\(\int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx\) [16]

Optimal result

Integrand size = 19, antiderivative size = 101 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cot (c+d x)}{d}-\frac {2 a \cot ^3(c+d x)}{3 d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^5(c+d x)}{5 d} \]

[Out]

a*arctanh(sin(d*x+c))/d-a*cot(d*x+c)/d-2/3*a*cot(d*x+c)^3/d-1/5*a*cot(d*x+c)^5/d-a*csc(d*x+c)/d-1/3*a*csc(d*x+
c)^3/d-1/5*a*csc(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3957, 2917, 2701, 308, 213, 3852} \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {2 a \cot ^3(c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}-\frac {a \csc ^5(c+d x)}{5 d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d} \]

[In]

Int[Csc[c + d*x]^6*(a + a*Sec[c + d*x]),x]

[Out]

(a*ArcTanh[Sin[c + d*x]])/d - (a*Cot[c + d*x])/d - (2*a*Cot[c + d*x]^3)/(3*d) - (a*Cot[c + d*x]^5)/(5*d) - (a*
Csc[c + d*x])/d - (a*Csc[c + d*x]^3)/(3*d) - (a*Csc[c + d*x]^5)/(5*d)

Rule 213

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

Rule 308

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 2701

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

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {8 a \cot (c+d x)}{15 d}-\frac {4 a \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac {a \cot (c+d x) \csc ^4(c+d x)}{5 d}-\frac {a \csc ^5(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\sin ^2(c+d x)\right )}{5 d} \]

[In]

Integrate[Csc[c + d*x]^6*(a + a*Sec[c + d*x]),x]

[Out]

(-8*a*Cot[c + d*x])/(15*d) - (4*a*Cot[c + d*x]*Csc[c + d*x]^2)/(15*d) - (a*Cot[c + d*x]*Csc[c + d*x]^4)/(5*d)
- (a*Csc[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, Sin[c + d*x]^2])/(5*d)

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {a \left (-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\)	\(83\)
default	\(\frac {a \left (-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\)	\(83\)
parallelrisch	\(-\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+10 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+80 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )\right ) a}{80 d}\)	\(95\)
norman	\(\frac {-\frac {a}{80 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{48 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(124\)
risch	\(-\frac {2 i a \left (15 \,{\mathrm e}^{7 i \left (d x +c \right )}-30 \,{\mathrm e}^{6 i \left (d x +c \right )}-35 \,{\mathrm e}^{5 i \left (d x +c \right )}+100 \,{\mathrm e}^{4 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}-46 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}+8\right )}{15 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(151\)

[In]

int(csc(d*x+c)^6*(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(-1/5/sin(d*x+c)^5-1/3/sin(d*x+c)^3-1/sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+a*(-8/15-1/5*csc(d*x+c)^4-4
/15*csc(d*x+c)^2)*cot(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (93) = 186\).

Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.88 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {16 \, a \cos \left (d x + c\right )^{4} + 14 \, a \cos \left (d x + c\right )^{3} - 54 \, a \cos \left (d x + c\right )^{2} - 15 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 15 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 16 \, a \cos \left (d x + c\right ) + 46 \, a}{30 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/30*(16*a*cos(d*x + c)^4 + 14*a*cos(d*x + c)^3 - 54*a*cos(d*x + c)^2 - 15*(a*cos(d*x + c)^3 - a*cos(d*x + c)
^2 - a*cos(d*x + c) + a)*log(sin(d*x + c) + 1)*sin(d*x + c) + 15*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(
d*x + c) + a)*log(-sin(d*x + c) + 1)*sin(d*x + c) - 16*a*cos(d*x + c) + 46*a)/((d*cos(d*x + c)^3 - d*cos(d*x +
c)^2 - d*cos(d*x + c) + d)*sin(d*x + c))

Sympy [F]

\[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=a \left (\int \csc ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{6}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(csc(d*x+c)**6*(a+a*sec(d*x+c)),x)

[Out]

a*(Integral(csc(c + d*x)**6*sec(c + d*x), x) + Integral(csc(c + d*x)**6, x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.95 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {a {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} + 3\right )}}{\sin \left (d x + c\right )^{5}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, {\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a}{\tan \left (d x + c\right )^{5}}}{30 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/30*(a*(2*(15*sin(d*x + c)^4 + 5*sin(d*x + c)^2 + 3)/sin(d*x + c)^5 - 15*log(sin(d*x + c) + 1) + 15*log(sin(
d*x + c) - 1)) + 2*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*a/tan(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {5 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 90 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, {\left (80 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{240 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/240*(5*a*tan(1/2*d*x + 1/2*c)^3 - 240*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 240*a*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) + 90*a*tan(1/2*d*x + 1/2*c) + 3*(80*a*tan(1/2*d*x + 1/2*c)^4 + 10*a*tan(1/2*d*x + 1/2*c)^2 + a)/tan
(1/2*d*x + 1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 14.53 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=\frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{5}\right )}{16\,d} \]

[In]

int((a + a/cos(c + d*x))/sin(c + d*x)^6,x)

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (3*a*tan(c/2 + (d*x)/2))/(8*d) - (a*tan(c/2 + (d*x)/2)^3)/(48*d) - (cot(c/
2 + (d*x)/2)^5*(a/5 + 2*a*tan(c/2 + (d*x)/2)^2 + 16*a*tan(c/2 + (d*x)/2)^4))/(16*d)