\(\int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx\) [690]

Optimal result

Integrand size = 21, antiderivative size = 68 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cot ^6(c+d x)}{6 a d}+\frac {\csc (c+d x)}{a d}-\frac {2 \csc ^3(c+d x)}{3 a d}+\frac {\csc ^5(c+d x)}{5 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2785, 2687, 30, 2686, 200} \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cot ^6(c+d x)}{6 a d}+\frac {\csc ^5(c+d x)}{5 a d}-\frac {2 \csc ^3(c+d x)}{3 a d}+\frac {\csc (c+d x)}{a d} \]

[In]

Int[Cot[c + d*x]^7/(a + a*Sin[c + d*x]),x]

[Out]

-1/6*Cot[c + d*x]^6/(a*d) + Csc[c + d*x]/(a*d) - (2*Csc[c + d*x]^3)/(3*a*d) + Csc[c + d*x]^5/(5*a*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 2785

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^6(c+d x) (-15 \cos (4 (c+d x))+78 \sin (c+d x)-5 (5+7 \sin (3 (c+d x))-3 \sin (5 (c+d x))))}{240 a d} \]

[In]

Integrate[Cot[c + d*x]^7/(a + a*Sin[c + d*x]),x]

[Out]

(Csc[c + d*x]^6*(-15*Cos[4*(c + d*x)] + 78*Sin[c + d*x] - 5*(5 + 7*Sin[3*(c + d*x)] - 3*Sin[5*(c + d*x)])))/(2
40*a*d)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {1}{2 \sin \left (d x +c \right )^{4}}-\frac {1}{6 \sin \left (d x +c \right )^{6}}-\frac {2}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{\sin \left (d x +c \right )}}{d a}\)	\(67\)
default	\(\frac {\frac {1}{2 \sin \left (d x +c \right )^{4}}-\frac {1}{6 \sin \left (d x +c \right )^{6}}-\frac {2}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{\sin \left (d x +c \right )}}{d a}\)	\(67\)
risch	\(\frac {2 i \left (-15 i {\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{11 i \left (d x +c \right )}-35 \,{\mathrm e}^{9 i \left (d x +c \right )}-50 i {\mathrm e}^{6 i \left (d x +c \right )}+78 \,{\mathrm e}^{7 i \left (d x +c \right )}-78 \,{\mathrm e}^{5 i \left (d x +c \right )}-15 i {\mathrm e}^{2 i \left (d x +c \right )}+35 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}\)	\(126\)
parallelrisch	\(\frac {-5 \left (\cot ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-100 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-100 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-75 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-75 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+600 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+600 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1920 a d}\)	\(162\)
norman	\(\frac {-\frac {1}{384 a d}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1920 d a}+\frac {7 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320 d a}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}-\frac {35 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d a}+\frac {35 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d a}+\frac {35 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d a}-\frac {35 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d a}-\frac {7 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{192 d a}+\frac {7 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{320 d a}+\frac {7 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1920 d a}-\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{384 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(242\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.41 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 \, \cos \left (d x + c\right )^{4} - 15 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 5}{30 \, {\left (a d \cos \left (d x + c\right )^{6} - 3 \, a d \cos \left (d x + c\right )^{4} + 3 \, a d \cos \left (d x + c\right )^{2} - a d\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 5}{30 \, a d \sin \left (d x + c\right )^{6}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 5}{30 \, a d \sin \left (d x + c\right )^{6}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^5-\frac {{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {{\sin \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )}{5}-\frac {1}{6}}{a\,d\,{\sin \left (c+d\,x\right )}^6} \]

[In]

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

[Out]

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