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]

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

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Rubi [A]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, 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 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])

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])

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.84, size = 37, normalized size = 0.79 \[ \frac {3 \, \cos \left (d x + c\right )^{4} + 4 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{12 \, a d} \]

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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giac [A]  time = 0.35, size = 47, normalized size = 1.00 \[ \frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} + 12 \, \sin \left (d x + c\right )}{12 \, a d} \]

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)

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maple [A]  time = 0.14, size = 45, normalized size = 0.96 \[ \frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )}{d a} \]

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 = 0.31, size = 47, normalized size = 1.00 \[ \frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} + 12 \, \sin \left (d x + c\right )}{12 \, a d} \]

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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mupad [B]  time = 4.66, size = 54, normalized size = 1.15 \[ \frac {\frac {\sin \left (c+d\,x\right )}{a}-\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 18.77, size = 530, normalized size = 11.28 \[ \begin {cases} \frac {6 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {10 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {10 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{5}{\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

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((6*tan(c/2 + d*x/2)**7/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*
x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*tan(c/2 + d*x/2)**6/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(
c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 10*tan(c/2 + d*x/2)**5/(3
*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**
2 + 3*a*d) + 10*tan(c/2 + d*x/2)**3/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 +
d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/2)**8 + 12*a*d*t
an(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*tan(c/2 + d*x/2)/(3*
a*d*tan(c/2 + d*x/2)**8 + 12*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 12*a*d*tan(c/2 + d*x/2)**2
+ 3*a*d), Ne(d, 0)), (x*cos(c)**5/(a*sin(c) + a), True))

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