3.232 \(\int \frac{\cos (c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=87 \[ \frac{\sin ^3(c+d x)}{3 a^2 d}-\frac{\sin ^2(c+d x)}{a^2 d}+\frac{3 \sin (c+d x)}{a^2 d}-\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{4 \log (\sin (c+d x)+1)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0885047, antiderivative size = 87, 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} \[ \frac{\sin ^3(c+d x)}{3 a^2 d}-\frac{\sin ^2(c+d x)}{a^2 d}+\frac{3 \sin (c+d x)}{a^2 d}-\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{4 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.564839, size = 73, normalized size = 0.84 \[ \frac{\sin ^3(c+d x)-3 \sin ^2(c+d x)+9 \sin (c+d x)-12 \log (\sin (c+d x)+1)-\frac{3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-12*Log[1 + Sin[c + d*x]] - 3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 9*Sin[c + d*x] - 3*Sin[c + d*x]^2 + S
in[c + d*x]^3)/(3*a^2*d)

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Maple [A]  time = 0.034, size = 83, normalized size = 1. \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,{a}^{2}d}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{{a}^{2}d}}+3\,{\frac{\sin \left ( dx+c \right ) }{{a}^{2}d}}-{\frac{1}{{a}^{2}d \left ( 1+\sin \left ( dx+c \right ) \right ) }}-4\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{{a}^{2}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))^2,x)

[Out]

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

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Maxima [A]  time = 1.12249, size = 95, normalized size = 1.09 \begin{align*} -\frac{\frac{3}{a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right )}{a^{2}} + \frac{12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}}}{3 \, 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))^2,x, algorithm="maxima")

[Out]

-1/3*(3/(a^2*sin(d*x + c) + a^2) - (sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 9*sin(d*x + c))/a^2 + 12*log(sin(d*x +
c) + 1)/a^2)/d

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Fricas [A]  time = 1.46774, size = 220, normalized size = 2.53 \begin{align*} \frac{2 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2} - 24 \,{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (4 \, \cos \left (d x + c\right )^{2} + 17\right )} \sin \left (d x + c\right ) + 11}{6 \,{\left (a^{2} d \sin \left (d x + c\right ) + a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/6*(2*cos(d*x + c)^4 - 16*cos(d*x + c)^2 - 24*(sin(d*x + c) + 1)*log(sin(d*x + c) + 1) + (4*cos(d*x + c)^2 +
17)*sin(d*x + c) + 11)/(a^2*d*sin(d*x + c) + a^2*d)

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Sympy [A]  time = 4.7202, size = 238, normalized size = 2.74 \begin{align*} \begin{cases} - \frac{12 \log{\left (\sin{\left (c + d x \right )} + 1 \right )} \sin{\left (c + d x \right )}}{3 a^{2} d \sin{\left (c + d x \right )} + 3 a^{2} d} - \frac{12 \log{\left (\sin{\left (c + d x \right )} + 1 \right )}}{3 a^{2} d \sin{\left (c + d x \right )} + 3 a^{2} d} + \frac{\sin ^{4}{\left (c + d x \right )}}{3 a^{2} d \sin{\left (c + d x \right )} + 3 a^{2} d} + \frac{8 \sin ^{2}{\left (c + d x \right )}}{3 a^{2} d \sin{\left (c + d x \right )} + 3 a^{2} d} + \frac{2 \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 a^{2} d \sin{\left (c + d x \right )} + 3 a^{2} d} + \frac{2 \cos ^{2}{\left (c + d x \right )}}{3 a^{2} d \sin{\left (c + d x \right )} + 3 a^{2} d} - \frac{12}{3 a^{2} d \sin{\left (c + d x \right )} + 3 a^{2} 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 )^{2}} & \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))**2,x)

[Out]

Piecewise((-12*log(sin(c + d*x) + 1)*sin(c + d*x)/(3*a**2*d*sin(c + d*x) + 3*a**2*d) - 12*log(sin(c + d*x) + 1
)/(3*a**2*d*sin(c + d*x) + 3*a**2*d) + sin(c + d*x)**4/(3*a**2*d*sin(c + d*x) + 3*a**2*d) + 8*sin(c + d*x)**2/
(3*a**2*d*sin(c + d*x) + 3*a**2*d) + 2*sin(c + d*x)*cos(c + d*x)**2/(3*a**2*d*sin(c + d*x) + 3*a**2*d) + 2*cos
(c + d*x)**2/(3*a**2*d*sin(c + d*x) + 3*a**2*d) - 12/(3*a**2*d*sin(c + d*x) + 3*a**2*d), Ne(d, 0)), (x*sin(c)*
*4*cos(c)/(a*sin(c) + a)**2, True))

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Giac [A]  time = 1.26145, size = 144, normalized size = 1.66 \begin{align*} -\frac{\frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}{\left (\frac{6 \, a}{a \sin \left (d x + c\right ) + a} - \frac{18 \, a^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} - 1\right )}}{a^{5}} - \frac{12 \, \log \left (\frac{{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left | a \right |}}\right )}{a^{2}} + \frac{3}{{\left (a \sin \left (d x + c\right ) + a\right )} a}}{3 \, 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))^2,x, algorithm="giac")

[Out]

-1/3*((a*sin(d*x + c) + a)^3*(6*a/(a*sin(d*x + c) + a) - 18*a^2/(a*sin(d*x + c) + a)^2 - 1)/a^5 - 12*log(abs(a
*sin(d*x + c) + a)/((a*sin(d*x + c) + a)^2*abs(a)))/a^2 + 3/((a*sin(d*x + c) + a)*a))/d