Integrand size = 21, antiderivative size = 73 \[ \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 x}{8 a}+\frac {\cos ^5(c+d x)}{5 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \]
Time = 0.54 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos ^7(c+d x) \left (30 \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {1+\sin (c+d x)} \left (-8-17 \sin (c+d x)+41 \sin ^2(c+d x)-6 \sin ^3(c+d x)-18 \sin ^4(c+d x)+8 \sin ^5(c+d x)\right )\right )}{40 a d (-1+\sin (c+d x))^4 (1+\sin (c+d x))^{7/2}} \]
-1/40*(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]]*(-8 - 17*Sin[c + d*x] + 41*Sin[c + d*x]^2 - 6*Sin[c + d*x]^3 - 18*Sin[c + d*x]^4 + 8*Sin[c + d*x]^5)))/(a*d*( -1 + Sin[c + d*x])^4*(1 + Sin[c + d*x])^(7/2))
Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3161, 3042, 3115, 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.
\(\displaystyle \int \frac {\cos ^6(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^6}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle \frac {\int \cos ^4(c+d x)dx}{a}+\frac {\cos ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx}{a}+\frac {\cos ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}}{a}+\frac {\cos ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}}{a}+\frac {\cos ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}}{a}+\frac {\cos ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\cos ^5(c+d x)}{5 a d}+\frac {\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )}{a}\) |
Cos[c + d*x]^5/(5*a*d) + ((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)/a
3.1.51.3.1 Defintions of rubi rules used
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 _)]), 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]
Time = 0.43 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {60 d x +2 \cos \left (5 d x +5 c \right )+5 \sin \left (4 d x +4 c \right )+10 \cos \left (3 d x +3 c \right )+40 \sin \left (2 d x +2 c \right )+20 \cos \left (d x +c \right )+32}{160 d a}\) | \(67\) |
risch | \(\frac {3 x}{8 a}+\frac {\cos \left (d x +c \right )}{8 a d}+\frac {\cos \left (5 d x +5 c \right )}{80 a d}+\frac {\sin \left (4 d x +4 c \right )}{32 d a}+\frac {\cos \left (3 d x +3 c \right )}{16 a d}+\frac {\sin \left (2 d x +2 c \right )}{4 d a}\) | \(90\) |
derivativedivides | \(\frac {\frac {2 \left (-\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {1}{5}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\) | \(114\) |
default | \(\frac {\frac {2 \left (-\frac {5 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {1}{5}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d a}\) | \(114\) |
norman | \(\frac {\frac {3 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {31 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {57 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {19 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {59 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {93 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {81 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {33 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {5 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {45 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {45 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {49 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {33}{20 a d}+\frac {9 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}+\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{10 d a}+\frac {9 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}+\frac {45 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {15 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {9 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}+\frac {9 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a}+\frac {183 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d a}+\frac {193 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d a}+\frac {15 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {3 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {45 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {3 x}{8 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(490\) |
1/160*(60*d*x+2*cos(5*d*x+5*c)+5*sin(4*d*x+4*c)+10*cos(3*d*x+3*c)+40*sin(2 *d*x+2*c)+20*cos(d*x+c)+32)/d/a
Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {8 \, \cos \left (d x + c\right )^{5} + 15 \, d x + 5 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40 \, a d} \]
Leaf count of result is larger than twice the leaf count of optimal. 1355 vs. \(2 (60) = 120\).
Time = 11.91 (sec) , antiderivative size = 1355, normalized size of antiderivative = 18.56 \[ \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]
Piecewise((15*d*x*tan(c/2 + d*x/2)**10/(40*a*d*tan(c/2 + d*x/2)**10 + 200* a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 75*d*x*tan(c/2 + d*x/2 )**8/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d* tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/ 2)**2 + 40*a*d) + 150*d*x*tan(c/2 + d*x/2)**6/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan (c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 150*d*x*tan(c/2 + d*x/2)**4/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/ 2 + d*x/2)**2 + 40*a*d) + 75*d*x*tan(c/2 + d*x/2)**2/(40*a*d*tan(c/2 + d*x /2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400* a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 15*d*x/( 40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/ 2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) - 50*tan(c/2 + d*x/2)**9/(40*a*d*tan(c/2 + d*x/2)**10 + 200*a*d* tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/ 2)**4 + 200*a*d*tan(c/2 + d*x/2)**2 + 40*a*d) + 80*tan(c/2 + d*x/2)**8/(40 *a*d*tan(c/2 + d*x/2)**10 + 200*a*d*tan(c/2 + d*x/2)**8 + 400*a*d*tan(c/2 + d*x/2)**6 + 400*a*d*tan(c/2 + d*x/2)**4 + 200*a*d*tan(c/2 + d*x/2)**2...
Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (65) = 130\).
Time = 0.27 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.53 \[ \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {25 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {80 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {10 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {40 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {25 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 8}{a + \frac {5 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{20 \, d} \]
1/20*((25*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3 + 80*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 10*sin(d*x + c)^7/(cos (d*x + c) + 1)^7 + 40*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 25*sin(d*x + c )^9/(cos(d*x + c) + 1)^9 + 8)/(a + 5*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a*sin(d*x + c)^10 /(cos(d*x + c) + 1)^10) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d
Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.56 \[ \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {15 \, {\left (d x + c\right )}}{a} - \frac {2 \, {\left (25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a}}{40 \, d} \]
1/40*(15*(d*x + c)/a - 2*(25*tan(1/2*d*x + 1/2*c)^9 - 40*tan(1/2*d*x + 1/2 *c)^8 + 10*tan(1/2*d*x + 1/2*c)^7 - 80*tan(1/2*d*x + 1/2*c)^4 - 10*tan(1/2 *d*x + 1/2*c)^3 - 25*tan(1/2*d*x + 1/2*c) - 8)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a))/d
Time = 9.96 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.47 \[ \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3\,x}{8\,a}+\frac {-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {2}{5}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]