3.556 \(\int \frac{\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)
Optimal. Leaf size=68 \[ \frac{\sin ^3(c+d x)}{3 a^3 d}-\frac{3 \sin ^2(c+d x)}{2 a^3 d}+\frac{4 \sin (c+d x)}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]
[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)
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Rubi [A] time = 0.0763898, antiderivative size = 68, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used =
{2836, 12, 77} \[ \frac{\sin ^3(c+d x)}{3 a^3 d}-\frac{3 \sin ^2(c+d x)}{2 a^3 d}+\frac{4 \sin (c+d x)}{a^3 d}-\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]
Antiderivative was successfully verified.
[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 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 77
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]))))
Rubi steps
\begin{align*} \int \frac{\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x}{a (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^6 d}\\ &=\frac{\operatorname{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] time = 0.341803, size = 51, normalized size = 0.75 \[ \frac{32 \sin ^3(c+d x)-144 \sin ^2(c+d x)+384 \sin (c+d x)-384 \log (\sin (c+d x)+1)+15}{96 a^3 d} \]
Antiderivative was successfully verified.
[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)
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Maple [A] time = 0.084, size = 65, 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}}-{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,{a}^{3}d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,{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)/(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-3/2*sin(d*x+c)^2/a^3/d+1/3*sin(d*x+c)^3/a^3/d
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Maxima [A] time = 1.0209, size = 72, normalized size = 1.06 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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
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Fricas [A] time = 1.0833, size = 132, normalized size = 1.94 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Sympy [A] time = 129.94, size = 1243, normalized size = 18.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**5*sin(d*x+c)/(a+a*sin(d*x+c))**3,x)
[Out]
Piecewise((-480*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**6/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(
c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 1440*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2
)**4/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**
3*d) - 1440*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2
+ d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 480*log(tan(c/2 + d*x/2) + 1)/(60*a**3*d*tan(c/2 +
d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 240*log(tan(c/2 +
d*x/2)**2 + 1)*tan(c/2 + d*x/2)**6/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*
d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 720*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 +
d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 720*log(tan(c/2 +
d*x/2)**2 + 1)*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*
d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 240*log(tan(c/2 + d*x/2)**2 + 1)/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**
3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 157*tan(c/2 + d*x/2)**6/(60*a**3*d*tan
(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 480*tan(c/2
+ d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 +
60*a**3*d) - 831*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a*
*3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 1120*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*t
an(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 831*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2
+ d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 480*tan(c/2 + d*x
/2)/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3
*d) - 157/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 6
0*a**3*d), Ne(d, 0)), (x*sin(c)*cos(c)**5/(a*sin(c) + a)**3, True))
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Giac [B] time = 1.33103, size = 190, normalized size = 2.79 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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