Integrand size = 28, antiderivative size = 231 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{b^2 d}+\frac {2 a^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {2 a^3 \left (a^2-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^2 (a+b)^{5/2} d}-\frac {\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}-\frac {\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}-\frac {a^4 \sin (c+d x)}{b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \]
arctanh(sin(d*x+c))/b^2/d+2*a^3*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+ b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d-2*a^3*(a^2-3*b^2)*arctanh((a-b)^(1/2)* tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/b^2/(a+b)^(5/2)/d-1/2*sin(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))-a^4*si n(d*x+c)/b/(a^2-b^2)^2/d/(b+a*cos(d*x+c))
Time = 2.68 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=-\frac {-\frac {4 \left (a^5-4 a^3 b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2}}+\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2}+\frac {2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^2}-\frac {2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^2}+\frac {2 a^4 \sin (c+d x)}{(a-b)^2 b (a+b)^2 (b+a \cos (c+d x))}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2}}{2 d} \]
-1/2*((-4*(a^5 - 4*a^3*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)) + Cot[(c + d*x)/2]/(a + b)^2 + (2*Log[Cos[ (c + d*x)/2] - Sin[(c + d*x)/2]])/b^2 - (2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/b^2 + (2*a^4*Sin[c + d*x])/((a - b)^2*b*(a + b)^2*(b + a*Cos[c + d*x])) + Tan[(c + d*x)/2]/(a - b)^2)/d
Time = 0.71 (sec) , antiderivative size = 231, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^3}{(a \sin (c+d x)+b \tan (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec (c+d x)}{(a \cos (c+d x)+b)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\sin \left (c+d x-\frac {\pi }{2}\right ) \cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \sin \left (\frac {1}{2} (2 c-\pi )+d x\right ) \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3376 |
| \(\displaystyle -\int \left (\frac {a^3}{b \left (a^2-b^2\right ) (-b-a \cos (c+d x))^2}-\frac {\sec (c+d x)}{b^2}-\frac {1}{2 (a-b)^2 (-\cos (c+d x)-1)}-\frac {1}{2 (a+b)^2 (1-\cos (c+d x))}+\frac {3 a^3 b^2-a^5}{b^2 \left (a^2-b^2\right )^2 (-b-a \cos (c+d x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {a^4 \sin (c+d x)}{b d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac {2 a^3 \left (a^2-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac {\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)}+\frac {\text {arctanh}(\sin (c+d x))}{b^2 d}\) |
ArcTanh[Sin[c + d*x]]/(b^2*d) + (2*a^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/ 2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*d) - (2*a^3*(a^2 - 3*b^2)*A rcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^2*(a + b)^(5/2)*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])) - (a^4*Sin[c + d*x])/(b*(a^2 - b^2 )^2*d*(b + a*Cos[c + d*x]))
3.3.63.3.1 Defintions of rubi rules used
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 = 17.05 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 a^{3} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b}-\frac {\left (a^{2}-4 b^{2}\right ) \operatorname {arctanh}\left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} b^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}}{d}\) | \(199\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 a^{3} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b}-\frac {\left (a^{2}-4 b^{2}\right ) \operatorname {arctanh}\left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} b^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}}{d}\) | \(199\) |
| risch | \(-\frac {2 i \left (2 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3 a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-a^{4}-2 a^{2} b^{2}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right ) b \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{2}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{2}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{2} d}+\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{b^{2} d}\) | \(529\) |
1/d*(-1/2*tan(1/2*d*x+1/2*c)/(a^2-2*a*b+b^2)+2*a^3/(a-b)^2/(a+b)^2/b^2*(a* b*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)-( a^2-4*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)))-1/2/(a+b)^2/tan(1/2*d*x+1/2*c)-1/b^2*ln(tan(1/2*d*x+1/2*c)-1)+ 1/b^2*ln(tan(1/2*d*x+1/2*c)+1))
Time = 0.77 (sec) , antiderivative size = 864, normalized size of antiderivative = 3.74 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\left [-\frac {2 \, a^{6} b - 2 \, b^{7} + {\left (a^{5} b - 4 \, a^{3} b^{3} + {\left (a^{6} - 4 \, a^{4} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{6} b + a^{4} b^{3} - 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{7} b^{2} - 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} - a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, a^{6} b - 2 \, b^{7} + 2 \, {\left (a^{5} b - 4 \, a^{3} b^{3} + {\left (a^{6} - 4 \, a^{4} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{6} b + a^{4} b^{3} - 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{7} b^{2} - 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} - a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
[-1/2*(2*a^6*b - 2*b^7 + (a^5*b - 4*a^3*b^3 + (a^6 - 4*a^4*b^2)*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*(a^6*b + a^4 *b^3 - 2*a^2*b^5)*cos(d*x + c)^2 - (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1)* sin(d*x + c) + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a^5*b^2 + 3 *a^3*b^4 - a*b^6)*cos(d*x + c))*log(-sin(d*x + c) + 1)*sin(d*x + c) + 2*(a ^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c))/(((a^7*b^2 - 3*a^5*b^4 + 3*a^3*b ^6 - a*b^8)*d*cos(d*x + c) + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d)*si n(d*x + c)), -1/2*(2*a^6*b - 2*b^7 + 2*(a^5*b - 4*a^3*b^3 + (a^6 - 4*a^4*b ^2)*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) - 2*(a^6*b + a^4*b^3 - 2*a ^2*b^5)*cos(d*x + c)^2 - (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a ^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1)*sin(d*x + c) + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(-sin(d*x + c) + 1)*sin(d*x + c) + 2*(a^5*b^2 - 2 *a^3*b^4 + a*b^6)*cos(d*x + c))/(((a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8 )*d*cos(d*x + c) + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d)*sin(d*x + c) )]
\[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.55 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.53 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (a^{5} - 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3} b - a^{2} b^{2} - a b^{3} + b^{4}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}} + \frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}}}{2 \, d} \]
1/2*(4*(a^5 - 4*a^3*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2 )))/((a^4*b^2 - 2*a^2*b^4 + b^6)*sqrt(-a^2 + b^2)) - tan(1/2*d*x + 1/2*c)/ (a^2 - 2*a*b + b^2) + (4*a^4*tan(1/2*d*x + 1/2*c)^2 - a^3*b*tan(1/2*d*x + 1/2*c)^2 + 3*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b^3*tan(1/2*d*x + 1/2*c) ^2 + b^4*tan(1/2*d*x + 1/2*c)^2 + a^3*b - a^2*b^2 - a*b^3 + b^4)/((a^4*b - 2*a^2*b^3 + b^5)*(a*tan(1/2*d*x + 1/2*c)^3 - b*tan(1/2*d*x + 1/2*c)^3 - a *tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))) + 2*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^2 - 2*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2)/d
Time = 27.60 (sec) , antiderivative size = 6056, normalized size of antiderivative = 26.22 \[ \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]
((a^2 - 2*a*b + b^2)/(a + b) + (tan(c/2 + (d*x)/2)^2*(4*a^4 - a^3*b - 3*a* b^3 + b^4 + 3*a^2*b^2))/(b*(a + b)^2))/(d*(tan(c/2 + (d*x)/2)^3*(6*a*b^2 - 6*a^2*b + 2*a^3 - 2*b^3) + tan(c/2 + (d*x)/2)*(2*a*b^2 + 2*a^2*b - 2*a^3 - 2*b^3))) - (atan(-(((tan(c/2 + (d*x)/2)*(32*b^26 - 96*a*b^25 - 224*a^2*b ^24 + 928*a^3*b^23 + 480*a^4*b^22 - 4000*a^5*b^21 + 992*a^6*b^20 + 9568*a^ 7*b^19 - 8128*a^8*b^18 - 12992*a^9*b^17 + 21344*a^10*b^16 + 8224*a^11*b^15 - 31744*a^12*b^14 + 2176*a^13*b^13 + 29600*a^14*b^12 - 8480*a^15*b^11 - 1 7632*a^16*b^10 + 7072*a^17*b^9 + 6528*a^18*b^8 - 3008*a^19*b^7 - 1376*a^20 *b^6 + 672*a^21*b^5 + 128*a^22*b^4 - 64*a^23*b^3) + (32*b^28 - 32*a*b^27 - 352*a^2*b^26 + 480*a^3*b^25 + 1504*a^4*b^24 - 2688*a^5*b^23 - 3168*a^6*b^ 22 + 8064*a^7*b^21 + 2880*a^8*b^20 - 14784*a^9*b^19 + 1344*a^10*b^18 + 174 72*a^11*b^17 - 6720*a^12*b^16 - 13440*a^13*b^15 + 8256*a^14*b^14 + 6528*a^ 15*b^13 - 5472*a^16*b^12 - 1824*a^17*b^11 + 2080*a^18*b^10 + 224*a^19*b^9 - 416*a^20*b^8 + 32*a^22*b^6 - (tan(c/2 + (d*x)/2)*(128*a^2*b^28 - 64*a*b^ 29 + 576*a^3*b^27 - 1280*a^4*b^26 - 2240*a^5*b^25 + 5760*a^6*b^24 + 4800*a ^7*b^23 - 15360*a^8*b^22 - 5760*a^9*b^21 + 26880*a^10*b^20 + 2688*a^11*b^1 9 - 32256*a^12*b^18 + 2688*a^13*b^17 + 26880*a^14*b^16 - 5760*a^15*b^15 - 15360*a^16*b^14 + 4800*a^17*b^13 + 5760*a^18*b^12 - 2240*a^19*b^11 - 1280* a^20*b^10 + 576*a^21*b^9 + 128*a^22*b^8 - 64*a^23*b^7))/b^2)/b^2)*1i)/b^2 + ((tan(c/2 + (d*x)/2)*(32*b^26 - 96*a*b^25 - 224*a^2*b^24 + 928*a^3*b^...