3.40 \(\int \frac{\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=58 \[ \frac{3 \sec (c+d x)}{2 a d}-\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\csc ^2(c+d x) \sec (c+d x)}{2 a d} \]

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(2*a*d) + (3*Sec[c + d*x])/(2*a*d) - (Csc[c + d*x]^2*Sec[c + d*x])/(2*a*d)

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Rubi [A]  time = 0.0959269, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3175, 2622, 288, 321, 207} \[ \frac{3 \sec (c+d x)}{2 a d}-\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\csc ^2(c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(2*a*d) + (3*Sec[c + d*x])/(2*a*d) - (Csc[c + d*x]^2*Sec[c + d*x])/(2*a*d)

Rule 3175

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 0.264649, size = 146, normalized size = 2.52 \[ \frac{\csc ^4(c+d x) \left (-6 \cos (2 (c+d x))+2 \cos (3 (c+d x))+3 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-3 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-2\right )+2\right )}{2 a d \left (\csc ^2\left (\frac{1}{2} (c+d x)\right )-\sec ^2\left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^4*(2 - 6*Cos[2*(c + d*x)] + 2*Cos[3*(c + d*x)] + 3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 3*Co
s[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-2 - 3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/2]])))
/(2*a*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))

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Maple [A]  time = 0.077, size = 87, normalized size = 1.5 \begin{align*}{\frac{1}{4\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{3\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{4\,da}}+{\frac{1}{4\,da \left ( 1+\cos \left ( dx+c \right ) \right ) }}-{\frac{3\,\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{4\,da}}+{\frac{1}{da\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/a/(-1+cos(d*x+c))+3/4/d/a*ln(-1+cos(d*x+c))+1/4/a/d/(1+cos(d*x+c))-3/4/d/a*ln(1+cos(d*x+c))+1/d/a/cos(d*
x+c)

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Maxima [A]  time = 0.968064, size = 95, normalized size = 1.64 \begin{align*} \frac{\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )} - \frac{3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac{3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84009, size = 266, normalized size = 4.59 \begin{align*} \frac{6 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 4}{4 \,{\left (a d \cos \left (d x + c\right )^{3} - a d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}}{\sin ^{2}{\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*sin(d*x+c)**2),x)

[Out]

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

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Giac [B]  time = 1.19287, size = 201, normalized size = 3.47 \begin{align*} \frac{\frac{6 \, \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{14 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{3 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1}{a{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + \frac{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac{\cos \left (d x + c\right ) - 1}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(6*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + (14*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*(co
s(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1)/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + (cos(d*x + c) - 1)^2/(
cos(d*x + c) + 1)^2)) - (cos(d*x + c) - 1)/(a*(cos(d*x + c) + 1)))/d