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]

-15/8*arctanh(cos(d*x+c))/a/d+15/8*sec(d*x+c)/a/d-5/8*csc(d*x+c)^2*sec(d*x+c)/a/d-1/4*csc(d*x+c)^4*sec(d*x+c)/
a/d

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Rubi [A]  time = 0.09, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 4.26, 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]

-1/64*(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))/(a*d)

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fricas [A]  time = 0.43, size = 135, normalized size = 1.65 \[ \frac {30 \, \cos \left (d x + c\right )^{4} - 50 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 16}{16 \, {\left (a d \cos \left (d x + c\right )^{5} - 2 \, a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )\right )}} \]

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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giac [B]  time = 0.17, size = 181, normalized size = 2.21 \[ \frac {\frac {{\left (\frac {16 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {90 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} + \frac {60 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} - \frac {\frac {16 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {128}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{64 \, d} \]

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

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maple [A]  time = 0.49, size = 123, normalized size = 1.50 \[ -\frac {1}{16 a d \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {7}{16 a d \left (\cos \left (d x +c \right )-1\right )}+\frac {15 \ln \left (\cos \left (d x +c \right )-1\right )}{16 a d}+\frac {1}{d a \cos \left (d x +c \right )}+\frac {1}{16 a d \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {7}{16 a d \left (1+\cos \left (d x +c \right )\right )}-\frac {15 \ln \left (1+\cos \left (d x +c \right )\right )}{16 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5/(a-a*sin(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.34, size = 90, normalized size = 1.10 \[ \frac {\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 25 \, \cos \left (d x + c\right )^{2} + 8\right )}}{a \cos \left (d x + c\right )^{5} - 2 \, a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )} - \frac {15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \]

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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mupad [B]  time = 0.10, size = 74, normalized size = 0.90 \[ \frac {\frac {15\,{\cos \left (c+d\,x\right )}^4}{8}-\frac {25\,{\cos \left (c+d\,x\right )}^2}{8}+1}{d\,\left (a\,{\cos \left (c+d\,x\right )}^5-2\,a\,{\cos \left (c+d\,x\right )}^3+a\,\cos \left (c+d\,x\right )\right )}-\frac {15\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{8\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^5*(a - a*sin(c + d*x)^2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\csc ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \]

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]

-Integral(csc(c + d*x)**5/(sin(c + d*x)**2 - 1), x)/a

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