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

Optimal. Leaf size=82 \[ -\frac{\csc ^4(c+d x)}{4 a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac{\cot (c+d x) \csc (c+d x)}{8 a d} \]

[Out]

-ArcTanh[Cos[c + d*x]]/(8*a*d) - (Cot[c + d*x]*Csc[c + d*x])/(8*a*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d) -
Csc[c + d*x]^4/(4*a*d)

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Rubi [A]  time = 0.158394, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2835, 2606, 30, 2611, 3768, 3770} \[ -\frac{\csc ^4(c+d x)}{4 a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac{\cot (c+d x) \csc (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[c + d*x]]/(8*a*d) - (Cot[c + d*x]*Csc[c + d*x])/(8*a*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a*d) -
Csc[c + d*x]^4/(4*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 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.368848, size = 91, normalized size = 1.11 \[ -\frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (2 \csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^4\left (\frac{1}{2} (c+d x)\right )-4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{16 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]^2*(2*Csc[(c + d*x)/2]^2 + 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)
/2]^4)*Sec[c + d*x])/(16*a*d*(1 + Sec[c + d*x]))

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Maple [A]  time = 0.06, size = 72, normalized size = 0.9 \begin{align*} -{\frac{1}{8\,da \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{16\,da}}+{\frac{1}{8\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{16\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02163, size = 116, normalized size = 1.41 \begin{align*} \frac{\frac{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 2\right )}}{a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a} - \frac{\log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac{\log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70858, size = 378, normalized size = 4.61 \begin{align*} \frac{2 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \, \cos \left (d x + c\right ) + 4}{16 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + (c
os(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) + 2*cos(d*x + c) + 4)/(a*d*cos
(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\csc ^{3}{\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)**3/(a+a*sec(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.29878, size = 174, normalized size = 2.12 \begin{align*} -\frac{\frac{2 \,{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{a{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{2 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} - \frac{\frac{2 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1)/(a*(cos(d*x + c) - 1)) - 2*log(abs(-co
s(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a - (2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*(cos(d*x + c) - 1)^2
/(cos(d*x + c) + 1)^2)/a^2)/d