\(\int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx\) [274]

Optimal result

Integrand size = 27, antiderivative size = 90 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d} \]

[Out]

1/8*a*arctanh(cos(d*x+c))/d-1/3*a*cot(d*x+c)^3/d-1/5*a*cot(d*x+c)^5/d+1/8*a*cot(d*x+c)*csc(d*x+c)/d-1/4*a*cot(
d*x+c)*csc(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2917, 2687, 14, 2691, 3853, 3855} \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {a \cot (c+d x) \csc (c+d x)}{8 d} \]

[In]

Int[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

(a*ArcTanh[Cos[c + d*x]])/(8*d) - (a*Cot[c + d*x]^3)/(3*d) - (a*Cot[c + d*x]^5)/(5*d) + (a*Cot[c + d*x]*Csc[c
+ d*x])/(8*d) - (a*Cot[c + d*x]*Csc[c + d*x]^3)/(4*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 2687

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 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 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 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]

Rubi steps \begin{align*} \text {integral}& = a \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx+a \int \cot ^2(c+d x) \csc ^4(c+d x) \, dx \\ & = -\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {1}{4} a \int \csc ^3(c+d x) \, dx+\frac {a \text {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{d} \\ & = \frac {a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {1}{8} a \int \csc (c+d x) \, dx+\frac {a \text {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{d} \\ & = \frac {a \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.97 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 a \cot (c+d x)}{15 d}+\frac {a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {a \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac {a \cot (c+d x) \csc ^4(c+d x)}{5 d}+\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {a \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d} \]

[In]

Integrate[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + a*Sin[c + d*x]),x]

[Out]

(2*a*Cot[c + d*x])/(15*d) + (a*Csc[(c + d*x)/2]^2)/(32*d) - (a*Csc[(c + d*x)/2]^4)/(64*d) + (a*Cot[c + d*x]*Cs
c[c + d*x]^2)/(15*d) - (a*Cot[c + d*x]*Csc[c + d*x]^4)/(5*d) + (a*Log[Cos[(c + d*x)/2]])/(8*d) - (a*Log[Sin[(c
+ d*x)/2]])/(8*d) - (a*Sec[(c + d*x)/2]^2)/(32*d) + (a*Sec[(c + d*x)/2]^4)/(64*d)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cos ^{3}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos ^{3}\left (d x +c \right )}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a \left (-\frac {\cos ^{3}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15 \sin \left (d x +c \right )^{3}}\right )}{d}\)	\(110\)
default	\(\frac {a \left (-\frac {\cos ^{3}\left (d x +c \right )}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos ^{3}\left (d x +c \right )}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )+a \left (-\frac {\cos ^{3}\left (d x +c \right )}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15 \sin \left (d x +c \right )^{3}}\right )}{d}\)	\(110\)
parallelrisch	\(-\frac {a \left (6 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-60 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{960 d}\)	\(120\)
risch	\(-\frac {a \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}+240 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}+80 i {\mathrm e}^{4 i \left (d x +c \right )}+80 i {\mathrm e}^{2 i \left (d x +c \right )}-90 \,{\mathrm e}^{3 i \left (d x +c \right )}-16 i-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}\)	\(140\)
norman	\(\frac {-\frac {a}{160 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d}-\frac {a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d}-\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {5 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {5 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}+\frac {a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d}+\frac {a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}\)	\(203\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (80) = 160\).

Time = 0.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.88 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {32 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 15 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(32*a*cos(d*x + c)^5 - 80*a*cos(d*x + c)^3 + 15*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*cos(
d*x + c) + 1/2)*sin(d*x + c) - 15*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(-1/2*cos(d*x + c) + 1/2)*sin
(d*x + c) - 30*(a*cos(d*x + c)^3 + a*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*
sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**6*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {15 \, a {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {16 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a}{\tan \left (d x + c\right )^{5}}}{240 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/240*(15*a*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) + 1
) + log(cos(d*x + c) - 1)) + 16*(5*tan(d*x + c)^2 + 3)*a/tan(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.60 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 60 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {274 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*a*tan(1/2*d*x + 1/2*c)^5 + 15*a*tan(1/2*d*x + 1/2*c)^4 + 10*a*tan(1/2*d*x + 1/2*c)^3 - 120*a*log(abs(
tan(1/2*d*x + 1/2*c))) - 60*a*tan(1/2*d*x + 1/2*c) + (274*a*tan(1/2*d*x + 1/2*c)^5 + 60*a*tan(1/2*d*x + 1/2*c)
^4 - 10*a*tan(1/2*d*x + 1/2*c)^2 - 15*a*tan(1/2*d*x + 1/2*c) - 6*a)/tan(1/2*d*x + 1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 9.87 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.59 \[ \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}\right )}{32\,d} \]

[In]

int((cos(c + d*x)^2*(a + a*sin(c + d*x)))/sin(c + d*x)^6,x)

[Out]

(a*tan(c/2 + (d*x)/2)^3)/(96*d) - (a*tan(c/2 + (d*x)/2))/(16*d) + (a*tan(c/2 + (d*x)/2)^4)/(64*d) + (a*tan(c/2
+ (d*x)/2)^5)/(160*d) - (a*log(tan(c/2 + (d*x)/2)))/(8*d) - (cot(c/2 + (d*x)/2)^5*(a/5 + (a*tan(c/2 + (d*x)/2
))/2 + (a*tan(c/2 + (d*x)/2)^2)/3 - 2*a*tan(c/2 + (d*x)/2)^4))/(32*d)