3.272 \(\int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx\)

Optimal. Leaf size=101 \[ -\frac {\left (a^2-b^2\right )^2}{4 b^5 (a+b \sin (x))^4}+\frac {4 a \left (a^2-b^2\right )}{3 b^5 (a+b \sin (x))^3}-\frac {3 a^2-b^2}{b^5 (a+b \sin (x))^2}+\frac {4 a}{b^5 (a+b \sin (x))}+\frac {\log (a+b \sin (x))}{b^5} \]

[Out]

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))

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4391, 2668, 697} \[ -\frac {\left (a^2-b^2\right )^2}{4 b^5 (a+b \sin (x))^4}+\frac {4 a \left (a^2-b^2\right )}{3 b^5 (a+b \sin (x))^3}-\frac {3 a^2-b^2}{b^5 (a+b \sin (x))^2}+\frac {4 a}{b^5 (a+b \sin (x))}+\frac {\log (a+b \sin (x))}{b^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x] + b*Tan[x])^(-5),x]

[Out]

Log[a + b*Sin[x]]/b^5 - (a^2 - b^2)^2/(4*b^5*(a + b*Sin[x])^4) + (4*a*(a^2 - b^2))/(3*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]))

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[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]

Rubi steps

\begin {align*} \int \frac {1}{(a \sec (x)+b \tan (x))^5} \, dx &=\int \frac {\cos ^5(x)}{(a+b \sin (x))^5} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{(a+x)^5} \, dx,x,b \sin (x)\right )}{b^5}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {\left (a^2-b^2\right )^2}{(a+x)^5}-\frac {4 \left (a^3-a b^2\right )}{(a+x)^4}+\frac {2 \left (3 a^2-b^2\right )}{(a+x)^3}-\frac {4 a}{(a+x)^2}+\frac {1}{a+x}\right ) \, dx,x,b \sin (x)\right )}{b^5}\\ &=\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))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 86, normalized size = 0.85 \[ \frac {\frac {25 a^4+12 b^2 \left (9 a^2+b^2\right ) \sin ^2(x)+8 a b \left (11 a^2+b^2\right ) \sin (x)+2 a^2 b^2+48 a b^3 \sin ^3(x)-3 b^4}{12 (a+b \sin (x))^4}+\log (a+b \sin (x))}{b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[x] + b*Tan[x])^(-5),x]

[Out]

(Log[a + b*Sin[x]] + (25*a^4 + 2*a^2*b^2 - 3*b^4 + 8*a*b*(11*a^2 + b^2)*Sin[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

________________________________________________________________________________________

fricas [B]  time = 1.06, size = 217, normalized size = 2.15 \[ \frac {25 \, a^{4} + 110 \, a^{2} b^{2} + 9 \, b^{4} - 12 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \relax (x)^{2} + 12 \, {\left (b^{4} \cos \relax (x)^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \relax (x)^{2} - 4 \, {\left (a b^{3} \cos \relax (x)^{2} - a^{3} b - a b^{3}\right )} \sin \relax (x)\right )} \log \left (b \sin \relax (x) + a\right ) - 8 \, {\left (6 \, a b^{3} \cos \relax (x)^{2} - 11 \, a^{3} b - 7 \, a b^{3}\right )} \sin \relax (x)}{12 \, {\left (b^{9} \cos \relax (x)^{4} + a^{4} b^{5} + 6 \, a^{2} b^{7} + b^{9} - 2 \, {\left (3 \, a^{2} b^{7} + b^{9}\right )} \cos \relax (x)^{2} - 4 \, {\left (a b^{8} \cos \relax (x)^{2} - a^{3} b^{6} - a b^{8}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="fricas")

[Out]

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)*sin(x))

________________________________________________________________________________________

giac [A]  time = 0.14, size = 91, normalized size = 0.90 \[ \frac {\log \left ({\left | b \sin \relax (x) + a \right |}\right )}{b^{5}} - \frac {25 \, b^{3} \sin \relax (x)^{4} + 52 \, a b^{2} \sin \relax (x)^{3} + 42 \, a^{2} b \sin \relax (x)^{2} - 12 \, b^{3} \sin \relax (x)^{2} + 12 \, a^{3} \sin \relax (x) - 8 \, a b^{2} \sin \relax (x) - 2 \, a^{2} b + 3 \, b^{3}}{12 \, {\left (b \sin \relax (x) + a\right )}^{4} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="giac")

[Out]

log(abs(b*sin(x) + a))/b^5 - 1/12*(25*b^3*sin(x)^4 + 52*a*b^2*sin(x)^3 + 42*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)

________________________________________________________________________________________

maple [A]  time = 0.21, size = 130, normalized size = 1.29 \[ \frac {4 a}{b^{5} \left (a +b \sin \relax (x )\right )}-\frac {a^{4}}{4 b^{5} \left (a +b \sin \relax (x )\right )^{4}}+\frac {a^{2}}{2 b^{3} \left (a +b \sin \relax (x )\right )^{4}}-\frac {1}{4 b \left (a +b \sin \relax (x )\right )^{4}}+\frac {4 a^{3}}{3 b^{5} \left (a +b \sin \relax (x )\right )^{3}}-\frac {4 a}{3 b^{3} \left (a +b \sin \relax (x )\right )^{3}}+\frac {\ln \left (a +b \sin \relax (x )\right )}{b^{5}}-\frac {3 a^{2}}{b^{5} \left (a +b \sin \relax (x )\right )^{2}}+\frac {1}{b^{3} \left (a +b \sin \relax (x )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)+b*tan(x))^5,x)

[Out]

4*a/b^5/(a+b*sin(x))-1/4/b^5/(a+b*sin(x))^4*a^4+1/2/b^3/(a+b*sin(x))^4*a^2-1/4/b/(a+b*sin(x))^4+4/3*a^3/b^5/(a
+b*sin(x))^3-4/3*a/b^3/(a+b*sin(x))^3+ln(a+b*sin(x))/b^5-3/b^5/(a+b*sin(x))^2*a^2+1/b^3/(a+b*sin(x))^2

________________________________________________________________________________________

maxima [B]  time = 0.51, size = 483, normalized size = 4.78 \[ -\frac {2 \, {\left (\frac {3 \, {\left (a^{7} - a^{3} b^{4}\right )} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {3 \, {\left (7 \, a^{6} b - 3 \, a^{2} b^{5}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {{\left (9 \, a^{7} + 52 \, a^{5} b^{2} - a^{3} b^{4} - 12 \, a b^{6}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {2 \, {\left (21 \, a^{6} b + 25 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 3 \, b^{7}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {{\left (9 \, a^{7} + 52 \, a^{5} b^{2} - a^{3} b^{4} - 12 \, a b^{6}\right )} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {3 \, {\left (7 \, a^{6} b - 3 \, a^{2} b^{5}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {3 \, {\left (a^{7} - a^{3} b^{4}\right )} \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}\right )}}{3 \, {\left (a^{8} b^{4} + \frac {8 \, a^{7} b^{5} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {8 \, a^{7} b^{5} \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}} + \frac {a^{8} b^{4} \sin \relax (x)^{8}}{{\left (\cos \relax (x) + 1\right )}^{8}} + \frac {4 \, {\left (a^{8} b^{4} + 6 \, a^{6} b^{6}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {8 \, {\left (3 \, a^{7} b^{5} + 4 \, a^{5} b^{7}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {2 \, {\left (3 \, a^{8} b^{4} + 24 \, a^{6} b^{6} + 8 \, a^{4} b^{8}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {8 \, {\left (3 \, a^{7} b^{5} + 4 \, a^{5} b^{7}\right )} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {4 \, {\left (a^{8} b^{4} + 6 \, a^{6} b^{6}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}}\right )}} + \frac {\log \left (a + \frac {2 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{b^{5}} - \frac {\log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="maxima")

[Out]

-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/(
cos(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)*si
n(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

________________________________________________________________________________________

mupad [B]  time = 3.84, size = 541, normalized size = 5.36 \[ \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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(x) + a/cos(x))^5,x)

[Out]

(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*ta
n(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 + 52*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))

________________________________________________________________________________________

sympy [A]  time = 15.19, size = 1719, normalized size = 17.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x))**5,x)

[Out]

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*sec(x) + 36*b**9*tan(x)**4) + 144
*a**3*b*log(a*sec(x)/b + tan(x))*tan(x)*sec(x)**3/(36*a**4*b**5*sec(x)**4 + 144*a**3*b**6*tan(x)*sec(x)**3 + 2
16*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 + 144*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*ta
n(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*log(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 + 21
6*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) + 6*a**2*b**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) + 144*a*b**3*log(a*sec(x)/b + tan(x))*tan(x)**3*sec(x)/(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*t
an(x)**4) - 72*a*b**3*log(tan(x)**2 + 1)*tan(x)**3*sec(x)/(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) - 52*a*b**3*tan(x)
**3*sec(x)/(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) + 24*a*b**3*tan(x)*sec(x)/(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) + 36*
b**4*log(a*sec(x)/b + tan(x))*tan(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*b**4*log(tan(x)**2 + 1)*tan(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*t
an(x)**3*sec(x) + 36*b**9*tan(x)**4) - 28*b**4*tan(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*b**4*tan(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) - 9*b**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), Ne(b, 0)), ((8*tan(x)**5/(15*se
c(x)**5) + 4*tan(x)**3/(3*sec(x)**5) + tan(x)/sec(x)**5)/a**5, True))

________________________________________________________________________________________