Integrand size = 21, antiderivative size = 49 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 x}{a^3}-\frac {3 \cos (c+d x)}{a^3 d}-\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2} \]
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\cos ^5(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5 \sqrt {2} a^3 d (1+\sin (c+d x))^{5/2}} \]
-1/5*(Cos[c + d*x]^5*Hypergeometric2F1[3/2, 5/2, 7/2, (1 - Sin[c + d*x])/2 ])/(Sqrt[2]*a^3*d*(1 + Sin[c + d*x])^(5/2))
Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3159, 3042, 3161, 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.
\(\displaystyle \int \frac {\cos ^4(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle -\frac {3 \int \frac {\cos ^2(c+d x)}{\sin (c+d x) a+a}dx}{a^2}-\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 \int \frac {\cos (c+d x)^2}{\sin (c+d x) a+a}dx}{a^2}-\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle -\frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (c+d x)}{a d}\right )}{a^2}-\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {3 \left (\frac {\cos (c+d x)}{a d}+\frac {x}{a}\right )}{a^2}-\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2}\) |
3.1.79.3.1 Defintions of rubi rules used
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] + Simp[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] && EqQ[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]
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 = 0.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a^{3} d}\) | \(54\) |
default | \(\frac {-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a^{3} d}\) | \(54\) |
parallelrisch | \(\frac {-6 \cos \left (d x +c \right ) d x +10 \cos \left (d x +c \right )+8 \sin \left (d x +c \right )-\cos \left (2 d x +2 c \right )-9}{2 a^{3} d \cos \left (d x +c \right )}\) | \(56\) |
risch | \(-\frac {3 x}{a^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}-\frac {8}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\) | \(64\) |
norman | \(\frac {-\frac {3 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {90 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {396 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {186 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {444 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {412 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {298 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {34 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {153 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {213 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {324 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {90 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {10}{a d}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {42 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {90 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {252 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {15 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {8 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {42 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {106 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {153 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {210 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {252 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {213 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(493\) |
Time = 0.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.59 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \, d x + {\left (3 \, d x + 5\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + {\left (3 \, d x + \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 4}{a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d} \]
-(3*d*x + (3*d*x + 5)*cos(d*x + c) + cos(d*x + c)^2 + (3*d*x + cos(d*x + c ) - 4)*sin(d*x + c) + 4)/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (46) = 92\).
Time = 13.89 (sec) , antiderivative size = 478, normalized size of antiderivative = 9.76 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} - \frac {3 d x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {3 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {3 d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {3 d x}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {8 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {10}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((-3*d*x*tan(c/2 + d*x/2)**3/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d *tan(c/2 + d*x/2)**2 + a**3*d*tan(c/2 + d*x/2) + a**3*d) - 3*d*x*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**3 *d*tan(c/2 + d*x/2) + a**3*d) - 3*d*x*tan(c/2 + d*x/2)/(a**3*d*tan(c/2 + d *x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**3*d*tan(c/2 + d*x/2) + a**3*d) - 3*d*x/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**3*d* tan(c/2 + d*x/2) + a**3*d) - 8*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2 )**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**3*d*tan(c/2 + d*x/2) + a**3*d) - 2* tan(c/2 + d*x/2)/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**3*d*tan(c/2 + d*x/2) + a**3*d) - 10/(a**3*d*tan(c/2 + d*x/2)**3 + a** 3*d*tan(c/2 + d*x/2)**2 + a**3*d*tan(c/2 + d*x/2) + a**3*d), Ne(d, 0)), (x *cos(c)**4/(a*sin(c) + a)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (49) = 98\).
Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 5}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{d} \]
-2*((sin(d*x + c)/(cos(d*x + c) + 1) + 4*sin(d*x + c)^2/(cos(d*x + c) + 1) ^2 + 5)/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1) + a^3*sin(d*x + c)^2/(c os(d*x + c) + 1)^2 + a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) + 3*arctan(s in(d*x + c)/(cos(d*x + c) + 1))/a^3)/d
Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.63 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} a^{3}}}{d} \]
-(3*(d*x + c)/a^3 + 2*(4*tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 5 )/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 1)*a^3))/d
Time = 6.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3\,x}{a^3}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+10}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]