\(\int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [560]

Optimal result

Integrand size = 27, antiderivative size = 83 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \csc (c+d x)}{a^3 d}+\frac {3 \csc ^2(c+d x)}{2 a^3 d}-\frac {\csc ^3(c+d x)}{3 a^3 d}-\frac {4 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (1+\sin (c+d x))}{a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 90} \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\csc ^3(c+d x)}{3 a^3 d}+\frac {3 \csc ^2(c+d x)}{2 a^3 d}-\frac {4 \csc (c+d x)}{a^3 d}-\frac {4 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-4*Csc[c + d*x])/(a^3*d) + (3*Csc[c + d*x]^2)/(2*a^3*d) - Csc[c + d*x]^3/(3*a^3*d) - (4*Log[Sin[c + d*x]])/(a
^3*d) + (4*Log[1 + Sin[c + d*x]])/(a^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2915

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

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.71 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {24 \csc (c+d x)-9 \csc ^2(c+d x)+2 \csc ^3(c+d x)+24 \log (\sin (c+d x))-24 \log (1+\sin (c+d x))}{6 a^3 d} \]

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/6*(24*Csc[c + d*x] - 9*Csc[c + d*x]^2 + 2*Csc[c + d*x]^3 + 24*Log[Sin[c + d*x]] - 24*Log[1 + Sin[c + d*x]])
/(a^3*d)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {3}{2 \sin \left (d x +c \right )^{2}}-\frac {4}{\sin \left (d x +c \right )}-4 \ln \left (\sin \left (d x +c \right )\right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(59\)
default	\(\frac {-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {3}{2 \sin \left (d x +c \right )^{2}}-\frac {4}{\sin \left (d x +c \right )}-4 \ln \left (\sin \left (d x +c \right )\right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(59\)
parallelrisch	\(\frac {-\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-96 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-51 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{3}}\)	\(110\)
risch	\(-\frac {2 i \left (12 \,{\mathrm e}^{5 i \left (d x +c \right )}-28 \,{\mathrm e}^{3 i \left (d x +c \right )}-9 i {\mathrm e}^{4 i \left (d x +c \right )}+12 \,{\mathrm e}^{i \left (d x +c \right )}+9 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\)	\(123\)
norman	\(\frac {-\frac {1}{24 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {17 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {17 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}+\frac {\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}+\frac {121 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {121 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {459 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {459 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {497 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}+\frac {497 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\)	\(299\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.28 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {24 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 24 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 24 \, \cos \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) - 26}{6 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/6*(24*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c))*sin(d*x + c) - 24*(cos(d*x + c)^2 - 1)*log(sin(d*x + c) +
1)*sin(d*x + c) + 24*cos(d*x + c)^2 + 9*sin(d*x + c) - 26)/((a^3*d*cos(d*x + c)^2 - a^3*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {24 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {24 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {24 \, \sin \left (d x + c\right )^{2} - 9 \, \sin \left (d x + c\right ) + 2}{a^{3} \sin \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/6*(24*log(sin(d*x + c) + 1)/a^3 - 24*log(sin(d*x + c))/a^3 - (24*sin(d*x + c)^2 - 9*sin(d*x + c) + 2)/(a^3*s
in(d*x + c)^3))/d

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.75 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {192 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {96 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {176 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 51 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 51 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{9}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/24*(192*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 96*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (176*tan(1/2*d*x +
1/2*c)^3 - 51*tan(1/2*d*x + 1/2*c)^2 + 9*tan(1/2*d*x + 1/2*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)^3) - (a^6*tan(1/2
*d*x + 1/2*c)^3 - 9*a^6*tan(1/2*d*x + 1/2*c)^2 + 51*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d

Mupad [B] (verification not implemented)

Time = 10.36 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.67 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}-\frac {17\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}\right )}{8\,a^3\,d} \]

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^4*(a + a*sin(c + d*x))^3),x)

[Out]

(3*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - tan(c/2 + (d*x)/2)^3/(24*a^3*d) - (4*log(tan(c/2 + (d*x)/2)))/(a^3*d) + (
8*log(tan(c/2 + (d*x)/2) + 1))/(a^3*d) - (17*tan(c/2 + (d*x)/2))/(8*a^3*d) - (cot(c/2 + (d*x)/2)^3*(17*tan(c/2
+ (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2) + 1/3))/(8*a^3*d)