\(\int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [505]

Optimal result

Integrand size = 27, antiderivative size = 61 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot ^6(c+d x)}{6 d}-\frac {a \csc (c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^5(c+d x)}{5 d} \]

[Out]

-1/6*a*cot(d*x+c)^6/d-a*csc(d*x+c)/d+2/3*a*csc(d*x+c)^3/d-1/5*a*csc(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2913, 2687, 30, 2686, 200} \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot ^6(c+d x)}{6 d}-\frac {a \csc ^5(c+d x)}{5 d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d} \]

[In]

Int[Cot[c + d*x]^5*Csc[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

-1/6*(a*Cot[c + d*x]^6)/d - (a*Csc[c + d*x])/d + (2*a*Csc[c + d*x]^3)/(3*d) - (a*Csc[c + d*x]^5)/(5*d)

Rule 30

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

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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 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 2913

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot ^6(c+d x)}{6 d}-\frac {a \csc (c+d x)}{d}+\frac {2 a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^5(c+d x)}{5 d} \]

[In]

Integrate[Cot[c + d*x]^5*Csc[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

-1/6*(a*Cot[c + d*x]^6)/d - (a*Csc[c + d*x])/d + (2*a*Csc[c + d*x]^3)/(3*d) - (a*Csc[c + d*x]^5)/(5*d)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(-\frac {a \left (\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}-\frac {2 \left (\csc ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+\csc \left (d x +c \right )\right )}{d}\)	\(64\)
default	\(-\frac {a \left (\frac {\left (\csc ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\csc ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{2}-\frac {2 \left (\csc ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+\csc \left (d x +c \right )\right )}{d}\)	\(64\)
risch	\(-\frac {2 i a \left (15 i {\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{11 i \left (d x +c \right )}-35 \,{\mathrm e}^{9 i \left (d x +c \right )}+50 i {\mathrm e}^{6 i \left (d x +c \right )}+78 \,{\mathrm e}^{7 i \left (d x +c \right )}-78 \,{\mathrm e}^{5 i \left (d x +c \right )}+15 i {\mathrm e}^{2 i \left (d x +c \right )}+35 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}\)	\(124\)
parallelrisch	\(-\frac {\left (\cot ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {12 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-6 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {12 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-6 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+120\right )\right ) a}{384 d}\)	\(155\)
norman	\(\frac {-\frac {a}{384 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d}+\frac {5 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d}+\frac {11 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}-\frac {3 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d}-\frac {25 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {5 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {25 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {3 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d}+\frac {11 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}+\frac {5 a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d}-\frac {a \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}-\frac {a \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(237\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-a/d*(1/6*csc(d*x+c)^6+1/5*csc(d*x+c)^5-1/2*csc(d*x+c)^4-2/3*csc(d*x+c)^3+1/2*csc(d*x+c)^2+csc(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.64 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {15 \, a \cos \left (d x + c\right )^{4} - 15 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, a \cos \left (d x + c\right )^{4} - 20 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right ) + 5 \, a}{30 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(15*a*cos(d*x + c)^4 - 15*a*cos(d*x + c)^2 + 2*(15*a*cos(d*x + c)^4 - 20*a*cos(d*x + c)^2 + 8*a)*sin(d*x
+ c) + 5*a)/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**7*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {30 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 15 \, a \sin \left (d x + c\right )^{2} + 6 \, a \sin \left (d x + c\right ) + 5 \, a}{30 \, d \sin \left (d x + c\right )^{6}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/30*(30*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 15*a*sin(d*x + c)^2 + 6*a*sin(d*x + c
) + 5*a)/(d*sin(d*x + c)^6)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {30 \, a \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} - 15 \, a \sin \left (d x + c\right )^{2} + 6 \, a \sin \left (d x + c\right ) + 5 \, a}{30 \, d \sin \left (d x + c\right )^{6}} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/30*(30*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 15*a*sin(d*x + c)^2 + 6*a*sin(d*x + c
) + 5*a)/(d*sin(d*x + c)^6)

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a\,{\sin \left (c+d\,x\right )}^5+\frac {a\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3}-\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {a\,\sin \left (c+d\,x\right )}{5}+\frac {a}{6}}{d\,{\sin \left (c+d\,x\right )}^6} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x)))/sin(c + d*x)^7,x)

[Out]

-(a/6 + (a*sin(c + d*x))/5 - (a*sin(c + d*x)^2)/2 - (2*a*sin(c + d*x)^3)/3 + (a*sin(c + d*x)^4)/2 + a*sin(c +
d*x)^5)/(d*sin(c + d*x)^6)