\(\int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [559]

Optimal result

Integrand size = 29, antiderivative size = 65 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a^3 d}+\frac {4 \log (\sin (c+d x))}{a^3 d}-\frac {4 \log (1+\sin (c+d x))}{a^3 d} \]

[Out]

3*csc(d*x+c)/a^3/d-1/2*csc(d*x+c)^2/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 = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 90} \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\csc ^2(c+d x)}{2 a^3 d}+\frac {3 \csc (c+d x)}{a^3 d}+\frac {4 \log (\sin (c+d x))}{a^3 d}-\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

(3*Csc[c + d*x])/(a^3*d) - Csc[c + d*x]^2/(2*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^3 (a-x)^2}{x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x)^2}{x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a}{x^3}-\frac {3}{x^2}+\frac {4}{a x}-\frac {4}{a (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {3 \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 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 = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6 \csc (c+d x)-\csc ^2(c+d x)+8 \log (\sin (c+d x))-8 \log (1+\sin (c+d x))}{2 a^3 d} \]

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^3,x]

[Out]

(6*Csc[c + d*x] - Csc[c + d*x]^2 + 8*Log[Sin[c + d*x]] - 8*Log[1 + Sin[c + d*x]])/(2*a^3*d)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {3}{\sin \left (d x +c \right )}+4 \ln \left (\sin \left (d x +c \right )\right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(49\)
default	\(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {3}{\sin \left (d x +c \right )}+4 \ln \left (\sin \left (d x +c \right )\right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(49\)
parallelrisch	\(\frac {-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-64 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}\)	\(84\)
risch	\(\frac {2 i \left (-i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}\)	\(100\)
norman	\(\frac {-\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {13 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {1}{8 a d}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {7 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {91 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {91 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {535 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {535 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {303 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {303 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}-\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}\)	\(299\)

[In]

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

[Out]

1/d/a^3*(-1/2/sin(d*x+c)^2+3/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 = 76, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \]

[In]

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

[Out]

1/2*(8*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c)) - 8*(cos(d*x + c)^2 - 1)*log(sin(d*x + c) + 1) - 6*sin(d*x +
c) + 1)/(a^3*d*cos(d*x + c)^2 - a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {8 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {8 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {6 \, \sin \left (d x + c\right ) - 1}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(8*log(sin(d*x + c) + 1)/a^3 - 8*log(sin(d*x + c))/a^3 - (6*sin(d*x + c) - 1)/(a^3*sin(d*x + c)^2))/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.77 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {64 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {32 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{8 \, d} \]

[In]

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

[Out]

-1/8*(64*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 32*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (48*tan(1/2*d*x + 1/
2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 1)/(a^3*tan(1/2*d*x + 1/2*c)^2) + (a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(
1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 10.44 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}\right )}{4\,a^3\,d} \]

[In]

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

[Out]

(4*log(tan(c/2 + (d*x)/2)))/(a^3*d) - tan(c/2 + (d*x)/2)^2/(8*a^3*d) - (8*log(tan(c/2 + (d*x)/2) + 1))/(a^3*d)
+ (3*tan(c/2 + (d*x)/2))/(2*a^3*d) + (cot(c/2 + (d*x)/2)^2*(6*tan(c/2 + (d*x)/2) - 1/2))/(4*a^3*d)