Integrand size = 28, antiderivative size = 113 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {b \cos (c+d x)}{a^2 d}+\frac {\cos ^2(c+d x)}{2 a d}+\frac {\log (1-\cos (c+d x))}{2 (a+b) d}+\frac {\log (1+\cos (c+d x))}{2 (a-b) d}-\frac {b^4 \log (b+a \cos (c+d x))}{a^3 \left (a^2-b^2\right ) d} \] Output:
-b*cos(d*x+c)/a^2/d+1/2*cos(d*x+c)^2/a/d+1/2*ln(1-cos(d*x+c))/(a+b)/d+1/2* ln(1+cos(d*x+c))/(a-b)/d-b^4*ln(b+a*cos(d*x+c))/a^3/(a^2-b^2)/d
Time = 0.49 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {-\frac {4 b \cos (c+d x)}{a^2}+\frac {\cos (2 (c+d x))}{a}+4 \left (\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a-b}+\frac {b^4 \log (b+a \cos (c+d x))}{a^3 \left (-a^2+b^2\right )}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}\right )}{4 d} \] Input:
Integrate[Cos[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]
Output:
((-4*b*Cos[c + d*x])/a^2 + Cos[2*(c + d*x)]/a + 4*(Log[Cos[(c + d*x)/2]]/( a - b) + (b^4*Log[b + a*Cos[c + d*x]])/(a^3*(-a^2 + b^2)) + Log[Sin[(c + d *x)/2]]/(a + b)))/(4*d)
Time = 0.63 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4897, 3042, 3316, 27, 604, 27, 2160, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^3}{a \sin (c+d x)+b \tan (c+d x)}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^4}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle -\frac {a \int \frac {\cos ^4(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {a^4 \cos ^4(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 604 |
| \(\displaystyle -\frac {-\frac {1}{2} \int -\frac {2 \left (-2 b \cos ^3(c+d x) a^3+2 b \cos (c+d x) a^3+b^2 a^2+\left (a^2-b^2\right ) \cos ^2(c+d x) a^2\right )}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))-\frac {1}{2} (a \cos (c+d x)+b)^2}{a^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {-2 b \cos ^3(c+d x) a^3+2 b \cos (c+d x) a^3+b^2 a^2+\left (a^2-b^2\right ) \cos ^2(c+d x) a^2}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))-\frac {1}{2} (a \cos (c+d x)+b)^2}{a^3 d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle -\frac {\int \left (\frac {b^4}{(a-b) (a+b) (b+a \cos (c+d x))}+2 b+\frac {a^3}{2 (a+b) (a-a \cos (c+d x))}-\frac {a^3}{2 (a-b) (\cos (c+d x) a+a)}\right )d(a \cos (c+d x))-\frac {1}{2} (a \cos (c+d x)+b)^2}{a^3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {a^3 \log (a-a \cos (c+d x))}{2 (a+b)}-\frac {a^3 \log (a \cos (c+d x)+a)}{2 (a-b)}+\frac {b^4 \log (a \cos (c+d x)+b)}{a^2-b^2}+2 a b \cos (c+d x)-\frac {1}{2} (a \cos (c+d x)+b)^2}{a^3 d}\) |
Input:
Int[Cos[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x]),x]
Output:
-((2*a*b*Cos[c + d*x] - (b + a*Cos[c + d*x])^2/2 - (a^3*Log[a - a*Cos[c + d*x]])/(2*(a + b)) - (a^3*Log[a + a*Cos[c + d*x]])/(2*(a - b)) + (b^4*Log[ b + a*Cos[c + d*x]])/(a^2 - b^2))/(a^3*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(b*d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b* x^2)^p*ExpandToSum[b*d^m*(m + n + 2*p + 1)*x^m - b*(m + n + 2*p + 1)*(c + d *x)^m - (c + d*x)^(m - 2)*(a*d^2*(m + n - 1) - b*c^2*(m + n + 2*p + 1) - 2* b*c*d*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && IntegerQ[2*p]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 2.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\cos \left (d x +c \right )^{2} a}{2}-b \cos \left (d x +c \right )}{a^{2}}+\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 a +2 b}-\frac {b^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{3} \left (a +b \right ) \left (a -b \right )}}{d}\) | \(100\) |
| default | \(\frac {\frac {\frac {\cos \left (d x +c \right )^{2} a}{2}-b \cos \left (d x +c \right )}{a^{2}}+\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 a -2 b}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 a +2 b}-\frac {b^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{3} \left (a +b \right ) \left (a -b \right )}}{d}\) | \(100\) |
| risch | \(\frac {i x}{a}+\frac {i x \,b^{2}}{a^{3}}+\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}-\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}-\frac {b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {2 i b^{4} x}{a^{3} \left (a^{2}-b^{2}\right )}+\frac {2 i b^{4} c}{a^{3} d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{3} d \left (a^{2}-b^{2}\right )}\) | \(273\) |
Input:
int(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/a^2*(1/2*cos(d*x+c)^2*a-b*cos(d*x+c))+1/(2*a-2*b)*ln(1+cos(d*x+c))+ 1/(2*a+2*b)*ln(-1+cos(d*x+c))-1/a^3*b^4/(a+b)/(a-b)*ln(b+a*cos(d*x+c)))
Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {2 \, b^{4} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) - {\left (a^{4} + a^{3} b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{4} - a^{3} b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d} \] Input:
integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="fricas")
Output:
-1/2*(2*b^4*log(a*cos(d*x + c) + b) - (a^4 - a^2*b^2)*cos(d*x + c)^2 + 2*( a^3*b - a*b^3)*cos(d*x + c) - (a^4 + a^3*b)*log(1/2*cos(d*x + c) + 1/2) - (a^4 - a^3*b)*log(-1/2*cos(d*x + c) + 1/2))/((a^5 - a^3*b^2)*d)
Timed out. \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c)),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {\frac {b^{4} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{5} - a^{3} b^{2}} + \frac {2 \, {\left (b + \frac {{\left (a + b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{3}}}{d} \] Input:
integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="maxima")
Output:
-(b^4*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^5 - a^3* b^2) + 2*(b + (a + b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^2 + 2*a^2*si n(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^ 4) - log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b) + (a^2 + b^2)*log(sin(d* x + c)^2/(cos(d*x + c) + 1)^2 + 1)/a^3)/d
Time = 0.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=-\frac {b^{4} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{5} d - a^{3} b^{2} d} + \frac {\log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a d - b d\right )}} + \frac {\log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a d + b d\right )}} + \frac {a d \cos \left (d x + c\right )^{2} - 2 \, b d \cos \left (d x + c\right )}{2 \, a^{2} d^{2}} \] Input:
integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x, algorithm="giac")
Output:
-b^4*log(abs(a*cos(d*x + c) + b))/(a^5*d - a^3*b^2*d) + 1/2*log(abs(cos(d* x + c) + 1))/(a*d - b*d) + 1/2*log(abs(cos(d*x + c) - 1))/(a*d + b*d) + 1/ 2*(a*d*cos(d*x + c)^2 - 2*b*d*cos(d*x + c))/(a^2*d^2)
Time = 16.80 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}-\frac {\frac {2\,b}{a^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a+b\right )}{a^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {b^4\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d\,\left (a^5-a^3\,b^2\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2+b^2\right )}{a^3\,d} \] Input:
int(cos(c + d*x)^3/(a*sin(c + d*x) + b*tan(c + d*x)),x)
Output:
log(tan(c/2 + (d*x)/2))/(d*(a + b)) - ((2*b)/a^2 + (2*tan(c/2 + (d*x)/2)^2 *(a + b))/a^2)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - ( b^4*log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2))/(d*(a^5 - a^3*b^2)) - (log(tan(c/2 + (d*x)/2)^2 + 1)*(a^2 + b^2))/(a^3*d)
Time = 0.16 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^3(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {-2 \cos \left (d x +c \right ) a^{3} b +2 \cos \left (d x +c \right ) a \,b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) b^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3} b -\sin \left (d x +c \right )^{2} a^{4}+\sin \left (d x +c \right )^{2} a^{2} b^{2}+2 a^{3} b -2 a \,b^{3}}{2 a^{3} d \left (a^{2}-b^{2}\right )} \] Input:
int(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c)),x)
Output:
( - 2*cos(c + d*x)*a**3*b + 2*cos(c + d*x)*a*b**3 - 2*log(tan((c + d*x)/2) **2 + 1)*a**4 + 2*log(tan((c + d*x)/2)**2 + 1)*b**4 - 2*log(tan((c + d*x)/ 2)**2*a - tan((c + d*x)/2)**2*b - a - b)*b**4 + 2*log(tan((c + d*x)/2))*a* *4 - 2*log(tan((c + d*x)/2))*a**3*b - sin(c + d*x)**2*a**4 + sin(c + d*x)* *2*a**2*b**2 + 2*a**3*b - 2*a*b**3)/(2*a**3*d*(a**2 - b**2))