3.1219 \(\int \frac {a+b \tan (e+f x)}{(c+d \tan (e+f x))^2} \, dx\)

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 2.15, size = 189, normalized size = 1.70 \[ \frac {(b c-a d) \left (\frac {2 d \left (\frac {c^2+d^2}{c+d \tan (e+f x)}-2 c \log (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^2}+\frac {i \log (-\tan (e+f x)+i)}{(c+i d)^2}-\frac {i \log (\tan (e+f x)+i)}{(c-i d)^2}\right )+\frac {b ((-d-i c) \log (-\tan (e+f x)+i)+i (c+i d) \log (\tan (e+f x)+i)+2 d \log (c+d \tan (e+f x)))}{c^2+d^2}}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 222, normalized size = 2.00 \[ \frac {2 \, b c d^{2} - 2 \, a d^{3} + 2 \, {\left (a c^{3} + 2 \, b c^{2} d - a c d^{2}\right )} f x - {\left (b c^{3} - 2 \, a c^{2} d - b c d^{2} + {\left (b c^{2} d - 2 \, a c d^{2} - b d^{3}\right )} \tan \left (f x + e\right )\right )} \log \left (\frac {d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, {\left (b c^{2} d - a c d^{2} - {\left (a c^{2} d + 2 \, b c d^{2} - a d^{3}\right )} f x\right )} \tan \left (f x + e\right )}{2 \, {\left ({\left (c^{4} d + 2 \, c^{2} d^{3} + d^{5}\right )} f \tan \left (f x + e\right ) + {\left (c^{5} + 2 \, c^{3} d^{2} + c d^{4}\right )} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.76, size = 241, normalized size = 2.17 \[ \frac {\frac {2 \, {\left (a c^{2} + 2 \, b c d - a d^{2}\right )} {\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {{\left (b c^{2} - 2 \, a c d - b d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} - \frac {2 \, {\left (b c^{2} d - 2 \, a c d^{2} - b d^{3}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{4} d + 2 \, c^{2} d^{3} + d^{5}} + \frac {2 \, {\left (b c^{2} d \tan \left (f x + e\right ) - 2 \, a c d^{2} \tan \left (f x + e\right ) - b d^{3} \tan \left (f x + e\right ) + 2 \, b c^{3} - 3 \, a c^{2} d - a d^{3}\right )}}{{\left (c^{4} + 2 \, c^{2} d^{2} + d^{4}\right )} {\left (d \tan \left (f x + e\right ) + c\right )}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.79, size = 177, normalized size = 1.59 \[ \frac {\frac {2 \, {\left (a c^{2} + 2 \, b c d - a d^{2}\right )} {\left (f x + e\right )}}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} - \frac {2 \, {\left (b c^{2} - 2 \, a c d - b d^{2}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {{\left (b c^{2} - 2 \, a c d - b d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{4} + 2 \, c^{2} d^{2} + d^{4}} + \frac {2 \, {\left (b c - a d\right )}}{c^{3} + c d^{2} + {\left (c^{2} d + d^{3}\right )} \tan \left (f x + e\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.61, size = 153, normalized size = 1.38 \[ \frac {\ln \left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (-b\,c^2+2\,a\,c\,d+b\,d^2\right )}{f\,{\left (c^2+d^2\right )}^2}-\frac {a\,d-b\,c}{f\,\left (c^2+d^2\right )\,\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )}{2\,f\,\left (-c^2+c\,d\,2{}\mathrm {i}+d^2\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,f\,\left (-c^2\,1{}\mathrm {i}+2\,c\,d+d^2\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.91, size = 2878, normalized size = 25.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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