Integrand size = 29, antiderivative size = 82 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \log (1+\sin (c+d x))}{a^3 d}-\frac {4 \sin (c+d x)}{a^3 d}+\frac {2 \sin ^2(c+d x)}{a^3 d}-\frac {\sin ^3(c+d x)}{a^3 d}+\frac {\sin ^4(c+d x)}{4 a^3 d} \] Output:
4*ln(1+sin(d*x+c))/a^3/d-4*sin(d*x+c)/a^3/d+2*sin(d*x+c)^2/a^3/d-sin(d*x+c )^3/a^3/d+1/4*sin(d*x+c)^4/a^3/d
Time = 0.88 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {16 \log (1+\sin (c+d x))-16 \sin (c+d x)+8 \sin ^2(c+d x)-4 \sin ^3(c+d x)+\sin ^4(c+d x)}{4 a^3 d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(16*Log[1 + Sin[c + d*x]] - 16*Sin[c + d*x] + 8*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + Sin[c + d*x]^4)/(4*a^3*d)
Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 2009}
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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^5(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^5}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {\sin ^2(c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^2 \sin ^2(c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {4 a^4}{\sin (c+d x) a+a}+\sin ^3(c+d x) a^3-3 \sin ^2(c+d x) a^3+4 \sin (c+d x) a^3-4 a^3\right )d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} a^4 \sin ^4(c+d x)-a^4 \sin ^3(c+d x)+2 a^4 \sin ^2(c+d x)-4 a^4 \sin (c+d x)+4 a^4 \log (a \sin (c+d x)+a)}{a^7 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(4*a^4*Log[a + a*Sin[c + d*x]] - 4*a^4*Sin[c + d*x] + 2*a^4*Sin[c + d*x]^2 - a^4*Sin[c + d*x]^3 + (a^4*Sin[c + d*x]^4)/4)/(a^7*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.66 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{4}}{4}-\sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )^{2}-4 \sin \left (d x +c \right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(58\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{4}}{4}-\sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )^{2}-4 \sin \left (d x +c \right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(58\) |
parallelrisch | \(\frac {-152 \sin \left (d x +c \right )-36 \cos \left (2 d x +2 c \right )+35+8 \sin \left (3 d x +3 c \right )+\cos \left (4 d x +4 c \right )-128 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+256 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32 d \,a^{3}}\) | \(78\) |
risch | \(-\frac {4 i x}{a^{3}}+\frac {19 i {\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {19 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {8 i c}{d \,a^{3}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {\cos \left (4 d x +4 c \right )}{32 d \,a^{3}}+\frac {\sin \left (3 d x +3 c \right )}{4 d \,a^{3}}-\frac {9 \cos \left (2 d x +2 c \right )}{8 d \,a^{3}}\) | \(127\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(1/4*sin(d*x+c)^4-sin(d*x+c)^3+2*sin(d*x+c)^2-4*sin(d*x+c)+4*ln(1+ sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{4 \, a^{3} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
1/4*(cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 5)*sin(d*x + c) + 16*log(sin(d*x + c) + 1))/(a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 1698 vs. \(2 (73) = 146\).
Time = 59.62 (sec) , antiderivative size = 1698, normalized size of antiderivative = 20.71 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**2/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((8*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**8/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 32*log(tan(c/2 + d*x/2) + 1)*t an(c/2 + d*x/2)**6/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2) **6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d ) + 48*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(a**3*d*tan(c/2 + d*x /2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a **3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 32*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 8* log(tan(c/2 + d*x/2) + 1)/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 4*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**8/(a**3*d*tan( c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2) **4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 16*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**6/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 24*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(a**3*d*tan( c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2) **4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 16*log(tan(c/2 + d*x/2)*...
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 8 \, \sin \left (d x + c\right )^{2} - 16 \, \sin \left (d x + c\right )}{a^{3}} + \frac {16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{4 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
1/4*((sin(d*x + c)^4 - 4*sin(d*x + c)^3 + 8*sin(d*x + c)^2 - 16*sin(d*x + c))/a^3 + 16*log(sin(d*x + c) + 1)/a^3)/d
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} + \frac {a^{9} d^{3} \sin \left (d x + c\right )^{4} - 4 \, a^{9} d^{3} \sin \left (d x + c\right )^{3} + 8 \, a^{9} d^{3} \sin \left (d x + c\right )^{2} - 16 \, a^{9} d^{3} \sin \left (d x + c\right )}{4 \, a^{12} d^{4}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
4*log(abs(sin(d*x + c) + 1))/(a^3*d) + 1/4*(a^9*d^3*sin(d*x + c)^4 - 4*a^9 *d^3*sin(d*x + c)^3 + 8*a^9*d^3*sin(d*x + c)^2 - 16*a^9*d^3*sin(d*x + c))/ (a^12*d^4)
Time = 32.70 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\sin \left (c+d\,x\right )}{a^3}+\frac {2\,{\sin \left (c+d\,x\right )}^2}{a^3}-\frac {{\sin \left (c+d\,x\right )}^3}{a^3}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a^3}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)
Output:
((4*log(sin(c + d*x) + 1))/a^3 - (4*sin(c + d*x))/a^3 + (2*sin(c + d*x)^2) /a^3 - sin(c + d*x)^3/a^3 + sin(c + d*x)^4/(4*a^3))/d
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right )^{4}-4 \sin \left (d x +c \right )^{3}+8 \sin \left (d x +c \right )^{2}-16 \sin \left (d x +c \right )-8}{4 a^{3} d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x)
Output:
( - 16*log(tan((c + d*x)/2)**2 + 1) + 32*log(tan((c + d*x)/2) + 1) + sin(c + d*x)**4 - 4*sin(c + d*x)**3 + 8*sin(c + d*x)**2 - 16*sin(c + d*x) - 8)/ (4*a**3*d)