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

Optimal. Leaf size=47 \[ \frac{(a-a \sin (c+d x))^4}{4 a^5 d}-\frac{2 (a-a \sin (c+d x))^3}{3 a^4 d} \]

[Out]

(-2*(a - a*Sin[c + d*x])^3)/(3*a^4*d) + (a - a*Sin[c + d*x])^4/(4*a^5*d)

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Rubi [A]  time = 0.0562803, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2667, 43} \[ \frac{(a-a \sin (c+d x))^4}{4 a^5 d}-\frac{2 (a-a \sin (c+d x))^3}{3 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - a*Sin[c + d*x])^3)/(3*a^4*d) + (a - a*Sin[c + d*x])^4/(4*a^5*d)

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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

Mathematica [A]  time = 0.0932865, size = 46, normalized size = 0.98 \[ \frac{\sin (c+d x) \left (3 \sin ^3(c+d x)-4 \sin ^2(c+d x)-6 \sin (c+d x)+12\right )}{12 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]*(12 - 6*Sin[c + d*x] - 4*Sin[c + d*x]^2 + 3*Sin[c + d*x]^3))/(12*a*d)

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Maple [A]  time = 0.05, size = 45, normalized size = 1. \begin{align*}{\frac{1}{da} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2}}+\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.930849, size = 63, normalized size = 1.34 \begin{align*} \frac{3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} + 12 \, \sin \left (d x + c\right )}{12 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67223, size = 93, normalized size = 1.98 \begin{align*} \frac{3 \, \cos \left (d x + c\right )^{4} + 4 \,{\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{12 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 126.403, size = 779, normalized size = 16.57 \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)**5/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((-7*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**8 + 60*a*d*tan(c/2 + d*x/2)**6 + 90*a*d*tan(c/2 +
d*x/2)**4 + 60*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*tan(c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**8 + 60*a*d
*tan(c/2 + d*x/2)**6 + 90*a*d*tan(c/2 + d*x/2)**4 + 60*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 58*tan(c/2 + d*x/2)
**6/(15*a*d*tan(c/2 + d*x/2)**8 + 60*a*d*tan(c/2 + d*x/2)**6 + 90*a*d*tan(c/2 + d*x/2)**4 + 60*a*d*tan(c/2 + d
*x/2)**2 + 15*a*d) + 50*tan(c/2 + d*x/2)**5/(15*a*d*tan(c/2 + d*x/2)**8 + 60*a*d*tan(c/2 + d*x/2)**6 + 90*a*d*
tan(c/2 + d*x/2)**4 + 60*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 42*tan(c/2 + d*x/2)**4/(15*a*d*tan(c/2 + d*x/2)**
8 + 60*a*d*tan(c/2 + d*x/2)**6 + 90*a*d*tan(c/2 + d*x/2)**4 + 60*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 50*tan(c/
2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**8 + 60*a*d*tan(c/2 + d*x/2)**6 + 90*a*d*tan(c/2 + d*x/2)**4 + 60*a*d*t
an(c/2 + d*x/2)**2 + 15*a*d) - 58*tan(c/2 + d*x/2)**2/(15*a*d*tan(c/2 + d*x/2)**8 + 60*a*d*tan(c/2 + d*x/2)**6
+ 90*a*d*tan(c/2 + d*x/2)**4 + 60*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*tan(c/2 + d*x/2)/(15*a*d*tan(c/2 + d
*x/2)**8 + 60*a*d*tan(c/2 + d*x/2)**6 + 90*a*d*tan(c/2 + d*x/2)**4 + 60*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 7/
(15*a*d*tan(c/2 + d*x/2)**8 + 60*a*d*tan(c/2 + d*x/2)**6 + 90*a*d*tan(c/2 + d*x/2)**4 + 60*a*d*tan(c/2 + d*x/2
)**2 + 15*a*d), Ne(d, 0)), (x*cos(c)**5/(a*sin(c) + a), True))

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Giac [A]  time = 1.14575, size = 63, normalized size = 1.34 \begin{align*} \frac{3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} + 12 \, \sin \left (d x + c\right )}{12 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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