\(\int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [517]

Optimal result

Integrand size = 27, antiderivative size = 110 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {4 a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x)}{d}+\frac {a^2 \sin ^2(c+d x)}{d}+\frac {a^2 \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 110, 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} \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \sin ^2(c+d x)}{d}-\frac {a^2 \sin (c+d x)}{d}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^2(c+d x)}{d}+\frac {a^2 \csc (c+d x)}{d}-\frac {4 a^2 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

(a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/d - (a^2*Csc[c + d*x]^3)/(3*d) - (4*a^2*Log[Sin[c + d*x]])/d - (a^
2*Sin[c + d*x])/d + (a^2*Sin[c + d*x]^2)/d + (a^2*Sin[c + d*x]^3)/(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 (a+x)^4}{x^4} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x)^2 (a+x)^4}{x^4} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \left (-a^2+\frac {a^6}{x^4}+\frac {2 a^5}{x^3}-\frac {a^4}{x^2}-\frac {4 a^3}{x}+2 a x+x^2\right ) \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {4 a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x)}{d}+\frac {a^2 \sin ^2(c+d x)}{d}+\frac {a^2 \sin ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.67 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (3 \csc (c+d x)-3 \csc ^2(c+d x)-\csc ^3(c+d x)-12 \log (\sin (c+d x))-3 \sin (c+d x)+3 \sin ^2(c+d x)+\sin ^3(c+d x)\right )}{3 d} \]

[In]

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

[Out]

(a^2*(3*Csc[c + d*x] - 3*Csc[c + d*x]^2 - Csc[c + d*x]^3 - 12*Log[Sin[c + d*x]] - 3*Sin[c + d*x] + 3*Sin[c + d
*x]^2 + Sin[c + d*x]^3))/(3*d)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.44

method	result	size
parallelrisch	\(\frac {\left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (48 \sin \left (d x +c \right )-16 \sin \left (3 d x +3 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-48 \sin \left (d x +c \right )+16 \sin \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin \left (5 d x +5 c \right )-\frac {5 \cos \left (2 d x +2 c \right )}{2}-\cos \left (4 d x +4 c \right )-\frac {\cos \left (6 d x +6 c \right )}{6}+9 \sin \left (d x +c \right )-10 \sin \left (3 d x +3 c \right )-\frac {5}{3}\right ) \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{128 d}\)	\(158\)
derivativedivides	\(\frac {a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+2 a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )}{d}\)	\(177\)
default	\(\frac {a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+2 a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{2}-\left (\cos ^{2}\left (d x +c \right )\right )-2 \ln \left (\sin \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )}{d}\)	\(177\)
risch	\(4 i a^{2} x +\frac {i a^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{4 d}+\frac {3 i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}-\frac {i a^{2} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {8 i a^{2} c}{d}+\frac {2 i a^{2} \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 \,{\mathrm e}^{3 i \left (d x +c \right )}-6 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}+6 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {4 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(225\)
norman	\(\frac {-\frac {a^{2}}{24 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {5 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {5 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {5 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {21 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {21 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {4 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(266\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/128*csc(1/2*d*x+1/2*c)^3*((48*sin(d*x+c)-16*sin(3*d*x+3*c))*ln(sec(1/2*d*x+1/2*c)^2)+(-48*sin(d*x+c)+16*sin(
3*d*x+3*c))*ln(tan(1/2*d*x+1/2*c))+sin(5*d*x+5*c)-5/2*cos(2*d*x+2*c)-cos(4*d*x+4*c)-1/6*cos(6*d*x+6*c)+9*sin(d
*x+c)-10*sin(3*d*x+3*c)-5/3)*sec(1/2*d*x+1/2*c)^3*a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \cos \left (d x + c\right )^{6} - 24 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(2*a^2*cos(d*x + c)^6 - 24*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d*x + c))*sin(d*x + c) - 3*(2*a^2*cos(d*
x + c)^4 - 3*a^2*cos(d*x + c)^2 - a^2)*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \sin \left (d x + c\right )^{3} + 3 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - 3 \, a^{2} \sin \left (d x + c\right ) + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, a^{2} \sin \left (d x + c\right ) - a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(a^2*sin(d*x + c)^3 + 3*a^2*sin(d*x + c)^2 - 12*a^2*log(sin(d*x + c)) - 3*a^2*sin(d*x + c) + (3*a^2*sin(d*
x + c)^2 - 3*a^2*sin(d*x + c) - a^2)/sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.97 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \sin \left (d x + c\right )^{3} + 3 \, a^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 3 \, a^{2} \sin \left (d x + c\right ) + \frac {22 \, a^{2} \sin \left (d x + c\right )^{3} + 3 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, a^{2} \sin \left (d x + c\right ) - a^{2}}{\sin \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(a^2*sin(d*x + c)^3 + 3*a^2*sin(d*x + c)^2 - 12*a^2*log(abs(sin(d*x + c))) - 3*a^2*sin(d*x + c) + (22*a^2*
sin(d*x + c)^3 + 3*a^2*sin(d*x + c)^2 - 3*a^2*sin(d*x + c) - a^2)/sin(d*x + c)^3)/d

Mupad [B] (verification not implemented)

Time = 9.63 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.62 \[ \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {4\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-30\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-26\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {4\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^2)/sin(c + d*x)^4,x)

[Out]

(3*a^2*tan(c/2 + (d*x)/2))/(8*d) - (a^2*tan(c/2 + (d*x)/2)^3)/(24*d) - (4*a^2*log(tan(c/2 + (d*x)/2)))/d - (a^
2*tan(c/2 + (d*x)/2)^2)/(4*d) - (6*a^2*tan(c/2 + (d*x)/2)^3 - 2*a^2*tan(c/2 + (d*x)/2)^2 + 8*a^2*tan(c/2 + (d*
x)/2)^4 - 26*a^2*tan(c/2 + (d*x)/2)^5 + 2*a^2*tan(c/2 + (d*x)/2)^6 - 30*a^2*tan(c/2 + (d*x)/2)^7 + 13*a^2*tan(
c/2 + (d*x)/2)^8 + a^2/3 + 2*a^2*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^3 + 24*tan(c/2 + (d*x)/2)^5 + 24
*tan(c/2 + (d*x)/2)^7 + 8*tan(c/2 + (d*x)/2)^9)) + (4*a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d