\(\int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [646]

Optimal result

Integrand size = 27, antiderivative size = 84 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {15 x}{8 a^3}-\frac {4 \cos (c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{a^3 d}+\frac {15 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d} \]

[Out]

-15/8*x/a^3-4*cos(d*x+c)/a^3/d+cos(d*x+c)^3/a^3/d+15/8*cos(d*x+c)*sin(d*x+c)/a^3/d+1/4*cos(d*x+c)*sin(d*x+c)^3
/a^3/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.25, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2938, 2758, 2761, 2715, 8} \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {5 \cos ^3(c+d x)}{4 a^3 d}-\frac {3 \cos ^5(c+d x)}{4 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {15 \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac {15 x}{8 a^3}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3} \]

[In]

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

[Out]

(-15*x)/(8*a^3) - (5*Cos[c + d*x]^3)/(4*a^3*d) - (15*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d) - Cos[c + d*x]^7/(d*
(a + a*Sin[c + d*x])^3) - (3*Cos[c + d*x]^5)/(4*d*(a^3 + a^3*Sin[c + d*x]))

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 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 ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac {3 \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx}{a} \\ & = -\frac {\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac {3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac {15 \int \frac {\cos ^4(c+d x)}{a+a \sin (c+d x)} \, dx}{4 a^2} \\ & = -\frac {5 \cos ^3(c+d x)}{4 a^3 d}-\frac {\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac {3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac {15 \int \cos ^2(c+d x) \, dx}{4 a^3} \\ & = -\frac {5 \cos ^3(c+d x)}{4 a^3 d}-\frac {15 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac {3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac {15 \int 1 \, dx}{8 a^3} \\ & = -\frac {15 x}{8 a^3}-\frac {5 \cos ^3(c+d x)}{4 a^3 d}-\frac {15 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {\cos ^7(c+d x)}{d (a+a \sin (c+d x))^3}-\frac {3 \cos ^5(c+d x)}{4 d \left (a^3+a^3 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(255\) vs. \(2(84)=168\).

Time = 0.98 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {(1+120 d x) \cos \left (\frac {c}{2}\right )+104 \cos \left (\frac {c}{2}+d x\right )+104 \cos \left (\frac {3 c}{2}+d x\right )-32 \cos \left (\frac {3 c}{2}+2 d x\right )+32 \cos \left (\frac {5 c}{2}+2 d x\right )-8 \cos \left (\frac {5 c}{2}+3 d x\right )-8 \cos \left (\frac {7 c}{2}+3 d x\right )+\cos \left (\frac {7 c}{2}+4 d x\right )-\cos \left (\frac {9 c}{2}+4 d x\right )-\sin \left (\frac {c}{2}\right )+120 d x \sin \left (\frac {c}{2}\right )-104 \sin \left (\frac {c}{2}+d x\right )+104 \sin \left (\frac {3 c}{2}+d x\right )-32 \sin \left (\frac {3 c}{2}+2 d x\right )-32 \sin \left (\frac {5 c}{2}+2 d x\right )+8 \sin \left (\frac {5 c}{2}+3 d x\right )-8 \sin \left (\frac {7 c}{2}+3 d x\right )+\sin \left (\frac {7 c}{2}+4 d x\right )+\sin \left (\frac {9 c}{2}+4 d x\right )}{64 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \]

[In]

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

[Out]

-1/64*((1 + 120*d*x)*Cos[c/2] + 104*Cos[c/2 + d*x] + 104*Cos[(3*c)/2 + d*x] - 32*Cos[(3*c)/2 + 2*d*x] + 32*Cos
[(5*c)/2 + 2*d*x] - 8*Cos[(5*c)/2 + 3*d*x] - 8*Cos[(7*c)/2 + 3*d*x] + Cos[(7*c)/2 + 4*d*x] - Cos[(9*c)/2 + 4*d
*x] - Sin[c/2] + 120*d*x*Sin[c/2] - 104*Sin[c/2 + d*x] + 104*Sin[(3*c)/2 + d*x] - 32*Sin[(3*c)/2 + 2*d*x] - 32
*Sin[(5*c)/2 + 2*d*x] + 8*Sin[(5*c)/2 + 3*d*x] - 8*Sin[(7*c)/2 + 3*d*x] + Sin[(7*c)/2 + 4*d*x] + Sin[(9*c)/2 +
4*d*x])/(a^3*d*(Cos[c/2] + Sin[c/2]))

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {-60 d x -104 \cos \left (d x +c \right )-\sin \left (4 d x +4 c \right )+8 \cos \left (3 d x +3 c \right )+32 \sin \left (2 d x +2 c \right )+96}{32 a^{3} d}\)	\(56\)
risch	\(-\frac {15 x}{8 a^{3}}-\frac {13 \cos \left (d x +c \right )}{4 a^{3} d}-\frac {\sin \left (4 d x +4 c \right )}{32 d \,a^{3}}+\frac {\cos \left (3 d x +3 c \right )}{4 d \,a^{3}}+\frac {\sin \left (2 d x +2 c \right )}{d \,a^{3}}\)	\(72\)
derivativedivides	\(\frac {\frac {4 \left (-\frac {15 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {23 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {9 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {11 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {3}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{3} d}\)	\(129\)
default	\(\frac {\frac {4 \left (-\frac {15 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {\left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {23 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {9 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {11 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {3}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{3} d}\)	\(129\)

[In]

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

[Out]

1/32*(-60*d*x-104*cos(d*x+c)-sin(4*d*x+4*c)+8*cos(3*d*x+3*c)+32*sin(2*d*x+2*c)+96)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8 \, \cos \left (d x + c\right )^{3} - 15 \, d x - {\left (2 \, \cos \left (d x + c\right )^{3} - 17 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 32 \, \cos \left (d x + c\right )}{8 \, a^{3} d} \]

[In]

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

[Out]

1/8*(8*cos(d*x + c)^3 - 15*d*x - (2*cos(d*x + c)^3 - 17*cos(d*x + c))*sin(d*x + c) - 32*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. 1246 vs. \(2 (78) = 156\).

Time = 58.30 (sec) , antiderivative size = 1246, normalized size of antiderivative = 14.83 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-15*d*x*tan(c/2 + d*x/2)**8/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3
*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) - 60*d*x*tan(c/2 + d*x/2)**6/(8*a**3*d*tan(
c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**
2 + 8*a**3*d) - 90*d*x*tan(c/2 + d*x/2)**4/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*
a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) - 60*d*x*tan(c/2 + d*x/2)**2/(8*a**3*d*
tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/
2)**2 + 8*a**3*d) - 15*d*x/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 +
d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) - 30*tan(c/2 + d*x/2)**7/(8*a**3*d*tan(c/2 + d*x/2)**8
+ 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) -
16*tan(c/2 + d*x/2)**6/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x
/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) - 46*tan(c/2 + d*x/2)**5/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32
*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) - 144*
tan(c/2 + d*x/2)**4/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)
**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) + 46*tan(c/2 + d*x/2)**3/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a*
*3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) - 176*tan
(c/2 + d*x/2)**2/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4
+ 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) + 30*tan(c/2 + d*x/2)/(8*a**3*d*tan(c/2 + d*x/2)**8 + 32*a**3*d*t
an(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/2)**2 + 8*a**3*d) - 48/(8*a**3*d*
tan(c/2 + d*x/2)**8 + 32*a**3*d*tan(c/2 + d*x/2)**6 + 48*a**3*d*tan(c/2 + d*x/2)**4 + 32*a**3*d*tan(c/2 + d*x/
2)**2 + 8*a**3*d), Ne(d, 0)), (x*sin(c)*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. 267 vs. \(2 (78) = 156\).

Time = 0.31 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.18 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {88 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {23 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {72 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {23 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {8 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 24}{a^{3} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \]

[In]

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

[Out]

1/4*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 88*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 23*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 72*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 23*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 8*sin(d*x + c
)^6/(cos(d*x + c) + 1)^6 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 24)/(a^3 + 4*a^3*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 6*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d
*x + c)^8/(cos(d*x + c) + 1)^8) - 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 23 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 23 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 88 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \]

[In]

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

[Out]

-1/8*(15*(d*x + c)/a^3 + 2*(15*tan(1/2*d*x + 1/2*c)^7 + 8*tan(1/2*d*x + 1/2*c)^6 + 23*tan(1/2*d*x + 1/2*c)^5 +
72*tan(1/2*d*x + 1/2*c)^4 - 23*tan(1/2*d*x + 1/2*c)^3 + 88*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) +
24)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d

Mupad [B] (verification not implemented)

Time = 10.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\cos \left (c+d\,x\right )}^3}{a^3\,d}-\frac {4\,\cos \left (c+d\,x\right )}{a^3\,d}-\frac {15\,x}{8\,a^3}-\frac {{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,a^3\,d}+\frac {17\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,a^3\,d} \]

[In]

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

[Out]

cos(c + d*x)^3/(a^3*d) - (4*cos(c + d*x))/(a^3*d) - (15*x)/(8*a^3) - (cos(c + d*x)^3*sin(c + d*x))/(4*a^3*d) +
(17*cos(c + d*x)*sin(c + d*x))/(8*a^3*d)