Integrand size = 28, antiderivative size = 92 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\cos (c+d x)}{a d}+\frac {\log (1-\cos (c+d x))}{2 (a+b) d}-\frac {\log (1+\cos (c+d x))}{2 (a-b) d}+\frac {b^3 \log (b+a \cos (c+d x))}{a^2 \left (a^2-b^2\right ) d} \]
cos(d*x+c)/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 ^3*ln(b+a*cos(d*x+c))/a^2/(a^2-b^2)/d
Time = 0.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {\cos (c+d x)}{a}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{-a+b}+\frac {b^3 \log (b+a \cos (c+d x))}{a^4-a^2 b^2}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a+b}}{d} \]
(Cos[c + d*x]/a + Log[Cos[(c + d*x)/2]]/(-a + b) + (b^3*Log[b + a*Cos[c + d*x]])/(a^4 - a^2*b^2) + Log[Sin[(c + d*x)/2]]/(a + b))/d
Time = 0.56 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4897, 3042, 25, 3316, 25, 27, 604, 25, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2}{a \sin (c+d x)+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a \cos (c+d x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )^3}{\cos \left (c+d x-\frac {\pi }{2}\right ) \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^3}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right ) \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {a \int -\frac {\cos ^3(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a \int \frac {\cos ^3(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a^3 \cos ^3(c+d x)}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 604 |
\(\displaystyle -\frac {-\int -\frac {\cos (c+d x) a^3-b \cos ^2(c+d x) a^2+b a^2}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))-a \cos (c+d x)-b}{a^2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\cos (c+d x) a^3-b \cos ^2(c+d x) a^2+b a^2}{(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))-a \cos (c+d x)-b}{a^2 d}\) |
\(\Big \downarrow \) 2160 |
\(\displaystyle -\frac {\int \left (\frac {b^3}{(b-a) (a+b) (b+a \cos (c+d x))}+\frac {a^2}{2 (a+b) (a-a \cos (c+d x))}+\frac {a^2}{2 (a-b) (\cos (c+d x) a+a)}\right )d(a \cos (c+d x))-a \cos (c+d x)-b}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {b^3 \log (a \cos (c+d x)+b)}{a^2-b^2}-\frac {a^2 \log (a-a \cos (c+d x))}{2 (a+b)}+\frac {a^2 \log (a \cos (c+d x)+a)}{2 (a-b)}-a \cos (c+d x)-b}{a^2 d}\) |
-((-b - a*Cos[c + d*x] - (a^2*Log[a - a*Cos[c + d*x]])/(2*(a + b)) + (a^2* Log[a + a*Cos[c + d*x]])/(2*(a - b)) - (b^3*Log[b + a*Cos[c + d*x]])/(a^2 - b^2))/(a^2*d))
3.3.51.3.1 Defintions of rubi rules used
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 = 1.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {\cos \left (d x +c \right )}{a}+\frac {b^{3} \ln \left (b +\cos \left (d x +c \right ) a \right )}{a^{2} \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(85\) |
default | \(\frac {\frac {\cos \left (d x +c \right )}{a}+\frac {b^{3} \ln \left (b +\cos \left (d x +c \right ) a \right )}{a^{2} \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\cos \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {\ln \left (\cos \left (d x +c \right )+1\right )}{2 a -2 b}}{d}\) | \(85\) |
risch | \(-\frac {i x b}{a^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}-\frac {i x}{a +b}-\frac {i c}{d \left (a +b \right )}-\frac {2 i b^{3} x}{a^{2} \left (a^{2}-b^{2}\right )}-\frac {2 i b^{3} c}{a^{2} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a +b \right )}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d \left (a^{2}-b^{2}\right )}\) | \(228\) |
1/d*(cos(d*x+c)/a+1/a^2*b^3/(a+b)/(a-b)*ln(b+cos(d*x+c)*a)+1/(2*a+2*b)*ln( cos(d*x+c)-1)-1/(2*a-2*b)*ln(cos(d*x+c)+1))
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {2 \, b^{3} \log \left (a \cos \left (d x + c\right ) + b\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) - {\left (a^{3} + a^{2} b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} - a^{2} b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d} \]
1/2*(2*b^3*log(a*cos(d*x + c) + b) + 2*(a^3 - a*b^2)*cos(d*x + c) - (a^3 + a^2*b)*log(1/2*cos(d*x + c) + 1/2) + (a^3 - a^2*b)*log(-1/2*cos(d*x + c) + 1/2))/((a^4 - a^2*b^2)*d)
\[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {b^{3} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{4} - a^{2} b^{2}} + \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a + b} + \frac {b \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{2}} + \frac {2}{a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]
(b^3*log(a + b - (a - b)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^4 - a^2*b ^2) + log(sin(d*x + c)/(cos(d*x + c) + 1))/(a + b) + b*log(sin(d*x + c)^2/ (cos(d*x + c) + 1)^2 + 1)/a^2 + 2/(a + a*sin(d*x + c)^2/(cos(d*x + c) + 1) ^2))/d
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (88) = 176\).
Time = 0.37 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {\frac {2 \, b^{3} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{4} - a^{2} b^{2}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a + b} + \frac {2 \, b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (2 \, a - b + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}}}{2 \, d} \]
1/2*(2*b^3*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(c os(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^4 - a^2*b^2) + log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a + b) + 2*b*log(abs(-(cos(d*x + c) - 1)/ (cos(d*x + c) + 1) + 1))/a^2 - 2*(2*a - b + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/(a^2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)))/d
Time = 23.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^2(c+d x)}{a \sin (c+d x)+b \tan (c+d x)} \, dx=\frac {2}{a\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d\,\left (a+b\right )}+\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}+\frac {b^3\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a^2\,d\,\left (a^2-b^2\right )} \]