\(\int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1217]

Optimal result

Integrand size = 27, antiderivative size = 130 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a b \sin (c+d x)}{d}-\frac {\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 d}+\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {b^2 \sin ^6(c+d x)}{6 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 130, 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, 962} \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 d}-\frac {\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 d}+\frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a b \sin ^5(c+d x)}{5 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin ^6(c+d x)}{6 d} \]

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(a^2*Log[Sin[c + d*x]])/d + (2*a*b*Sin[c + d*x])/d - ((2*a^2 - b^2)*Sin[c + d*x]^2)/(2*d) - (4*a*b*Sin[c + d*x
]^3)/(3*d) + ((a^2 - 2*b^2)*Sin[c + d*x]^4)/(4*d) + (2*a*b*Sin[c + d*x]^5)/(5*d) + (b^2*Sin[c + d*x]^6)/(6*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 962

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 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 (a+x)^2 \left (b^2-x^2\right )^2}{x} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a+x)^2 \left (b^2-x^2\right )^2}{x} \, dx,x,b \sin (c+d x)\right )}{b^4 d} \\ & = \frac {\text {Subst}\left (\int \left (2 a b^4+\frac {a^2 b^4}{x}-b^2 \left (2 a^2-b^2\right ) x-4 a b^2 x^2+\left (a^2-2 b^2\right ) x^3+2 a x^4+x^5\right ) \, dx,x,b \sin (c+d x)\right )}{b^4 d} \\ & = \frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a b \sin (c+d x)}{d}-\frac {\left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 d}+\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {b^2 \sin ^6(c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {60 a^2 \log (\sin (c+d x))+120 a b \sin (c+d x)+30 \left (-2 a^2+b^2\right ) \sin ^2(c+d x)-80 a b \sin ^3(c+d x)+15 \left (a^2-2 b^2\right ) \sin ^4(c+d x)+24 a b \sin ^5(c+d x)+10 b^2 \sin ^6(c+d x)}{60 d} \]

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]*(a + b*Sin[c + d*x])^2,x]

[Out]

(60*a^2*Log[Sin[c + d*x]] + 120*a*b*Sin[c + d*x] + 30*(-2*a^2 + b^2)*Sin[c + d*x]^2 - 80*a*b*Sin[c + d*x]^3 +
15*(a^2 - 2*b^2)*Sin[c + d*x]^4 + 24*a*b*Sin[c + d*x]^5 + 10*b^2*Sin[c + d*x]^6)/(60*d)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.62

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+\frac {2 a b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {b^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{6}}{d}\)	\(81\)
default	\(\frac {a^{2} \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+\frac {2 a b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {b^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{6}}{d}\)	\(81\)
parallelrisch	\(\frac {30 \cos \left (4 d x +4 c \right ) a^{2}-30 b^{2} \cos \left (4 d x +4 c \right )-5 b^{2} \cos \left (6 d x +6 c \right )+24 a b \sin \left (5 d x +5 c \right )+200 a b \sin \left (3 d x +3 c \right )+360 a^{2} \cos \left (2 d x +2 c \right )-75 b^{2} \cos \left (2 d x +2 c \right )+1200 a b \sin \left (d x +c \right )+960 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-960 a^{2} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-390 a^{2}+110 b^{2}}{960 d}\)	\(155\)
risch	\(-i a^{2} x +\frac {3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}-\frac {5 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{128 d}+\frac {3 a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {5 \,{\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{128 d}-\frac {2 i a^{2} c}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {5 a b \sin \left (d x +c \right )}{4 d}-\frac {\cos \left (6 d x +6 c \right ) b^{2}}{192 d}+\frac {a b \sin \left (5 d x +5 c \right )}{40 d}+\frac {a^{2} \cos \left (4 d x +4 c \right )}{32 d}-\frac {\cos \left (4 d x +4 c \right ) b^{2}}{32 d}+\frac {5 a b \sin \left (3 d x +3 c \right )}{24 d}\)	\(202\)
norman	\(\frac {-\frac {12 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (2 a^{2}-b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (2 a^{2}-b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 \left (12 a^{2}-5 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {28 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {104 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {104 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {28 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(283\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {10 \, b^{2} \cos \left (d x + c\right )^{6} - 15 \, a^{2} \cos \left (d x + c\right )^{4} - 30 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, a^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (3 \, a b \cos \left (d x + c\right )^{4} + 4 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/60*(10*b^2*cos(d*x + c)^6 - 15*a^2*cos(d*x + c)^4 - 30*a^2*cos(d*x + c)^2 - 60*a^2*log(1/2*sin(d*x + c)) -
8*(3*a*b*cos(d*x + c)^4 + 4*a*b*cos(d*x + c)^2 + 8*a*b)*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {10 \, b^{2} \sin \left (d x + c\right )^{6} + 24 \, a b \sin \left (d x + c\right )^{5} - 80 \, a b \sin \left (d x + c\right )^{3} + 15 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{4} + 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 120 \, a b \sin \left (d x + c\right ) - 30 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/60*(10*b^2*sin(d*x + c)^6 + 24*a*b*sin(d*x + c)^5 - 80*a*b*sin(d*x + c)^3 + 15*(a^2 - 2*b^2)*sin(d*x + c)^4
+ 60*a^2*log(sin(d*x + c)) + 120*a*b*sin(d*x + c) - 30*(2*a^2 - b^2)*sin(d*x + c)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {10 \, b^{2} \sin \left (d x + c\right )^{6} + 24 \, a b \sin \left (d x + c\right )^{5} + 15 \, a^{2} \sin \left (d x + c\right )^{4} - 30 \, b^{2} \sin \left (d x + c\right )^{4} - 80 \, a b \sin \left (d x + c\right )^{3} - 60 \, a^{2} \sin \left (d x + c\right )^{2} + 30 \, b^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 120 \, a b \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(10*b^2*sin(d*x + c)^6 + 24*a*b*sin(d*x + c)^5 + 15*a^2*sin(d*x + c)^4 - 30*b^2*sin(d*x + c)^4 - 80*a*b*s
in(d*x + c)^3 - 60*a^2*sin(d*x + c)^2 + 30*b^2*sin(d*x + c)^2 + 60*a^2*log(abs(sin(d*x + c))) + 120*a*b*sin(d*
x + c))/d

Mupad [B] (verification not implemented)

Time = 11.62 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.18 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^4}{4\,d}-\frac {b^2\,{\cos \left (c+d\,x\right )}^6}{6\,d}+\frac {16\,a\,b\,\sin \left (c+d\,x\right )}{15\,d}+\frac {8\,a\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,d}+\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]

[In]

int((cos(c + d*x)^5*(a + b*sin(c + d*x))^2)/sin(c + d*x),x)

[Out]

(a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a^2*log(1/cos(c/2 + (d*x)/2)^2))/d + (a^2*cos(c + d*x)^2
)/(2*d) + (a^2*cos(c + d*x)^4)/(4*d) - (b^2*cos(c + d*x)^6)/(6*d) + (16*a*b*sin(c + d*x))/(15*d) + (8*a*b*cos(
c + d*x)^2*sin(c + d*x))/(15*d) + (2*a*b*cos(c + d*x)^4*sin(c + d*x))/(5*d)