Integrand size = 13, antiderivative size = 1 \[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx=0 \] Output:
0
Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
Time = 1.93 (sec) , antiderivative size = 400, normalized size of antiderivative = 400.00 \[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx=\frac {\csc (x) \left (-8 a b \left (7 a^2+6 b^2\right )-96 (a-i b) (a+i b) b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )-6 \left (3 a^4+12 a^2 b^2+8 b^4\right ) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+6 \left (3 a^4+12 a^2 b^2+8 b^4\right ) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {3 a^4}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^4}-\frac {8 a^3 b \sin \left (\frac {x}{2}\right )}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^3}-\frac {8 a b \left (7 a^2+6 b^2\right ) \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )}-\frac {3 a^4}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^4}+\frac {8 a^3 b \sin \left (\frac {x}{2}\right )}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3}-\frac {a^2 \left (9 a^2+4 a b+12 b^2\right )}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}+\frac {8 a b \left (7 a^2+6 b^2\right ) \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}-\frac {a^2 \left (9 a^2-4 a b+12 b^2\right )}{-1+\sin (x)}\right ) (b \cos (x)+a \sin (x))}{48 a^5 (a+b \cot (x))} \] Input:
Integrate[Sec[x]^5/(a + b*Cot[x]),x]
Output:
(Csc[x]*(-8*a*b*(7*a^2 + 6*b^2) - 96*(a - I*b)*(a + I*b)*b*Sqrt[a^2 + b^2] *ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]] - 6*(3*a^4 + 12*a^2*b^2 + 8*b^ 4)*Log[Cos[x/2] - Sin[x/2]] + 6*(3*a^4 + 12*a^2*b^2 + 8*b^4)*Log[Cos[x/2] + Sin[x/2]] + (3*a^4)/(Cos[x/2] - Sin[x/2])^4 - (8*a^3*b*Sin[x/2])/(Cos[x/ 2] - Sin[x/2])^3 - (8*a*b*(7*a^2 + 6*b^2)*Sin[x/2])/(Cos[x/2] - Sin[x/2]) - (3*a^4)/(Cos[x/2] + Sin[x/2])^4 + (8*a^3*b*Sin[x/2])/(Cos[x/2] + Sin[x/2 ])^3 - (a^2*(9*a^2 + 4*a*b + 12*b^2))/(Cos[x/2] + Sin[x/2])^2 + (8*a*b*(7* a^2 + 6*b^2)*Sin[x/2])/(Cos[x/2] + Sin[x/2]) - (a^2*(9*a^2 - 4*a*b + 12*b^ 2))/(-1 + Sin[x]))*(b*Cos[x] + a*Sin[x]))/(48*a^5*(a + b*Cot[x]))
Result contains higher order function than in optimal. Order 3 vs. order 1 in optimal.
Time = 0.55 (sec) , antiderivative size = 145, normalized size of antiderivative = 145.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3042, 4001, 25, 25, 3042, 3589, 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.
| \(\displaystyle \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right )^5 \left (a-b \tan \left (x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4001 |
| \(\displaystyle \int -\frac {\tan (x) \sec ^4(x)}{-a \sin (x)-b \cos (x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\sec ^4(x) \tan (x)}{b \cos (x)+a \sin (x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\tan (x) \sec ^4(x)}{a \sin (x)+b \cos (x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (x)}{\cos (x)^5 (a \sin (x)+b \cos (x))}dx\) |
| \(\Big \downarrow \) 3589 |
| \(\displaystyle \int \left (\frac {\sec ^5(x)}{a}-\frac {b \sec ^4(x)}{a (a \sin (x)+b \cos (x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \text {arctanh}(\sin (x))}{2 a^3}+\frac {b^2 \tan (x) \sec (x)}{2 a^3}-\frac {b \sec ^3(x)}{3 a^2}+\frac {b^2 \left (a^2+b^2\right ) \text {arctanh}(\sin (x))}{a^5}+\frac {b \left (a^2+b^2\right )^{3/2} \text {arctanh}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^5}-\frac {b \left (a^2+b^2\right ) \sec (x)}{a^4}+\frac {3 \text {arctanh}(\sin (x))}{8 a}+\frac {\tan (x) \sec ^3(x)}{4 a}+\frac {3 \tan (x) \sec (x)}{8 a}\) |
Input:
Int[Sec[x]^5/(a + b*Cot[x]),x]
Output:
(3*ArcTanh[Sin[x]])/(8*a) + (b^2*ArcTanh[Sin[x]])/(2*a^3) + (b^2*(a^2 + b^ 2)*ArcTanh[Sin[x]])/a^5 + (b*(a^2 + b^2)^(3/2)*ArcTanh[(a*Cos[x] - b*Sin[x ])/Sqrt[a^2 + b^2]])/a^5 - (b*(a^2 + b^2)*Sec[x])/a^4 - (b*Sec[x]^3)/(3*a^ 2) + (3*Sec[x]*Tan[x])/(8*a) + (b^2*Sec[x]*Tan[x])/(2*a^3) + (Sec[x]^3*Tan [x])/(4*a)
Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_. ) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[Ex pandTrig[cos[c + d*x]^m*(sin[c + d*x]^n/(a*cos[c + d*x] + b*sin[c + d*x])), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Int[Sin[e + f*x]^m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/C os[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && ILtQ [n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 12.03 (sec) , antiderivative size = 311, normalized size of antiderivative = 311.00
| method | result | size |
| default | \(\frac {2 b \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \operatorname {arctanh}\left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{5} \sqrt {a^{2}+b^{2}}}+\frac {1}{4 a \left (\tan \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {-3 a -2 b}{6 a^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {-7 a^{2}-4 a b -4 b^{2}}{8 a^{3} \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {\left (-3 a^{4}-12 a^{2} b^{2}-8 b^{4}\right ) \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{8 a^{5}}-\frac {-5 a^{3}-12 a^{2} b -4 a \,b^{2}-8 b^{3}}{8 a^{4} \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {1}{4 a \left (\tan \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {-3 a +2 b}{6 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {7 a^{2}-4 a b +4 b^{2}}{8 a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\left (3 a^{4}+12 a^{2} b^{2}+8 b^{4}\right ) \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{8 a^{5}}-\frac {-5 a^{3}+12 a^{2} b -4 a \,b^{2}+8 b^{3}}{8 a^{4} \left (\tan \left (\frac {x}{2}\right )+1\right )}\) | \(311\) |
| risch | \(-\frac {{\mathrm e}^{i x} \left (-12 i a \,b^{2} {\mathrm e}^{2 i x}+12 i a \,b^{2} {\mathrm e}^{6 i x}+24 a^{2} b \,{\mathrm e}^{6 i x}+24 b^{3} {\mathrm e}^{6 i x}-9 i a^{3}-12 i a \,b^{2}+104 a^{2} b \,{\mathrm e}^{4 i x}+72 b^{3} {\mathrm e}^{4 i x}+9 i a^{3} {\mathrm e}^{6 i x}+33 i a^{3} {\mathrm e}^{4 i x}+104 a^{2} b \,{\mathrm e}^{2 i x}+72 b^{3} {\mathrm e}^{2 i x}+12 i a \,b^{2} {\mathrm e}^{4 i x}-33 i a^{3} {\mathrm e}^{2 i x}+24 a^{2} b +24 b^{3}\right )}{12 a^{4} \left ({\mathrm e}^{2 i x}+1\right )^{4}}+\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b \ln \left ({\mathrm e}^{i x}-\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \left (i b -a \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\right )}{a^{5}}-\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b \ln \left ({\mathrm e}^{i x}+\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \left (i b -a \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\right )}{a^{5}}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right )}{8 a}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right ) b^{2}}{2 a^{3}}-\frac {\ln \left ({\mathrm e}^{i x}-i\right ) b^{4}}{a^{5}}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right )}{8 a}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right ) b^{2}}{2 a^{3}}+\frac {\ln \left ({\mathrm e}^{i x}+i\right ) b^{4}}{a^{5}}\) | \(413\) |
Input:
int(sec(x)^5/(a+b*cot(x)),x,method=_RETURNVERBOSE)
Output:
2*b*(a^4+2*a^2*b^2+b^4)/a^5/(a^2+b^2)^(1/2)*arctanh(1/2*(-2*b*tan(1/2*x)+2 *a)/(a^2+b^2)^(1/2))+1/4/a/(tan(1/2*x)-1)^4-1/6*(-3*a-2*b)/a^2/(tan(1/2*x) -1)^3-1/8*(-7*a^2-4*a*b-4*b^2)/a^3/(tan(1/2*x)-1)^2+1/8/a^5*(-3*a^4-12*a^2 *b^2-8*b^4)*ln(tan(1/2*x)-1)-1/8*(-5*a^3-12*a^2*b-4*a*b^2-8*b^3)/a^4/(tan( 1/2*x)-1)-1/4/a/(tan(1/2*x)+1)^4-1/6*(-3*a+2*b)/a^2/(tan(1/2*x)+1)^3-1/8*( 7*a^2-4*a*b+4*b^2)/a^3/(tan(1/2*x)+1)^2+1/8*(3*a^4+12*a^2*b^2+8*b^4)/a^5*l n(tan(1/2*x)+1)-1/8*(-5*a^3+12*a^2*b-4*a*b^2+8*b^3)/a^4/(tan(1/2*x)+1)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.21 (sec) , antiderivative size = 236, normalized size of antiderivative = 236.00 \[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx=\frac {24 \, {\left (a^{2} b + b^{3}\right )} \sqrt {a^{2} + b^{2}} \cos \left (x\right )^{4} \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) + 3 \, {\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - 3 \, {\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) - 16 \, a^{3} b \cos \left (x\right ) - 48 \, {\left (a^{3} b + a b^{3}\right )} \cos \left (x\right )^{3} + 6 \, {\left (2 \, a^{4} + {\left (3 \, a^{4} + 4 \, a^{2} b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{48 \, a^{5} \cos \left (x\right )^{4}} \] Input:
integrate(sec(x)^5/(a+b*cot(x)),x, algorithm="fricas")
Output:
1/48*(24*(a^2*b + b^3)*sqrt(a^2 + b^2)*cos(x)^4*log((2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 - a^2 - 2*b^2 - 2*sqrt(a^2 + b^2)*(a*cos(x) - b*sin( x)))/(2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 + a^2)) + 3*(3*a^4 + 12*a ^2*b^2 + 8*b^4)*cos(x)^4*log(sin(x) + 1) - 3*(3*a^4 + 12*a^2*b^2 + 8*b^4)* cos(x)^4*log(-sin(x) + 1) - 16*a^3*b*cos(x) - 48*(a^3*b + a*b^3)*cos(x)^3 + 6*(2*a^4 + (3*a^4 + 4*a^2*b^2)*cos(x)^2)*sin(x))/(a^5*cos(x)^4)
\[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx=\int \frac {\sec ^{5}{\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \] Input:
integrate(sec(x)**5/(a+b*cot(x)),x)
Output:
Integral(sec(x)**5/(a + b*cot(x)), x)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.12 (sec) , antiderivative size = 396, normalized size of antiderivative = 396.00 \[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx=-\frac {32 \, a^{2} b + 24 \, b^{3} - \frac {3 \, {\left (5 \, a^{3} + 4 \, a b^{2}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {8 \, {\left (10 \, a^{2} b + 9 \, b^{3}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {3 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {24 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {3 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} - \frac {24 \, {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} - \frac {3 \, {\left (5 \, a^{3} + 4 \, a b^{2}\right )} \sin \left (x\right )^{7}}{{\left (\cos \left (x\right ) + 1\right )}^{7}}}{12 \, {\left (a^{4} - \frac {4 \, a^{4} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {6 \, a^{4} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {4 \, a^{4} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac {a^{4} \sin \left (x\right )^{8}}{{\left (\cos \left (x\right ) + 1\right )}^{8}}\right )}} + \frac {{\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{8 \, a^{5}} - \frac {{\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{8 \, a^{5}} + \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{5}} \] Input:
integrate(sec(x)^5/(a+b*cot(x)),x, algorithm="maxima")
Output:
-1/12*(32*a^2*b + 24*b^3 - 3*(5*a^3 + 4*a*b^2)*sin(x)/(cos(x) + 1) - 8*(10 *a^2*b + 9*b^3)*sin(x)^2/(cos(x) + 1)^2 - 3*(3*a^3 - 4*a*b^2)*sin(x)^3/(co s(x) + 1)^3 + 24*(4*a^2*b + 3*b^3)*sin(x)^4/(cos(x) + 1)^4 - 3*(3*a^3 - 4* a*b^2)*sin(x)^5/(cos(x) + 1)^5 - 24*(2*a^2*b + b^3)*sin(x)^6/(cos(x) + 1)^ 6 - 3*(5*a^3 + 4*a*b^2)*sin(x)^7/(cos(x) + 1)^7)/(a^4 - 4*a^4*sin(x)^2/(co s(x) + 1)^2 + 6*a^4*sin(x)^4/(cos(x) + 1)^4 - 4*a^4*sin(x)^6/(cos(x) + 1)^ 6 + a^4*sin(x)^8/(cos(x) + 1)^8) + 1/8*(3*a^4 + 12*a^2*b^2 + 8*b^4)*log(si n(x)/(cos(x) + 1) + 1)/a^5 - 1/8*(3*a^4 + 12*a^2*b^2 + 8*b^4)*log(sin(x)/( cos(x) + 1) - 1)/a^5 + (a^4*b + 2*a^2*b^3 + b^5)*log((a - b*sin(x)/(cos(x) + 1) + sqrt(a^2 + b^2))/(a - b*sin(x)/(cos(x) + 1) - sqrt(a^2 + b^2)))/(s qrt(a^2 + b^2)*a^5)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.14 (sec) , antiderivative size = 329, normalized size of antiderivative = 329.00 \[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx=\frac {{\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{8 \, a^{5}} - \frac {{\left (3 \, a^{4} + 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{8 \, a^{5}} + \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{5}} + \frac {15 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{7} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{7} + 48 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{6} + 24 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{6} + 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} - 96 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{4} - 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} + 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 80 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, a^{3} \tan \left (\frac {1}{2} \, x\right ) + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - 32 \, a^{2} b - 24 \, b^{3}}{12 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{4} a^{4}} \] Input:
integrate(sec(x)^5/(a+b*cot(x)),x, algorithm="giac")
Output:
1/8*(3*a^4 + 12*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*x) + 1))/a^5 - 1/8*(3*a^4 + 12*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*x) - 1))/a^5 + (a^4*b + 2*a^2*b^3 + b^5)*log(abs(2*b*tan(1/2*x) - 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*tan(1/2*x) - 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^5) + 1/12*(15*a^3*tan(1/2* x)^7 + 12*a*b^2*tan(1/2*x)^7 + 48*a^2*b*tan(1/2*x)^6 + 24*b^3*tan(1/2*x)^6 + 9*a^3*tan(1/2*x)^5 - 12*a*b^2*tan(1/2*x)^5 - 96*a^2*b*tan(1/2*x)^4 - 72 *b^3*tan(1/2*x)^4 + 9*a^3*tan(1/2*x)^3 - 12*a*b^2*tan(1/2*x)^3 + 80*a^2*b* tan(1/2*x)^2 + 72*b^3*tan(1/2*x)^2 + 15*a^3*tan(1/2*x) + 12*a*b^2*tan(1/2* x) - 32*a^2*b - 24*b^3)/((tan(1/2*x)^2 - 1)^4*a^4)
Time = 10.03 (sec) , antiderivative size = 1971, normalized size of antiderivative = 1971.00 \[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(x)^5*(a + b*cot(x))),x)
Output:
((tan(x/2)*(5*a^2 + 4*b^2))/(4*a^3) - (2*(4*a^2*b + 3*b^3))/(3*a^4) + (tan (x/2)^3*(3*a^2 - 4*b^2))/(4*a^3) + (tan(x/2)^5*(3*a^2 - 4*b^2))/(4*a^3) + (tan(x/2)^7*(5*a^2 + 4*b^2))/(4*a^3) + (2*tan(x/2)^6*(2*a^2*b + b^3))/a^4 - (2*tan(x/2)^4*(4*a^2*b + 3*b^3))/a^4 + (2*tan(x/2)^2*(10*a^2*b + 9*b^3)) /(3*a^4))/(6*tan(x/2)^4 - 4*tan(x/2)^2 - 4*tan(x/2)^6 + tan(x/2)^8 + 1) - (atan(((((3*a^4)/8 + b^4 + (3*a^2*b^2)/2)*((32*a^4*b^10 + 96*a^6*b^8 + 96* a^8*b^6 + 36*a^10*b^4 + (9*a^12*b^2)/2)/a^11 - (((3*a^4)/8 + b^4 + (3*a^2* b^2)/2)*((12*a^14*b + 16*a^10*b^5 + 28*a^12*b^3)/a^11 - (tan(x/2)*(128*a^1 0*b^6 + 256*a^12*b^4 + 128*a^14*b^2))/(2*a^12) + ((32*a^3*b^2 + (tan(x/2)* (192*a^16*b + 128*a^14*b^3))/(2*a^12))*((3*a^4)/8 + b^4 + (3*a^2*b^2)/2))/ a^5))/a^5 + (tan(x/2)*(18*a^14*b + 128*a^4*b^11 + 576*a^6*b^9 + 960*a^8*b^ 7 + 712*a^10*b^5 + 217*a^12*b^3))/(2*a^12))*1i)/a^5 + (((3*a^4)/8 + b^4 + (3*a^2*b^2)/2)*((32*a^4*b^10 + 96*a^6*b^8 + 96*a^8*b^6 + 36*a^10*b^4 + (9* a^12*b^2)/2)/a^11 - (((3*a^4)/8 + b^4 + (3*a^2*b^2)/2)*((tan(x/2)*(128*a^1 0*b^6 + 256*a^12*b^4 + 128*a^14*b^2))/(2*a^12) - (12*a^14*b + 16*a^10*b^5 + 28*a^12*b^3)/a^11 + ((32*a^3*b^2 + (tan(x/2)*(192*a^16*b + 128*a^14*b^3) )/(2*a^12))*((3*a^4)/8 + b^4 + (3*a^2*b^2)/2))/a^5))/a^5 + (tan(x/2)*(18*a ^14*b + 128*a^4*b^11 + 576*a^6*b^9 + 960*a^8*b^7 + 712*a^10*b^5 + 217*a^12 *b^3))/(2*a^12))*1i)/a^5)/((32*b^13 + 152*a^2*b^11 + 280*a^4*b^9 + 247*a^6 *b^7 + 102*a^8*b^5 + 15*a^10*b^3)/a^11 - (tan(x/2)*(128*b^14 + 640*a^2*...
Time = 0.18 (sec) , antiderivative size = 679, normalized size of antiderivative = 679.00 \[ \int \frac {\sec ^5(x)}{a+b \cot (x)} \, dx =\text {Too large to display} \] Input:
int(sec(x)^5/(a+b*cot(x)),x)
Output:
(48*sqrt(a**2 + b**2)*atan((tan(x/2)*b*i - a*i)/sqrt(a**2 + b**2))*sin(x)* *4*a**2*b*i + 48*sqrt(a**2 + b**2)*atan((tan(x/2)*b*i - a*i)/sqrt(a**2 + b **2))*sin(x)**4*b**3*i - 96*sqrt(a**2 + b**2)*atan((tan(x/2)*b*i - a*i)/sq rt(a**2 + b**2))*sin(x)**2*a**2*b*i - 96*sqrt(a**2 + b**2)*atan((tan(x/2)* b*i - a*i)/sqrt(a**2 + b**2))*sin(x)**2*b**3*i + 48*sqrt(a**2 + b**2)*atan ((tan(x/2)*b*i - a*i)/sqrt(a**2 + b**2))*a**2*b*i + 48*sqrt(a**2 + b**2)*a tan((tan(x/2)*b*i - a*i)/sqrt(a**2 + b**2))*b**3*i + 24*cos(x)*sin(x)**2*a **3*b + 24*cos(x)*sin(x)**2*a*b**3 - 32*cos(x)*a**3*b - 24*cos(x)*a*b**3 - 9*log(tan(x/2) - 1)*sin(x)**4*a**4 - 36*log(tan(x/2) - 1)*sin(x)**4*a**2* b**2 - 24*log(tan(x/2) - 1)*sin(x)**4*b**4 + 18*log(tan(x/2) - 1)*sin(x)** 2*a**4 + 72*log(tan(x/2) - 1)*sin(x)**2*a**2*b**2 + 48*log(tan(x/2) - 1)*s in(x)**2*b**4 - 9*log(tan(x/2) - 1)*a**4 - 36*log(tan(x/2) - 1)*a**2*b**2 - 24*log(tan(x/2) - 1)*b**4 + 9*log(tan(x/2) + 1)*sin(x)**4*a**4 + 36*log( tan(x/2) + 1)*sin(x)**4*a**2*b**2 + 24*log(tan(x/2) + 1)*sin(x)**4*b**4 - 18*log(tan(x/2) + 1)*sin(x)**2*a**4 - 72*log(tan(x/2) + 1)*sin(x)**2*a**2* b**2 - 48*log(tan(x/2) + 1)*sin(x)**2*b**4 + 9*log(tan(x/2) + 1)*a**4 + 36 *log(tan(x/2) + 1)*a**2*b**2 + 24*log(tan(x/2) + 1)*b**4 - 8*sin(x)**4*a** 3*b - 12*sin(x)**4*a*b**3 - 9*sin(x)**3*a**4 - 12*sin(x)**3*a**2*b**2 + 16 *sin(x)**2*a**3*b + 24*sin(x)**2*a*b**3 + 15*sin(x)*a**4 + 12*sin(x)*a**2* b**2 - 8*a**3*b - 12*a*b**3)/(24*a**5*(sin(x)**4 - 2*sin(x)**2 + 1))