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

Optimal. Leaf size=70 \[ -\frac{2 \cos ^3(c+d x)}{3 a^2 d}-\frac{\sin (c+d x) \cos (c+d x)}{a^2 d}-\frac{x}{a^2}-\frac{\cos ^5(c+d x)}{d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.109522, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2859, 2682, 2635, 8} \[ -\frac{2 \cos ^3(c+d x)}{3 a^2 d}-\frac{\sin (c+d x) \cos (c+d x)}{a^2 d}-\frac{x}{a^2}-\frac{\cos ^5(c+d x)}{d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [B]  time = 0.747531, size = 204, normalized size = 2.91 \[ \frac{-24 d x \sin \left (\frac{c}{2}\right )+21 \sin \left (\frac{c}{2}+d x\right )-21 \sin \left (\frac{3 c}{2}+d x\right )+6 \sin \left (\frac{3 c}{2}+2 d x\right )+6 \sin \left (\frac{5 c}{2}+2 d x\right )-\sin \left (\frac{5 c}{2}+3 d x\right )+\sin \left (\frac{7 c}{2}+3 d x\right )-2 \cos \left (\frac{c}{2}\right ) (12 d x+1)-21 \cos \left (\frac{c}{2}+d x\right )-21 \cos \left (\frac{3 c}{2}+d x\right )+6 \cos \left (\frac{3 c}{2}+2 d x\right )-6 \cos \left (\frac{5 c}{2}+2 d x\right )+\cos \left (\frac{5 c}{2}+3 d x\right )+\cos \left (\frac{7 c}{2}+3 d x\right )+2 \sin \left (\frac{c}{2}\right )}{24 a^2 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + 12*d*x)*Cos[c/2] - 21*Cos[c/2 + d*x] - 21*Cos[(3*c)/2 + d*x] + 6*Cos[(3*c)/2 + 2*d*x] - 6*Cos[(5*c)/2
+ 2*d*x] + Cos[(5*c)/2 + 3*d*x] + Cos[(7*c)/2 + 3*d*x] + 2*Sin[c/2] - 24*d*x*Sin[c/2] + 21*Sin[c/2 + d*x] - 2
1*Sin[(3*c)/2 + d*x] + 6*Sin[(3*c)/2 + 2*d*x] + 6*Sin[(5*c)/2 + 2*d*x] - Sin[(5*c)/2 + 3*d*x] + Sin[(7*c)/2 +
3*d*x])/(24*a^2*d*(Cos[c/2] + Sin[c/2]))

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Maple [B]  time = 0.089, size = 177, normalized size = 2.5 \begin{align*} -2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-8\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{10}{3\,d{a}^{2}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^2,x)

[Out]

-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)
^4-8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*
c)-10/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3-2/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.67655, size = 248, normalized size = 3.54 \begin{align*} \frac{2 \,{\left (\frac{\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 5}{a^{2} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{3 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

2/3*((3*sin(d*x + c)/(cos(d*x + c) + 1) - 12*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 3*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a
^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) - 3*arctan(sin(d*x + c)/(cos
(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.09594, size = 115, normalized size = 1.64 \begin{align*} \frac{\cos \left (d x + c\right )^{3} - 3 \, d x + 3 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right )}{3 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 44.2726, size = 694, normalized size = 9.91 \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)**4*sin(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((-3*d*x*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*
tan(c/2 + d*x/2)**2 + 3*a**2*d) - 9*d*x*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 +
d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 9*d*x*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**
6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 3*d*x/(3*a**2*d*tan(c/2 + d*x/2)
**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 6*tan(c/2 + d*x/2)**5/(3*a**2*
d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 6*tan(c/2 +
d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2
*d) - 24*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 +
d*x/2)**2 + 3*a**2*d) + 6*tan(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a
**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 10/(3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a*
*2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d), Ne(d, 0)), (x*sin(c)*cos(c)**4/(a*sin(c) + a)**2, True))

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Giac [A]  time = 1.33507, size = 119, normalized size = 1.7 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}}{a^{2}} + \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(3*(d*x + c)/a^2 + 2*(3*tan(1/2*d*x + 1/2*c)^5 + 3*tan(1/2*d*x + 1/2*c)^4 + 12*tan(1/2*d*x + 1/2*c)^2 - 3
*tan(1/2*d*x + 1/2*c) + 5)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^2))/d