3.81 \(\int \frac {\csc ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=42 \[ -\frac {2 a \cos (c+d x)+a}{6 d (1-\cos (c+d x)) (a \cos (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3872, 2836, 12, 81} \[ -\frac {2 a \cos (c+d x)+a}{6 d (1-\cos (c+d x)) (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 81

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*
x)^(n + 1)*(e + f*x)^(p + 1)*(2*a*d*f*(n + p + 3) - b*(d*e*(n + 2) + c*f*(p + 2)) + b*d*f*(n + p + 2)*x))/(d^2
*f^2*(n + p + 2)*(n + p + 3)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && NeQ[n + p + 3,
0] && EqQ[d*f*(n + p + 2)*(a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1)))) - b*(d*e*(n + 1)
+ c*f*(p + 1))*(a*d*f*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2))), 0]

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 38, normalized size = 0.90 \[ -\frac {(2 \cos (c+d x)+1) \csc ^2(c+d x)}{6 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

-1/6*((1 + 2*Cos[c + d*x])*Csc[c + d*x]^2)/(a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.66, size = 60, normalized size = 1.43 \[ \frac {2 \, \cos \left (d x + c\right ) + 1}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.30, size = 82, normalized size = 1.95 \[ \frac {\frac {3 \, {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}} + \frac {\frac {6 \, a^{4} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a^{4} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{6}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/96*(3*(cos(d*x + c) + 1)/(a^2*(cos(d*x + c) - 1)) + (6*a^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a^4*(cos(
d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/a^6)/d

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maple [A]  time = 0.72, size = 57, normalized size = 1.36 \[ \frac {\frac {1}{-16+16 \cos \left (d x +c \right )}+\frac {1}{12 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {1}{8 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {1}{16 \left (1+\cos \left (d x +c \right )\right )}}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3/(a+a*sec(d*x+c))^2,x)

[Out]

1/d/a^2*(1/16/(-1+cos(d*x+c))+1/12/(1+cos(d*x+c))^3-1/8/(1+cos(d*x+c))^2-1/16/(1+cos(d*x+c)))

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maxima [A]  time = 0.58, size = 59, normalized size = 1.40 \[ \frac {2 \, \cos \left (d x + c\right ) + 1}{6 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right ) - a^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.94, size = 58, normalized size = 1.38 \[ -\frac {\frac {\cos \left (c+d\,x\right )}{3}+\frac {1}{6}}{d\,\left (-a^2\,{\cos \left (c+d\,x\right )}^4-2\,a^2\,{\cos \left (c+d\,x\right )}^3+2\,a^2\,\cos \left (c+d\,x\right )+a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^3*(a + a/cos(c + d*x))^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\csc ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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