3.240 \(\int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=111 \[ \frac {\sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {6 \sin (c+d x)}{a^3 d}-\frac {5}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {10 \log (\sin (c+d x)+1)}{a^3 d}+\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

-10*ln(1+sin(d*x+c))/a^3/d+6*sin(d*x+c)/a^3/d-3/2*sin(d*x+c)^2/a^3/d+1/3*sin(d*x+c)^3/a^3/d+1/2/a/d/(a+a*sin(d
*x+c))^2-5/d/(a^3+a^3*sin(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 111, normalized size of antiderivative = 1.00, 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} \[ \frac {\sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {6 \sin (c+d x)}{a^3 d}-\frac {5}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {10 \log (\sin (c+d x)+1)}{a^3 d}+\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Log[1 + Sin[c + d*x]])/(a^3*d) + (6*Sin[c + d*x])/(a^3*d) - (3*Sin[c + d*x]^2)/(2*a^3*d) + Sin[c + d*x]^3
/(3*a^3*d) + 1/(2*a*d*(a + a*Sin[c + d*x])^2) - 5/(d*(a^3 + a^3*Sin[c + d*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])

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]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{a^5 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{(a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a^6 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (6 a^2-3 a x+x^2-\frac {a^5}{(a+x)^3}+\frac {5 a^4}{(a+x)^2}-\frac {10 a^3}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^6 d}\\ &=-\frac {10 \log (1+\sin (c+d x))}{a^3 d}+\frac {6 \sin (c+d x)}{a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {\sin ^3(c+d x)}{3 a^3 d}+\frac {1}{2 a d (a+a \sin (c+d x))^2}-\frac {5}{d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.70, size = 106, normalized size = 0.95 \[ \frac {32 \sin ^5(c+d x)-80 \sin ^4(c+d x)+320 \sin ^3(c+d x)+\sin ^2(c+d x) (1023-960 \log (\sin (c+d x)+1))-6 \sin (c+d x) (320 \log (\sin (c+d x)+1)-21)-960 \log (\sin (c+d x)+1)-417}{96 a^3 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-417 - 960*Log[1 + Sin[c + d*x]] - 6*(-21 + 320*Log[1 + Sin[c + d*x]])*Sin[c + d*x] + (1023 - 960*Log[1 + Sin
[c + d*x]])*Sin[c + d*x]^2 + 320*Sin[c + d*x]^3 - 80*Sin[c + d*x]^4 + 32*Sin[c + d*x]^5)/(96*a^3*d*(1 + Sin[c
+ d*x])^2)

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fricas [A]  time = 0.52, size = 117, normalized size = 1.05 \[ \frac {10 \, \cos \left (d x + c\right )^{4} + 115 \, \cos \left (d x + c\right )^{2} - 120 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (d x + c\right )^{4} - 24 \, \cos \left (d x + c\right )^{2} + 37\right )} \sin \left (d x + c\right ) - 80}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(10*cos(d*x + c)^4 + 115*cos(d*x + c)^2 - 120*(cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d*x + c) + 1)
- 2*(2*cos(d*x + c)^4 - 24*cos(d*x + c)^2 + 37)*sin(d*x + c) - 80)/(a^3*d*cos(d*x + c)^2 - 2*a^3*d*sin(d*x +
c) - 2*a^3*d)

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giac [A]  time = 0.22, size = 89, normalized size = 0.80 \[ -\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac {3 \, {\left (10 \, \sin \left (d x + c\right ) + 9\right )}}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a^{6} \sin \left (d x + c\right )^{3} - 9 \, a^{6} \sin \left (d x + c\right )^{2} + 36 \, a^{6} \sin \left (d x + c\right )}{a^{9}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(60*log(abs(sin(d*x + c) + 1))/a^3 + 3*(10*sin(d*x + c) + 9)/(a^3*(sin(d*x + c) + 1)^2) - (2*a^6*sin(d*x
+ c)^3 - 9*a^6*sin(d*x + c)^2 + 36*a^6*sin(d*x + c))/a^9)/d

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maple [A]  time = 0.24, size = 101, normalized size = 0.91 \[ \frac {\sin ^{3}\left (d x +c \right )}{3 a^{3} d}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2 a^{3} d}+\frac {6 \sin \left (d x +c \right )}{a^{3} d}+\frac {1}{2 d \,a^{3} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {10 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}-\frac {5}{d \,a^{3} \left (1+\sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x)

[Out]

1/3*sin(d*x+c)^3/a^3/d-3/2*sin(d*x+c)^2/a^3/d+6*sin(d*x+c)/a^3/d+1/2/d/a^3/(1+sin(d*x+c))^2-10*ln(1+sin(d*x+c)
)/a^3/d-5/d/a^3/(1+sin(d*x+c))

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maxima [A]  time = 0.32, size = 95, normalized size = 0.86 \[ -\frac {\frac {3 \, {\left (10 \, \sin \left (d x + c\right ) + 9\right )}}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {2 \, \sin \left (d x + c\right )^{3} - 9 \, \sin \left (d x + c\right )^{2} + 36 \, \sin \left (d x + c\right )}{a^{3}} + \frac {60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/6*(3*(10*sin(d*x + c) + 9)/(a^3*sin(d*x + c)^2 + 2*a^3*sin(d*x + c) + a^3) - (2*sin(d*x + c)^3 - 9*sin(d*x
+ c)^2 + 36*sin(d*x + c))/a^3 + 60*log(sin(d*x + c) + 1)/a^3)/d

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mupad [B]  time = 0.13, size = 108, normalized size = 0.97 \[ \frac {6\,\sin \left (c+d\,x\right )}{a^3\,d}-\frac {5\,\sin \left (c+d\,x\right )+\frac {9}{2}}{d\,\left (a^3\,{\sin \left (c+d\,x\right )}^2+2\,a^3\,\sin \left (c+d\,x\right )+a^3\right )}-\frac {10\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d}-\frac {3\,{\sin \left (c+d\,x\right )}^2}{2\,a^3\,d}+\frac {{\sin \left (c+d\,x\right )}^3}{3\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*sin(c + d*x)^5)/(a + a*sin(c + d*x))^3,x)

[Out]

(6*sin(c + d*x))/(a^3*d) - (5*sin(c + d*x) + 9/2)/(d*(2*a^3*sin(c + d*x) + a^3 + a^3*sin(c + d*x)^2)) - (10*lo
g(sin(c + d*x) + 1))/(a^3*d) - (3*sin(c + d*x)^2)/(2*a^3*d) + sin(c + d*x)^3/(3*a^3*d)

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sympy [A]  time = 8.06, size = 394, normalized size = 3.55 \[ \begin {cases} - \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {120 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} + \frac {2 \sin ^{5}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {5 \sin ^{4}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} + \frac {20 \sin ^{3}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {120 \sin {\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {90}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{5}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((-60*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(6*a**3*d*sin(c + d*x)**2 + 12*a**3*d*sin(c + d*x) + 6*a*
*3*d) - 120*log(sin(c + d*x) + 1)*sin(c + d*x)/(6*a**3*d*sin(c + d*x)**2 + 12*a**3*d*sin(c + d*x) + 6*a**3*d)
- 60*log(sin(c + d*x) + 1)/(6*a**3*d*sin(c + d*x)**2 + 12*a**3*d*sin(c + d*x) + 6*a**3*d) + 2*sin(c + d*x)**5/
(6*a**3*d*sin(c + d*x)**2 + 12*a**3*d*sin(c + d*x) + 6*a**3*d) - 5*sin(c + d*x)**4/(6*a**3*d*sin(c + d*x)**2 +
12*a**3*d*sin(c + d*x) + 6*a**3*d) + 20*sin(c + d*x)**3/(6*a**3*d*sin(c + d*x)**2 + 12*a**3*d*sin(c + d*x) +
6*a**3*d) - 120*sin(c + d*x)/(6*a**3*d*sin(c + d*x)**2 + 12*a**3*d*sin(c + d*x) + 6*a**3*d) - 90/(6*a**3*d*sin
(c + d*x)**2 + 12*a**3*d*sin(c + d*x) + 6*a**3*d), Ne(d, 0)), (x*sin(c)**5*cos(c)/(a*sin(c) + a)**3, True))

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