\(\int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx\) [223]

Optimal result

Integrand size = 27, antiderivative size = 80 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {4 a^4 \csc (c+d x)}{d}-\frac {a^4 \csc ^2(c+d x)}{2 d}+\frac {6 a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {a^4 \sin ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, 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 = {2912, 12, 45} \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^4 \sin ^2(c+d x)}{2 d}+\frac {4 a^4 \sin (c+d x)}{d}-\frac {a^4 \csc ^2(c+d x)}{2 d}-\frac {4 a^4 \csc (c+d x)}{d}+\frac {6 a^4 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

(-4*a^4*Csc[c + d*x])/d - (a^4*Csc[c + d*x]^2)/(2*d) + (6*a^4*Log[Sin[c + d*x]])/d + (4*a^4*Sin[c + d*x])/d +
(a^4*Sin[c + d*x]^2)/(2*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a^3 (a+x)^4}{x^3} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^2 \text {Subst}\left (\int \frac {(a+x)^4}{x^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \text {Subst}\left (\int \left (4 a+\frac {a^4}{x^3}+\frac {4 a^3}{x^2}+\frac {6 a^2}{x}+x\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {4 a^4 \csc (c+d x)}{d}-\frac {a^4 \csc ^2(c+d x)}{2 d}+\frac {6 a^4 \log (\sin (c+d x))}{d}+\frac {4 a^4 \sin (c+d x)}{d}+\frac {a^4 \sin ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {1}{2} a^4 \left (-\frac {8 \csc (c+d x)}{d}-\frac {\csc ^2(c+d x)}{d}+\frac {12 \log (\sin (c+d x))}{d}+\frac {8 \sin (c+d x)}{d}+\frac {\sin ^2(c+d x)}{d}\right ) \]

[In]

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

[Out]

(a^4*((-8*Csc[c + d*x])/d - Csc[c + d*x]^2/d + (12*Log[Sin[c + d*x]])/d + (8*Sin[c + d*x])/d + Sin[c + d*x]^2/
d))/2

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(-\frac {a^{4} \left (\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+4 \csc \left (d x +c \right )+6 \ln \left (\csc \left (d x +c \right )\right )-\frac {4}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(57\)
default	\(-\frac {a^{4} \left (\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+4 \csc \left (d x +c \right )+6 \ln \left (\csc \left (d x +c \right )\right )-\frac {4}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(57\)
risch	\(-6 i a^{4} x -\frac {a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {2 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {12 i a^{4} c}{d}-\frac {2 i a^{4} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {6 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(166\)
parallelrisch	\(\frac {a^{4} \left (48 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-48 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (2 d x +2 c \right )-8 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (d x +c \right )+2 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-32 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (d x +c \right )-16 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}\)	\(175\)
norman	\(\frac {-\frac {a^{4}}{8 d}-\frac {2 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{4} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{4} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {6 a^{4} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {25 a^{4} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {25 a^{4} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {6 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a^{4} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(266\)

[In]

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

[Out]

-1/d*a^4*(1/2*csc(d*x+c)^2+4*csc(d*x+c)+6*ln(csc(d*x+c))-4/csc(d*x+c)-1/2/csc(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.22 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=-\frac {2 \, a^{4} \cos \left (d x + c\right )^{4} - 16 \, a^{4} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 3 \, a^{4} \cos \left (d x + c\right )^{2} - a^{4} - 24 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - a^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

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

[Out]

-1/4*(2*a^4*cos(d*x + c)^4 - 16*a^4*cos(d*x + c)^2*sin(d*x + c) - 3*a^4*cos(d*x + c)^2 - a^4 - 24*(a^4*cos(d*x
+ c)^2 - a^4)*log(1/2*sin(d*x + c)))/(d*cos(d*x + c)^2 - d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \sin \left (d x + c\right )^{2} + 12 \, a^{4} \log \left (\sin \left (d x + c\right )\right ) + 8 \, a^{4} \sin \left (d x + c\right ) - \frac {8 \, a^{4} \sin \left (d x + c\right ) + a^{4}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {a^{4} \sin \left (d x + c\right )^{2} + 12 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 8 \, a^{4} \sin \left (d x + c\right ) - \frac {8 \, a^{4} \sin \left (d x + c\right ) + a^{4}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.59 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^4 \, dx=\frac {6\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {-24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {15\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-16\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^4}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {6\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

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

[Out]

(6*a^4*log(tan(c/2 + (d*x)/2)))/d - (a^4*tan(c/2 + (d*x)/2)^2)/(8*d) - (a^4*tan(c/2 + (d*x)/2)^2 - 16*a^4*tan(
c/2 + (d*x)/2)^3 - (15*a^4*tan(c/2 + (d*x)/2)^4)/2 - 24*a^4*tan(c/2 + (d*x)/2)^5 + a^4/2 + 8*a^4*tan(c/2 + (d*
x)/2))/(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)) - (2*a^4*tan(c/2 + (d*x)
/2))/d - (6*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d