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

Optimal. Leaf size=95 \[ \frac {\sin (c+d x)}{a^4 d}-\frac {6}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {4 \log (\sin (c+d x)+1)}{a^4 d}+\frac {2}{d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 95, 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} \[ \frac {\sin (c+d x)}{a^4 d}-\frac {6}{d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {2}{d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {4 \log (\sin (c+d x)+1)}{a^4 d}-\frac {1}{3 a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 6.56, size = 127, normalized size = 1.34 \[ -\frac {3 (2 \sin (c+d x)+1)^2}{16 a^4 d (\sin (c+d x)+1)^3}-\frac {\frac {252 \sin ^2(c+d x)+444 \sin (c+d x)+197}{(\sin (c+d x)+1)^3}-48 \sin (c+d x)+192 \log (\sin (c+d x)+1)}{48 a^4 d}-\frac {1}{24 a^4 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*1/(a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6) - (3*(1 + 2*Sin[c + d*x])^2)/(16*a^4*d*(1 + Sin[c + d*
x])^3) - (192*Log[1 + Sin[c + d*x]] - 48*Sin[c + d*x] + (197 + 444*Sin[c + d*x] + 252*Sin[c + d*x]^2)/(1 + Sin
[c + d*x])^3)/(48*a^4*d)

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fricas [A]  time = 0.54, size = 132, normalized size = 1.39 \[ -\frac {3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} + 12 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 19}{3 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 + 12*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(
sin(d*x + c) + 1) - 9*(cos(d*x + c)^2 + 2)*sin(d*x + c) - 19)/(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))

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giac [A]  time = 0.21, size = 66, normalized size = 0.69 \[ -\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, \sin \left (d x + c\right )}{a^{4}} + \frac {18 \, \sin \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) + 13}{a^{4} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{3 \, d} \]

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))^4,x, algorithm="giac")

[Out]

-1/3*(12*log(abs(sin(d*x + c) + 1))/a^4 - 3*sin(d*x + c)/a^4 + (18*sin(d*x + c)^2 + 30*sin(d*x + c) + 13)/(a^4
*(sin(d*x + c) + 1)^3))/d

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maple [A]  time = 0.24, size = 86, normalized size = 0.91 \[ \frac {\sin \left (d x +c \right )}{a^{4} d}-\frac {1}{3 d \,a^{4} \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {4 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{4} d}-\frac {6}{d \,a^{4} \left (1+\sin \left (d x +c \right )\right )}+\frac {2}{d \,a^{4} \left (1+\sin \left (d x +c \right )\right )^{2}} \]

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))^4,x)

[Out]

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

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maxima [A]  time = 0.38, size = 94, normalized size = 0.99 \[ -\frac {\frac {18 \, \sin \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right ) + 13}{a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} - \frac {3 \, \sin \left (d x + c\right )}{a^{4}}}{3 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

-1/3*((18*sin(d*x + c)^2 + 30*sin(d*x + c) + 13)/(a^4*sin(d*x + c)^3 + 3*a^4*sin(d*x + c)^2 + 3*a^4*sin(d*x +
c) + a^4) + 12*log(sin(d*x + c) + 1)/a^4 - 3*sin(d*x + c)/a^4)/d

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mupad [B]  time = 0.12, size = 69, normalized size = 0.73 \[ \frac {\sin \left (c+d\,x\right )}{a^4\,d}-\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^4\,d}-\frac {6\,{\sin \left (c+d\,x\right )}^2+10\,\sin \left (c+d\,x\right )+\frac {13}{3}}{a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^3} \]

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))^4,x)

[Out]

sin(c + d*x)/(a^4*d) - (4*log(sin(c + d*x) + 1))/(a^4*d) - (10*sin(c + d*x) + 6*sin(c + d*x)^2 + 13/3)/(a^4*d*
(sin(c + d*x) + 1)^3)

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sympy [A]  time = 5.57, size = 527, normalized size = 5.55 \[ \begin {cases} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{3}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {12 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} + \frac {3 \sin ^{4}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {36 \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {54 \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {22}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-12*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*
a**4*d*sin(c + d*x) + 3*a**4*d) - 36*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*
d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 36*log(sin(c + d*x) + 1)*sin(c + d*x)/(3*a**4*d*sin(c
+ d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 12*log(sin(c + d*x) + 1)/(3*a**4*d*
sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) + 3*sin(c + d*x)**4/(3*a**4*d*s
in(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 36*sin(c + d*x)**2/(3*a**4*d*s
in(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 54*sin(c + d*x)/(3*a**4*d*sin(
c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d) - 22/(3*a**4*d*sin(c + d*x)**3 + 9*
a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d), Ne(d, 0)), (x*sin(c)**4*cos(c)/(a*sin(c) + a)**4,
True))

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