Integrand size = 27, antiderivative size = 84 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {15 x}{8 a^3}-\frac {4 \cos (c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{a^3 d}+\frac {15 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d} \] Output:
-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
Leaf count is larger than twice the leaf count of optimal. \(255\) vs. \(2(84)=168\).
Time = 1.62 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {(1+120 d x) \cos \left (\frac {c}{2}\right )+104 \cos \left (\frac {c}{2}+d x\right )+104 \cos \left (\frac {3 c}{2}+d x\right )-32 \cos \left (\frac {3 c}{2}+2 d x\right )+32 \cos \left (\frac {5 c}{2}+2 d x\right )-8 \cos \left (\frac {5 c}{2}+3 d x\right )-8 \cos \left (\frac {7 c}{2}+3 d x\right )+\cos \left (\frac {7 c}{2}+4 d x\right )-\cos \left (\frac {9 c}{2}+4 d x\right )-\sin \left (\frac {c}{2}\right )+120 d x \sin \left (\frac {c}{2}\right )-104 \sin \left (\frac {c}{2}+d x\right )+104 \sin \left (\frac {3 c}{2}+d x\right )-32 \sin \left (\frac {3 c}{2}+2 d x\right )-32 \sin \left (\frac {5 c}{2}+2 d x\right )+8 \sin \left (\frac {5 c}{2}+3 d x\right )-8 \sin \left (\frac {7 c}{2}+3 d x\right )+\sin \left (\frac {7 c}{2}+4 d x\right )+\sin \left (\frac {9 c}{2}+4 d x\right )}{64 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \] Input:
Integrate[(Cos[c + d*x]^6*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]
Output:
-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]))
Time = 0.59 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.40, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3338, 3042, 3158, 3042, 3161, 3042, 3115, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sin (c+d x) \cos ^6(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^6}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle -\frac {3 \int \frac {\cos ^6(c+d x)}{(\sin (c+d x) a+a)^2}dx}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \int \frac {\cos (c+d x)^6}{(\sin (c+d x) a+a)^2}dx}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle -\frac {3 \left (\frac {5 \int \frac {\cos ^4(c+d x)}{\sin (c+d x) a+a}dx}{4 a}+\frac {\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \left (\frac {5 \int \frac {\cos (c+d x)^4}{\sin (c+d x) a+a}dx}{4 a}+\frac {\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle -\frac {3 \left (\frac {5 \left (\frac {\int \cos ^2(c+d x)dx}{a}+\frac {\cos ^3(c+d x)}{3 a d}\right )}{4 a}+\frac {\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \left (\frac {5 \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{a}+\frac {\cos ^3(c+d x)}{3 a d}\right )}{4 a}+\frac {\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {3 \left (\frac {5 \left (\frac {\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}}{a}+\frac {\cos ^3(c+d x)}{3 a d}\right )}{4 a}+\frac {\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {3 \left (\frac {\cos ^5(c+d x)}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {5 \left (\frac {\cos ^3(c+d x)}{3 a d}+\frac {\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}}{a}\right )}{4 a}\right )}{a}-\frac {\cos ^7(c+d x)}{d (a \sin (c+d x)+a)^3}\) |
Input:
Int[(Cos[c + d*x]^6*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]
Output:
-(Cos[c + d*x]^7/(d*(a + a*Sin[c + d*x])^3)) - (3*(Cos[c + d*x]^5/(4*d*(a^ 2 + a^2*Sin[c + d*x])) + (5*(Cos[c + d*x]^3/(3*a*d) + (x/2 + (Cos[c + d*x] *Sin[c + d*x])/(2*d))/a))/(4*a)))/a
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[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] && In tegersQ[2*m, 2*p]
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] + Si mp[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]
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] + Simp[(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]
Time = 0.84 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67
| method | result | size |
| parallelrisch | \(\frac {-60 d x +8 \cos \left (3 d x +3 c \right )-104 \cos \left (d x +c \right )-\sin \left (4 d x +4 c \right )+32 \sin \left (2 d x +2 c \right )+96}{32 d \,a^{3}}\) | \(56\) |
| risch | \(-\frac {15 x}{8 a^{3}}-\frac {13 \cos \left (d x +c \right )}{4 a^{3} d}-\frac {\sin \left (4 d x +4 c \right )}{32 d \,a^{3}}+\frac {\cos \left (3 d x +3 c \right )}{4 a^{3} d}+\frac {\sin \left (2 d x +2 c \right )}{d \,a^{3}}\) | \(72\) |
| derivativedivides | \(\frac {\frac {4 \left (-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {3}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{3}}\) | \(129\) |
| default | \(\frac {\frac {4 \left (-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {3}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {15 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{3}}\) | \(129\) |
Input:
int(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/32*(-60*d*x+8*cos(3*d*x+3*c)-104*cos(d*x+c)-sin(4*d*x+4*c)+32*sin(2*d*x+ 2*c)+96)/d/a^3
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8 \, \cos \left (d x + c\right )^{3} - 15 \, d x - {\left (2 \, \cos \left (d x + c\right )^{3} - 17 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 32 \, \cos \left (d x + c\right )}{8 \, a^{3} d} \] Input:
integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 1246 vs. \(2 (78) = 156\).
Time = 61.12 (sec) , antiderivative size = 1246, normalized size of antiderivative = 14.83 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**6*sin(d*x+c)/(a+a*sin(d*x+c))**3,x)
Output:
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 + 3 2*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)...
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (78) = 156\).
Time = 0.12 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.18 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {88 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {23 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {72 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {23 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {8 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 24}{a^{3} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \] Input:
integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
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
Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 23 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 23 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 88 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \] Input:
integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-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) + 2 4)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d
Time = 32.76 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\cos \left (c+d\,x\right )}^3}{a^3\,d}-\frac {4\,\cos \left (c+d\,x\right )}{a^3\,d}-\frac {15\,x}{8\,a^3}-\frac {{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,a^3\,d}+\frac {17\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,a^3\,d} \] Input:
int((cos(c + d*x)^6*sin(c + d*x))/(a + a*sin(c + d*x))^3,x)
Output:
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)
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right )-24 \cos \left (d x +c \right )-15 d x +24}{8 a^{3} d} \] Input:
int(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^3,x)
Output:
(2*cos(c + d*x)*sin(c + d*x)**3 - 8*cos(c + d*x)*sin(c + d*x)**2 + 15*cos( c + d*x)*sin(c + d*x) - 24*cos(c + d*x) - 15*d*x + 24)/(8*a**3*d)