Integrand size = 21, antiderivative size = 61 \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^3 d}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\sin ^2(c+d x)}{2 b d} \] Output:
-(a^2-b^2)*ln(a+b*sin(d*x+c))/b^3/d+a*sin(d*x+c)/b^2/d-1/2*sin(d*x+c)^2/b/ d
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\left (\left (a^2-b^2\right ) \log (a+b \sin (c+d x))\right )+a b \sin (c+d x)-\frac {1}{2} b^2 \sin ^2(c+d x)}{b^3 d} \] Input:
Integrate[Cos[c + d*x]^3/(a + b*Sin[c + d*x]),x]
Output:
(-((a^2 - b^2)*Log[a + b*Sin[c + d*x]]) + a*b*Sin[c + d*x] - (b^2*Sin[c + d*x]^2)/2)/(b^3*d)
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {b^2-b^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (a-b \sin (c+d x)+\frac {b^2-a^2}{a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\left (a^2-b^2\right ) \log (a+b \sin (c+d x))+a b \sin (c+d x)-\frac {1}{2} b^2 \sin ^2(c+d x)}{b^3 d}\) |
Input:
Int[Cos[c + d*x]^3/(a + b*Sin[c + d*x]),x]
Output:
(-((a^2 - b^2)*Log[a + b*Sin[c + d*x]]) + a*b*Sin[c + d*x] - (b^2*Sin[c + d*x]^2)/2)/(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 = 0.60 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {b \sin \left (d x +c \right )^{2}}{2}+a \sin \left (d x +c \right )}{b^{2}}+\frac {\left (-a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(54\) |
default | \(\frac {\frac {-\frac {b \sin \left (d x +c \right )^{2}}{2}+a \sin \left (d x +c \right )}{b^{2}}+\frac {\left (-a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3}}}{d}\) | \(54\) |
parallelrisch | \(\frac {4 a^{2} \left (\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \right )\right )-4 b^{2} \left (\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \right )\right )-b^{2}+b^{2} \cos \left (2 d x +2 c \right )+4 a b \sin \left (d x +c \right )}{4 b^{3} d}\) | \(132\) |
risch | \(\frac {i x \,a^{2}}{b^{3}}-\frac {i x}{b}+\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}+\frac {2 i a^{2} c}{b^{3} d}-\frac {2 i c}{b d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b d}\) | \(188\) |
norman | \(\frac {-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b d}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2} d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{2} d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{b^{2} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{b^{3} d}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}{b^{3} d}\) | \(190\) |
Input:
int(cos(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/b^2*(-1/2*b*sin(d*x+c)^2+a*sin(d*x+c))+(-a^2+b^2)/b^3*ln(a+b*sin(d* x+c)))
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \sin \left (d x + c\right ) - 2 \, {\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{2 \, b^{3} d} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
1/2*(b^2*cos(d*x + c)^2 + 2*a*b*sin(d*x + c) - 2*(a^2 - b^2)*log(b*sin(d*x + c) + a))/(b^3*d)
Timed out. \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3/(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )}{b^{2}} + \frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2*((b*sin(d*x + c)^2 - 2*a*sin(d*x + c))/b^2 + 2*(a^2 - b^2)*log(b*sin( d*x + c) + a)/b^3)/d
Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{3} d} - \frac {b d \sin \left (d x + c\right )^{2} - 2 \, a d \sin \left (d x + c\right )}{2 \, b^{2} d^{2}} \] Input:
integrate(cos(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-(a^2 - b^2)*log(abs(b*sin(d*x + c) + a))/(b^3*d) - 1/2*(b*d*sin(d*x + c)^ 2 - 2*a*d*sin(d*x + c))/(b^2*d^2)
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^2}{2\,b}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{b^3}-\frac {a\,\sin \left (c+d\,x\right )}{b^2}}{d} \] Input:
int(cos(c + d*x)^3/(a + b*sin(c + d*x)),x)
Output:
-(sin(c + d*x)^2/(2*b) + (log(a + b*sin(c + d*x))*(a^2 - b^2))/b^3 - (a*si n(c + d*x))/b^2)/d
Time = 0.17 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.31 \[ \int \frac {\cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b^{2}-\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +2 b^{2}}{2 b^{3} d} \] Input:
int(cos(d*x+c)^3/(a+b*sin(d*x+c)),x)
Output:
(2*log(tan((c + d*x)/2)**2 + 1)*a**2 - 2*log(tan((c + d*x)/2)**2 + 1)*b**2 - 2*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**2 + 2*log(ta n((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b**2 - sin(c + d*x)**2*b** 2 + 2*sin(c + d*x)*a*b + 2*b**2)/(2*b**3*d)