3.41 \(\int \frac{\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=82 \[ \frac{15 \sec (c+d x)}{8 a d}-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac{\csc ^4(c+d x) \sec (c+d x)}{4 a d}-\frac{5 \csc ^2(c+d x) \sec (c+d x)}{8 a d} \]

[Out]

(-15*ArcTanh[Cos[c + d*x]])/(8*a*d) + (15*Sec[c + d*x])/(8*a*d) - (5*Csc[c + d*x]^2*Sec[c + d*x])/(8*a*d) - (C
sc[c + d*x]^4*Sec[c + d*x])/(4*a*d)

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Rubi [A]  time = 0.0945551, antiderivative size = 82, 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, 321, 207} \[ \frac{15 \sec (c+d x)}{8 a d}-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac{\csc ^4(c+d x) \sec (c+d x)}{4 a d}-\frac{5 \csc ^2(c+d x) \sec (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5/(a - a*Sin[c + d*x]^2),x]

[Out]

(-15*ArcTanh[Cos[c + d*x]])/(8*a*d) + (15*Sec[c + d*x])/(8*a*d) - (5*Csc[c + d*x]^2*Sec[c + d*x])/(8*a*d) - (C
sc[c + d*x]^4*Sec[c + d*x])/(4*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 ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx &=\frac{\int \csc ^5(c+d x) \sec ^2(c+d x) \, dx}{a}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^3} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac{\csc ^4(c+d x) \sec (c+d x)}{4 a d}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{4 a d}\\ &=-\frac{5 \csc ^2(c+d x) \sec (c+d x)}{8 a d}-\frac{\csc ^4(c+d x) \sec (c+d x)}{4 a d}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{8 a d}\\ &=\frac{15 \sec (c+d x)}{8 a d}-\frac{5 \csc ^2(c+d x) \sec (c+d x)}{8 a d}-\frac{\csc ^4(c+d x) \sec (c+d x)}{4 a d}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{8 a d}\\ &=-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac{15 \sec (c+d x)}{8 a d}-\frac{5 \csc ^2(c+d x) \sec (c+d x)}{8 a d}-\frac{\csc ^4(c+d x) \sec (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 4.47197, size = 132, normalized size = 1.61 \[ -\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right )+14 \csc ^2\left (\frac{1}{2} (c+d x)\right )+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (-14 \tan ^2\left (\frac{1}{2} (c+d x)\right )+\cos (c+d x) \left (\sec ^4\left (\frac{1}{2} (c+d x)\right )-8 \left (-15 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+8\right )\right )+78\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )-1}}{64 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5/(a - a*Sin[c + d*x]^2),x]

[Out]

-(14*Csc[(c + d*x)/2]^2 + Csc[(c + d*x)/2]^4 + (Sec[(c + d*x)/2]^2*(78 + Cos[c + d*x]*(-8*(8 + 15*Log[Cos[(c +
d*x)/2]] - 15*Log[Sin[(c + d*x)/2]]) + Sec[(c + d*x)/2]^4) - 14*Tan[(c + d*x)/2]^2))/(-1 + Tan[(c + d*x)/2]^2
))/(64*a*d)

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Maple [A]  time = 0.08, size = 123, normalized size = 1.5 \begin{align*} -{\frac{1}{16\,da \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{7}{16\,da \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{15\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{16\,da}}+{\frac{1}{16\,da \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{7}{16\,da \left ( 1+\cos \left ( dx+c \right ) \right ) }}-{\frac{15\,\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{16\,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)^5/(a-sin(d*x+c)^2*a),x)

[Out]

-1/16/d/a/(-1+cos(d*x+c))^2+7/16/d/a/(-1+cos(d*x+c))+15/16/d/a*ln(-1+cos(d*x+c))+1/16/a/d/(1+cos(d*x+c))^2+7/1
6/a/d/(1+cos(d*x+c))-15/16/d/a*ln(1+cos(d*x+c))+1/d/a/cos(d*x+c)

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Maxima [A]  time = 0.963531, size = 122, normalized size = 1.49 \begin{align*} \frac{\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{4} - 25 \, \cos \left (d x + c\right )^{2} + 8\right )}}{a \cos \left (d x + c\right )^{5} - 2 \, a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )} - \frac{15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac{15 \, \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)^5/(a-a*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

1/16*(2*(15*cos(d*x + c)^4 - 25*cos(d*x + c)^2 + 8)/(a*cos(d*x + c)^5 - 2*a*cos(d*x + c)^3 + a*cos(d*x + c)) -
15*log(cos(d*x + c) + 1)/a + 15*log(cos(d*x + c) - 1)/a)/d

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Fricas [A]  time = 1.79891, size = 382, normalized size = 4.66 \begin{align*} \frac{30 \, \cos \left (d x + c\right )^{4} - 50 \, \cos \left (d x + c\right )^{2} - 15 \,{\left (\cos \left (d x + c\right )^{5} - 2 \, \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 ) + 15 \,{\left (\cos \left (d x + c\right )^{5} - 2 \, \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 ) + 16}{16 \,{\left (a d \cos \left (d x + c\right )^{5} - 2 \, 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)^5/(a-a*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

1/16*(30*cos(d*x + c)^4 - 50*cos(d*x + c)^2 - 15*(cos(d*x + c)^5 - 2*cos(d*x + c)^3 + cos(d*x + c))*log(1/2*co
s(d*x + c) + 1/2) + 15*(cos(d*x + c)^5 - 2*cos(d*x + c)^3 + cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 16)/(
a*d*cos(d*x + c)^5 - 2*a*d*cos(d*x + c)^3 + a*d*cos(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)**5/(a-a*sin(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.20958, size = 244, normalized size = 2.98 \begin{align*} \frac{\frac{{\left (\frac{16 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{90 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} + \frac{60 \, \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{16 \, 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}} + \frac{128}{a{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a-a*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/64*((16*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 90*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x +
c) + 1)^2/(a*(cos(d*x + c) - 1)^2) + 60*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a - (16*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 + 128/(a*((cos(d*x + c) - 1)/(
cos(d*x + c) + 1) + 1)))/d