3.241 \(\int \frac{\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=93 \[ \frac{\sin ^2(c+d x)}{2 a^3 d}-\frac{3 \sin (c+d x)}{a^3 d}+\frac{4}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{6 \log (\sin (c+d x)+1)}{a^3 d}-\frac{1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.0963241, antiderivative size = 93, 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 = {2833, 12, 43} \[ \frac{\sin ^2(c+d x)}{2 a^3 d}-\frac{3 \sin (c+d x)}{a^3 d}+\frac{4}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac{6 \log (\sin (c+d x)+1)}{a^3 d}-\frac{1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

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])

Rubi steps

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

Mathematica [A]  time = 2.13367, size = 78, normalized size = 0.84 \[ \frac{8 \sin ^2(c+d x)+\left (\frac{64}{(\sin (c+d x)+1)^2}-48\right ) \sin (c+d x)+96 \log (\sin (c+d x)+1)+\frac{56}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}}{16 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]

[Out]

(96*Log[1 + Sin[c + d*x]] + 56/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 8*Sin[c + d*x]^2 + Sin[c + d*x]*(-48
+ 64/(1 + Sin[c + d*x])^2))/(16*a^3*d)

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Maple [A]  time = 0.039, size = 85, normalized size = 0.9 \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,{a}^{3}d}}-3\,{\frac{\sin \left ( dx+c \right ) }{{a}^{3}d}}-{\frac{1}{2\,{a}^{3}d \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+4\,{\frac{1}{{a}^{3}d \left ( 1+\sin \left ( dx+c \right ) \right ) }}+6\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{{a}^{3}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x)

[Out]

1/2*sin(d*x+c)^2/a^3/d-3*sin(d*x+c)/a^3/d-1/2/d/a^3/(1+sin(d*x+c))^2+4/d/a^3/(1+sin(d*x+c))+6*ln(1+sin(d*x+c))
/a^3/d

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Maxima [A]  time = 1.13208, size = 109, normalized size = 1.17 \begin{align*} \frac{\frac{8 \, \sin \left (d x + c\right ) + 7}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac{\sin \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right )}{a^{3}} + \frac{12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*((8*sin(d*x + c) + 7)/(a^3*sin(d*x + c)^2 + 2*a^3*sin(d*x + c) + a^3) + (sin(d*x + c)^2 - 6*sin(d*x + c))/
a^3 + 12*log(sin(d*x + c) + 1)/a^3)/d

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Fricas [A]  time = 1.50938, size = 284, normalized size = 3.05 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{4} + 19 \, \cos \left (d x + c\right )^{2} - 24 \,{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (4 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) - 8}{4 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(2*cos(d*x + c)^4 + 19*cos(d*x + c)^2 - 24*(cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d*x + c) + 1) +
2*(4*cos(d*x + c)^2 - 3)*sin(d*x + c) - 8)/(a^3*d*cos(d*x + c)^2 - 2*a^3*d*sin(d*x + c) - 2*a^3*d)

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Sympy [A]  time = 7.04268, size = 456, normalized size = 4.9 \begin{align*} \begin{cases} \frac{12 \log{\left (\sin{\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{24 \log{\left (\sin{\left (c + d x \right )} + 1 \right )} \sin{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{12 \log{\left (\sin{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{3 \sin ^{4}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{2 \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{4 \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{20 \sin{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{2 \cos ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} + \frac{16}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin{\left (c + d x \right )} + 2 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{4}{\left (c \right )} \cos{\left (c \right )}}{\left (a \sin{\left (c \right )} + a\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((12*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3
*d) + 24*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 12
*log(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 3*sin(c + d*x)**4/(2*a*
*3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 2*sin(c + d*x)**2*cos(c + d*x)**2/(2*a**3*d*sin(c +
d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 4*sin(c + d*x)*cos(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a*
*3*d*sin(c + d*x) + 2*a**3*d) + 20*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d)
+ 2*cos(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 16/(2*a**3*d*sin(c + d*x)*
*2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Ne(d, 0)), (x*sin(c)**4*cos(c)/(a*sin(c) + a)**3, True))

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Giac [A]  time = 1.19862, size = 99, normalized size = 1.06 \begin{align*} \frac{\frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac{8 \, \sin \left (d x + c\right ) + 7}{a^{3}{\left (\sin \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{2} - 6 \, a^{3} \sin \left (d x + c\right )}{a^{6}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(12*log(abs(sin(d*x + c) + 1))/a^3 + (8*sin(d*x + c) + 7)/(a^3*(sin(d*x + c) + 1)^2) + (a^3*sin(d*x + c)^2
- 6*a^3*sin(d*x + c))/a^6)/d