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

Optimal. Leaf size=78 \[ \frac{5 \sec ^3(c+d x)}{6 a^2 d}+\frac{5 \sec (c+d x)}{2 a^2 d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\csc ^2(c+d x) \sec ^3(c+d x)}{2 a^2 d} \]

[Out]

(-5*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (5*Sec[c + d*x])/(2*a^2*d) + (5*Sec[c + d*x]^3)/(6*a^2*d) - (Csc[c + d*
x]^2*Sec[c + d*x]^3)/(2*a^2*d)

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Rubi [A]  time = 0.0847661, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3175, 2622, 288, 302, 207} \[ \frac{5 \sec ^3(c+d x)}{6 a^2 d}+\frac{5 \sec (c+d x)}{2 a^2 d}-\frac{5 \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{\csc ^2(c+d x) \sec ^3(c+d x)}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTanh[Cos[c + d*x]])/(2*a^2*d) + (5*Sec[c + d*x])/(2*a^2*d) + (5*Sec[c + d*x]^3)/(6*a^2*d) - (Csc[c + d*
x]^2*Sec[c + d*x]^3)/(2*a^2*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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [B]  time = 0.491932, size = 208, normalized size = 2.67 \[ \frac{2 \csc ^8(c+d x) \left (-40 \cos (2 (c+d x))+13 \cos (3 (c+d x))-30 \cos (4 (c+d x))+13 \cos (5 (c+d x))+15 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+15 \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-15 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-15 \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (30 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-30 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-26\right )+22\right )}{3 a^2 d \left (\csc ^2\left (\frac{1}{2} (c+d x)\right )-\sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Csc[c + d*x]^8*(22 - 40*Cos[2*(c + d*x)] + 13*Cos[3*(c + d*x)] - 30*Cos[4*(c + d*x)] + 13*Cos[5*(c + d*x)]
+ 15*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 15*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 15*Cos[3*(c + d*x)]*
Log[Sin[(c + d*x)/2]] - 15*Cos[5*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-26 - 30*Log[Cos[(c + d*x)/2
]] + 30*Log[Sin[(c + d*x)/2]])))/(3*a^2*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2)^3)

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Maple [A]  time = 0.086, size = 104, normalized size = 1.3 \begin{align*}{\frac{1}{4\,{a}^{2}d \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{5\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{4\,{a}^{2}d}}+{\frac{1}{4\,{a}^{2}d \left ( 1+\cos \left ( dx+c \right ) \right ) }}-{\frac{5\,\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{4\,{a}^{2}d}}+{\frac{1}{3\,{a}^{2}d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+2\,{\frac{1}{{a}^{2}d\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)^2,x)

[Out]

1/4/d/a^2/(-1+cos(d*x+c))+5/4/d/a^2*ln(-1+cos(d*x+c))+1/4/d/a^2/(1+cos(d*x+c))-5/4/d/a^2*ln(1+cos(d*x+c))+1/3/
d/a^2/cos(d*x+c)^3+2/d/a^2/cos(d*x+c)

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Maxima [A]  time = 0.956819, size = 116, normalized size = 1.49 \begin{align*} \frac{\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 2\right )}}{a^{2} \cos \left (d x + c\right )^{5} - a^{2} \cos \left (d x + c\right )^{3}} - \frac{15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{15 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{12 \, 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)^2,x, algorithm="maxima")

[Out]

1/12*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 2)/(a^2*cos(d*x + c)^5 - a^2*cos(d*x + c)^3) - 15*log(cos(d*x
+ c) + 1)/a^2 + 15*log(cos(d*x + c) - 1)/a^2)/d

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Fricas [A]  time = 1.76094, size = 312, normalized size = 4. \begin{align*} \frac{30 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 15 \,{\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 4}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{3}\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)^2,x, algorithm="fricas")

[Out]

1/12*(30*cos(d*x + c)^4 - 20*cos(d*x + c)^2 - 15*(cos(d*x + c)^5 - cos(d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2)
+ 15*(cos(d*x + c)^5 - cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) - 4)/(a^2*d*cos(d*x + c)^5 - a^2*d*cos(d*
x + c)^3)

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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 ^{4}{\left (c + d x \right )} - 2 \sin ^{2}{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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)**2,x)

[Out]

Integral(csc(c + d*x)**3/(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1), x)/a**2

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Giac [B]  time = 1.21886, size = 236, normalized size = 3.03 \begin{align*} -\frac{\frac{3 \,{\left (\frac{10 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{30 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac{3 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{a^{2}{\left (\cos \left (d x + c\right ) + 1\right )}} - \frac{16 \,{\left (\frac{12 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{9 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 7\right )}}{a^{2}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{24 \, 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)^2,x, algorithm="giac")

[Out]

-1/24*(3*(10*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1)/(a^2*(cos(d*x + c) - 1)) - 30*log(a
bs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + 3*(cos(d*x + c) - 1)/(a^2*(cos(d*x + c) + 1)) - 16*(12*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 7)/(a^2*((cos(d*x + c) - 1)/
(cos(d*x + c) + 1) + 1)^3))/d