Integrand size = 21, antiderivative size = 72 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\log (a+b \sin (c+d x))}{b^3 d}+\frac {a^2-b^2}{2 b^3 d (a+b \sin (c+d x))^2}-\frac {2 a}{b^3 d (a+b \sin (c+d x))} \]
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\log (a+b \sin (c+d x))+\frac {3 a^2+b^2+4 a b \sin (c+d x)}{2 (a+b \sin (c+d x))^2}}{b^3 d} \]
-((Log[a + b*Sin[c + d*x]] + (3*a^2 + b^2 + 4*a*b*Sin[c + d*x])/(2*(a + b* Sin[c + d*x])^2))/(b^3*d))
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 476, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{(a+b \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {b^2-b^2 \sin ^2(c+d x)}{(a+b \sin (c+d x))^3}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (\frac {2 a}{(a+b \sin (c+d x))^2}+\frac {1}{-a-b \sin (c+d x)}+\frac {b^2-a^2}{(a+b \sin (c+d x))^3}\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^2-b^2}{2 (a+b \sin (c+d x))^2}-\frac {2 a}{a+b \sin (c+d x)}-\log (a+b \sin (c+d x))}{b^3 d}\) |
(-Log[a + b*Sin[c + d*x]] + (a^2 - b^2)/(2*(a + b*Sin[c + d*x])^2) - (2*a) /(a + b*Sin[c + d*x]))/(b^3*d)
3.5.51.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.97 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {-\frac {-a^{2}+b^{2}}{2 b^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3}}-\frac {2 a}{b^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(66\) |
default | \(\frac {-\frac {-a^{2}+b^{2}}{2 b^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3}}-\frac {2 a}{b^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(66\) |
risch | \(\frac {i x}{b^{3}}+\frac {2 i c}{b^{3} d}-\frac {2 i \left (3 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d \,b^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{3} d}\) | \(151\) |
parallelrisch | \(\frac {\left (-8 a^{3} b \sin \left (d x +c \right )+2 a^{2} b^{2} \cos \left (2 d x +2 c \right )-4 a^{4}-2 a^{2} b^{2}\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+\left (8 a^{3} b \sin \left (d x +c \right )-2 a^{2} b^{2} \cos \left (2 d x +2 c \right )+4 a^{4}+2 a^{2} b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 a^{2} b^{2}-b^{4}\right ) \cos \left (2 d x +2 c \right )+\left (4 a^{3} b +4 a \,b^{3}\right ) \sin \left (d x +c \right )+3 a^{2} b^{2}+b^{4}}{4 \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 \sin \left (d x +c \right ) a b +a^{2}\right ) a^{2} d \,b^{3}}\) | \(230\) |
norman | \(\frac {\frac {\left (6 a^{2}+2 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} b d}+\frac {\left (6 a^{2}+2 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} b d}+\frac {2 \left (9 a^{2}+3 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d \,a^{2}}+\frac {2 \left (9 a^{2}+3 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d \,a^{2}}+\frac {2 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,b^{2} d}+\frac {8 \left (a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{2} d}+\frac {12 \left (a^{2}+b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{2} d}+\frac {8 \left (a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{2} d}+\frac {2 \left (a^{2}+b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{2} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} {\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}-\frac {\ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{3} d}\) | \(378\) |
1/d*(-1/2*(-a^2+b^2)/b^3/(a+b*sin(d*x+c))^2-1/b^3*ln(a+b*sin(d*x+c))-2*a/b ^3/(a+b*sin(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.53 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {4 \, a b \sin \left (d x + c\right ) + 3 \, a^{2} + b^{2} - 2 \, {\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{2 \, {\left (b^{5} d \cos \left (d x + c\right )^{2} - 2 \, a b^{4} d \sin \left (d x + c\right ) - {\left (a^{2} b^{3} + b^{5}\right )} d\right )}} \]
1/2*(4*a*b*sin(d*x + c) + 3*a^2 + b^2 - 2*(b^2*cos(d*x + c)^2 - 2*a*b*sin( d*x + c) - a^2 - b^2)*log(b*sin(d*x + c) + a))/(b^5*d*cos(d*x + c)^2 - 2*a *b^4*d*sin(d*x + c) - (a^2*b^3 + b^5)*d)
Leaf count of result is larger than twice the leaf count of optimal. 398 vs. \(2 (61) = 122\).
Time = 0.76 (sec) , antiderivative size = 398, normalized size of antiderivative = 5.53 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\begin {cases} \frac {x \cos ^{3}{\left (c \right )}}{a^{3}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{3}} & \text {for}\: b = 0 \\\frac {x \cos ^{3}{\left (c \right )}}{\left (a + b \sin {\left (c \right )}\right )^{3}} & \text {for}\: d = 0 \\- \frac {2 a^{2} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {2 a^{2}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {4 a b \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )} \sin {\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {2 a b \sin {\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {2 b^{2} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} - \frac {b^{2} \cos ^{2}{\left (c + d x \right )}}{2 a^{2} b^{3} d + 4 a b^{4} d \sin {\left (c + d x \right )} + 2 b^{5} d \sin ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]
Piecewise((x*cos(c)**3/a**3, Eq(b, 0) & Eq(d, 0)), ((2*sin(c + d*x)**3/(3* d) + sin(c + d*x)*cos(c + d*x)**2/d)/a**3, Eq(b, 0)), (x*cos(c)**3/(a + b* sin(c))**3, Eq(d, 0)), (-2*a**2*log(a/b + sin(c + d*x))/(2*a**2*b**3*d + 4 *a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d*x)**2) - 2*a**2/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d*x)**2) - 4*a*b*log(a/b + s in(c + d*x))*sin(c + d*x)/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b** 5*d*sin(c + d*x)**2) - 2*a*b*sin(c + d*x)/(2*a**2*b**3*d + 4*a*b**4*d*sin( c + d*x) + 2*b**5*d*sin(c + d*x)**2) - 2*b**2*log(a/b + sin(c + d*x))*sin( c + d*x)**2/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d* x)**2) - b**2*cos(c + d*x)**2/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2 *b**5*d*sin(c + d*x)**2), True))
Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {4 \, a b \sin \left (d x + c\right ) + 3 \, a^{2} + b^{2}}{b^{5} \sin \left (d x + c\right )^{2} + 2 \, a b^{4} \sin \left (d x + c\right ) + a^{2} b^{3}} + \frac {2 \, \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}}}{2 \, d} \]
-1/2*((4*a*b*sin(d*x + c) + 3*a^2 + b^2)/(b^5*sin(d*x + c)^2 + 2*a*b^4*sin (d*x + c) + a^2*b^3) + 2*log(b*sin(d*x + c) + a)/b^3)/d
Time = 0.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {2 \, \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{3}} + \frac {4 \, a \sin \left (d x + c\right ) + \frac {3 \, a^{2} + b^{2}}{b}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{2}}}{2 \, d} \]
-1/2*(2*log(abs(b*sin(d*x + c) + a))/b^3 + (4*a*sin(d*x + c) + (3*a^2 + b^ 2)/b)/((b*sin(d*x + c) + a)^2*b^2))/d
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{b^3\,d}-\frac {\frac {3\,a^2+b^2}{2\,b^3}+\frac {2\,a\,\sin \left (c+d\,x\right )}{b^2}}{d\,\left (a^2+2\,a\,b\,\sin \left (c+d\,x\right )+b^2\,{\sin \left (c+d\,x\right )}^2\right )} \]