\(\int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [434]

Optimal result

Integrand size = 27, antiderivative size = 80 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {9 x}{2 a^3}+\frac {9 \cos (c+d x)}{2 a^3 d}+\frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3+a^3 \sin (c+d x)\right )} \]

[Out]

9/2*x/a^3+9/2*cos(d*x+c)/a^3/d+cos(d*x+c)^5/d/(a+a*sin(d*x+c))^3+3/2*cos(d*x+c)^3/d/(a^3+a^3*sin(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2938, 2758, 2761, 8} \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {9 \cos (c+d x)}{2 a^3 d}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {9 x}{2 a^3}+\frac {\cos ^5(c+d x)}{d (a \sin (c+d x)+a)^3} \]

[In]

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

[Out]

(9*x)/(2*a^3) + (9*Cos[c + d*x])/(2*a^3*d) + Cos[c + d*x]^5/(d*(a + a*Sin[c + d*x])^3) + (3*Cos[c + d*x]^3)/(2
*d*(a^3 + a^3*Sin[c + d*x]))

Rule 8

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

Rule 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(a*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2761

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

Rule 2938

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p
+ 1))), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx}{a} \\ & = \frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {9 \int \frac {\cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{2 a^2} \\ & = \frac {9 \cos (c+d x)}{2 a^3 d}+\frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {9 \int 1 \, dx}{2 a^3} \\ & = \frac {9 x}{2 a^3}+\frac {9 \cos (c+d x)}{2 a^3 d}+\frac {\cos ^5(c+d x)}{d (a+a \sin (c+d x))^3}+\frac {3 \cos ^3(c+d x)}{2 d \left (a^3+a^3 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.79 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {180 d x \cos \left (\frac {d x}{2}\right )+59 \cos \left (c+\frac {d x}{2}\right )+55 \cos \left (c+\frac {3 d x}{2}\right )+5 \cos \left (3 c+\frac {5 d x}{2}\right )-381 \sin \left (\frac {d x}{2}\right )+180 d x \sin \left (c+\frac {d x}{2}\right )+55 \sin \left (2 c+\frac {3 d x}{2}\right )-5 \sin \left (2 c+\frac {5 d x}{2}\right )}{40 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(180*d*x*Cos[(d*x)/2] + 59*Cos[c + (d*x)/2] + 55*Cos[c + (3*d*x)/2] + 5*Cos[3*c + (5*d*x)/2] - 381*Sin[(d*x)/2
] + 180*d*x*Sin[c + (d*x)/2] + 55*Sin[2*c + (3*d*x)/2] - 5*Sin[2*c + (5*d*x)/2])/(40*a^3*d*(Cos[c/2] + Sin[c/2
])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\(\frac {36 d x \cos \left (d x +c \right )+56 \cos \left (d x +c \right )-33 \sin \left (d x +c \right )+12 \cos \left (2 d x +2 c \right )-\sin \left (3 d x +3 c \right )+44}{8 d \,a^{3} \cos \left (d x +c \right )}\)	\(67\)
risch	\(\frac {9 x}{2 a^{3}}+\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {8}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}-\frac {\sin \left (2 d x +2 c \right )}{4 d \,a^{3}}\)	\(81\)
derivativedivides	\(\frac {\frac {4 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{3}}\)	\(92\)
default	\(\frac {\frac {4 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d \,a^{3}}\)	\(92\)
norman	\(\frac {\frac {909 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {315 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {1395 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {135 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {1215 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {585 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {1395 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {1215 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {909 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {585 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {14}{a d}+\frac {315 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {135 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {9 x}{2 a}+\frac {296 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {167 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {45 x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {9 x \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {43 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {61 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {1136 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {45 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {807 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {360 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {1034 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {881 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {1083 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {614 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {9 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {533 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {130 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(565\)

[In]

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

[Out]

1/8/d/a^3*(36*d*x*cos(d*x+c)+56*cos(d*x+c)-33*sin(d*x+c)+12*cos(2*d*x+2*c)-sin(3*d*x+3*c)+44)/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\cos \left (d x + c\right )^{3} + 9 \, d x + {\left (9 \, d x + 13\right )} \cos \left (d x + c\right ) + 6 \, \cos \left (d x + c\right )^{2} + {\left (9 \, d x - \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) + 8}{2 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/2*(cos(d*x + c)^3 + 9*d*x + (9*d*x + 13)*cos(d*x + c) + 6*cos(d*x + c)^2 + (9*d*x - cos(d*x + c)^2 + 5*cos(d
*x + c) - 8)*sin(d*x + c) + 8)/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1244 vs. \(2 (71) = 142\).

Time = 23.08 (sec) , antiderivative size = 1244, normalized size of antiderivative = 15.55 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((9*d*x*tan(c/2 + d*x/2)**5/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*t
an(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 9*d*x*tan(c/2 + d*
x/2)**4/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d
*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 18*d*x*tan(c/2 + d*x/2)**3/(2*a**3*d*tan(c/2 +
d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**
3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 18*d*x*tan(c/2 + d*x/2)**2/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2
+ d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**
3*d) + 9*d*x*tan(c/2 + d*x/2)/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2
+ d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 9*d*x/(2*a**3*d*tan(c/2 +
d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a*
*3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 18*tan(c/2 + d*x/2)**4/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 +
d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d
) + 14*tan(c/2 + d*x/2)**3/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d
*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 42*tan(c/2 + d*x/2)**2/(2*a*
*3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*
x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d) + 10*tan(c/2 + d*x/2)/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d
*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3 + 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2)
+ 2*a**3*d) + 28/(2*a**3*d*tan(c/2 + d*x/2)**5 + 2*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**3
+ 4*a**3*d*tan(c/2 + d*x/2)**2 + 2*a**3*d*tan(c/2 + d*x/2) + 2*a**3*d), Ne(d, 0)), (x*sin(c)*cos(c)**4/(a*sin(
c) + a)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (74) = 148\).

Time = 0.30 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.81 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 14}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{d} \]

[In]

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

[Out]

((5*sin(d*x + c)/(cos(d*x + c) + 1) + 21*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 7*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 9*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14)/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a^3*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^3*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {9 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} + \frac {16}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{2 \, d} \]

[In]

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

[Out]

1/2*(9*(d*x + c)/a^3 + 2*(tan(1/2*d*x + 1/2*c)^3 + 6*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 6)/((tan(
1/2*d*x + 1/2*c)^2 + 1)^2*a^3) + 16/(a^3*(tan(1/2*d*x + 1/2*c) + 1)))/d

Mupad [B] (verification not implemented)

Time = 11.78 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^4(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {9\,x}{2\,a^3}+\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+14}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

[In]

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

[Out]

(9*x)/(2*a^3) + (5*tan(c/2 + (d*x)/2) + 21*tan(c/2 + (d*x)/2)^2 + 7*tan(c/2 + (d*x)/2)^3 + 9*tan(c/2 + (d*x)/2
)^4 + 14)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)