^3(2 (c+d x))}{5760 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + a*Sec[c + d*x])^3,x]

[Out]

-(Csc[c]*Csc[2*(c + d*x)]^3*(5376*Sin[c] - 1152*Sin[d*x] - 1764*Sin[c + d*x] - 1323*Sin[2*(c + d*x)] + 98*Sin[
3*(c + d*x)] + 588*Sin[4*(c + d*x)] + 294*Sin[5*(c + d*x)] + 49*Sin[6*(c + d*x)] + 3456*Sin[2*c + d*x] - 1152*
Sin[c + 2*d*x] + 2880*Sin[3*c + 2*d*x] - 128*Sin[2*c + 3*d*x] - 768*Sin[3*c + 4*d*x] - 384*Sin[4*c + 5*d*x] -
64*Sin[5*c + 6*d*x]))/(5760*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.074, size = 60, normalized size = 0.6 \begin{align*}{\frac{1}{64\,d{a}^{3}} \left ( -{\frac{1}{9} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}}+{\frac{3}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-3\,\tan \left ( 1/2\,dx+c/2 \right ) -{\frac{1}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}} \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))^3,x)

[Out]

1/64/d/a^3*(-1/9*tan(1/2*d*x+1/2*c)^9+3/5*tan(1/2*d*x+1/2*c)^5-3*tan(1/2*d*x+1/2*c)-1/3/tan(1/2*d*x+1/2*c)^3)

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Maxima [A]  time = 1.12351, size = 124, normalized size = 1.2 \begin{align*} -\frac{\frac{\frac{135 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{27 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{5 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{3}} + \frac{15 \,{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{3} \sin \left (d x + c\right )^{3}}}{2880 \, 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))^3,x, algorithm="maxima")

[Out]

-1/2880*((135*sin(d*x + c)/(cos(d*x + c) + 1) - 27*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^9/(cos
(d*x + c) + 1)^9)/a^3 + 15*(cos(d*x + c) + 1)^3/(a^3*sin(d*x + c)^3))/d

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Fricas [A]  time = 1.89216, size = 359, normalized size = 3.49 \begin{align*} \frac{2 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2}{45 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} - 3 \, a^{3} d \cos \left (d x + c\right ) - a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/45*(2*cos(d*x + c)^6 + 6*cos(d*x + c)^5 + 3*cos(d*x + c)^4 - 7*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 6*cos(d*x
+ c) + 2)/((a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 2*a^3*d*cos(d*x + c)^3 - 2*a^3*d*cos(d*x + c)^2 -
3*a^3*d*cos(d*x + c) - a^3*d)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28664, size = 99, normalized size = 0.96 \begin{align*} -\frac{\frac{15}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}} + \frac{5 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 27 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 135 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{27}}}{2880 \, 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))^3,x, algorithm="giac")

[Out]

-1/2880*(15/(a^3*tan(1/2*d*x + 1/2*c)^3) + (5*a^24*tan(1/2*d*x + 1/2*c)^9 - 27*a^24*tan(1/2*d*x + 1/2*c)^5 + 1
35*a^24*tan(1/2*d*x + 1/2*c))/a^27)/d