\(\int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx\) [412]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 73 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{8 a}-\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2918, 2645, 30, 2648, 2715, 8} \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos ^3(c+d x)}{3 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac {\sin (c+d x) \cos (c+d x)}{8 a d}-\frac {x}{8 a} \]

[In]

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

[Out]

-1/8*x/a - Cos[c + d*x]^3/(3*a*d) - (Cos[c + d*x]*Sin[c + d*x])/(8*a*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d
)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
 &&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
 + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*n]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \cos ^2(c+d x) \sin (c+d x) \, dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{a} \\ & = \frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac {\int \cos ^2(c+d x) \, dx}{4 a}-\frac {\text {Subst}\left (\int x^2 \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac {\int 1 \, dx}{8 a} \\ & = -\frac {x}{8 a}-\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(219\) vs. \(2(73)=146\).

Time = 1.12 (sec) , antiderivative size = 219, normalized size of antiderivative = 3.00 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {-24 (c-d x) \cos \left (\frac {c}{2}\right )+24 \cos \left (\frac {c}{2}+d x\right )+24 \cos \left (\frac {3 c}{2}+d x\right )+8 \cos \left (\frac {5 c}{2}+3 d x\right )+8 \cos \left (\frac {7 c}{2}+3 d x\right )-3 \cos \left (\frac {7 c}{2}+4 d x\right )+3 \cos \left (\frac {9 c}{2}+4 d x\right )+48 \sin \left (\frac {c}{2}\right )-24 c \sin \left (\frac {c}{2}\right )+24 d x \sin \left (\frac {c}{2}\right )-24 \sin \left (\frac {c}{2}+d x\right )+24 \sin \left (\frac {3 c}{2}+d x\right )-8 \sin \left (\frac {5 c}{2}+3 d x\right )+8 \sin \left (\frac {7 c}{2}+3 d x\right )-3 \sin \left (\frac {7 c}{2}+4 d x\right )-3 \sin \left (\frac {9 c}{2}+4 d x\right )}{192 a d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \]

[In]

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

[Out]

-1/192*(-24*(c - d*x)*Cos[c/2] + 24*Cos[c/2 + d*x] + 24*Cos[(3*c)/2 + d*x] + 8*Cos[(5*c)/2 + 3*d*x] + 8*Cos[(7
*c)/2 + 3*d*x] - 3*Cos[(7*c)/2 + 4*d*x] + 3*Cos[(9*c)/2 + 4*d*x] + 48*Sin[c/2] - 24*c*Sin[c/2] + 24*d*x*Sin[c/
2] - 24*Sin[c/2 + d*x] + 24*Sin[(3*c)/2 + d*x] - 8*Sin[(5*c)/2 + 3*d*x] + 8*Sin[(7*c)/2 + 3*d*x] - 3*Sin[(7*c)
/2 + 4*d*x] - 3*Sin[(9*c)/2 + 4*d*x])/(a*d*(Cos[c/2] + Sin[c/2]))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.62

method result size
parallelrisch \(\frac {-12 d x -8 \cos \left (3 d x +3 c \right )-24 \cos \left (d x +c \right )+3 \sin \left (4 d x +4 c \right )-32}{96 d a}\) \(45\)
risch \(-\frac {x}{8 a}-\frac {\cos \left (d x +c \right )}{4 a d}+\frac {\sin \left (4 d x +4 c \right )}{32 d a}-\frac {\cos \left (3 d x +3 c \right )}{12 a d}\) \(56\)
derivativedivides \(\frac {\frac {4 \left (-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {1}{6}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\) \(129\)
default \(\frac {\frac {4 \left (-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {1}{6}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\) \(129\)
norman \(\frac {-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {x}{8 a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}-\frac {5 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {5 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {5 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {5 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {5 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {5 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}-\frac {5 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {5 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {5}{12 a d}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {5 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {3 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}-\frac {19 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) \(399\)

[In]

int(cos(d*x+c)^4*sin(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/96*(-12*d*x-8*cos(3*d*x+3*c)-24*cos(d*x+c)+3*sin(4*d*x+4*c)-32)/d/a

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {8 \, \cos \left (d x + c\right )^{3} + 3 \, d x - 3 \, {\left (2 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (56) = 112\).

Time = 6.42 (sec) , antiderivative size = 1134, normalized size of antiderivative = 15.53 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-3*d*x*tan(c/2 + d*x/2)**8/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c
/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 12*d*x*tan(c/2 + d*x/2)**6/(24*a*d*tan(c/2 + d*x/2)**8
 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 18*d*x*ta
n(c/2 + d*x/2)**4/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*
a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 12*d*x*tan(c/2 + d*x/2)**2/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 +
d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 3*d*x/(24*a*d*tan(c/2 + d*x/2
)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 6*tan
(c/2 + d*x/2)**7/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a
*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 48*tan(c/2 + d*x/2)**6/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2
)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 42*tan(c/2 + d*x/2)**5/(24*a*d*tan
(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*
a*d) - 48*tan(c/2 + d*x/2)**4/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x
/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 42*tan(c/2 + d*x/2)**3/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*ta
n(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 16*tan(c/2 + d*x/2)**
2/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*
x/2)**2 + 24*a*d) + 6*tan(c/2 + d*x/2)/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(
c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 16/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x
/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d), Ne(d, 0)), (x*sin(c)*cos(c)**4/(a
*sin(c) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (65) = 130\).

Time = 0.32 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.52 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {21 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {24 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {24 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 8}{a + \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{12 \, d} \]

[In]

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

[Out]

1/12*((3*sin(d*x + c)/(cos(d*x + c) + 1) - 8*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 21*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 24*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 24*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 8)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(c
os(d*x + c) + 1)^8) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.74 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]

[In]

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

[Out]

-1/24*(3*(d*x + c)/a + 2*(3*tan(1/2*d*x + 1/2*c)^7 + 24*tan(1/2*d*x + 1/2*c)^6 - 21*tan(1/2*d*x + 1/2*c)^5 + 2
4*tan(1/2*d*x + 1/2*c)^4 + 21*tan(1/2*d*x + 1/2*c)^3 + 8*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 8)/
((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d

Mupad [B] (verification not implemented)

Time = 9.61 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.59 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6\,\cos \left (c+d\,x\right )+2\,\cos \left (3\,c+3\,d\,x\right )-\frac {3\,\sin \left (4\,c+4\,d\,x\right )}{4}+3\,d\,x}{24\,a\,d} \]

[In]

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

[Out]

-(6*cos(c + d*x) + 2*cos(3*c + 3*d*x) - (3*sin(4*c + 4*d*x))/4 + 3*d*x)/(24*a*d)