3.18 \(\int \csc ^{10}(c+d x) (a+a \sec (c+d x)) \, dx\)

Optimal. Leaf size=165 \[ -\frac {a \cot ^9(c+d x)}{9 d}-\frac {4 a \cot ^7(c+d x)}{7 d}-\frac {6 a \cot ^5(c+d x)}{5 d}-\frac {4 a \cot ^3(c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}-\frac {a \csc ^9(c+d x)}{9 d}-\frac {a \csc ^7(c+d x)}{7 d}-\frac {a \csc ^5(c+d x)}{5 d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d}+\frac {a \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

a*arctanh(sin(d*x+c))/d-a*cot(d*x+c)/d-4/3*a*cot(d*x+c)^3/d-6/5*a*cot(d*x+c)^5/d-4/7*a*cot(d*x+c)^7/d-1/9*a*co
t(d*x+c)^9/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-1/7*a*csc(d*x+c)^7/d-1/9*a*csc(d*x+c)^9/
d

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3872, 2838, 2621, 302, 207, 3767} \[ -\frac {a \cot ^9(c+d x)}{9 d}-\frac {4 a \cot ^7(c+d x)}{7 d}-\frac {6 a \cot ^5(c+d x)}{5 d}-\frac {4 a \cot ^3(c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}-\frac {a \csc ^9(c+d x)}{9 d}-\frac {a \csc ^7(c+d x)}{7 d}-\frac {a \csc ^5(c+d x)}{5 d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d}+\frac {a \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^10*(a + a*Sec[c + d*x]),x]

[Out]

(a*ArcTanh[Sin[c + d*x]])/d - (a*Cot[c + d*x])/d - (4*a*Cot[c + d*x]^3)/(3*d) - (6*a*Cot[c + d*x]^5)/(5*d) - (
4*a*Cot[c + d*x]^7)/(7*d) - (a*Cot[c + d*x]^9)/(9*d) - (a*Csc[c + d*x])/d - (a*Csc[c + d*x]^3)/(3*d) - (a*Csc[
c + d*x]^5)/(5*d) - (a*Csc[c + d*x]^7)/(7*d) - (a*Csc[c + d*x]^9)/(9*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 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 2621

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 2838

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 3767

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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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 \csc ^{10}(c+d x) (a+a \sec (c+d x)) \, dx &=-\int (-a-a \cos (c+d x)) \csc ^{10}(c+d x) \sec (c+d x) \, dx\\ &=a \int \csc ^{10}(c+d x) \, dx+a \int \csc ^{10}(c+d x) \sec (c+d x) \, dx\\ &=-\frac {a \operatorname {Subst}\left (\int \frac {x^{10}}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}-\frac {a \operatorname {Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {a \cot (c+d x)}{d}-\frac {4 a \cot ^3(c+d x)}{3 d}-\frac {6 a \cot ^5(c+d x)}{5 d}-\frac {4 a \cot ^7(c+d x)}{7 d}-\frac {a \cot ^9(c+d x)}{9 d}-\frac {a \operatorname {Subst}\left (\int \left (1+x^2+x^4+x^6+x^8+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac {a \cot (c+d x)}{d}-\frac {4 a \cot ^3(c+d x)}{3 d}-\frac {6 a \cot ^5(c+d x)}{5 d}-\frac {4 a \cot ^7(c+d x)}{7 d}-\frac {a \cot ^9(c+d x)}{9 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 \csc ^7(c+d x)}{7 d}-\frac {a \csc ^9(c+d x)}{9 d}-\frac {a \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac {a \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a \cot (c+d x)}{d}-\frac {4 a \cot ^3(c+d x)}{3 d}-\frac {6 a \cot ^5(c+d x)}{5 d}-\frac {4 a \cot ^7(c+d x)}{7 d}-\frac {a \cot ^9(c+d x)}{9 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 \csc ^7(c+d x)}{7 d}-\frac {a \csc ^9(c+d x)}{9 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 135, normalized size = 0.82 \[ -\frac {a \csc ^9(c+d x) \, _2F_1\left (-\frac {9}{2},1;-\frac {7}{2};\sin ^2(c+d x)\right )}{9 d}-\frac {128 a \cot (c+d x)}{315 d}-\frac {a \cot (c+d x) \csc ^8(c+d x)}{9 d}-\frac {8 a \cot (c+d x) \csc ^6(c+d x)}{63 d}-\frac {16 a \cot (c+d x) \csc ^4(c+d x)}{105 d}-\frac {64 a \cot (c+d x) \csc ^2(c+d x)}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^10*(a + a*Sec[c + d*x]),x]

[Out]

(-128*a*Cot[c + d*x])/(315*d) - (64*a*Cot[c + d*x]*Csc[c + d*x]^2)/(315*d) - (16*a*Cot[c + d*x]*Csc[c + d*x]^4
)/(105*d) - (8*a*Cot[c + d*x]*Csc[c + d*x]^6)/(63*d) - (a*Cot[c + d*x]*Csc[c + d*x]^8)/(9*d) - (a*Csc[c + d*x]
^9*Hypergeometric2F1[-9/2, 1, -7/2, Sin[c + d*x]^2])/(9*d)

________________________________________________________________________________________

fricas [B]  time = 0.69, size = 366, normalized size = 2.22 \[ -\frac {256 \, a \cos \left (d x + c\right )^{8} + 374 \, a \cos \left (d x + c\right )^{7} - 1526 \, a \cos \left (d x + c\right )^{6} - 1204 \, a \cos \left (d x + c\right )^{5} + 3220 \, a \cos \left (d x + c\right )^{4} + 1316 \, a \cos \left (d x + c\right )^{3} - 2996 \, a \cos \left (d x + c\right )^{2} - 315 \, {\left (a \cos \left (d x + c\right )^{7} - a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{5} + 3 \, a \cos \left (d x + c\right )^{4} + 3 \, a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 315 \, {\left (a \cos \left (d x + c\right )^{7} - a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{5} + 3 \, a \cos \left (d x + c\right )^{4} + 3 \, a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 496 \, a \cos \left (d x + c\right ) + 1126 \, a}{630 \, {\left (d \cos \left (d x + c\right )^{7} - d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{5} + 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{3} - 3 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^10*(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/630*(256*a*cos(d*x + c)^8 + 374*a*cos(d*x + c)^7 - 1526*a*cos(d*x + c)^6 - 1204*a*cos(d*x + c)^5 + 3220*a*c
os(d*x + c)^4 + 1316*a*cos(d*x + c)^3 - 2996*a*cos(d*x + c)^2 - 315*(a*cos(d*x + c)^7 - a*cos(d*x + c)^6 - 3*a
*cos(d*x + c)^5 + 3*a*cos(d*x + c)^4 + 3*a*cos(d*x + c)^3 - 3*a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*log(sin(d
*x + c) + 1)*sin(d*x + c) + 315*(a*cos(d*x + c)^7 - a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^5 + 3*a*cos(d*x + c)^4
+ 3*a*cos(d*x + c)^3 - 3*a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*log(-sin(d*x + c) + 1)*sin(d*x + c) - 496*a*c
os(d*x + c) + 1126*a)/((d*cos(d*x + c)^7 - d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^5 + 3*d*cos(d*x + c)^4 + 3*d*co
s(d*x + c)^3 - 3*d*cos(d*x + c)^2 - d*cos(d*x + c) + d)*sin(d*x + c))

________________________________________________________________________________________

giac [A]  time = 0.27, size = 164, normalized size = 0.99 \[ -\frac {45 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 630 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4830 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80640 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 80640 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 40950 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {80640 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 13650 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 2898 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 450 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}}}{80640 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^10*(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/80640*(45*a*tan(1/2*d*x + 1/2*c)^7 + 630*a*tan(1/2*d*x + 1/2*c)^5 + 4830*a*tan(1/2*d*x + 1/2*c)^3 - 80640*a
*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 80640*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 40950*a*tan(1/2*d*x + 1/2*c
) + (80640*a*tan(1/2*d*x + 1/2*c)^8 + 13650*a*tan(1/2*d*x + 1/2*c)^6 + 2898*a*tan(1/2*d*x + 1/2*c)^4 + 450*a*t
an(1/2*d*x + 1/2*c)^2 + 35*a)/tan(1/2*d*x + 1/2*c)^9)/d

________________________________________________________________________________________

maple [A]  time = 0.81, size = 183, normalized size = 1.11 \[ -\frac {128 a \cot \left (d x +c \right )}{315 d}-\frac {a \cot \left (d x +c \right ) \left (\csc ^{8}\left (d x +c \right )\right )}{9 d}-\frac {8 a \cot \left (d x +c \right ) \left (\csc ^{6}\left (d x +c \right )\right )}{63 d}-\frac {16 a \cot \left (d x +c \right ) \left (\csc ^{4}\left (d x +c \right )\right )}{105 d}-\frac {64 a \cot \left (d x +c \right ) \left (\csc ^{2}\left (d x +c \right )\right )}{315 d}-\frac {a}{9 d \sin \left (d x +c \right )^{9}}-\frac {a}{7 d \sin \left (d x +c \right )^{7}}-\frac {a}{5 d \sin \left (d x +c \right )^{5}}-\frac {a}{3 d \sin \left (d x +c \right )^{3}}-\frac {a}{d \sin \left (d x +c \right )}+\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^10*(a+a*sec(d*x+c)),x)

[Out]

-128/315*a*cot(d*x+c)/d-1/9/d*a*cot(d*x+c)*csc(d*x+c)^8-8/63/d*a*cot(d*x+c)*csc(d*x+c)^6-16/105/d*a*cot(d*x+c)
*csc(d*x+c)^4-64/315/d*a*cot(d*x+c)*csc(d*x+c)^2-1/9/d*a/sin(d*x+c)^9-1/7/d*a/sin(d*x+c)^7-1/5/d*a/sin(d*x+c)^
5-1/3/d*a/sin(d*x+c)^3-1/d*a/sin(d*x+c)+1/d*a*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 136, normalized size = 0.82 \[ -\frac {a {\left (\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{8} + 105 \, \sin \left (d x + c\right )^{6} + 63 \, \sin \left (d x + c\right )^{4} + 45 \, \sin \left (d x + c\right )^{2} + 35\right )}}{\sin \left (d x + c\right )^{9}} - 315 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, {\left (315 \, \tan \left (d x + c\right )^{8} + 420 \, \tan \left (d x + c\right )^{6} + 378 \, \tan \left (d x + c\right )^{4} + 180 \, \tan \left (d x + c\right )^{2} + 35\right )} a}{\tan \left (d x + c\right )^{9}}}{630 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^10*(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/630*(a*(2*(315*sin(d*x + c)^8 + 105*sin(d*x + c)^6 + 63*sin(d*x + c)^4 + 45*sin(d*x + c)^2 + 35)/sin(d*x +
c)^9 - 315*log(sin(d*x + c) + 1) + 315*log(sin(d*x + c) - 1)) + 2*(315*tan(d*x + c)^8 + 420*tan(d*x + c)^6 + 3
78*tan(d*x + c)^4 + 180*tan(d*x + c)^2 + 35)*a/tan(d*x + c)^9)/d

________________________________________________________________________________________

mupad [B]  time = 1.64, size = 159, normalized size = 0.96 \[ \frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {65\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,d}-\frac {23\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{384\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{128\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{1792\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (256\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {130\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {46\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{7}+\frac {a}{9}\right )}{256\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))/sin(c + d*x)^10,x)

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (65*a*tan(c/2 + (d*x)/2))/(128*d) - (23*a*tan(c/2 + (d*x)/2)^3)/(384*d) -
(a*tan(c/2 + (d*x)/2)^5)/(128*d) - (a*tan(c/2 + (d*x)/2)^7)/(1792*d) - (cot(c/2 + (d*x)/2)^9*(a/9 + (10*a*tan(
c/2 + (d*x)/2)^2)/7 + (46*a*tan(c/2 + (d*x)/2)^4)/5 + (130*a*tan(c/2 + (d*x)/2)^6)/3 + 256*a*tan(c/2 + (d*x)/2
)^8))/(256*d)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**10*(a+a*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________