3.1.64 \(\int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx\) [64]

Optimal. Leaf size=106 \[ -\frac {\tanh ^{-1}(\cos (c+d x))}{16 a d}-\frac {\cot (c+d x) \csc (c+d x)}{16 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\csc ^6(c+d x)}{6 a d} \]

[Out]

-1/16*arctanh(cos(d*x+c))/a/d-1/16*cot(d*x+c)*csc(d*x+c)/a/d-1/24*cot(d*x+c)*csc(d*x+c)^3/a/d+1/6*cot(d*x+c)*c
sc(d*x+c)^5/a/d-1/6*csc(d*x+c)^6/a/d

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Rubi [A]
time = 0.13, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3957, 2914, 2686, 30, 2691, 3853, 3855} \begin {gather*} -\frac {\csc ^6(c+d x)}{6 a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{16 a d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}-\frac {\cot (c+d x) \csc (c+d x)}{16 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*ArcTanh[Cos[c + d*x]]/(a*d) - (Cot[c + d*x]*Csc[c + d*x])/(16*a*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(24*a
*d) + (Cot[c + d*x]*Csc[c + d*x]^5)/(6*a*d) - Csc[c + d*x]^6/(6*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

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 2691

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 2914

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 3853

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 3855

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

Rubi steps

\begin {align*} \int \frac {\csc ^5(c+d x)}{a+a \sec (c+d x)} \, dx &=-\int \frac {\cot (c+d x) \csc ^4(c+d x)}{-a-a \cos (c+d x)} \, dx\\ &=-\frac {\int \cot ^2(c+d x) \csc ^5(c+d x) \, dx}{a}+\frac {\int \cot (c+d x) \csc ^6(c+d x) \, dx}{a}\\ &=\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}+\frac {\int \csc ^5(c+d x) \, dx}{6 a}-\frac {\text {Subst}\left (\int x^5 \, dx,x,\csc (c+d x)\right )}{a d}\\ &=-\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\csc ^6(c+d x)}{6 a d}+\frac {\int \csc ^3(c+d x) \, dx}{8 a}\\ &=-\frac {\cot (c+d x) \csc (c+d x)}{16 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\csc ^6(c+d x)}{6 a d}+\frac {\int \csc (c+d x) \, dx}{16 a}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{16 a d}-\frac {\cot (c+d x) \csc (c+d x)}{16 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{24 a d}+\frac {\cot (c+d x) \csc ^5(c+d x)}{6 a d}-\frac {\csc ^6(c+d x)}{6 a d}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 122, normalized size = 1.15 \begin {gather*} -\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (12 \csc ^2\left (\frac {1}{2} (c+d x)\right )+3 \csc ^4\left (\frac {1}{2} (c+d x)\right )+24 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 \sec ^4\left (\frac {1}{2} (c+d x)\right )+2 \sec ^6\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x)}{192 a d (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(Cos[(c + d*x)/2]^2*(12*Csc[(c + d*x)/2]^2 + 3*Csc[(c + d*x)/2]^4 + 24*(Log[Cos[(c + d*x)/2]] - Log[Sin
[(c + d*x)/2]]) + 3*Sec[(c + d*x)/2]^4 + 2*Sec[(c + d*x)/2]^6)*Sec[c + d*x])/(a*d*(1 + Sec[c + d*x]))

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Maple [A]
time = 0.12, size = 79, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {-\frac {1}{32 \left (-1+\cos \left (d x +c \right )\right )^{2}}+\frac {1}{-16+16 \cos \left (d x +c \right )}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{32}-\frac {1}{24 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {1}{32 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{32}}{d a}\)	\(79\)
default	\(\frac {-\frac {1}{32 \left (-1+\cos \left (d x +c \right )\right )^{2}}+\frac {1}{-16+16 \cos \left (d x +c \right )}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{32}-\frac {1}{24 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {1}{32 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{32}}{d a}\)	\(79\)
norman	\(\frac {-\frac {1}{128 a d}-\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a d}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{32 a d}-\frac {3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 a d}-\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{192 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 a d}\)	\(117\)
risch	\(\frac {3 \,{\mathrm e}^{9 i \left (d x +c \right )}+6 \,{\mathrm e}^{8 i \left (d x +c \right )}-8 \,{\mathrm e}^{7 i \left (d x +c \right )}-22 \,{\mathrm e}^{6 i \left (d x +c \right )}-150 \,{\mathrm e}^{5 i \left (d x +c \right )}-22 \,{\mathrm e}^{4 i \left (d x +c \right )}-8 \,{\mathrm e}^{3 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}}{24 a d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{16 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{16 a d}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/32/(-1+cos(d*x+c))^2+1/16/(-1+cos(d*x+c))+1/32*ln(-1+cos(d*x+c))-1/24/(1+cos(d*x+c))^3-1/32/(1+cos(d
*x+c))^2-1/32*ln(1+cos(d*x+c)))

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Maxima [A]
time = 0.27, size = 130, normalized size = 1.23 \begin {gather*} \frac {\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )^{2} - 5 \, \cos \left (d x + c\right ) - 8\right )}}{a \cos \left (d x + c\right )^{5} + a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) + a} - \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}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/96*(2*(3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 - 5*cos(d*x + c)^2 - 5*cos(d*x + c) - 8)/(a*cos(d*x + c)^5 + a*co
s(d*x + c)^4 - 2*a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 + a*cos(d*x + c) + a) - 3*log(cos(d*x + c) + 1)/a + 3*l
og(cos(d*x + c) - 1)/a)/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (96) = 192\).
time = 2.67, size = 217, normalized size = 2.05 \begin {gather*} \frac {6 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{3} - 10 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 ) + 3 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 ) - 10 \, \cos \left (d x + c\right ) - 16}{96 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{3} - 2 \, a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/96*(6*cos(d*x + c)^4 + 6*cos(d*x + c)^3 - 10*cos(d*x + c)^2 - 3*(cos(d*x + c)^5 + cos(d*x + c)^4 - 2*cos(d*x
+ c)^3 - 2*cos(d*x + c)^2 + cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 3*(cos(d*x + c)^5 + cos(d*x + c)^
4 - 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) - 10*cos(d*x + c) - 1
6)/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^3 - 2*a*d*cos(d*x + c)^2 + a*d*cos(d*x + c) +
a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\csc ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**5/(a+a*sec(d*x+c)),x)

[Out]

Integral(csc(c + d*x)**5/(sec(c + d*x) + 1), x)/a

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Giac [A]
time = 0.48, size = 182, normalized size = 1.72 \begin {gather*} \frac {\frac {3 \, {\left (\frac {6 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {6 \, {\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 {12 \, \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 {12 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {9 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}}}{384 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*(6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 6*(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) + 12*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + (12*a^2*(cos(d
*x + c) - 1)/(cos(d*x + c) + 1) - 9*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 2*a^2*(cos(d*x + c) - 1)^3
/(cos(d*x + c) + 1)^3)/a^3)/d

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Mupad [B]
time = 1.01, size = 115, normalized size = 1.08 \begin {gather*} -\frac {\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{16\,a\,d}-\frac {-\frac {{\cos \left (c+d\,x\right )}^4}{16}-\frac {{\cos \left (c+d\,x\right )}^3}{16}+\frac {5\,{\cos \left (c+d\,x\right )}^2}{48}+\frac {5\,\cos \left (c+d\,x\right )}{48}+\frac {1}{6}}{d\,\left (a\,{\cos \left (c+d\,x\right )}^5+a\,{\cos \left (c+d\,x\right )}^4-2\,a\,{\cos \left (c+d\,x\right )}^3-2\,a\,{\cos \left (c+d\,x\right )}^2+a\,\cos \left (c+d\,x\right )+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- atanh(cos(c + d*x))/(16*a*d) - ((5*cos(c + d*x))/48 + (5*cos(c + d*x)^2)/48 - cos(c + d*x)^3/16 - cos(c + d*
x)^4/16 + 1/6)/(d*(a + a*cos(c + d*x) - 2*a*cos(c + d*x)^2 - 2*a*cos(c + d*x)^3 + a*cos(c + d*x)^4 + a*cos(c +
d*x)^5))

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