Integrand size = 28, antiderivative size = 243 \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\frac {2 b x}{a^3}+\frac {2 b^6 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {2 b^4 \left (5 a^2-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}-\frac {\sin (c+d x)}{a^2 d}-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}-\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}-\frac {b^5 \sin (c+d x)}{a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \] Output:
2*b*x/a^3+2*b^6*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a -b)^(5/2)/(a+b)^(5/2)/d+2*b^4*(5*a^2-3*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d* x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2)/d-sin(d*x+c)/a^2/d-1/2*s in(d*x+c)/(a+b)^2/d/(1-cos(d*x+c))-1/2*sin(d*x+c)/(a-b)^2/d/(1+cos(d*x+c)) -b^5*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))
Time = 2.59 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=-\frac {-\frac {4 b (c+d x)}{a^3}+\frac {4 b^4 \left (5 a^2-2 b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}+\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2}+\frac {2 \sin (c+d x)}{a^2}+\frac {2 b^5 \sin (c+d x)}{a^2 (a-b)^2 (a+b)^2 (b+a \cos (c+d x))}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2}}{2 d} \] Input:
Integrate[Cos[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]
Output:
-1/2*((-4*b*(c + d*x))/a^3 + (4*b^4*(5*a^2 - 2*b^2)*ArcTanh[((-a + b)*Tan[ (c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2)) + Cot[(c + d*x)/2] /(a + b)^2 + (2*Sin[c + d*x])/a^2 + (2*b^5*Sin[c + d*x])/(a^2*(a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])) + Tan[(c + d*x)/2]/(a - b)^2)/d
Time = 0.92 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 4897, 3042, 25, 3376, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^3}{(a \sin (c+d x)+b \tan (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a \cos (c+d x)+b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )^5}{\cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^5}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3376 |
| \(\displaystyle -\int \left (\frac {b^5}{a^3 \left (a^2-b^2\right ) (-b-a \cos (c+d x))^2}-\frac {2 b}{a^3}+\frac {\cos (c+d x)}{a^2}-\frac {1}{2 (a-b)^2 (-\cos (c+d x)-1)}-\frac {1}{2 (a+b)^2 (1-\cos (c+d x))}+\frac {5 a^2 b^4-3 b^6}{a^3 \left (a^2-b^2\right )^2 (-b-a \cos (c+d x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b^6 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac {2 b x}{a^3}-\frac {b^5 \sin (c+d x)}{a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac {\sin (c+d x)}{a^2 d}+\frac {2 b^4 \left (5 a^2-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac {\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)}\) |
Input:
Int[Cos[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]
Output:
(2*b*x)/a^3 + (2*b^6*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/ (a^3*(a - b)^(5/2)*(a + b)^(5/2)*d) + (2*b^4*(5*a^2 - 3*b^2)*ArcTanh[(Sqrt [a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(5/2)*(a + b)^(5/2)*d ) - Sin[c + d*x]/(a^2*d) - Sin[c + d*x]/(2*(a + b)^2*d*(1 - Cos[c + d*x])) - Sin[c + d*x]/(2*(a - b)^2*d*(1 + Cos[c + d*x])) - (b^5*Sin[c + d*x])/(a ^2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
Time = 4.80 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b^{4} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b}-\frac {\left (5 a^{2}-2 b^{2}\right ) \operatorname {arctanh}\left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{3}}}{d}\) | \(215\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b^{4} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a -b}-\frac {\left (5 a^{2}-2 b^{2}\right ) \operatorname {arctanh}\left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{3}}}{d}\) | \(215\) |
| risch | \(\frac {2 b x}{a^{3}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}-\frac {2 i \left (a^{6} {\mathrm e}^{3 i \left (d x +c \right )}+a^{4} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+b^{6} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{3} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a \,b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+a^{6} {\mathrm e}^{i \left (d x +c \right )}-3 a^{4} b^{2} {\mathrm e}^{i \left (d x +c \right )}-b^{6} {\mathrm e}^{i \left (d x +c \right )}-2 a^{5} b -a \,b^{5}\right )}{d \left (a^{2}-b^{2}\right )^{2} a^{3} \left (a \,{\mathrm e}^{4 i \left (d x +c \right )}+2 b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 b \,{\mathrm e}^{i \left (d x +c \right )}-a \right )}+\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {2 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}-\frac {5 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}+\frac {2 b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}\) | \(580\) |
Input:
int(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2*tan(1/2*d*x+1/2*c)/(a^2-2*a*b+b^2)+2/a^3*(-a*tan(1/2*d*x+1/2*c)/ (1+tan(1/2*d*x+1/2*c)^2)+2*b*arctan(tan(1/2*d*x+1/2*c)))-1/2/(a+b)^2/tan(1 /2*d*x+1/2*c)-2*b^4/(a+b)^2/(a-b)^2/a^3*(-a*b*tan(1/2*d*x+1/2*c)/(tan(1/2* d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2-a-b)-(5*a^2-2*b^2)/((a+b)*(a-b))^(1/ 2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a+b)*(a-b))^(1/2))))
Time = 0.13 (sec) , antiderivative size = 857, normalized size of antiderivative = 3.53 \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="fricas" )
Output:
[-1/2*(4*a^7*b - 6*a^5*b^3 + 6*a^3*b^5 - 4*a*b^7 - 2*(a^8 - 3*a^6*b^2 + 3* a^4*b^4 - a^2*b^6)*cos(d*x + c)^3 + (5*a^2*b^5 - 2*b^7 + (5*a^3*b^4 - 2*a* b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2) *cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2* a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 2*(3*a^7*b - 5*a^5*b^3 + 4*a^3*b^5 - 2*a*b^7)*cos(d*x + c)^2 + 2*(2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 - a^2*b^6)*cos(d*x + c) - 4*((a^7*b - 3*a^5*b^3 + 3 *a^3*b^5 - a*b^7)*d*x*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^ 8)*d*x)*sin(d*x + c))/(((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*d*cos(d*x + c) + (a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d)*sin(d*x + c)), -(2*a^ 7*b - 3*a^5*b^3 + 3*a^3*b^5 - 2*a*b^7 - (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2 *b^6)*cos(d*x + c)^3 - (5*a^2*b^5 - 2*b^7 + (5*a^3*b^4 - 2*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - (3*a^7*b - 5*a^5*b^3 + 4*a^3*b^5 - 2 *a*b^7)*cos(d*x + c)^2 + (2*a^8 - 5*a^6*b^2 + 4*a^4*b^4 - a^2*b^6)*cos(d*x + c) - 2*((a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x*cos(d*x + c) + (a^6 *b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*x)*sin(d*x + c))/(((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*d*cos(d*x + c) + (a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d)*sin(d*x + c))]
Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="maxima" )
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 1362 vs. \(2 (219) = 438\).
Time = 0.45 (sec) , antiderivative size = 1362, normalized size of antiderivative = 5.60 \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
-1/2*(2*((2*a^4*b - 2*a^3*b^2 - 4*a^2*b^3 - a*b^4 + 2*b^5)*sqrt(-a^2 + b^2 )*abs(a^7 - 2*a^5*b^2 + a^3*b^4)*abs(a - b) - (2*a^11*b - 2*a^10*b^2 - 8*a ^9*b^3 + 13*a^8*b^4 + 12*a^7*b^5 - 24*a^6*b^6 - 8*a^5*b^7 + 17*a^4*b^8 + 2 *a^3*b^9 - 4*a^2*b^10)*sqrt(-a^2 + b^2)*abs(a - b))*(pi*floor(1/2*(d*x + c )/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a^6*b - 2*a^4*b^3 + a^2*b ^5 + sqrt((a^7 + a^6*b - 2*a^5*b^2 - 2*a^4*b^3 + a^3*b^4 + a^2*b^5)*(a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^3*b^4 - a^2*b^5) + (a^6*b - 2*a^4*b^3 + a^2*b^5)^2))/(a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^3*b^4 - a^2*b^5)))) /((a^7 - 2*a^5*b^2 + a^3*b^4)^2*(a^2 - 2*a*b + b^2) + (a^8*b - 2*a^7*b^2 - a^6*b^3 + 4*a^5*b^4 - a^4*b^5 - 2*a^3*b^6 + a^2*b^7)*abs(a^7 - 2*a^5*b^2 + a^3*b^4)) + 2*(2*a^11*b - 2*a^10*b^2 - 8*a^9*b^3 + 13*a^8*b^4 + 12*a^7*b ^5 - 24*a^6*b^6 - 8*a^5*b^7 + 17*a^4*b^8 + 2*a^3*b^9 - 4*a^2*b^10 + 2*a^4* b*abs(a^7 - 2*a^5*b^2 + a^3*b^4) - 2*a^3*b^2*abs(a^7 - 2*a^5*b^2 + a^3*b^4 ) - 4*a^2*b^3*abs(a^7 - 2*a^5*b^2 + a^3*b^4) - a*b^4*abs(a^7 - 2*a^5*b^2 + a^3*b^4) + 2*b^5*abs(a^7 - 2*a^5*b^2 + a^3*b^4))*(pi*floor(1/2*(d*x + c)/ pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a^6*b - 2*a^4*b^3 + a^2*b^5 - sqrt((a^7 + a^6*b - 2*a^5*b^2 - 2*a^4*b^3 + a^3*b^4 + a^2*b^5)*(a^7 - a ^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^3*b^4 - a^2*b^5) + (a^6*b - 2*a^4*b^3 + a ^2*b^5)^2))/(a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^3*b^4 - a^2*b^5))))/( a^6*b*abs(a^7 - 2*a^5*b^2 + a^3*b^4) - 2*a^4*b^3*abs(a^7 - 2*a^5*b^2 + ...
Time = 21.13 (sec) , antiderivative size = 7329, normalized size of antiderivative = 30.16 \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int(cos(c + d*x)^3/(a*sin(c + d*x) + b*tan(c + d*x))^2,x)
Output:
((a^2 - 2*a*b + b^2)/(a + b) + (2*tan(c/2 + (d*x)/2)^2*(2*a*b^4 + 3*a^4*b + 2*a^5 + 4*b^5 - 3*a^2*b^3 - 6*a^3*b^2))/(a^2*(a + b)^2) - (tan(c/2 + (d* x)/2)^4*(4*a*b^4 - 7*a^4*b + 5*a^5 - 8*b^5 + 7*a^2*b^3 - 5*a^3*b^2))/(a^2* (a + b)^2))/(d*(tan(c/2 + (d*x)/2)^5*(6*a*b^2 - 6*a^2*b + 2*a^3 - 2*b^3) - tan(c/2 + (d*x)/2)^3*(4*a^2*b - 8*a*b^2 + 4*b^3) + tan(c/2 + (d*x)/2)*(2* a*b^2 + 2*a^2*b - 2*a^3 - 2*b^3))) - tan(c/2 + (d*x)/2)/(2*d*(a - b)^2) + (4*b*atan((1920*a^7*b^22*tan(c/2 + (d*x)/2))/(1920*a^7*b^22 - 1920*a^8*b^2 1 - 16640*a^9*b^20 + 16640*a^10*b^19 + 62080*a^11*b^18 - 62080*a^12*b^17 - 131072*a^13*b^16 + 131072*a^14*b^15 + 172672*a^15*b^14 - 172672*a^16*b^13 - 147200*a^17*b^12 + 147200*a^18*b^11 + 81280*a^19*b^10 - 81280*a^20*b^9 - 28160*a^21*b^8 + 28160*a^22*b^7 + 5632*a^23*b^6 - 5632*a^24*b^5 - 512*a^ 25*b^4 + 512*a^26*b^3) - (1920*a^8*b^21*tan(c/2 + (d*x)/2))/(1920*a^7*b^22 - 1920*a^8*b^21 - 16640*a^9*b^20 + 16640*a^10*b^19 + 62080*a^11*b^18 - 62 080*a^12*b^17 - 131072*a^13*b^16 + 131072*a^14*b^15 + 172672*a^15*b^14 - 1 72672*a^16*b^13 - 147200*a^17*b^12 + 147200*a^18*b^11 + 81280*a^19*b^10 - 81280*a^20*b^9 - 28160*a^21*b^8 + 28160*a^22*b^7 + 5632*a^23*b^6 - 5632*a^ 24*b^5 - 512*a^25*b^4 + 512*a^26*b^3) - (16640*a^9*b^20*tan(c/2 + (d*x)/2) )/(1920*a^7*b^22 - 1920*a^8*b^21 - 16640*a^9*b^20 + 16640*a^10*b^19 + 6208 0*a^11*b^18 - 62080*a^12*b^17 - 131072*a^13*b^16 + 131072*a^14*b^15 + 1726 72*a^15*b^14 - 172672*a^16*b^13 - 147200*a^17*b^12 + 147200*a^18*b^11 +...
Time = 0.17 (sec) , antiderivative size = 698, normalized size of antiderivative = 2.87 \[ \int \frac {\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x)
Output:
(10*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt( - a**2 + b**2))*cos(c + d*x)*sin(c + d*x)*a**3*b**4 - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2) )*cos(c + d*x)*sin(c + d*x)*a*b**6 + 10*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)*a**2* b**5 - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)* b)/sqrt( - a**2 + b**2))*sin(c + d*x)*b**7 - cos(c + d*x)*sin(c + d*x)**2* a**8 + 3*cos(c + d*x)*sin(c + d*x)**2*a**6*b**2 - 3*cos(c + d*x)*sin(c + d *x)**2*a**4*b**4 + cos(c + d*x)*sin(c + d*x)**2*a**2*b**6 + 2*cos(c + d*x) *sin(c + d*x)*a**7*b*d*x - 6*cos(c + d*x)*sin(c + d*x)*a**5*b**3*d*x + 6*c os(c + d*x)*sin(c + d*x)*a**3*b**5*d*x - 2*cos(c + d*x)*sin(c + d*x)*a*b** 7*d*x - cos(c + d*x)*a**8 + 2*cos(c + d*x)*a**6*b**2 - cos(c + d*x)*a**4*b **4 - 3*sin(c + d*x)**2*a**7*b + 5*sin(c + d*x)**2*a**5*b**3 - 4*sin(c + d *x)**2*a**3*b**5 + 2*sin(c + d*x)**2*a*b**7 + 2*sin(c + d*x)*a**6*b**2*d*x - 6*sin(c + d*x)*a**4*b**4*d*x + 6*sin(c + d*x)*a**2*b**6*d*x - 2*sin(c + d*x)*b**8*d*x + a**7*b - 2*a**5*b**3 + a**3*b**5)/(sin(c + d*x)*a**3*d*(c os(c + d*x)*a**7 - 3*cos(c + d*x)*a**5*b**2 + 3*cos(c + d*x)*a**3*b**4 - c os(c + d*x)*a*b**6 + a**6*b - 3*a**4*b**3 + 3*a**2*b**5 - b**7))