\(\int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [77]

Optimal result

Integrand size = 21, antiderivative size = 77 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5 x}{2 a^3}+\frac {5 \cos ^3(c+d x)}{3 a^3 d}+\frac {5 \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac {2 \cos ^5(c+d x)}{a d (a+a \sin (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2759, 2761, 2715, 8} \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5 \cos ^3(c+d x)}{3 a^3 d}+\frac {5 \sin (c+d x) \cos (c+d x)}{2 a^3 d}+\frac {5 x}{2 a^3}+\frac {2 \cos ^5(c+d x)}{a d (a \sin (c+d x)+a)^2} \]

[In]

Int[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 8

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

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 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 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]

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.57 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\cos ^7(c+d x) \left (30 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (-22+31 \sin (c+d x)-11 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )\right )}{6 a^3 d (-1+\sin (c+d x))^4 (1+\sin (c+d x))^{7/2}} \]

[In]

Integrate[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/6*(Cos[c + d*x]^7*(30*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[1 - Sin[c + d*x]] + Sqrt[1 + Sin[c + d*x]
]*(-22 + 31*Sin[c + d*x] - 11*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3)))/(a^3*d*(-1 + Sin[c + d*x])^4*(1 + Sin[c + d
*x])^(7/2))

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.58

method	result	size
parallelrisch	\(\frac {30 d x -\cos \left (3 d x +3 c \right )-9 \sin \left (2 d x +2 c \right )+45 \cos \left (d x +c \right )+44}{12 a^{3} d}\)	\(45\)
risch	\(\frac {5 x}{2 a^{3}}+\frac {15 \cos \left (d x +c \right )}{4 a^{3} d}-\frac {\cos \left (3 d x +3 c \right )}{12 a^{3} d}-\frac {3 \sin \left (2 d x +2 c \right )}{4 a^{3} d}\)	\(56\)
derivativedivides	\(\frac {\frac {2 \left (\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {11}{3}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\)	\(90\)
default	\(\frac {\frac {2 \left (\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {11}{3}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\)	\(90\)
norman	\(\frac {\frac {200 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2435 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {3197 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {3055 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2701 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1969 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {3325 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {1699 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {200 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {340 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {1253 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {100 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {21 \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {22}{3 a d}+\frac {25 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+\frac {101 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}+\frac {500 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {640 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {169 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {5 x \left (\tan ^{17}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {3 \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {640 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {25 x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {340 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {341 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {40 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {289 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {725 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {100 x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {673 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {40 x \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {500 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {69 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {725 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {5 x}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(637\)

[In]

int(cos(d*x+c)^6/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/12*(30*d*x-cos(3*d*x+3*c)-9*sin(2*d*x+2*c)+45*cos(d*x+c)+44)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.58 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, \cos \left (d x + c\right )^{3} - 15 \, d x + 9 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 24 \, \cos \left (d x + c\right )}{6 \, a^{3} d} \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/6*(2*cos(d*x + c)^3 - 15*d*x + 9*cos(d*x + c)*sin(d*x + c) - 24*cos(d*x + c))/(a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 690 vs. \(2 (71) = 142\).

Time = 37.03 (sec) , antiderivative size = 690, normalized size of antiderivative = 8.96 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} \frac {15 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {45 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {45 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {15 d x}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {18 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {36 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {96 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} - \frac {18 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} + \frac {44}{6 a^{3} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{6}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**6/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((15*d*x*tan(c/2 + d*x/2)**6/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*
d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 45*d*x*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c
/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 45*d*x*tan(c/2 + d*x/2)**2/(6*a**3*d*tan(c/2 + d*
x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 15*d*x/(6*a**3*d*tan(c/2
+ d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 18*tan(c/2 + d*x/2)
**5/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d)
+ 36*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d
*x/2)**2 + 6*a**3*d) + 96*tan(c/2 + d*x/2)**2/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**4 +
18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) - 18*tan(c/2 + d*x/2)/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(
c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d) + 44/(6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan
(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d), Ne(d, 0)), (x*cos(c)**6/(a*sin(c) + a)**3, True)
)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (71) = 142\).

Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.39 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {9 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 22}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/3*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 18*sin(d*x + c)^4/(cos(d*x
+ c) + 1)^4 - 9*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 22)/(a^3 + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) - 15*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 = 88, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {15 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(15*(d*x + c)/a^3 + 2*(9*tan(1/2*d*x + 1/2*c)^5 + 18*tan(1/2*d*x + 1/2*c)^4 + 48*tan(1/2*d*x + 1/2*c)^2 -
9*tan(1/2*d*x + 1/2*c) + 22)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d

Mupad [B] (verification not implemented)

Time = 6.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5\,x}{2\,a^3}+\frac {4\,\cos \left (c+d\,x\right )}{a^3\,d}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^3\,d}-\frac {3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,a^3\,d} \]

[In]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^3,x)

[Out]

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