3.376 \(\int \cot ^4(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=114 \[ -\frac {a \cot ^7(c+d x)}{7 d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {a \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}+\frac {a \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac {a \cot (c+d x) \csc (c+d x)}{16 d} \]

[Out]

-1/16*a*arctanh(cos(d*x+c))/d-1/5*a*cot(d*x+c)^5/d-1/7*a*cot(d*x+c)^7/d-1/16*a*cot(d*x+c)*csc(d*x+c)/d+1/8*a*c
ot(d*x+c)*csc(d*x+c)^3/d-1/6*a*cot(d*x+c)^3*csc(d*x+c)^3/d

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Rubi [A]  time = 0.15, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2838, 2607, 14, 2611, 3768, 3770} \[ -\frac {a \cot ^7(c+d x)}{7 d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {a \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}+\frac {a \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac {a \cot (c+d x) \csc (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

-(a*ArcTanh[Cos[c + d*x]])/(16*d) - (a*Cot[c + d*x]^5)/(5*d) - (a*Cot[c + d*x]^7)/(7*d) - (a*Cot[c + d*x]*Csc[
c + d*x])/(16*d) + (a*Cot[c + d*x]*Csc[c + d*x]^3)/(8*d) - (a*Cot[c + d*x]^3*Csc[c + d*x]^3)/(6*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

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 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 3768

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 3770

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

Rubi steps

\begin {align*} \int \cot ^4(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cot ^4(c+d x) \csc ^3(c+d x) \, dx+a \int \cot ^4(c+d x) \csc ^4(c+d x) \, dx\\ &=-\frac {a \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}-\frac {1}{2} a \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx+\frac {a \operatorname {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{d}\\ &=\frac {a \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac {a \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}+\frac {1}{8} a \int \csc ^3(c+d x) \, dx+\frac {a \operatorname {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,-\cot (c+d x)\right )}{d}\\ &=-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \cot ^7(c+d x)}{7 d}-\frac {a \cot (c+d x) \csc (c+d x)}{16 d}+\frac {a \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac {a \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}+\frac {1}{16} a \int \csc (c+d x) \, dx\\ &=-\frac {a \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \cot ^7(c+d x)}{7 d}-\frac {a \cot (c+d x) \csc (c+d x)}{16 d}+\frac {a \cot (c+d x) \csc ^3(c+d x)}{8 d}-\frac {a \cot ^3(c+d x) \csc ^3(c+d x)}{6 d}\\ \end {align*}

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Mathematica [B]  time = 0.08, size = 239, normalized size = 2.10 \[ -\frac {2 a \cot (c+d x)}{35 d}-\frac {a \csc ^6\left (\frac {1}{2} (c+d x)\right )}{384 d}+\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a \sec ^6\left (\frac {1}{2} (c+d x)\right )}{384 d}-\frac {a \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{16 d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{16 d}-\frac {a \cot (c+d x) \csc ^6(c+d x)}{7 d}+\frac {8 a \cot (c+d x) \csc ^4(c+d x)}{35 d}-\frac {a \cot (c+d x) \csc ^2(c+d x)}{35 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

(-2*a*Cot[c + d*x])/(35*d) - (a*Csc[(c + d*x)/2]^2)/(64*d) + (a*Csc[(c + d*x)/2]^4)/(64*d) - (a*Csc[(c + d*x)/
2]^6)/(384*d) - (a*Cot[c + d*x]*Csc[c + d*x]^2)/(35*d) + (8*a*Cot[c + d*x]*Csc[c + d*x]^4)/(35*d) - (a*Cot[c +
d*x]*Csc[c + d*x]^6)/(7*d) - (a*Log[Cos[(c + d*x)/2]])/(16*d) + (a*Log[Sin[(c + d*x)/2]])/(16*d) + (a*Sec[(c
+ d*x)/2]^2)/(64*d) - (a*Sec[(c + d*x)/2]^4)/(64*d) + (a*Sec[(c + d*x)/2]^6)/(384*d)

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fricas [B]  time = 0.53, size = 221, normalized size = 1.94 \[ -\frac {192 \, a \cos \left (d x + c\right )^{7} - 672 \, a \cos \left (d x + c\right )^{5} + 105 \, {\left (a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} + 3 \, a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 105 \, {\left (a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} + 3 \, a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 70 \, {\left (3 \, a \cos \left (d x + c\right )^{5} + 8 \, a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3360 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^8*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/3360*(192*a*cos(d*x + c)^7 - 672*a*cos(d*x + c)^5 + 105*(a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^4 + 3*a*cos(d*
x + c)^2 - a)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 105*(a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^4 + 3*a*cos(
d*x + c)^2 - a)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 70*(3*a*cos(d*x + c)^5 + 8*a*cos(d*x + c)^3 - 3*a*
cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [B]  time = 0.22, size = 229, normalized size = 2.01 \[ \frac {15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 35 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 21 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 105 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 840 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 315 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {2178 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 315 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 105 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 105 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^8*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/13440*(15*a*tan(1/2*d*x + 1/2*c)^7 + 35*a*tan(1/2*d*x + 1/2*c)^6 - 21*a*tan(1/2*d*x + 1/2*c)^5 - 105*a*tan(1
/2*d*x + 1/2*c)^4 - 105*a*tan(1/2*d*x + 1/2*c)^3 - 105*a*tan(1/2*d*x + 1/2*c)^2 + 840*a*log(abs(tan(1/2*d*x +
1/2*c))) + 315*a*tan(1/2*d*x + 1/2*c) - (2178*a*tan(1/2*d*x + 1/2*c)^7 + 315*a*tan(1/2*d*x + 1/2*c)^6 - 105*a*
tan(1/2*d*x + 1/2*c)^5 - 105*a*tan(1/2*d*x + 1/2*c)^4 - 105*a*tan(1/2*d*x + 1/2*c)^3 - 21*a*tan(1/2*d*x + 1/2*
c)^2 + 35*a*tan(1/2*d*x + 1/2*c) + 15*a)/tan(1/2*d*x + 1/2*c)^7)/d

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maple [A]  time = 0.27, size = 160, normalized size = 1.40 \[ -\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{6 d \sin \left (d x +c \right )^{6}}-\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{24 d \sin \left (d x +c \right )^{4}}+\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{48 d \sin \left (d x +c \right )^{2}}+\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{48 d}+\frac {a \cos \left (d x +c \right )}{16 d}+\frac {a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16 d}-\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{7 d \sin \left (d x +c \right )^{7}}-\frac {2 a \left (\cos ^{5}\left (d x +c \right )\right )}{35 d \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^8*(a+a*sin(d*x+c)),x)

[Out]

-1/6/d*a/sin(d*x+c)^6*cos(d*x+c)^5-1/24/d*a/sin(d*x+c)^4*cos(d*x+c)^5+1/48/d*a/sin(d*x+c)^2*cos(d*x+c)^5+1/48*
a*cos(d*x+c)^3/d+1/16*a*cos(d*x+c)/d+1/16/d*a*ln(csc(d*x+c)-cot(d*x+c))-1/7/d*a/sin(d*x+c)^7*cos(d*x+c)^5-2/35
/d*a/sin(d*x+c)^5*cos(d*x+c)^5

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maxima [A]  time = 0.32, size = 118, normalized size = 1.04 \[ \frac {35 \, a {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac {96 \, {\left (7 \, \tan \left (d x + c\right )^{2} + 5\right )} a}{\tan \left (d x + c\right )^{7}}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^8*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/3360*(35*a*(2*(3*cos(d*x + c)^5 + 8*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*
cos(d*x + c)^2 - 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 96*(7*tan(d*x + c)^2 + 5)*a/tan(d*x
+ c)^7)/d

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mupad [B]  time = 9.95, size = 385, normalized size = 3.38 \[ \frac {a\,\left (15\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+35\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-35\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-21\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+105\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+21\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+840\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\right )}{13440\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + a*sin(c + d*x)))/sin(c + d*x)^8,x)

[Out]

(a*(15*sin(c/2 + (d*x)/2)^14 - 15*cos(c/2 + (d*x)/2)^14 + 35*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^13 - 35*cos
(c/2 + (d*x)/2)^13*sin(c/2 + (d*x)/2) - 21*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^12 - 105*cos(c/2 + (d*x)/2)
^3*sin(c/2 + (d*x)/2)^11 - 105*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^10 - 105*cos(c/2 + (d*x)/2)^5*sin(c/2 +
(d*x)/2)^9 + 315*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8 - 315*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^6 +
105*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^5 + 105*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^4 + 105*cos(c/2 +
(d*x)/2)^11*sin(c/2 + (d*x)/2)^3 + 21*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^2 + 840*log(sin(c/2 + (d*x)/2)
/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^7))/(13440*d*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)
/2)^7)

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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(cos(d*x+c)**4*csc(d*x+c)**8*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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