3.1195 \(\int \frac {c+d \tan (e+f x)}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=111 \[ -\frac {b c-a d}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {\left (a^2 (-d)+2 a b c+b^2 d\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2}+\frac {x \left (a^2 c+2 a b d-b^2 c\right )}{\left (a^2+b^2\right )^2} \]

[Out]

(a^2*c+2*a*b*d-b^2*c)*x/(a^2+b^2)^2+(-a^2*d+2*a*b*c+b^2*d)*ln(a*cos(f*x+e)+b*sin(f*x+e))/(a^2+b^2)^2/f+(a*d-b*
c)/(a^2+b^2)/f/(a+b*tan(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3529, 3531, 3530} \[ -\frac {b c-a d}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {\left (a^2 (-d)+2 a b c+b^2 d\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2}+\frac {x \left (a^2 c+2 a b d-b^2 c\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2*c - b^2*c + 2*a*b*d)*x)/(a^2 + b^2)^2 + ((2*a*b*c - a^2*d + b^2*d)*Log[a*Cos[e + f*x] + b*Sin[e + f*x]])
/((a^2 + b^2)^2*f) - (b*c - a*d)/((a^2 + b^2)*f*(a + b*Tan[e + f*x]))

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 2.22, size = 190, normalized size = 1.71 \[ \frac {\frac {d ((-b-i a) \log (-\tan (e+f x)+i)+i (a+i b) \log (\tan (e+f x)+i)+2 b \log (a+b \tan (e+f x)))}{a^2+b^2}-(b c-a d) \left (\frac {2 b \left (\frac {a^2+b^2}{a+b \tan (e+f x)}-2 a \log (a+b \tan (e+f x))\right )}{\left (a^2+b^2\right )^2}+\frac {i \log (-\tan (e+f x)+i)}{(a+i b)^2}-\frac {i \log (\tan (e+f x)+i)}{(a-i b)^2}\right )}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*(((-I)*a - b)*Log[I - Tan[e + f*x]] + I*(a + I*b)*Log[I + Tan[e + f*x]] + 2*b*Log[a + b*Tan[e + f*x]]))/(a
^2 + b^2) - (b*c - a*d)*((I*Log[I - Tan[e + f*x]])/(a + I*b)^2 - (I*Log[I + Tan[e + f*x]])/(a - I*b)^2 + (2*b*
(-2*a*Log[a + b*Tan[e + f*x]] + (a^2 + b^2)/(a + b*Tan[e + f*x])))/(a^2 + b^2)^2))/(2*b*f)

________________________________________________________________________________________

fricas [B]  time = 0.48, size = 225, normalized size = 2.03 \[ -\frac {2 \, b^{3} c - 2 \, a b^{2} d - 2 \, {\left (2 \, a^{2} b d + {\left (a^{3} - a b^{2}\right )} c\right )} f x - {\left (2 \, a^{2} b c - {\left (a^{3} - a b^{2}\right )} d + {\left (2 \, a b^{2} c - {\left (a^{2} b - b^{3}\right )} d\right )} \tan \left (f x + e\right )\right )} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, {\left (a b^{2} c - a^{2} b d + {\left (2 \, a b^{2} d + {\left (a^{2} b - b^{3}\right )} c\right )} f x\right )} \tan \left (f x + e\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} f \tan \left (f x + e\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.75, size = 241, normalized size = 2.17 \[ \frac {\frac {2 \, {\left (a^{2} c - b^{2} c + 2 \, a b d\right )} {\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a b c - a^{2} d + b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (2 \, a b^{2} c - a^{2} b d + b^{3} d\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {2 \, {\left (2 \, a b^{2} c \tan \left (f x + e\right ) - a^{2} b d \tan \left (f x + e\right ) + b^{3} d \tan \left (f x + e\right ) + 3 \, a^{2} b c + b^{3} c - 2 \, a^{3} d\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \tan \left (f x + e\right ) + a\right )}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^2*c - b^2*c + 2*a*b*d)*(f*x + e)/(a^4 + 2*a^2*b^2 + b^4) - (2*a*b*c - a^2*d + b^2*d)*log(tan(f*x + e
)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) + 2*(2*a*b^2*c - a^2*b*d + b^3*d)*log(abs(b*tan(f*x + e) + a))/(a^4*b + 2*a^2
*b^3 + b^5) - 2*(2*a*b^2*c*tan(f*x + e) - a^2*b*d*tan(f*x + e) + b^3*d*tan(f*x + e) + 3*a^2*b*c + b^3*c - 2*a^
3*d)/((a^4 + 2*a^2*b^2 + b^4)*(b*tan(f*x + e) + a)))/f

________________________________________________________________________________________

maple [B]  time = 0.24, size = 301, normalized size = 2.71 \[ \frac {d a}{f \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (f x +e \right )\right )}-\frac {c b}{f \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (f x +e \right )\right )}-\frac {\ln \left (a +b \tan \left (f x +e \right )\right ) a^{2} d}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {2 \ln \left (a +b \tan \left (f x +e \right )\right ) a b c}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (a +b \tan \left (f x +e \right )\right ) b^{2} d}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a^{2} d}{2 f \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a b c}{f \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) b^{2} d}{2 f \left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) a^{2} c}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {2 \arctan \left (\tan \left (f x +e \right )\right ) a b d}{f \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{2} c}{f \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f/(a^2+b^2)/(a+b*tan(f*x+e))*d*a-1/f/(a^2+b^2)/(a+b*tan(f*x+e))*c*b-1/f/(a^2+b^2)^2*ln(a+b*tan(f*x+e))*a^2*d
+2/f/(a^2+b^2)^2*ln(a+b*tan(f*x+e))*a*b*c+1/f/(a^2+b^2)^2*ln(a+b*tan(f*x+e))*b^2*d+1/2/f/(a^2+b^2)^2*ln(1+tan(
f*x+e)^2)*a^2*d-1/f/(a^2+b^2)^2*ln(1+tan(f*x+e)^2)*a*b*c-1/2/f/(a^2+b^2)^2*ln(1+tan(f*x+e)^2)*b^2*d+1/f/(a^2+b
^2)^2*arctan(tan(f*x+e))*a^2*c+2/f/(a^2+b^2)^2*arctan(tan(f*x+e))*a*b*d-1/f/(a^2+b^2)^2*arctan(tan(f*x+e))*b^2
*c

________________________________________________________________________________________

maxima [A]  time = 0.76, size = 180, normalized size = 1.62 \[ \frac {\frac {2 \, {\left (2 \, a b d + {\left (a^{2} - b^{2}\right )} c\right )} {\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (2 \, a b c - {\left (a^{2} - b^{2}\right )} d\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a b c - {\left (a^{2} - b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (b c - a d\right )}}{a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \tan \left (f x + e\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.48, size = 152, normalized size = 1.37 \[ \frac {a\,d-b\,c}{f\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (c+d\,1{}\mathrm {i}\right )}{2\,f\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (d+c\,1{}\mathrm {i}\right )}{2\,f\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (-d\,a^2+2\,c\,a\,b+d\,b^2\right )}{f\,{\left (a^2+b^2\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*d - b*c)/(f*(a^2 + b^2)*(a + b*tan(e + f*x))) - (log(tan(e + f*x) - 1i)*(c + d*1i))/(2*f*(2*a*b - a^2*1i +
b^2*1i)) - (log(tan(e + f*x) + 1i)*(c*1i + d))/(2*f*(a*b*2i - a^2 + b^2)) + (log(a + b*tan(e + f*x))*(b^2*d -
a^2*d + 2*a*b*c))/(f*(a^2 + b^2)^2)

________________________________________________________________________________________

sympy [A]  time = 1.90, size = 2878, normalized size = 25.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________