3.544 \(\int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=55 \[ \frac {\sin ^6(c+d x)}{6 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {\sin ^4(c+d x)}{4 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 55, normalized size of antiderivative = 1.00, 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, 43} \[ \frac {\sin ^6(c+d x)}{6 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {\sin ^4(c+d x)}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]^4/(4*a^2*d) - (2*Sin[c + d*x]^5)/(5*a^2*d) + Sin[c + d*x]^6/(6*a^2*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 43

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 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]

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a-x)^2 x^3}{a^3} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int (a-x)^2 x^3 \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 x^3-2 a x^4+x^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=\frac {\sin ^4(c+d x)}{4 a^2 d}-\frac {2 \sin ^5(c+d x)}{5 a^2 d}+\frac {\sin ^6(c+d x)}{6 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 38, normalized size = 0.69 \[ \frac {\sin ^4(c+d x) \left (10 \sin ^2(c+d x)-24 \sin (c+d x)+15\right )}{60 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(Sin[c + d*x]^4*(15 - 24*Sin[c + d*x] + 10*Sin[c + d*x]^2))/(60*a^2*d)

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fricas [A]  time = 0.75, size = 67, normalized size = 1.22 \[ -\frac {10 \, \cos \left (d x + c\right )^{6} - 45 \, \cos \left (d x + c\right )^{4} + 60 \, \cos \left (d x + c\right )^{2} + 24 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right )}{60 \, a^{2} d} \]

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))^2,x, algorithm="fricas")

[Out]

-1/60*(10*cos(d*x + c)^6 - 45*cos(d*x + c)^4 + 60*cos(d*x + c)^2 + 24*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*
sin(d*x + c))/(a^2*d)

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giac [A]  time = 0.23, size = 39, normalized size = 0.71 \[ \frac {10 \, \sin \left (d x + c\right )^{6} - 24 \, \sin \left (d x + c\right )^{5} + 15 \, \sin \left (d x + c\right )^{4}}{60 \, a^{2} d} \]

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))^2,x, algorithm="giac")

[Out]

1/60*(10*sin(d*x + c)^6 - 24*sin(d*x + c)^5 + 15*sin(d*x + c)^4)/(a^2*d)

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maple [A]  time = 0.38, size = 39, normalized size = 0.71 \[ \frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d \,a^{2}} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.72, size = 39, normalized size = 0.71 \[ \frac {10 \, \sin \left (d x + c\right )^{6} - 24 \, \sin \left (d x + c\right )^{5} + 15 \, \sin \left (d x + c\right )^{4}}{60 \, a^{2} d} \]

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))^2,x, algorithm="maxima")

[Out]

1/60*(10*sin(d*x + c)^6 - 24*sin(d*x + c)^5 + 15*sin(d*x + c)^4)/(a^2*d)

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mupad [B]  time = 0.06, size = 36, normalized size = 0.65 \[ \frac {{\sin \left (c+d\,x\right )}^4\,\left (10\,{\sin \left (c+d\,x\right )}^2-24\,\sin \left (c+d\,x\right )+15\right )}{60\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)

[Out]

(sin(c + d*x)^4*(10*sin(c + d*x)^2 - 24*sin(c + d*x) + 15))/(60*a^2*d)

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sympy [A]  time = 128.34, size = 682, normalized size = 12.40 \[ \begin {cases} \frac {60 \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {192 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {280 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} - \frac {192 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} + \frac {60 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{15 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 300 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 225 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 90 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 15 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\relax (c )} \cos ^{5}{\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

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))**2,x)

[Out]

Piecewise((60*tan(c/2 + d*x/2)**8/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*
d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 +
d*x/2)**2 + 15*a**2*d) - 192*tan(c/2 + d*x/2)**7/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)*
*10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a*
*2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 280*tan(c/2 + d*x/2)**6/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*ta
n(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*
x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 192*tan(c/2 + d*x/2)**5/(15*a**2*d*tan(c/2 + d*x/2)**12
+ 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2
*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 60*tan(c/2 + d*x/2)**4/(15*a**2*d*tan(c/
2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)
**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d), Ne(d, 0)), (x*sin(c)**3*cos
(c)**5/(a*sin(c) + a)**2, True))

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