\(\int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx\) [523]

Optimal result

Integrand size = 27, antiderivative size = 137 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}-\frac {5 a^3 \sin ^4(c+d x)}{4 d}+\frac {a^3 \sin ^5(c+d x)}{5 d}+\frac {a^3 \sin ^6(c+d x)}{2 d}+\frac {a^3 \sin ^7(c+d x)}{7 d} \]

[Out]

a^3*ln(sin(d*x+c))/d+3*a^3*sin(d*x+c)/d+1/2*a^3*sin(d*x+c)^2/d-5/3*a^3*sin(d*x+c)^3/d-5/4*a^3*sin(d*x+c)^4/d+1
/5*a^3*sin(d*x+c)^5/d+1/2*a^3*sin(d*x+c)^6/d+1/7*a^3*sin(d*x+c)^7/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 137, 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} \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^7(c+d x)}{7 d}+\frac {a^3 \sin ^6(c+d x)}{2 d}+\frac {a^3 \sin ^5(c+d x)}{5 d}-\frac {5 a^3 \sin ^4(c+d x)}{4 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}+\frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*Log[Sin[c + d*x]])/d + (3*a^3*Sin[c + d*x])/d + (a^3*Sin[c + d*x]^2)/(2*d) - (5*a^3*Sin[c + d*x]^3)/(3*d)
- (5*a^3*Sin[c + d*x]^4)/(4*d) + (a^3*Sin[c + d*x]^5)/(5*d) + (a^3*Sin[c + d*x]^6)/(2*d) + (a^3*Sin[c + d*x]^
7)/(7*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 (a-x)^2 (a+x)^5}{x} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x)^2 (a+x)^5}{x} \, dx,x,a \sin (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (3 a^6+\frac {a^7}{x}+a^5 x-5 a^4 x^2-5 a^3 x^3+a^2 x^4+3 a x^5+x^6\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d} \\ & = \frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d}-\frac {5 a^3 \sin ^3(c+d x)}{3 d}-\frac {5 a^3 \sin ^4(c+d x)}{4 d}+\frac {a^3 \sin ^5(c+d x)}{5 d}+\frac {a^3 \sin ^6(c+d x)}{2 d}+\frac {a^3 \sin ^7(c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.64 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (420 \log (\sin (c+d x))+1260 \sin (c+d x)+210 \sin ^2(c+d x)-700 \sin ^3(c+d x)-525 \sin ^4(c+d x)+84 \sin ^5(c+d x)+210 \sin ^6(c+d x)+60 \sin ^7(c+d x)\right )}{420 d} \]

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(420*Log[Sin[c + d*x]] + 1260*Sin[c + d*x] + 210*Sin[c + d*x]^2 - 700*Sin[c + d*x]^3 - 525*Sin[c + d*x]^4
+ 84*Sin[c + d*x]^5 + 210*Sin[c + d*x]^6 + 60*Sin[c + d*x]^7))/(420*d)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\frac {a^{3} \left (6720 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6720 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-420+945 \cos \left (2 d x +2 c \right )-15 \sin \left (7 d x +7 c \right )+189 \sin \left (5 d x +5 c \right )-105 \cos \left (6 d x +6 c \right )+13125 \sin \left (d x +c \right )+2065 \sin \left (3 d x +3 c \right )-420 \cos \left (4 d x +4 c \right )\right )}{6720 d}\)	\(111\)
derivativedivides	\(\frac {a^{3} \left (-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2}+\frac {3 a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{3} \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(131\)
default	\(\frac {a^{3} \left (-\frac {\left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{2}+\frac {3 a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{3} \left (\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(131\)
risch	\(-i a^{3} x -\frac {2 i a^{3} c}{d}+\frac {9 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}+\frac {9 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {125 a^{3} \sin \left (d x +c \right )}{64 d}-\frac {a^{3} \sin \left (7 d x +7 c \right )}{448 d}-\frac {a^{3} \cos \left (6 d x +6 c \right )}{64 d}+\frac {9 a^{3} \sin \left (5 d x +5 c \right )}{320 d}-\frac {a^{3} \cos \left (4 d x +4 c \right )}{16 d}+\frac {59 a^{3} \sin \left (3 d x +3 c \right )}{192 d}\)	\(171\)
norman	\(\frac {\frac {2 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {68 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {646 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2488 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {646 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {68 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {6 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {10 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {10 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(303\)

[In]

int(cos(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/6720*a^3*(6720*ln(tan(1/2*d*x+1/2*c))-6720*ln(sec(1/2*d*x+1/2*c)^2)-420+945*cos(2*d*x+2*c)-15*sin(7*d*x+7*c)
+189*sin(5*d*x+5*c)-105*cos(6*d*x+6*c)+13125*sin(d*x+c)+2065*sin(3*d*x+3*c)-420*cos(4*d*x+4*c))/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {210 \, a^{3} \cos \left (d x + c\right )^{6} - 105 \, a^{3} \cos \left (d x + c\right )^{4} - 210 \, a^{3} \cos \left (d x + c\right )^{2} - 420 \, a^{3} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (15 \, a^{3} \cos \left (d x + c\right )^{6} - 66 \, a^{3} \cos \left (d x + c\right )^{4} - 88 \, a^{3} \cos \left (d x + c\right )^{2} - 176 \, a^{3}\right )} \sin \left (d x + c\right )}{420 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/420*(210*a^3*cos(d*x + c)^6 - 105*a^3*cos(d*x + c)^4 - 210*a^3*cos(d*x + c)^2 - 420*a^3*log(1/2*sin(d*x + c
)) + 4*(15*a^3*cos(d*x + c)^6 - 66*a^3*cos(d*x + c)^4 - 88*a^3*cos(d*x + c)^2 - 176*a^3)*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.78 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {60 \, a^{3} \sin \left (d x + c\right )^{7} + 210 \, a^{3} \sin \left (d x + c\right )^{6} + 84 \, a^{3} \sin \left (d x + c\right )^{5} - 525 \, a^{3} \sin \left (d x + c\right )^{4} - 700 \, a^{3} \sin \left (d x + c\right )^{3} + 210 \, a^{3} \sin \left (d x + c\right )^{2} + 420 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 1260 \, a^{3} \sin \left (d x + c\right )}{420 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/420*(60*a^3*sin(d*x + c)^7 + 210*a^3*sin(d*x + c)^6 + 84*a^3*sin(d*x + c)^5 - 525*a^3*sin(d*x + c)^4 - 700*a
^3*sin(d*x + c)^3 + 210*a^3*sin(d*x + c)^2 + 420*a^3*log(sin(d*x + c)) + 1260*a^3*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {60 \, a^{3} \sin \left (d x + c\right )^{7} + 210 \, a^{3} \sin \left (d x + c\right )^{6} + 84 \, a^{3} \sin \left (d x + c\right )^{5} - 525 \, a^{3} \sin \left (d x + c\right )^{4} - 700 \, a^{3} \sin \left (d x + c\right )^{3} + 210 \, a^{3} \sin \left (d x + c\right )^{2} + 420 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 1260 \, a^{3} \sin \left (d x + c\right )}{420 \, d} \]

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/420*(60*a^3*sin(d*x + c)^7 + 210*a^3*sin(d*x + c)^6 + 84*a^3*sin(d*x + c)^5 - 525*a^3*sin(d*x + c)^4 - 700*a
^3*sin(d*x + c)^3 + 210*a^3*sin(d*x + c)^2 + 420*a^3*log(abs(sin(d*x + c))) + 1260*a^3*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 10.01 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.30 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {176\,a^3\,\sin \left (c+d\,x\right )}{105\,d}-\frac {a^3\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^4}{4\,d}-\frac {a^3\,{\cos \left (c+d\,x\right )}^6}{2\,d}+\frac {88\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{105\,d}+\frac {22\,a^3\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{35\,d}-\frac {a^3\,{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )}{7\,d} \]

[In]

int((cos(c + d*x)^5*(a + a*sin(c + d*x))^3)/sin(c + d*x),x)

[Out]

(176*a^3*sin(c + d*x))/(105*d) - (a^3*log(1/cos(c/2 + (d*x)/2)^2))/d + (a^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2)))/d + (a^3*cos(c + d*x)^2)/(2*d) + (a^3*cos(c + d*x)^4)/(4*d) - (a^3*cos(c + d*x)^6)/(2*d) + (88*a^3*c
os(c + d*x)^2*sin(c + d*x))/(105*d) + (22*a^3*cos(c + d*x)^4*sin(c + d*x))/(35*d) - (a^3*cos(c + d*x)^6*sin(c
+ d*x))/(7*d)