\(\int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [556]

Optimal result

Integrand size = 27, antiderivative size = 68 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \log (1+\sin (c+d x))}{a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {\sin ^3(c+d x)}{3 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 68, 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, 78} \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-4*Log[1 + Sin[c + d*x]])/(a^3*d) + (4*Sin[c + d*x])/(a^3*d) - (3*Sin[c + d*x]^2)/(2*a^3*d) + Sin[c + d*x]^3/
(3*a^3*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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {15-384 \log (1+\sin (c+d x))+384 \sin (c+d x)-144 \sin ^2(c+d x)+32 \sin ^3(c+d x)}{96 a^3 d} \]

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(15 - 384*Log[1 + Sin[c + d*x]] + 384*Sin[c + d*x] - 144*Sin[c + d*x]^2 + 32*Sin[c + d*x]^3)/(96*a^3*d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(48\)
default	\(\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(48\)
parallelrisch	\(\frac {-96 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+48 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9+51 \sin \left (d x +c \right )-\sin \left (3 d x +3 c \right )+9 \cos \left (2 d x +2 c \right )}{12 d \,a^{3}}\)	\(69\)
risch	\(\frac {4 i x}{a^{3}}-\frac {17 i {\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {17 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {8 i c}{d \,a^{3}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}-\frac {\sin \left (3 d x +3 c \right )}{12 d \,a^{3}}+\frac {3 \cos \left (2 d x +2 c \right )}{4 d \,a^{3}}\)	\(110\)
norman	\(\frac {\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {8 \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {34 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {34 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {278 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {278 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {628 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {628 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1124 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1124 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1706 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1706 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2266 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2266 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2576 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2576 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}+\frac {4 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\)	\(379\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {9 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 13\right )} \sin \left (d x + c\right ) - 24 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{6 \, a^{3} d} \]

[In]

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

[Out]

1/6*(9*cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 13)*sin(d*x + c) - 24*log(sin(d*x + c) + 1))/(a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1102 vs. \(2 (61) = 122\).

Time = 35.16 (sec) , antiderivative size = 1102, normalized size of antiderivative = 16.21 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-24*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**6/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2
+ d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) - 72*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(3*a
**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) - 72*log(t
an(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3
*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) - 24*log(tan(c/2 + d*x/2) + 1)/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan
(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 12*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)
**6/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) +
36*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)*
*4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 36*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**2/(3*a**3*d*
tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 12*log(tan(c/2
+ d*x/2)**2 + 1)/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2
+ 3*a**3*d) + 24*tan(c/2 + d*x/2)**5/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*t
an(c/2 + d*x/2)**2 + 3*a**3*d) - 18*tan(c/2 + d*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x
/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 56*tan(c/2 + d*x/2)**3/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a
**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) - 18*tan(c/2 + d*x/2)**2/(3*a**3*d*tan(c/
2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 24*tan(c/2 + d*x/2)/
(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d), Ne(d,
0)), (x*sin(c)*cos(c)**5/(a*sin(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {2 \, \sin \left (d x + c\right )^{3} - 9 \, \sin \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right )}{a^{3}} - \frac {24 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*((2*sin(d*x + c)^3 - 9*sin(d*x + c)^2 + 24*sin(d*x + c))/a^3 - 24*log(sin(d*x + c) + 1)/a^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (64) = 128\).

Time = 0.35 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {6 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac {12 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {11 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 28 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}\right )}}{3 \, d} \]

[In]

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

[Out]

2/3*(6*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 12*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (11*tan(1/2*d*x + 1/2
*c)^6 - 12*tan(1/2*d*x + 1/2*c)^5 + 42*tan(1/2*d*x + 1/2*c)^4 - 28*tan(1/2*d*x + 1/2*c)^3 + 42*tan(1/2*d*x + 1
/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 11)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\sin \left (c+d\,x\right )}{a^3}+\frac {3\,{\sin \left (c+d\,x\right )}^2}{2\,a^3}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a^3}}{d} \]

[In]

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

[Out]

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