\(\int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [250]

Optimal result

Integrand size = 27, antiderivative size = 95 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {4 \log (1+\sin (c+d x))}{a^4 d}+\frac {\sin (c+d x)}{a^4 d}-\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {2}{d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {6}{d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 95, 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} \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin (c+d x)}{a^4 d}-\frac {6}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {4 \log (\sin (c+d x)+1)}{a^4 d}+\frac {2}{d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

[In]

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

[Out]

(-4*Log[1 + Sin[c + d*x]])/(a^4*d) + Sin[c + d*x]/(a^4*d) - 1/(3*a*d*(a + a*Sin[c + d*x])^3) + 2/(d*(a^2 + a^2
*Sin[c + d*x])^2) - 6/(d*(a^4 + a^4*Sin[c + d*x]))

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 {x^4}{a^4 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \frac {x^4}{(a+x)^4} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {a^4}{(a+x)^4}-\frac {4 a^3}{(a+x)^3}+\frac {6 a^2}{(a+x)^2}-\frac {4 a}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = -\frac {4 \log (1+\sin (c+d x))}{a^4 d}+\frac {\sin (c+d x)}{a^4 d}-\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {2}{d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {6}{d \left (a^4+a^4 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.15 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-13-12 \log (1+\sin (c+d x))-9 (3+4 \log (1+\sin (c+d x))) \sin (c+d x)-9 (1+4 \log (1+\sin (c+d x))) \sin ^2(c+d x)+(9-12 \log (1+\sin (c+d x))) \sin ^3(c+d x)+3 \sin ^4(c+d x)}{3 a^4 d (1+\sin (c+d x))^3} \]

[In]

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

[Out]

(-13 - 12*Log[1 + Sin[c + d*x]] - 9*(3 + 4*Log[1 + Sin[c + d*x]])*Sin[c + d*x] - 9*(1 + 4*Log[1 + Sin[c + d*x]
])*Sin[c + d*x]^2 + (9 - 12*Log[1 + Sin[c + d*x]])*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4)/(3*a^4*d*(1 + Sin[c + d*
x])^3)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {\sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {6}{1+\sin \left (d x +c \right )}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {2}{\left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\)	\(62\)
default	\(\frac {\sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {6}{1+\sin \left (d x +c \right )}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {2}{\left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\)	\(62\)
risch	\(\frac {4 i x}{a^{4}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{4}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{4}}+\frac {8 i c}{d \,a^{4}}-\frac {4 i \left (30 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}-44 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\)	\(157\)
parallelrisch	\(\frac {\left (144 \cos \left (2 d x +2 c \right )-360 \sin \left (d x +c \right )+24 \sin \left (3 d x +3 c \right )-240\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-288 \cos \left (2 d x +2 c \right )+720 \sin \left (d x +c \right )-48 \sin \left (3 d x +3 c \right )+480\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+132 \cos \left (2 d x +2 c \right )-3 \cos \left (4 d x +4 c \right )-228 \sin \left (d x +c \right )+44 \sin \left (3 d x +3 c \right )-129}{6 d \,a^{4} \left (-10+6 \cos \left (2 d x +2 c \right )+\sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )\right )}\)	\(174\)
norman	\(\frac {\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {8 \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {48 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {48 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {464 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {464 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1112 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1112 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {5288 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {5288 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2152 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2152 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {3376 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {3376 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4592 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4592 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}+\frac {4 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}\)	\(379\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.39 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} + 12 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 19}{3 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/3*(3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 + 12*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(
sin(d*x + c) + 1) - 9*(cos(d*x + c)^2 + 2)*sin(d*x + c) - 19)/(3*a^4*d*cos(d*x + c)^2 - 4*a^4*d + (a^4*d*cos(d
*x + c)^2 - 4*a^4*d)*sin(d*x + c))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (82) = 164\).

Time = 1.25 (sec) , antiderivative size = 527, normalized size of antiderivative = 5.55 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\begin {cases} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{3}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} + \frac {3 \sin ^{4}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {54 \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {22}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-12*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*
a**4*d*sin(c + d*x) + 3*a**4*d) - 36*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*
d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 36*log(sin(c + d*x) + 1)*sin(c + d*x)/(3*a**4*d*sin(c
+ d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 12*log(sin(c + d*x) + 1)/(3*a**4*d*
sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) + 3*sin(c + d*x)**4/(3*a**4*d*s
in(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 36*sin(c + d*x)**2/(3*a**4*d*s
in(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 54*sin(c + d*x)/(3*a**4*d*sin(
c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 22/(3*a**4*d*sin(c + d*x)**3 + 9*
a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d), Ne(d, 0)), (x*sin(c)**4*cos(c)/(a*sin(c) + a)**4,
True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {18 \, \sin \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) + 13}{a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} - \frac {3 \, \sin \left (d x + c\right )}{a^{4}}}{3 \, d} \]

[In]

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

[Out]

-1/3*((18*sin(d*x + c)^2 + 30*sin(d*x + c) + 13)/(a^4*sin(d*x + c)^3 + 3*a^4*sin(d*x + c)^2 + 3*a^4*sin(d*x +
c) + a^4) + 12*log(sin(d*x + c) + 1)/a^4 - 3*sin(d*x + c)/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, \sin \left (d x + c\right )}{a^{4}} + \frac {18 \, \sin \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) + 13}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

-1/3*(12*log(abs(sin(d*x + c) + 1))/a^4 - 3*sin(d*x + c)/a^4 + (18*sin(d*x + c)^2 + 30*sin(d*x + c) + 13)/(a^4
*(sin(d*x + c) + 1)^3))/d

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin \left (c+d\,x\right )}{a^4\,d}-\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^4\,d}-\frac {6\,{\sin \left (c+d\,x\right )}^2+10\,\sin \left (c+d\,x\right )+\frac {13}{3}}{a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^3} \]

[In]

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

[Out]

sin(c + d*x)/(a^4*d) - (4*log(sin(c + d*x) + 1))/(a^4*d) - (10*sin(c + d*x) + 6*sin(c + d*x)^2 + 13/3)/(a^4*d*
(sin(c + d*x) + 1)^3)