\(\int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx\) [371]

Optimal result

Integrand size = 25, antiderivative size = 94 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {3 a x}{2}+\frac {3 a \text {arctanh}(\cos (c+d x))}{2 d}-\frac {3 a \cos (c+d x)}{2 d}-\frac {3 a \cot (c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {a \cos (c+d x) \cot ^2(c+d x)}{2 d} \]

[Out]

-3/2*a*x+3/2*a*arctanh(cos(d*x+c))/d-3/2*a*cos(d*x+c)/d-3/2*a*cot(d*x+c)/d+1/2*a*cos(d*x+c)^2*cot(d*x+c)/d-1/2
*a*cos(d*x+c)*cot(d*x+c)^2/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2917, 2672, 294, 327, 212, 2671, 209} \[ \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\cos (c+d x))}{2 d}-\frac {3 a \cos (c+d x)}{2 d}-\frac {3 a \cot (c+d x)}{2 d}+\frac {a \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac {a \cos (c+d x) \cot ^2(c+d x)}{2 d}-\frac {3 a x}{2} \]

[In]

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

[Out]

(-3*a*x)/2 + (3*a*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a*Cos[c + d*x])/(2*d) - (3*a*Cot[c + d*x])/(2*d) + (a*Cos[
c + d*x]^2*Cot[c + d*x])/(2*d) - (a*Cos[c + d*x]*Cot[c + d*x]^2)/(2*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \left (12 c+12 d x+8 \cos (c+d x)+8 \cot (c+d x)+\csc ^2\left (\frac {1}{2} (c+d x)\right )-12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )+2 \sin (2 (c+d x))\right )}{8 d} \]

[In]

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

[Out]

-1/8*(a*(12*c + 12*d*x + 8*Cos[c + d*x] + 8*Cot[c + d*x] + Csc[(c + d*x)/2]^2 - 12*Log[Cos[(c + d*x)/2]] + 12*
Log[Sin[(c + d*x)/2]] - Sec[(c + d*x)/2]^2 + 2*Sin[2*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cos ^{5}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+a \left (-\frac {\cos ^{5}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(116\)
default	\(\frac {a \left (-\frac {\cos ^{5}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+a \left (-\frac {\cos ^{5}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{2}-\frac {3 \cos \left (d x +c \right )}{2}-\frac {3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(116\)
parallelrisch	\(\frac {\left (-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\left (\frac {\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {1}{2}\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {9 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-\frac {\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )}{9}-\frac {\cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )}{9}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}+\left (\cos \left (d x +c \right )-\frac {3}{4}\right ) \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 d x}{2}\right ) a}{d}\)	\(133\)
risch	\(-\frac {3 a x}{2}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {a \left ({\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}-2 i {\mathrm e}^{2 i \left (d x +c \right )}+2 i\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}\)	\(155\)
norman	\(\frac {-\frac {a}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {3 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {3 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {3 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-3 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {5 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {5 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(214\)

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(-1/sin(d*x+c)*cos(d*x+c)^5-(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)-3/2*d*x-3/2*c)+a*(-1/2/sin(d*x+c)^
2*cos(d*x+c)^5-1/2*cos(d*x+c)^3-3/2*cos(d*x+c)-3/2*ln(csc(d*x+c)-cot(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.48 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {6 \, a d x \cos \left (d x + c\right )^{2} + 4 \, a \cos \left (d x + c\right )^{3} - 6 \, a d x - 6 \, a \cos \left (d x + c\right ) - 3 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (a \cos \left (d x + c\right )^{3} - 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**3*(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a - a {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.73 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, {\left (d x + c\right )} a - 12 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2}}}{8 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(a*tan(1/2*d*x + 1/2*c)^2 - 12*(d*x + c)*a - 12*a*log(abs(tan(1/2*d*x + 1/2*c))) + 4*a*tan(1/2*d*x + 1/2*c
) + (6*a*tan(1/2*d*x + 1/2*c)^6 + 4*a*tan(1/2*d*x + 1/2*c)^5 - 5*a*tan(1/2*d*x + 1/2*c)^4 - 16*a*tan(1/2*d*x +
1/2*c)^3 - 12*a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c) - a)/(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x
+ 1/2*c))^2)/d

Mupad [B] (verification not implemented)

Time = 10.43 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.54 \[ \int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {17\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {3\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {3\,a\,\mathrm {atan}\left (\frac {9\,a^2}{9\,a^2-9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,a^2-9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

[In]

int((cos(c + d*x)^4*(a + a*sin(c + d*x)))/sin(c + d*x)^3,x)

[Out]

(a*tan(c/2 + (d*x)/2))/(2*d) - (a/2 + 2*a*tan(c/2 + (d*x)/2) + 9*a*tan(c/2 + (d*x)/2)^2 + 8*a*tan(c/2 + (d*x)/
2)^3 + (17*a*tan(c/2 + (d*x)/2)^4)/2 - 2*a*tan(c/2 + (d*x)/2)^5)/(d*(4*tan(c/2 + (d*x)/2)^2 + 8*tan(c/2 + (d*x
)/2)^4 + 4*tan(c/2 + (d*x)/2)^6)) + (a*tan(c/2 + (d*x)/2)^2)/(8*d) - (3*a*log(tan(c/2 + (d*x)/2)))/(2*d) - (3*
a*atan((9*a^2)/(9*a^2 - 9*a^2*tan(c/2 + (d*x)/2)) + (9*a^2*tan(c/2 + (d*x)/2))/(9*a^2 - 9*a^2*tan(c/2 + (d*x)/
2))))/d