\(\int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [1313]

Optimal result

Integrand size = 27, antiderivative size = 107 \[ \int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\log (\sin (c+d x))}{a d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{b^3 d}-\frac {a \sin ^2(c+d x)}{2 b^2 d}+\frac {\sin ^3(c+d x)}{3 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 107, 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 = {2916, 12, 908} \[ \int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{b^3 d}-\frac {a \sin ^2(c+d x)}{2 b^2 d}+\frac {\log (\sin (c+d x))}{a d}+\frac {\sin ^3(c+d x)}{3 b d} \]

[In]

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

[Out]

Log[Sin[c + d*x]]/(a*d) - ((a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(a*b^4*d) + ((a^2 - 2*b^2)*Sin[c + d*x])/(b^
3*d) - (a*Sin[c + d*x]^2)/(2*b^2*d) + Sin[c + d*x]^3/(3*b*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 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {6 \left (b^4 \log (\sin (c+d x))-\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))\right )+6 a b \left (a^2-2 b^2\right ) \sin (c+d x)-3 a^2 b^2 \sin ^2(c+d x)+2 a b^3 \sin ^3(c+d x)}{6 a b^4 d} \]

[In]

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

[Out]

(6*(b^4*Log[Sin[c + d*x]] - (a^2 - b^2)^2*Log[a + b*Sin[c + d*x]]) + 6*a*b*(a^2 - 2*b^2)*Sin[c + d*x] - 3*a^2*
b^2*Sin[c + d*x]^2 + 2*a*b^3*Sin[c + d*x]^3)/(6*a*b^4*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {\left (-a^{4}+2 a^{2} b^{2}-b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a \,b^{4}}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}}{d}\)	\(105\)
default	\(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {\left (-a^{4}+2 a^{2} b^{2}-b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a \,b^{4}}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}}{d}\)	\(105\)
parallelrisch	\(\frac {-12 \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+12 \left (a^{4}-2 a^{2} b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{4}+3 a^{2} b^{2} \cos \left (2 d x +2 c \right )-a \sin \left (3 d x +3 c \right ) b^{3}+3 \left (4 a^{3} b -7 a \,b^{3}\right ) \sin \left (d x +c \right )-3 a^{2} b^{2}}{12 a \,b^{4} d}\)	\(156\)
norman	\(\frac {\frac {2 \left (9 a^{2}-14 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {2 \left (9 a^{2}-14 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}-\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}+\frac {2 \left (a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{3} d}+\frac {2 \left (a^{2}-2 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {a \left (a^{2}-2 b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a \,b^{4} d}\)	\(295\)
risch	\(-\frac {4 i a c}{b^{2} d}-\frac {2 i a x}{b^{2}}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {2 i a^{3} c}{b^{4} d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )}}{8 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,b^{2}}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {i a^{3} x}{b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}+\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}-\frac {\sin \left (3 d x +3 c \right )}{12 b d}\)	\(306\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 6 \, b^{4} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} b + 5 \, a b^{3}\right )} \sin \left (d x + c\right )}{6 \, a b^{4} d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a} + \frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a b^{4}}}{6 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} + \frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right ) - 12 \, b^{2} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a b^{4}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*log(abs(sin(d*x + c)))/a + (2*b^2*sin(d*x + c)^3 - 3*a*b*sin(d*x + c)^2 + 6*a^2*sin(d*x + c) - 12*b^2*s
in(d*x + c))/b^3 - 6*(a^4 - 2*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/(a*b^4))/d

Mupad [B] (verification not implemented)

Time = 11.95 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.37 \[ \int \frac {\cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-2\,b^2\right )}{b^3}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^2-2\,b^2\right )}{b^3}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{b^2}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{b^2}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2-4\,b^2\right )}{3\,b^3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2-2\,b^2\right )}{b^4\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,{\left (a^2-b^2\right )}^2}{a\,b^4\,d} \]

[In]

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

[Out]

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