\(\int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx\) [503]

Optimal result

Integrand size = 19, antiderivative size = 81 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 a \csc (c+d x)}{d}+\frac {a \csc ^2(c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2786, 90} \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin (c+d x)}{d}-\frac {a \csc ^4(c+d x)}{4 d}-\frac {a \csc ^3(c+d x)}{3 d}+\frac {a \csc ^2(c+d x)}{d}+\frac {2 a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d} \]

[In]

Int[Cot[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

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

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 2786

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \cot ^4(c+d x)}{4 d}+\frac {2 a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\tan (c+d x))}{d}+\frac {a \sin (c+d x)}{d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(101\)
default	\(\frac {a \left (-\frac {\cos ^{6}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {\cos ^{6}\left (d x +c \right )}{\sin \left (d x +c \right )}+\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )\right )+a \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(101\)
risch	\(-i a x -\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 i a c}{d}+\frac {4 i a \left (3 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-3 i {\mathrm e}^{4 i \left (d x +c \right )}-7 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(165\)
parallelrisch	\(-\frac {19 \left (\frac {32 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{19}-\frac {32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{19}+\frac {\left (27+\frac {\left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-8 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\csc ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{19}+\frac {\left (-\frac {25 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{6}+3 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-\frac {\cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )}{4}-\frac {\cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )}{4}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{19}-2 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{32 d}\)	\(190\)
norman	\(\frac {-\frac {a}{64 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d}+\frac {11 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}+\frac {5 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {15 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {11 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}-\frac {a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(205\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.36 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {12 \, a \cos \left (d x + c\right )^{2} - 12 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (3 \, a \cos \left (d x + c\right )^{4} - 12 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right ) - 9 \, a}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

[In]

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

[Out]

-1/12*(12*a*cos(d*x + c)^2 - 12*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(1/2*sin(d*x + c)) - 4*(3*a*cos
(d*x + c)^4 - 12*a*cos(d*x + c)^2 + 8*a)*sin(d*x + c) - 9*a)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \log \left (\sin \left (d x + c\right )\right ) + 12 \, a \sin \left (d x + c\right ) + \frac {24 \, a \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 4 \, a \sin \left (d x + c\right ) - 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

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

[Out]

1/12*(12*a*log(sin(d*x + c)) + 12*a*sin(d*x + c) + (24*a*sin(d*x + c)^3 + 12*a*sin(d*x + c)^2 - 4*a*sin(d*x +
c) - 3*a)/sin(d*x + c)^4)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, a \sin \left (d x + c\right ) - \frac {25 \, a \sin \left (d x + c\right )^{4} - 24 \, a \sin \left (d x + c\right )^{3} - 12 \, a \sin \left (d x + c\right )^{2} + 4 \, a \sin \left (d x + c\right ) + 3 \, a}{\sin \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

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

[Out]

1/12*(12*a*log(abs(sin(d*x + c))) + 12*a*sin(d*x + c) - (25*a*sin(d*x + c)^4 - 24*a*sin(d*x + c)^3 - 12*a*sin(
d*x + c)^2 + 4*a*sin(d*x + c) + 3*a)/sin(d*x + c)^4)/d

Mupad [B] (verification not implemented)

Time = 10.33 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.56 \[ \int \cot ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {7\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {46\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

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

[Out]

(7*a*tan(c/2 + (d*x)/2))/(8*d) + ((11*a*tan(c/2 + (d*x)/2)^2)/4 - (2*a*tan(c/2 + (d*x)/2))/3 - a/4 + (40*a*tan
(c/2 + (d*x)/2)^3)/3 + 3*a*tan(c/2 + (d*x)/2)^4 + 46*a*tan(c/2 + (d*x)/2)^5)/(d*(16*tan(c/2 + (d*x)/2)^4 + 16*
tan(c/2 + (d*x)/2)^6)) - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d + (3*a*tan(c/2 + (d*x)/2)^2)/(16*d) - (a*tan(c/2
+ (d*x)/2)^3)/(24*d) - (a*tan(c/2 + (d*x)/2)^4)/(64*d) + (a*log(tan(c/2 + (d*x)/2)))/d