Integrand size = 11, antiderivative size = 101 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx=\frac {\log (a+b \sin (x))}{b^5}-\frac {\left (a^2-b^2\right )^2}{4 b^5 (a+b \sin (x))^4}+\frac {4 a \left (a^2-b^2\right )}{3 b^5 (a+b \sin (x))^3}-\frac {3 a^2-b^2}{b^5 (a+b \sin (x))^2}+\frac {4 a}{b^5 (a+b \sin (x))} \] Output:
ln(a+b*sin(x))/b^5-1/4*(a^2-b^2)^2/b^5/(a+b*sin(x))^4+4/3*a*(a^2-b^2)/b^5/ (a+b*sin(x))^3-(3*a^2-b^2)/b^5/(a+b*sin(x))^2+4*a/b^5/(a+b*sin(x))
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx=\frac {\log (a+b \sin (x))+\frac {25 a^4+2 a^2 b^2-3 b^4+8 a b \left (11 a^2+b^2\right ) \sin (x)+12 b^2 \left (9 a^2+b^2\right ) \sin ^2(x)+48 a b^3 \sin ^3(x)}{12 (a+b \sin (x))^4}}{b^5} \] Input:
Integrate[(a*Sec[x] + b*Tan[x])^(-5),x]
Output:
(Log[a + b*Sin[x]] + (25*a^4 + 2*a^2*b^2 - 3*b^4 + 8*a*b*(11*a^2 + b^2)*Si n[x] + 12*b^2*(9*a^2 + b^2)*Sin[x]^2 + 48*a*b^3*Sin[x]^3)/(12*(a + b*Sin[x ])^4))/b^5
Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4891, 3042, 3147, 476, 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.
| \(\displaystyle \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sec (x)+b \tan (x))^5}dx\) |
| \(\Big \downarrow \) 4891 |
| \(\displaystyle \int \frac {\cos ^5(x)}{(a+b \sin (x))^5}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (x)^5}{(a+b \sin (x))^5}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {\int \frac {\left (b^2-b^2 \sin ^2(x)\right )^2}{(a+b \sin (x))^5}d(b \sin (x))}{b^5}\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \frac {\int \left (\frac {\left (a^2-b^2\right )^2}{(a+b \sin (x))^5}+\frac {1}{a+b \sin (x)}-\frac {4 a}{(a+b \sin (x))^2}+\frac {2 \left (3 a^2-b^2\right )}{(a+b \sin (x))^3}-\frac {4 \left (a^3-a b^2\right )}{(a+b \sin (x))^4}\right )d(b \sin (x))}{b^5}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\left (a^2-b^2\right )^2}{4 (a+b \sin (x))^4}+\frac {4 a \left (a^2-b^2\right )}{3 (a+b \sin (x))^3}-\frac {3 a^2-b^2}{(a+b \sin (x))^2}+\frac {4 a}{a+b \sin (x)}+\log (a+b \sin (x))}{b^5}\) |
Input:
Int[(a*Sec[x] + b*Tan[x])^(-5),x]
Output:
(Log[a + b*Sin[x]] - (a^2 - b^2)^2/(4*(a + b*Sin[x])^4) + (4*a*(a^2 - b^2) )/(3*(a + b*Sin[x])^3) - (3*a^2 - b^2)/(a + b*Sin[x])^2 + (4*a)/(a + b*Sin [x]))/b^5
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x _)]^(n_.))^(p_), x_Symbol] :> Int[ActivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a *Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]
Time = 176.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(\frac {4 a \left (a^{2}-b^{2}\right )}{3 b^{5} \left (a +b \sin \left (x \right )\right )^{3}}-\frac {6 a^{2}-2 b^{2}}{2 b^{5} \left (a +b \sin \left (x \right )\right )^{2}}+\frac {\ln \left (a +b \sin \left (x \right )\right )}{b^{5}}+\frac {4 a}{b^{5} \left (a +b \sin \left (x \right )\right )}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{4 b^{5} \left (a +b \sin \left (x \right )\right )^{4}}\) | \(102\) |
| risch | \(-\frac {i x}{b^{5}}+\frac {8 i a \,b^{3} {\mathrm e}^{7 i x}-\frac {176 i a^{3} b \,{\mathrm e}^{5 i x}}{3}-\frac {88 i a \,b^{3} {\mathrm e}^{5 i x}}{3}-36 a^{2} b^{2} {\mathrm e}^{6 i x}-4 b^{4} {\mathrm e}^{6 i x}+\frac {176 i a^{3} b \,{\mathrm e}^{3 i x}}{3}+\frac {88 i a \,b^{3} {\mathrm e}^{3 i x}}{3}+\frac {100 a^{4} {\mathrm e}^{4 i x}}{3}+\frac {224 a^{2} b^{2} {\mathrm e}^{4 i x}}{3}+4 b^{4} {\mathrm e}^{4 i x}-8 i {\mathrm e}^{i x} a \,b^{3}-36 a^{2} b^{2} {\mathrm e}^{2 i x}-4 b^{4} {\mathrm e}^{2 i x}}{\left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{4} b^{5}}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{b}-1\right )}{b^{5}}\) | \(214\) |
Input:
int(1/(a*sec(x)+b*tan(x))^5,x,method=_RETURNVERBOSE)
Output:
4/3*a*(a^2-b^2)/b^5/(a+b*sin(x))^3-1/2*(6*a^2-2*b^2)/b^5/(a+b*sin(x))^2+ln (a+b*sin(x))/b^5+4*a/b^5/(a+b*sin(x))-1/4*(a^4-2*a^2*b^2+b^4)/b^5/(a+b*sin (x))^4
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (97) = 194\).
Time = 0.10 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.15 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx=\frac {25 \, a^{4} + 110 \, a^{2} b^{2} + 9 \, b^{4} - 12 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} + 12 \, {\left (b^{4} \cos \left (x\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} - 4 \, {\left (a b^{3} \cos \left (x\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (x\right )\right )} \log \left (b \sin \left (x\right ) + a\right ) - 8 \, {\left (6 \, a b^{3} \cos \left (x\right )^{2} - 11 \, a^{3} b - 7 \, a b^{3}\right )} \sin \left (x\right )}{12 \, {\left (b^{9} \cos \left (x\right )^{4} + a^{4} b^{5} + 6 \, a^{2} b^{7} + b^{9} - 2 \, {\left (3 \, a^{2} b^{7} + b^{9}\right )} \cos \left (x\right )^{2} - 4 \, {\left (a b^{8} \cos \left (x\right )^{2} - a^{3} b^{6} - a b^{8}\right )} \sin \left (x\right )\right )}} \] Input:
integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="fricas")
Output:
1/12*(25*a^4 + 110*a^2*b^2 + 9*b^4 - 12*(9*a^2*b^2 + b^4)*cos(x)^2 + 12*(b ^4*cos(x)^4 + a^4 + 6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)*cos(x)^2 - 4*(a* b^3*cos(x)^2 - a^3*b - a*b^3)*sin(x))*log(b*sin(x) + a) - 8*(6*a*b^3*cos(x )^2 - 11*a^3*b - 7*a*b^3)*sin(x))/(b^9*cos(x)^4 + a^4*b^5 + 6*a^2*b^7 + b^ 9 - 2*(3*a^2*b^7 + b^9)*cos(x)^2 - 4*(a*b^8*cos(x)^2 - a^3*b^6 - a*b^8)*si n(x))
Leaf count of result is larger than twice the leaf count of optimal. 1719 vs. \(2 (90) = 180\).
Time = 5.91 (sec) , antiderivative size = 1719, normalized size of antiderivative = 17.02 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx=\text {Too large to display} \] Input:
integrate(1/(a*sec(x)+b*tan(x))**5,x)
Output:
Piecewise((36*a**4*log(a*sec(x)/b + tan(x))*sec(x)**4/(36*a**4*b**5*sec(x) **4 + 144*a**3*b**6*tan(x)*sec(x)**3 + 216*a**2*b**7*tan(x)**2*sec(x)**2 + 144*a*b**8*tan(x)**3*sec(x) + 36*b**9*tan(x)**4) - 18*a**4*log(tan(x)**2 + 1)*sec(x)**4/(36*a**4*b**5*sec(x)**4 + 144*a**3*b**6*tan(x)*sec(x)**3 + 216*a**2*b**7*tan(x)**2*sec(x)**2 + 144*a*b**8*tan(x)**3*sec(x) + 36*b**9* tan(x)**4) + 20*a**4*sec(x)**4/(36*a**4*b**5*sec(x)**4 + 144*a**3*b**6*tan (x)*sec(x)**3 + 216*a**2*b**7*tan(x)**2*sec(x)**2 + 144*a*b**8*tan(x)**3*s ec(x) + 36*b**9*tan(x)**4) + 144*a**3*b*log(a*sec(x)/b + tan(x))*tan(x)*se c(x)**3/(36*a**4*b**5*sec(x)**4 + 144*a**3*b**6*tan(x)*sec(x)**3 + 216*a** 2*b**7*tan(x)**2*sec(x)**2 + 144*a*b**8*tan(x)**3*sec(x) + 36*b**9*tan(x)* *4) - 72*a**3*b*log(tan(x)**2 + 1)*tan(x)*sec(x)**3/(36*a**4*b**5*sec(x)** 4 + 144*a**3*b**6*tan(x)*sec(x)**3 + 216*a**2*b**7*tan(x)**2*sec(x)**2 + 1 44*a*b**8*tan(x)**3*sec(x) + 36*b**9*tan(x)**4) + 44*a**3*b*tan(x)*sec(x)* *3/(36*a**4*b**5*sec(x)**4 + 144*a**3*b**6*tan(x)*sec(x)**3 + 216*a**2*b** 7*tan(x)**2*sec(x)**2 + 144*a*b**8*tan(x)**3*sec(x) + 36*b**9*tan(x)**4) + 216*a**2*b**2*log(a*sec(x)/b + tan(x))*tan(x)**2*sec(x)**2/(36*a**4*b**5* sec(x)**4 + 144*a**3*b**6*tan(x)*sec(x)**3 + 216*a**2*b**7*tan(x)**2*sec(x )**2 + 144*a*b**8*tan(x)**3*sec(x) + 36*b**9*tan(x)**4) - 108*a**2*b**2*lo g(tan(x)**2 + 1)*tan(x)**2*sec(x)**2/(36*a**4*b**5*sec(x)**4 + 144*a**3*b* *6*tan(x)*sec(x)**3 + 216*a**2*b**7*tan(x)**2*sec(x)**2 + 144*a*b**8*ta...
Leaf count of result is larger than twice the leaf count of optimal. 483 vs. \(2 (97) = 194\).
Time = 0.14 (sec) , antiderivative size = 483, normalized size of antiderivative = 4.78 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx =\text {Too large to display} \] Input:
integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="maxima")
Output:
-2/3*(3*(a^7 - a^3*b^4)*sin(x)/(cos(x) + 1) + 3*(7*a^6*b - 3*a^2*b^5)*sin( x)^2/(cos(x) + 1)^2 + (9*a^7 + 52*a^5*b^2 - a^3*b^4 - 12*a*b^6)*sin(x)^3/( cos(x) + 1)^3 + 2*(21*a^6*b + 25*a^4*b^3 - 7*a^2*b^5 - 3*b^7)*sin(x)^4/(co s(x) + 1)^4 + (9*a^7 + 52*a^5*b^2 - a^3*b^4 - 12*a*b^6)*sin(x)^5/(cos(x) + 1)^5 + 3*(7*a^6*b - 3*a^2*b^5)*sin(x)^6/(cos(x) + 1)^6 + 3*(a^7 - a^3*b^4 )*sin(x)^7/(cos(x) + 1)^7)/(a^8*b^4 + 8*a^7*b^5*sin(x)/(cos(x) + 1) + 8*a^ 7*b^5*sin(x)^7/(cos(x) + 1)^7 + a^8*b^4*sin(x)^8/(cos(x) + 1)^8 + 4*(a^8*b ^4 + 6*a^6*b^6)*sin(x)^2/(cos(x) + 1)^2 + 8*(3*a^7*b^5 + 4*a^5*b^7)*sin(x) ^3/(cos(x) + 1)^3 + 2*(3*a^8*b^4 + 24*a^6*b^6 + 8*a^4*b^8)*sin(x)^4/(cos(x ) + 1)^4 + 8*(3*a^7*b^5 + 4*a^5*b^7)*sin(x)^5/(cos(x) + 1)^5 + 4*(a^8*b^4 + 6*a^6*b^6)*sin(x)^6/(cos(x) + 1)^6) + log(a + 2*b*sin(x)/(cos(x) + 1) + a*sin(x)^2/(cos(x) + 1)^2)/b^5 - log(sin(x)^2/(cos(x) + 1)^2 + 1)/b^5
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx=\frac {\log \left ({\left | b \sin \left (x\right ) + a \right |}\right )}{b^{5}} - \frac {25 \, b^{3} \sin \left (x\right )^{4} + 52 \, a b^{2} \sin \left (x\right )^{3} + 42 \, a^{2} b \sin \left (x\right )^{2} - 12 \, b^{3} \sin \left (x\right )^{2} + 12 \, a^{3} \sin \left (x\right ) - 8 \, a b^{2} \sin \left (x\right ) - 2 \, a^{2} b + 3 \, b^{3}}{12 \, {\left (b \sin \left (x\right ) + a\right )}^{4} b^{4}} \] Input:
integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="giac")
Output:
log(abs(b*sin(x) + a))/b^5 - 1/12*(25*b^3*sin(x)^4 + 52*a*b^2*sin(x)^3 + 4 2*a^2*b*sin(x)^2 - 12*b^3*sin(x)^2 + 12*a^3*sin(x) - 8*a*b^2*sin(x) - 2*a^ 2*b + 3*b^3)/((b*sin(x) + a)^4*b^4)
Time = 16.90 (sec) , antiderivative size = 541, normalized size of antiderivative = 5.36 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {16\,a}{\frac {32\,a^3}{b^2}-16\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-16\,a+\frac {32\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {32\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{b^2}}+\frac {16\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{\frac {32\,a^3}{b^2}-16\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-16\,a+\frac {32\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {32\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{b^2}}+\frac {32\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{32\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )-16\,a\,b+\frac {32\,a^3}{b}+\frac {32\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{b}-16\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}\right )}{b^5}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (7\,a^4-3\,b^4\right )}{a^2\,b^3}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6\,\left (7\,a^4-3\,b^4\right )}{a^2\,b^3}+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4-b^4\right )}{a\,b^4}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7\,\left (a^4-b^4\right )}{a\,b^4}+\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (21\,a^6+25\,a^4\,b^2-7\,a^2\,b^4-3\,b^6\right )}{3\,a^4\,b^3}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (9\,a^6+52\,a^4\,b^2-a^2\,b^4-12\,b^6\right )}{3\,a^3\,b^4}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (9\,a^6+52\,a^4\,b^2-a^2\,b^4-12\,b^6\right )}{3\,a^3\,b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (4\,a^4+24\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^6\,\left (4\,a^4+24\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (24\,a^3\,b+32\,a\,b^3\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (24\,a^3\,b+32\,a\,b^3\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (6\,a^4+48\,a^2\,b^2+16\,b^4\right )+a^4+a^4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^8+8\,a^3\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7+8\,a^3\,b\,\mathrm {tan}\left (\frac {x}{2}\right )} \] Input:
int(1/(b*tan(x) + a/cos(x))^5,x)
Output:
(2*atanh((16*a)/((32*a^3)/b^2 - 16*a*tan(x/2)^2 - 16*a + (32*a^2*tan(x/2)) /b + (32*a^3*tan(x/2)^2)/b^2) + (16*a*tan(x/2)^2)/((32*a^3)/b^2 - 16*a*tan (x/2)^2 - 16*a + (32*a^2*tan(x/2))/b + (32*a^3*tan(x/2)^2)/b^2) + (32*a^2* tan(x/2))/(32*a^2*tan(x/2) - 16*a*b + (32*a^3)/b + (32*a^3*tan(x/2)^2)/b - 16*a*b*tan(x/2)^2)))/b^5 - ((2*tan(x/2)^2*(7*a^4 - 3*b^4))/(a^2*b^3) + (2 *tan(x/2)^6*(7*a^4 - 3*b^4))/(a^2*b^3) + (2*tan(x/2)*(a^4 - b^4))/(a*b^4) + (2*tan(x/2)^7*(a^4 - b^4))/(a*b^4) + (4*tan(x/2)^4*(21*a^6 - 3*b^6 - 7*a ^2*b^4 + 25*a^4*b^2))/(3*a^4*b^3) + (2*tan(x/2)^3*(9*a^6 - 12*b^6 - a^2*b^ 4 + 52*a^4*b^2))/(3*a^3*b^4) + (2*tan(x/2)^5*(9*a^6 - 12*b^6 - a^2*b^4 + 5 2*a^4*b^2))/(3*a^3*b^4))/(tan(x/2)^2*(4*a^4 + 24*a^2*b^2) + tan(x/2)^6*(4* a^4 + 24*a^2*b^2) + tan(x/2)^3*(32*a*b^3 + 24*a^3*b) + tan(x/2)^5*(32*a*b^ 3 + 24*a^3*b) + tan(x/2)^4*(6*a^4 + 16*b^4 + 48*a^2*b^2) + a^4 + a^4*tan(x /2)^8 + 8*a^3*b*tan(x/2)^7 + 8*a^3*b*tan(x/2))
Time = 0.16 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.45 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx=\frac {-6 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right )^{4} a^{4}-24 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right )^{3} \tan \left (x \right ) a^{3} b -36 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right )^{2} \tan \left (x \right )^{2} a^{2} b^{2}-24 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \sec \left (x \right ) \tan \left (x \right )^{3} a \,b^{3}-6 \,\mathrm {log}\left (\tan \left (x \right )^{2}+1\right ) \tan \left (x \right )^{4} b^{4}+12 \,\mathrm {log}\left (\sec \left (x \right ) a +\tan \left (x \right ) b \right ) \sec \left (x \right )^{4} a^{4}+48 \,\mathrm {log}\left (\sec \left (x \right ) a +\tan \left (x \right ) b \right ) \sec \left (x \right )^{3} \tan \left (x \right ) a^{3} b +72 \,\mathrm {log}\left (\sec \left (x \right ) a +\tan \left (x \right ) b \right ) \sec \left (x \right )^{2} \tan \left (x \right )^{2} a^{2} b^{2}+48 \,\mathrm {log}\left (\sec \left (x \right ) a +\tan \left (x \right ) b \right ) \sec \left (x \right ) \tan \left (x \right )^{3} a \,b^{3}+12 \,\mathrm {log}\left (\sec \left (x \right ) a +\tan \left (x \right ) b \right ) \tan \left (x \right )^{4} b^{4}+3 \sec \left (x \right )^{4} a^{4}-22 \sec \left (x \right )^{2} \tan \left (x \right )^{2} a^{2} b^{2}+2 \sec \left (x \right )^{2} a^{2} b^{2}-32 \sec \left (x \right ) \tan \left (x \right )^{3} a \,b^{3}+8 \sec \left (x \right ) \tan \left (x \right ) a \,b^{3}-13 \tan \left (x \right )^{4} b^{4}+6 \tan \left (x \right )^{2} b^{4}-3 b^{4}}{12 b^{5} \left (\sec \left (x \right )^{4} a^{4}+4 \sec \left (x \right )^{3} \tan \left (x \right ) a^{3} b +6 \sec \left (x \right )^{2} \tan \left (x \right )^{2} a^{2} b^{2}+4 \sec \left (x \right ) \tan \left (x \right )^{3} a \,b^{3}+\tan \left (x \right )^{4} b^{4}\right )} \] Input:
int(1/(a*sec(x)+b*tan(x))^5,x)
Output:
( - 6*log(tan(x)**2 + 1)*sec(x)**4*a**4 - 24*log(tan(x)**2 + 1)*sec(x)**3* tan(x)*a**3*b - 36*log(tan(x)**2 + 1)*sec(x)**2*tan(x)**2*a**2*b**2 - 24*l og(tan(x)**2 + 1)*sec(x)*tan(x)**3*a*b**3 - 6*log(tan(x)**2 + 1)*tan(x)**4 *b**4 + 12*log(sec(x)*a + tan(x)*b)*sec(x)**4*a**4 + 48*log(sec(x)*a + tan (x)*b)*sec(x)**3*tan(x)*a**3*b + 72*log(sec(x)*a + tan(x)*b)*sec(x)**2*tan (x)**2*a**2*b**2 + 48*log(sec(x)*a + tan(x)*b)*sec(x)*tan(x)**3*a*b**3 + 1 2*log(sec(x)*a + tan(x)*b)*tan(x)**4*b**4 + 3*sec(x)**4*a**4 - 22*sec(x)** 2*tan(x)**2*a**2*b**2 + 2*sec(x)**2*a**2*b**2 - 32*sec(x)*tan(x)**3*a*b**3 + 8*sec(x)*tan(x)*a*b**3 - 13*tan(x)**4*b**4 + 6*tan(x)**2*b**4 - 3*b**4) /(12*b**5*(sec(x)**4*a**4 + 4*sec(x)**3*tan(x)*a**3*b + 6*sec(x)**2*tan(x) **2*a**2*b**2 + 4*sec(x)*tan(x)**3*a*b**3 + tan(x)**4*b**4))