3.239 \(\int \frac {\tan (e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=93 \[ \frac {1}{2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac {1}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)^3} \]

[Out]

-1/2*ln(a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(a-b)^3/f+1/4/(a-b)/f/(a+b*tan(f*x+e)^2)^2+1/2/(a-b)^2/f/(a+b*tan(f*x+e
)^2)

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Rubi [A]  time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3670, 444, 44} \[ \frac {1}{2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac {1}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

-Log[a*Cos[e + f*x]^2 + b*Sin[e + f*x]^2]/(2*(a - b)^3*f) + 1/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) + 1/(2*(a
- b)^2*f*(a + b*Tan[e + f*x]^2))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.75, size = 82, normalized size = 0.88 \[ \frac {\frac {(a-b)^2}{\left (a+b \tan ^2(e+f x)\right )^2}+\frac {2 (a-b)}{a+b \tan ^2(e+f x)}-2 \log \left (a+b \tan ^2(e+f x)\right )-4 \log (\cos (e+f x))}{4 f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(-4*Log[Cos[e + f*x]] - 2*Log[a + b*Tan[e + f*x]^2] + (a - b)^2/(a + b*Tan[e + f*x]^2)^2 + (2*(a - b))/(a + b*
Tan[e + f*x]^2))/(4*(a - b)^3*f)

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fricas [B]  time = 0.45, size = 206, normalized size = 2.22 \[ -\frac {3 \, b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, a b - b^{2} + 2 \, {\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

-1/4*(3*b^2*tan(f*x + e)^4 + 2*(2*a*b + b^2)*tan(f*x + e)^2 + 4*a*b - b^2 + 2*(b^2*tan(f*x + e)^4 + 2*a*b*tan(
f*x + e)^2 + a^2)*log((b*tan(f*x + e)^2 + a)/(tan(f*x + e)^2 + 1)))/((a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*t
an(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*tan(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*
b^3)*f)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)2/f*(1/(2*a^3-6*a^2*b+6*a*b^2-2*b^3)*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1))-1/(4*a^3-12*a^2*b+
12*a*b^2-4*b^3)*ln(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+4
*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a)+(3*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^4-12*((1-cos(f*
x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^4+8*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^3*b+24*((1-cos(f*x+exp(1
)))/(1+cos(f*x+exp(1))))^3*a^2*b^2-8*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a*b^3+18*((1-cos(f*x+exp(1)))
/(1+cos(f*x+exp(1))))^2*a^4-16*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^3*b-48*((1-cos(f*x+exp(1)))/(1+co
s(f*x+exp(1))))^2*a^2*b^2+80*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b^3-16*((1-cos(f*x+exp(1)))/(1+cos(
f*x+exp(1))))^2*b^4-12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^4+8*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a
^3*b+24*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2*b^2-8*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b^3+3*a^4)
/(8*a^5-24*a^4*b+24*a^3*b^2-8*a^2*b^3)/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a-2*(1-cos(f*x+exp(1)))/(1
+cos(f*x+exp(1)))*a+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+a)^2)

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maple [B]  time = 0.24, size = 190, normalized size = 2.04 \[ -\frac {\ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 f \left (a -b \right )^{3}}+\frac {a}{2 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {b}{2 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {a^{2}}{4 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {a b}{2 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{2}}{4 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)/(a+b*tan(f*x+e)^2)^3,x)

[Out]

-1/2/f/(a-b)^3*ln(a+b*tan(f*x+e)^2)+1/2/f/(a-b)^3*a/(a+b*tan(f*x+e)^2)-1/2/f/(a-b)^3/(a+b*tan(f*x+e)^2)*b+1/4/
f/(a-b)^3*a^2/(a+b*tan(f*x+e)^2)^2-1/2/f/(a-b)^3*a*b/(a+b*tan(f*x+e)^2)^2+1/4/f*b^2/(a-b)^3/(a+b*tan(f*x+e)^2)
^2+1/2/f/(a-b)^3*ln(1+tan(f*x+e)^2)

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maxima [B]  time = 0.53, size = 192, normalized size = 2.06 \[ \frac {\frac {4 \, {\left (a b - b^{2}\right )} \sin \left (f x + e\right )^{2} - 4 \, a b + b^{2}}{a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (f x + e\right )^{2}} - \frac {2 \, \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/4*((4*(a*b - b^2)*sin(f*x + e)^2 - 4*a*b + b^2)/(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^5 - 5*a^4*b + 10*a
^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sin(f*x + e)^4 - 2*(a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*sin(f*
x + e)^2) - 2*log(-(a - b)*sin(f*x + e)^2 + a)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3))/f

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mupad [B]  time = 12.46, size = 375, normalized size = 4.03 \[ \frac {a^2\,\left (-3+\mathrm {atan}\left (\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}-b\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}}{2\,a+a\,{\mathrm {tan}\left (e+f\,x\right )}^2+b\,{\mathrm {tan}\left (e+f\,x\right )}^2}\right )\,4{}\mathrm {i}\right )+b^2\,\left (2\,{\mathrm {tan}\left (e+f\,x\right )}^2-1+{\mathrm {tan}\left (e+f\,x\right )}^4\,\mathrm {atan}\left (\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}-b\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}}{2\,a+a\,{\mathrm {tan}\left (e+f\,x\right )}^2+b\,{\mathrm {tan}\left (e+f\,x\right )}^2}\right )\,4{}\mathrm {i}\right )+a\,b\,\left (4-2\,{\mathrm {tan}\left (e+f\,x\right )}^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}-b\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}}{2\,a+a\,{\mathrm {tan}\left (e+f\,x\right )}^2+b\,{\mathrm {tan}\left (e+f\,x\right )}^2}\right )\,8{}\mathrm {i}\right )}{f\,\left (-4\,a^5-8\,a^4\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2+12\,a^4\,b-4\,a^3\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4+24\,a^3\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2-12\,a^3\,b^2+12\,a^2\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^4-24\,a^2\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2+4\,a^2\,b^3-12\,a\,b^4\,{\mathrm {tan}\left (e+f\,x\right )}^4+8\,a\,b^4\,{\mathrm {tan}\left (e+f\,x\right )}^2+4\,b^5\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)/(a + b*tan(e + f*x)^2)^3,x)

[Out]

(a^2*(atan((a*tan(e + f*x)^2*1i - b*tan(e + f*x)^2*1i)/(2*a + a*tan(e + f*x)^2 + b*tan(e + f*x)^2))*4i - 3) +
b^2*(tan(e + f*x)^4*atan((a*tan(e + f*x)^2*1i - b*tan(e + f*x)^2*1i)/(2*a + a*tan(e + f*x)^2 + b*tan(e + f*x)^
2))*4i + 2*tan(e + f*x)^2 - 1) + a*b*(tan(e + f*x)^2*atan((a*tan(e + f*x)^2*1i - b*tan(e + f*x)^2*1i)/(2*a + a
*tan(e + f*x)^2 + b*tan(e + f*x)^2))*8i - 2*tan(e + f*x)^2 + 4))/(f*(12*a^4*b - 4*a^5 + 4*a^2*b^3 - 12*a^3*b^2
+ 4*b^5*tan(e + f*x)^4 + 8*a*b^4*tan(e + f*x)^2 - 8*a^4*b*tan(e + f*x)^2 - 12*a*b^4*tan(e + f*x)^4 - 24*a^2*b
^3*tan(e + f*x)^2 + 24*a^3*b^2*tan(e + f*x)^2 + 12*a^2*b^3*tan(e + f*x)^4 - 4*a^3*b^2*tan(e + f*x)^4))

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sympy [A]  time = 133.35, size = 2876, normalized size = 30.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Piecewise((zoo*x/tan(e)**5, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (log(tan(e + f*x)**2 + 1)/(2*a**3*f), Eq(b, 0)),
(-1/(6*b**3*f*tan(e + f*x)**6 + 18*b**3*f*tan(e + f*x)**4 + 18*b**3*f*tan(e + f*x)**2 + 6*b**3*f), Eq(a, b)),
(x*tan(e)/(a + b*tan(e)**2)**3, Eq(f, 0)), (-2*a**2*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*a**5*f + 8*a**
4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3
*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e
+ f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*a**2*log(I*sqrt(a)*sqrt(1/b) + tan(e +
f*x))/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*ta
n(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3
*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 2*a**2*log(tan(e +
f*x)**2 + 1)/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b
**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a
**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 3*a**2/(4*
a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x
)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a
*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 4*a*b*log(-I*sqrt(a)*sqrt(1
/b) + tan(e + f*x))*tan(e + f*x)**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e
+ f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f
*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e +
f*x)**4) - 4*a*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)*
*2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b
**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**
4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 4*a*b*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(4*a**5*f + 8
*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*
a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*ta
n(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 2*a*b*tan(e + f*x)**2/(4*a**5*f + 8*a
**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a*
*3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(
e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 4*a*b/(4*a**5*f + 8*a**4*b*f*tan(e + f*
x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**
2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*
b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*b**2*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f
*x)**4/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*t
an(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**
3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*b**2*log(I*sqrt
(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**4/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b*
*2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a
**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5
*f*tan(e + f*x)**4) + 2*b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**4/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 -
12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*
f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*
tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*b**2*tan(e + f*x)**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 -
12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f
*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*t
an(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + b**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**
3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 +
24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*
b**5*f*tan(e + f*x)**4), True))

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