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

Optimal. Leaf size=80 \[ \frac {9 \cos (c+d x)}{2 a^3 d}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {9 x}{2 a^3}+\frac {\cos ^5(c+d x)}{d (a \sin (c+d x)+a)^3} \]

[Out]

9/2*x/a^3+9/2*cos(d*x+c)/a^3/d+cos(d*x+c)^5/d/(a+a*sin(d*x+c))^3+3/2*cos(d*x+c)^3/d/(a^3+a^3*sin(d*x+c))

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Rubi [A]  time = 0.14, antiderivative size = 80, normalized size of antiderivative = 1.00, 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, 2679, 2682, 8} \[ \frac {9 \cos (c+d x)}{2 a^3 d}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {9 x}{2 a^3}+\frac {\cos ^5(c+d x)}{d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*x)/(2*a^3) + (9*Cos[c + d*x])/(2*a^3*d) + Cos[c + d*x]^5/(d*(a + a*Sin[c + d*x])^3) + (3*Cos[c + d*x]^3)/(2
*d*(a^3 + a^3*Sin[c + d*x]))

Rule 8

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

Rule 2679

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

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 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]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx}{a}\\ &=\frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {9 \int \frac {\cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{2 a^2}\\ &=\frac {9 \cos (c+d x)}{2 a^3 d}+\frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {9 \int 1 \, dx}{2 a^3}\\ &=\frac {9 x}{2 a^3}+\frac {9 \cos (c+d x)}{2 a^3 d}+\frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.64, size = 143, normalized size = 1.79 \[ \frac {180 d x \sin \left (c+\frac {d x}{2}\right )+55 \sin \left (2 c+\frac {3 d x}{2}\right )-5 \sin \left (2 c+\frac {5 d x}{2}\right )+59 \cos \left (c+\frac {d x}{2}\right )+55 \cos \left (c+\frac {3 d x}{2}\right )+5 \cos \left (3 c+\frac {5 d x}{2}\right )-381 \sin \left (\frac {d x}{2}\right )+180 d x \cos \left (\frac {d x}{2}\right )}{40 a^3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(180*d*x*Cos[(d*x)/2] + 59*Cos[c + (d*x)/2] + 55*Cos[c + (3*d*x)/2] + 5*Cos[3*c + (5*d*x)/2] - 381*Sin[(d*x)/2
] + 180*d*x*Sin[c + (d*x)/2] + 55*Sin[2*c + (3*d*x)/2] - 5*Sin[2*c + (5*d*x)/2])/(40*a^3*d*(Cos[c/2] + Sin[c/2
])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.44, size = 100, normalized size = 1.25 \[ \frac {\cos \left (d x + c\right )^{3} + 9 \, d x + {\left (9 \, d x + 13\right )} \cos \left (d x + c\right ) + 6 \, \cos \left (d x + c\right )^{2} + {\left (9 \, d x - \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) + 8}{2 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/2*(cos(d*x + c)^3 + 9*d*x + (9*d*x + 13)*cos(d*x + c) + 6*cos(d*x + c)^2 + (9*d*x - cos(d*x + c)^2 + 5*cos(d
*x + c) - 8)*sin(d*x + c) + 8)/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

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giac [A]  time = 0.20, size = 91, normalized size = 1.14 \[ \frac {\frac {9 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} + \frac {16}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{2 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/2*(9*(d*x + c)/a^3 + 2*(tan(1/2*d*x + 1/2*c)^3 + 6*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 6)/((tan(
1/2*d*x + 1/2*c)^2 + 1)^2*a^3) + 16/(a^3*(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [B]  time = 0.40, size = 163, normalized size = 2.04 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {6}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {8}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

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))^3,x)

[Out]

1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^
2-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2+9/d/a^3*arctan(tan(
1/2*d*x+1/2*c))+8/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.42, size = 225, normalized size = 2.81 \[ \frac {\frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 14}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{d} \]

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))^3,x, algorithm="maxima")

[Out]

((5*sin(d*x + c)/(cos(d*x + c) + 1) + 21*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 7*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14)/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a^3*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^3*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 10.80, size = 94, normalized size = 1.18 \[ \frac {9\,x}{2\,a^3}+\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+14}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*x)/(2*a^3) + (5*tan(c/2 + (d*x)/2) + 21*tan(c/2 + (d*x)/2)^2 + 7*tan(c/2 + (d*x)/2)^3 + 9*tan(c/2 + (d*x)/2
)^4 + 14)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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sympy [A]  time = 61.27, size = 1244, normalized size = 15.55 \[ \text {result too large to display} \]

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))**3,x)

[Out]

Piecewise((9*d*x*tan(c/2 + d*x/2)**5/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*t
an(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 9*d*x*tan(c/2 + d*
x/2)**4/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d
*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 18*d*x*tan(c/2 + d*x/2)**3/(2*a**3*d*tan(c/2 +
d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**
3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 18*d*x*tan(c/2 + d*x/2)**2/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2
+ d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**
3*d) + 9*d*x*tan(c/2 + d*x/2)/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2
+ d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 9*d*x/(2*a**3*d*tan(c/2 +
d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a*
*3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 18*tan(c/2 + d*x/2)**4/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 +
d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d
) + 14*tan(c/2 + d*x/2)**3/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d
*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 42*tan(c/2 + d*x/2)**2/(2*a*
*3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*
x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 10*tan(c/2 + d*x/2)/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d
*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2)
+ 2*a**3*d) + 28/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3
+ 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d), Ne(d, 0)), (x*sin(c)*cos(c)**4/(a*sin(
c) + a)**3, True))

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