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

Optimal. Leaf size=93 \[ \frac {\sin ^2(c+d x)}{2 a^3 d}-\frac {3 \sin (c+d x)}{a^3 d}+\frac {4}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {6 \log (\sin (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 93, 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} \[ \frac {\sin ^2(c+d x)}{2 a^3 d}-\frac {3 \sin (c+d x)}{a^3 d}+\frac {4}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {6 \log (\sin (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 2.08, size = 78, normalized size = 0.84 \[ \frac {8 \sin ^2(c+d x)+\left (\frac {64}{(\sin (c+d x)+1)^2}-48\right ) \sin (c+d x)+96 \log (\sin (c+d x)+1)+\frac {56}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}}{16 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(96*Log[1 + Sin[c + d*x]] + 56/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 8*Sin[c + d*x]^2 + Sin[c + d*x]*(-48
+ 64/(1 + Sin[c + d*x])^2))/(16*a^3*d)

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 107, normalized size = 1.15 \[ -\frac {2 \, \cos \left (d x + c\right )^{4} + 19 \, \cos \left (d x + c\right )^{2} - 24 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) - 8}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(2*cos(d*x + c)^4 + 19*cos(d*x + c)^2 - 24*(cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d*x + c) + 1) +
2*(4*cos(d*x + c)^2 - 3)*sin(d*x + c) - 8)/(a^3*d*cos(d*x + c)^2 - 2*a^3*d*sin(d*x + c) - 2*a^3*d)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 73, normalized size = 0.78 \[ \frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac {8 \, \sin \left (d x + c\right ) + 7}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{2} - 6 \, a^{3} \sin \left (d x + c\right )}{a^{6}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(12*log(abs(sin(d*x + c) + 1))/a^3 + (8*sin(d*x + c) + 7)/(a^3*(sin(d*x + c) + 1)^2) + (a^3*sin(d*x + c)^2
- 6*a^3*sin(d*x + c))/a^6)/d

________________________________________________________________________________________

maple [A]  time = 0.24, size = 85, normalized size = 0.91 \[ \frac {\sin ^{2}\left (d x +c \right )}{2 a^{3} d}-\frac {3 \sin \left (d x +c \right )}{a^{3} d}-\frac {1}{2 d \,a^{3} \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {6 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {4}{d \,a^{3} \left (1+\sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 81, normalized size = 0.87 \[ \frac {\frac {8 \, \sin \left (d x + c\right ) + 7}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac {\sin \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right )}{a^{3}} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*((8*sin(d*x + c) + 7)/(a^3*sin(d*x + c)^2 + 2*a^3*sin(d*x + c) + a^3) + (sin(d*x + c)^2 - 6*sin(d*x + c))/
a^3 + 12*log(sin(d*x + c) + 1)/a^3)/d

________________________________________________________________________________________

mupad [B]  time = 8.49, size = 91, normalized size = 0.98 \[ \frac {6\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d}+\frac {4\,\sin \left (c+d\,x\right )+\frac {7}{2}}{d\,\left (a^3\,{\sin \left (c+d\,x\right )}^2+2\,a^3\,\sin \left (c+d\,x\right )+a^3\right )}-\frac {3\,\sin \left (c+d\,x\right )}{a^3\,d}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*log(sin(c + d*x) + 1))/(a^3*d) + (4*sin(c + d*x) + 7/2)/(d*(2*a^3*sin(c + d*x) + a^3 + a^3*sin(c + d*x)^2))
- (3*sin(c + d*x))/(a^3*d) + sin(c + d*x)^2/(2*a^3*d)

________________________________________________________________________________________

sympy [A]  time = 5.47, size = 347, normalized size = 3.73 \[ \begin {cases} \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {24 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {\sin ^{4}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {4 \sin ^{3}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {24 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {18}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((12*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3
*d) + 24*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 12
*log(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + sin(c + d*x)**4/(2*a**3
*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 4*sin(c + d*x)**3/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*
d*sin(c + d*x) + 2*a**3*d) + 24*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 1
8/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Ne(d, 0)), (x*sin(c)**4*cos(c)/(a*sin(c) + a)
**3, True))

________________________________________________________________________________________