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

Optimal. Leaf size=82 \[ \frac{\sin ^4(c+d x)}{4 a^3 d}-\frac{\sin ^3(c+d x)}{a^3 d}+\frac{2 \sin ^2(c+d x)}{a^3 d}-\frac{4 \sin (c+d x)}{a^3 d}+\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

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

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Rubi [A]  time = 0.118477, antiderivative size = 82, 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} \[ \frac{\sin ^4(c+d x)}{4 a^3 d}-\frac{\sin ^3(c+d x)}{a^3 d}+\frac{2 \sin ^2(c+d x)}{a^3 d}-\frac{4 \sin (c+d x)}{a^3 d}+\frac{4 \log (\sin (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(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) - Sin[c + d*x]^3/(a^
3*d) + Sin[c + d*x]^4/(4*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 ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^2}{a^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^2}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-4 a^3+4 a^2 x-3 a x^2+x^3+\frac{4 a^4}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 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{\sin ^3(c+d x)}{a^3 d}+\frac{\sin ^4(c+d x)}{4 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.932708, size = 59, normalized size = 0.72 \[ \frac{-152 \sin (c+d x)+8 \sin (3 (c+d x))-36 \cos (2 (c+d x))+\cos (4 (c+d x))+128 \log (\sin (c+d x)+1)+35}{32 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35 - 36*Cos[2*(c + d*x)] + Cos[4*(c + d*x)] + 128*Log[1 + Sin[c + d*x]] - 152*Sin[c + d*x] + 8*Sin[3*(c + d*x
)])/(32*a^3*d)

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Maple [A]  time = 0.102, size = 81, 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{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{{a}^{3}d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,{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)^2/(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-sin(d*x+c)^3/a^3/d+1/4*sin(d*x+c)^4/a^3/d

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Maxima [A]  time = 1.07935, size = 82, normalized size = 1. \begin{align*} \frac{\frac{\sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 8 \, \sin \left (d x + c\right )^{2} - 16 \, \sin \left (d x + c\right )}{a^{3}} + \frac{16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((sin(d*x + c)^4 - 4*sin(d*x + c)^3 + 8*sin(d*x + c)^2 - 16*sin(d*x + c))/a^3 + 16*log(sin(d*x + c) + 1)/a
^3)/d

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Fricas [A]  time = 1.1258, size = 155, normalized size = 1.89 \begin{align*} \frac{\cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 4 \,{\left (\cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{4 \, 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)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 5)*sin(d*x + c) + 16*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)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.32593, size = 225, normalized size = 2.74 \begin{align*} -\frac{\frac{12 \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac{24 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{25 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 124 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 96 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 210 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 96 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 124 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 25}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{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)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/3*(12*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 24*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (25*tan(1/2*d*x + 1
/2*c)^8 - 24*tan(1/2*d*x + 1/2*c)^7 + 124*tan(1/2*d*x + 1/2*c)^6 - 96*tan(1/2*d*x + 1/2*c)^5 + 210*tan(1/2*d*x
+ 1/2*c)^4 - 96*tan(1/2*d*x + 1/2*c)^3 + 124*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 25)/((tan(1/2
*d*x + 1/2*c)^2 + 1)^4*a^3))/d