Integrand size = 21, antiderivative size = 63 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 a \log (a+b \sin (c+d x))}{b^3 d}-\frac {\sin (c+d x)}{b^2 d}+\frac {a^2-b^2}{b^3 d (a+b \sin (c+d x))} \] Output:
2*a*ln(a+b*sin(d*x+c))/b^3/d-sin(d*x+c)/b^2/d+(a^2-b^2)/b^3/d/(a+b*sin(d*x +c))
Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 a \log (a+b \sin (c+d x))-b \sin (c+d x)+\frac {(a-b) (a+b)}{a+b \sin (c+d x)}}{b^3 d} \] Input:
Integrate[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]
Output:
(2*a*Log[a + b*Sin[c + d*x]] - b*Sin[c + d*x] + ((a - b)*(a + b))/(a + b*S in[c + d*x]))/(b^3*d)
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{(a+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {b^2-b^2 \sin ^2(c+d x)}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (\frac {2 a}{a+b \sin (c+d x)}+\frac {b^2-a^2}{(a+b \sin (c+d x))^2}-1\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^2-b^2}{a+b \sin (c+d x)}+2 a \log (a+b \sin (c+d x))-b \sin (c+d x)}{b^3 d}\) |
Input:
Int[Cos[c + d*x]^3/(a + b*Sin[c + d*x])^2,x]
Output:
(2*a*Log[a + b*Sin[c + d*x]] - b*Sin[c + d*x] + (a^2 - b^2)/(a + b*Sin[c + d*x]))/(b^3*d)
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 = 0.41 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )}{b^{2}}+\frac {2 a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3}}-\frac {-a^{2}+b^{2}}{b^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(60\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )}{b^{2}}+\frac {2 a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3}}-\frac {-a^{2}+b^{2}}{b^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(60\) |
risch | \(-\frac {2 i a x}{b^{3}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}-\frac {4 i a c}{b^{3} d}+\frac {2 \left (a^{2}-b^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{d \,b^{3} \left (-i {\mathrm e}^{2 i \left (d x +c \right )} b +i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{3} d}\) | \(152\) |
parallelrisch | \(-\frac {-4 a^{2}+3 b^{2}+4 a^{2} \left (-\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )\right )+4 \sin \left (d x +c \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a b -4 \sin \left (d x +c \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a b -b^{2} \cos \left (2 d x +2 c \right )}{d \,b^{3} \left (2 a +2 b \sin \left (d x +c \right )\right )}\) | \(155\) |
norman | \(\frac {-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b d}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{b d}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b d}-\frac {2 \left (2 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,b^{2} d}-\frac {6 \left (2 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a \,b^{2} d}-\frac {6 \left (2 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a \,b^{2} d}-\frac {2 \left (2 a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a \,b^{2} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}-\frac {2 a \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{b^{3} d}+\frac {2 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{3} d}\) | \(295\) |
Input:
int(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b^2*sin(d*x+c)+2/b^3*a*ln(a+b*sin(d*x+c))-1/b^3*(-a^2+b^2)/(a+b*si n(d*x+c)))
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {b^{2} \cos \left (d x + c\right )^{2} - a b \sin \left (d x + c\right ) + a^{2} - 2 \, b^{2} + 2 \, {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4} d \sin \left (d x + c\right ) + a b^{3} d} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
(b^2*cos(d*x + c)^2 - a*b*sin(d*x + c) + a^2 - 2*b^2 + 2*(a*b*sin(d*x + c) + a^2)*log(b*sin(d*x + c) + a))/(b^4*d*sin(d*x + c) + a*b^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (53) = 106\).
Time = 0.75 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.51 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\begin {cases} \frac {x \cos ^{3}{\left (c \right )}}{a^{2}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}}{a^{2}} & \text {for}\: b = 0 \\\frac {x \cos ^{3}{\left (c \right )}}{\left (a + b \sin {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\\frac {2 a^{2} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} + \frac {2 a^{2}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} + \frac {2 a b \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )} \sin {\left (c + d x \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} - \frac {2 b^{2} \sin ^{2}{\left (c + d x \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} - \frac {b^{2} \cos ^{2}{\left (c + d x \right )}}{a b^{3} d + b^{4} d \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3/(a+b*sin(d*x+c))**2,x)
Output:
Piecewise((x*cos(c)**3/a**2, 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**2, Eq(b, 0)), (x*cos(c)**3/(a + b* sin(c))**2, Eq(d, 0)), (2*a**2*log(a/b + sin(c + d*x))/(a*b**3*d + b**4*d* sin(c + d*x)) + 2*a**2/(a*b**3*d + b**4*d*sin(c + d*x)) + 2*a*b*log(a/b + sin(c + d*x))*sin(c + d*x)/(a*b**3*d + b**4*d*sin(c + d*x)) - 2*b**2*sin(c + d*x)**2/(a*b**3*d + b**4*d*sin(c + d*x)) - b**2*cos(c + d*x)**2/(a*b**3 *d + b**4*d*sin(c + d*x)), True))
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {a^{2} - b^{2}}{b^{4} \sin \left (d x + c\right ) + a b^{3}} + \frac {2 \, a \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}} - \frac {\sin \left (d x + c\right )}{b^{2}}}{d} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
((a^2 - b^2)/(b^4*sin(d*x + c) + a*b^3) + 2*a*log(b*sin(d*x + c) + a)/b^3 - sin(d*x + c)/b^2)/d
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, a \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{3} d} - \frac {\sin \left (d x + c\right )}{b^{2} d} + \frac {a^{2} - b^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{3} d} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
2*a*log(abs(b*sin(d*x + c) + a))/(b^3*d) - sin(d*x + c)/(b^2*d) + (a^2 - b ^2)/((b*sin(d*x + c) + a)*b^3*d)
Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2\,a\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{b^3\,d}-\frac {\sin \left (c+d\,x\right )}{b^2\,d}+\frac {a^2-b^2}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^3+a\,b^2\right )} \] Input:
int(cos(c + d*x)^3/(a + b*sin(c + d*x))^2,x)
Output:
(2*a*log(a + b*sin(c + d*x)))/(b^3*d) - sin(c + d*x)/(b^2*d) + (a^2 - b^2) /(b*d*(a*b^2 + b^3*sin(c + d*x)))
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\cos \left (d x +c \right )^{2} b^{2}+2 \,\mathrm {log}\left (\sin \left (d x +c \right ) b +a \right ) \sin \left (d x +c \right ) a b +2 \,\mathrm {log}\left (\sin \left (d x +c \right ) b +a \right ) a^{2}-2 \sin \left (d x +c \right )^{2} b^{2}-2 \sin \left (d x +c \right ) a b}{b^{3} d \left (\sin \left (d x +c \right ) b +a \right )} \] Input:
int(cos(d*x+c)^3/(a+b*sin(d*x+c))^2,x)
Output:
( - cos(c + d*x)**2*b**2 + 2*log(sin(c + d*x)*b + a)*sin(c + d*x)*a*b + 2* log(sin(c + d*x)*b + a)*a**2 - 2*sin(c + d*x)**2*b**2 - 2*sin(c + d*x)*a*b )/(b**3*d*(sin(c + d*x)*b + a))