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

Optimal. Leaf size=102 \[ \frac{\sin ^5(c+d x)}{5 a^3 d}-\frac{3 \sin ^4(c+d x)}{4 a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}-\frac{2 \sin ^2(c+d x)}{a^3 d}+\frac{4 \sin (c+d x)}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

[Out]

(-4*Log[1 + Sin[c + d*x]])/(a^3*d) + (4*Sin[c + d*x])/(a^3*d) - (2*Sin[c + d*x]^2)/(a^3*d) + (4*Sin[c + d*x]^3
)/(3*a^3*d) - (3*Sin[c + d*x]^4)/(4*a^3*d) + Sin[c + d*x]^5/(5*a^3*d)

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Rubi [A]  time = 0.125569, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac{\sin ^5(c+d x)}{5 a^3 d}-\frac{3 \sin ^4(c+d x)}{4 a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}-\frac{2 \sin ^2(c+d x)}{a^3 d}+\frac{4 \sin (c+d x)}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(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) - (2*Sin[c + d*x]^2)/(a^3*d) + (4*Sin[c + d*x]^3
)/(3*a^3*d) - (3*Sin[c + d*x]^4)/(4*a^3*d) + Sin[c + d*x]^5/(5*a^3*d)

Rule 2836

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*x)/b
)^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]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

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]))

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^3}{a^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^3}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (4 a^4-4 a^3 x+4 a^2 x^2-3 a x^3+x^4-\frac{4 a^5}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=-\frac{4 \log (1+\sin (c+d x))}{a^3 d}+\frac{4 \sin (c+d x)}{a^3 d}-\frac{2 \sin ^2(c+d x)}{a^3 d}+\frac{4 \sin ^3(c+d x)}{3 a^3 d}-\frac{3 \sin ^4(c+d x)}{4 a^3 d}+\frac{\sin ^5(c+d x)}{5 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.962285, size = 71, normalized size = 0.7 \[ \frac{192 \sin ^5(c+d x)-720 \sin ^4(c+d x)+1280 \sin ^3(c+d x)-1920 \sin ^2(c+d x)+3840 \sin (c+d x)-3840 \log (\sin (c+d x)+1)+45}{960 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(45 - 3840*Log[1 + Sin[c + d*x]] + 3840*Sin[c + d*x] - 1920*Sin[c + d*x]^2 + 1280*Sin[c + d*x]^3 - 720*Sin[c +
d*x]^4 + 192*Sin[c + d*x]^5)/(960*a^3*d)

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Maple [A]  time = 0.119, size = 97, normalized size = 1. \begin{align*} -4\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{{a}^{3}d}}+4\,{\frac{\sin \left ( dx+c \right ) }{{a}^{3}d}}-2\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{{a}^{3}d}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,{a}^{3}d}}-{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,{a}^{3}d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,{a}^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1279, size = 99, normalized size = 0.97 \begin{align*} \frac{\frac{12 \, \sin \left (d x + c\right )^{5} - 45 \, \sin \left (d x + c\right )^{4} + 80 \, \sin \left (d x + c\right )^{3} - 120 \, \sin \left (d x + c\right )^{2} + 240 \, \sin \left (d x + c\right )}{a^{3}} - \frac{240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*((12*sin(d*x + c)^5 - 45*sin(d*x + c)^4 + 80*sin(d*x + c)^3 - 120*sin(d*x + c)^2 + 240*sin(d*x + c))/a^3
- 240*log(sin(d*x + c) + 1)/a^3)/d

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Fricas [A]  time = 1.13642, size = 196, normalized size = 1.92 \begin{align*} -\frac{45 \, \cos \left (d x + c\right )^{4} - 210 \, \cos \left (d x + c\right )^{2} - 4 \,{\left (3 \, \cos \left (d x + c\right )^{4} - 26 \, \cos \left (d x + c\right )^{2} + 83\right )} \sin \left (d x + c\right ) + 240 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(45*cos(d*x + c)^4 - 210*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^4 - 26*cos(d*x + c)^2 + 83)*sin(d*x + c) + 2
40*log(sin(d*x + c) + 1))/(a^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.34583, size = 261, normalized size = 2.56 \begin{align*} \frac{\frac{60 \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac{120 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{137 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} - 120 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 805 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 640 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1910 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1136 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1910 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 640 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 805 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 120 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 137}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5} a^{3}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(60*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 120*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (137*tan(1/2*d*x +
1/2*c)^10 - 120*tan(1/2*d*x + 1/2*c)^9 + 805*tan(1/2*d*x + 1/2*c)^8 - 640*tan(1/2*d*x + 1/2*c)^7 + 1910*tan(1
/2*d*x + 1/2*c)^6 - 1136*tan(1/2*d*x + 1/2*c)^5 + 1910*tan(1/2*d*x + 1/2*c)^4 - 640*tan(1/2*d*x + 1/2*c)^3 + 8
05*tan(1/2*d*x + 1/2*c)^2 - 120*tan(1/2*d*x + 1/2*c) + 137)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a^3))/d