3.249 \(\int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=116 \[ \frac {\sin ^2(c+d x)}{2 a^4 d}-\frac {4 \sin (c+d x)}{a^4 d}+\frac {10}{d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {10 \log (\sin (c+d x)+1)}{a^4 d}-\frac {5}{2 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

[Out]

10*ln(1+sin(d*x+c))/a^4/d-4*sin(d*x+c)/a^4/d+1/2*sin(d*x+c)^2/a^4/d+1/3/a/d/(a+a*sin(d*x+c))^3-5/2/d/(a^2+a^2*
sin(d*x+c))^2+10/d/(a^4+a^4*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 116, 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} \[ \frac {\sin ^2(c+d x)}{2 a^4 d}-\frac {4 \sin (c+d x)}{a^4 d}+\frac {10}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {5}{2 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac {10 \log (\sin (c+d x)+1)}{a^4 d}+\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^5)/(a + a*Sin[c + d*x])^4,x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.88, size = 119, normalized size = 1.03 \[ \frac {3 \sin ^5(c+d x)-15 \sin ^4(c+d x)+\sin ^3(c+d x) (60 \log (\sin (c+d x)+1)-63)+9 \sin ^2(c+d x) (20 \log (\sin (c+d x)+1)-1)+9 \sin (c+d x) (20 \log (\sin (c+d x)+1)+9)+60 \log (\sin (c+d x)+1)+47}{6 a^4 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^5)/(a + a*Sin[c + d*x])^4,x]

[Out]

(47 + 60*Log[1 + Sin[c + d*x]] + 9*(9 + 20*Log[1 + Sin[c + d*x]])*Sin[c + d*x] + 9*(-1 + 20*Log[1 + Sin[c + d*
x]])*Sin[c + d*x]^2 + (-63 + 60*Log[1 + Sin[c + d*x]])*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 3*Sin[c + d*x]^5)/
(6*a^4*d*(1 + Sin[c + d*x])^3)

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 144, normalized size = 1.24 \[ \frac {30 \, \cos \left (d x + c\right )^{4} - 87 \, \cos \left (d x + c\right )^{2} + 120 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, \cos \left (d x + c\right )^{4} + 39 \, \cos \left (d x + c\right )^{2} + 10\right )} \sin \left (d x + c\right ) - 34}{12 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

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))^4,x, algorithm="fricas")

[Out]

1/12*(30*cos(d*x + c)^4 - 87*cos(d*x + c)^2 + 120*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*l
og(sin(d*x + c) + 1) - 3*(2*cos(d*x + c)^4 + 39*cos(d*x + c)^2 + 10)*sin(d*x + c) - 34)/(3*a^4*d*cos(d*x + c)^
2 - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c))

________________________________________________________________________________________

giac [A]  time = 0.22, size = 84, normalized size = 0.72 \[ \frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} + \frac {60 \, \sin \left (d x + c\right )^{2} + 105 \, \sin \left (d x + c\right ) + 47}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (a^{4} \sin \left (d x + c\right )^{2} - 8 \, a^{4} \sin \left (d x + c\right )\right )}}{a^{8}}}{6 \, d} \]

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))^4,x, algorithm="giac")

[Out]

1/6*(60*log(abs(sin(d*x + c) + 1))/a^4 + (60*sin(d*x + c)^2 + 105*sin(d*x + c) + 47)/(a^4*(sin(d*x + c) + 1)^3
) + 3*(a^4*sin(d*x + c)^2 - 8*a^4*sin(d*x + c))/a^8)/d

________________________________________________________________________________________

maple [A]  time = 0.23, size = 103, normalized size = 0.89 \[ \frac {\sin ^{2}\left (d x +c \right )}{2 a^{4} d}-\frac {4 \sin \left (d x +c \right )}{a^{4} d}+\frac {1}{3 d \,a^{4} \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{2 d \,a^{4} \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {10 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{4} d}+\frac {10}{d \,a^{4} \left (1+\sin \left (d x +c \right )\right )} \]

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))^4,x)

[Out]

1/2*sin(d*x+c)^2/a^4/d-4*sin(d*x+c)/a^4/d+1/3/d/a^4/(1+sin(d*x+c))^3-5/2/d/a^4/(1+sin(d*x+c))^2+10*ln(1+sin(d*
x+c))/a^4/d+10/d/a^4/(1+sin(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 105, normalized size = 0.91 \[ \frac {\frac {60 \, \sin \left (d x + c\right )^{2} + 105 \, \sin \left (d x + c\right ) + 47}{a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}} + \frac {3 \, {\left (\sin \left (d x + c\right )^{2} - 8 \, \sin \left (d x + c\right )\right )}}{a^{4}} + \frac {60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}}}{6 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

1/6*((60*sin(d*x + c)^2 + 105*sin(d*x + c) + 47)/(a^4*sin(d*x + c)^3 + 3*a^4*sin(d*x + c)^2 + 3*a^4*sin(d*x +
c) + a^4) + 3*(sin(d*x + c)^2 - 8*sin(d*x + c))/a^4 + 60*log(sin(d*x + c) + 1)/a^4)/d

________________________________________________________________________________________

mupad [B]  time = 8.53, size = 114, normalized size = 0.98 \[ \frac {10\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^4\,d}-\frac {4\,\sin \left (c+d\,x\right )}{a^4\,d}+\frac {10\,{\sin \left (c+d\,x\right )}^2+\frac {35\,\sin \left (c+d\,x\right )}{2}+\frac {47}{6}}{d\,\left (a^4\,{\sin \left (c+d\,x\right )}^3+3\,a^4\,{\sin \left (c+d\,x\right )}^2+3\,a^4\,\sin \left (c+d\,x\right )+a^4\right )}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a^4\,d} \]

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))^4,x)

[Out]

(10*log(sin(c + d*x) + 1))/(a^4*d) - (4*sin(c + d*x))/(a^4*d) + ((35*sin(c + d*x))/2 + 10*sin(c + d*x)^2 + 47/
6)/(d*(3*a^4*sin(c + d*x) + a^4 + 3*a^4*sin(c + d*x)^2 + a^4*sin(c + d*x)^3)) + sin(c + d*x)^2/(2*a^4*d)

________________________________________________________________________________________

sympy [A]  time = 10.05, size = 588, normalized size = 5.07 \[ \begin {cases} \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{3}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {180 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {180 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {3 \sin ^{5}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} - \frac {15 \sin ^{4}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {180 \sin ^{2}{\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {270 \sin {\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} + \frac {110}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{5}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{4}} & \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))**4,x)

[Out]

Piecewise((60*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18
*a**4*d*sin(c + d*x) + 6*a**4*d) + 180*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(6*a**4*d*sin(c + d*x)**3 + 18*a*
*4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 180*log(sin(c + d*x) + 1)*sin(c + d*x)/(6*a**4*d*s
in(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 60*log(sin(c + d*x) + 1)/(6*
a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 3*sin(c + d*x)**5/(6
*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) - 15*sin(c + d*x)**4/
(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 180*sin(c + d*x)*
*2/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 270*sin(c + d*
x)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 110/(6*a**4*d*
sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d), Ne(d, 0)), (x*sin(c)**5*cos(
c)/(a*sin(c) + a)**4, True))

________________________________________________________________________________________