3.510 \(\int (a+b \tan (d+e x)) (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x))^2 \, dx\)

Optimal. Leaf size=144 \[ \frac{\left (a^2+b^2\right ) (a \tan (d+e x)+b)^3}{3 e}+\frac{b \left (a^2+b^2\right ) (a \tan (d+e x)+b)^2}{2 e}-\frac{a \left (a^4-b^4\right ) \tan (d+e x)}{e}+\frac{b \left (3 a^2-b^2\right ) \left (a^2+b^2\right ) \log (\cos (d+e x))}{e}+a x \left (a^2-3 b^2\right ) \left (a^2+b^2\right )+\frac{b (a \tan (d+e x)+b)^4}{4 e} \]

[Out]

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

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Rubi [A]  time = 0.268532, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3708, 3528, 12, 3525, 3475} \[ \frac{\left (a^2+b^2\right ) (a \tan (d+e x)+b)^3}{3 e}+\frac{b \left (a^2+b^2\right ) (a \tan (d+e x)+b)^2}{2 e}-\frac{a \left (a^4-b^4\right ) \tan (d+e x)}{e}+\frac{b \left (3 a^2-b^2\right ) \left (a^2+b^2\right ) \log (\cos (d+e x))}{e}+a x \left (a^2-3 b^2\right ) \left (a^2+b^2\right )+\frac{b (a \tan (d+e x)+b)^4}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3708

Int[((A_) + (B_.)*tan[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*tan[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Tan[d + e*x])*(b + 2*c*Tan[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rule 3528

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3525

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^2 \, dx &=\frac{\int \left (2 a b+2 a^2 \tan (d+e x)\right )^4 (a+b \tan (d+e x)) \, dx}{16 a^4}\\ &=\frac{b (b+a \tan (d+e x))^4}{4 e}+\frac{\int 2 a \left (a^2+b^2\right ) \tan (d+e x) \left (2 a b+2 a^2 \tan (d+e x)\right )^3 \, dx}{16 a^4}\\ &=\frac{b (b+a \tan (d+e x))^4}{4 e}+\frac{\left (a^2+b^2\right ) \int \tan (d+e x) \left (2 a b+2 a^2 \tan (d+e x)\right )^3 \, dx}{8 a^3}\\ &=\frac{\left (a^2+b^2\right ) (b+a \tan (d+e x))^3}{3 e}+\frac{b (b+a \tan (d+e x))^4}{4 e}+\frac{\left (a^2+b^2\right ) \int \left (2 a b+2 a^2 \tan (d+e x)\right )^2 \left (-2 a^2+2 a b \tan (d+e x)\right ) \, dx}{8 a^3}\\ &=\frac{b \left (a^2+b^2\right ) (b+a \tan (d+e x))^2}{2 e}+\frac{\left (a^2+b^2\right ) (b+a \tan (d+e x))^3}{3 e}+\frac{b (b+a \tan (d+e x))^4}{4 e}+\frac{\left (a^2+b^2\right ) \int \left (2 a b+2 a^2 \tan (d+e x)\right ) \left (-8 a^3 b-4 a^2 \left (a^2-b^2\right ) \tan (d+e x)\right ) \, dx}{8 a^3}\\ &=a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) x-\frac{a \left (a^4-b^4\right ) \tan (d+e x)}{e}+\frac{b \left (a^2+b^2\right ) (b+a \tan (d+e x))^2}{2 e}+\frac{\left (a^2+b^2\right ) (b+a \tan (d+e x))^3}{3 e}+\frac{b (b+a \tan (d+e x))^4}{4 e}-\left (b \left (3 a^2-b^2\right ) \left (a^2+b^2\right )\right ) \int \tan (d+e x) \, dx\\ &=a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) x+\frac{b \left (3 a^2-b^2\right ) \left (a^2+b^2\right ) \log (\cos (d+e x))}{e}-\frac{a \left (a^4-b^4\right ) \tan (d+e x)}{e}+\frac{b \left (a^2+b^2\right ) (b+a \tan (d+e x))^2}{2 e}+\frac{\left (a^2+b^2\right ) (b+a \tan (d+e x))^3}{3 e}+\frac{b (b+a \tan (d+e x))^4}{4 e}\\ \end{align*}

Mathematica [C]  time = 2.32042, size = 153, normalized size = 1.06 \[ \frac{4 a^3 \left (a^2+4 b^2\right ) \tan ^3(d+e x)+18 a^2 b \left (a^2+2 b^2\right ) \tan ^2(d+e x)-12 a \left (-2 a^2 b^2+a^4-4 b^4\right ) \tan (d+e x)+6 \left (a^2+b^2\right ) \left (i (a+i b)^3 \log (\tan (d+e x)+i)-i (a-i b)^3 \log (-\tan (d+e x)+i)\right )+3 a^4 b \tan ^4(d+e x)}{12 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(a^2 + b^2)*((-I)*(a - I*b)^3*Log[I - Tan[d + e*x]] + I*(a + I*b)^3*Log[I + Tan[d + e*x]]) - 12*a*(a^4 - 2*
a^2*b^2 - 4*b^4)*Tan[d + e*x] + 18*a^2*b*(a^2 + 2*b^2)*Tan[d + e*x]^2 + 4*a^3*(a^2 + 4*b^2)*Tan[d + e*x]^3 + 3
*a^4*b*Tan[d + e*x]^4)/(12*e)

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Maple [A]  time = 0.008, size = 245, normalized size = 1.7 \begin{align*}{\frac{{a}^{4}b \left ( \tan \left ( ex+d \right ) \right ) ^{4}}{4\,e}}+{\frac{ \left ( \tan \left ( ex+d \right ) \right ) ^{3}{a}^{5}}{3\,e}}+{\frac{4\, \left ( \tan \left ( ex+d \right ) \right ) ^{3}{a}^{3}{b}^{2}}{3\,e}}+{\frac{3\, \left ( \tan \left ( ex+d \right ) \right ) ^{2}{a}^{4}b}{2\,e}}+3\,{\frac{{a}^{2} \left ( \tan \left ( ex+d \right ) \right ) ^{2}{b}^{3}}{e}}-{\frac{{a}^{5}\tan \left ( ex+d \right ) }{e}}+2\,{\frac{{a}^{3}{b}^{2}\tan \left ( ex+d \right ) }{e}}+4\,{\frac{a{b}^{4}\tan \left ( ex+d \right ) }{e}}-{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){a}^{4}b}{2\,e}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){a}^{2}{b}^{3}}{e}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ){b}^{5}}{2\,e}}+{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ){a}^{5}}{e}}-2\,{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ){a}^{3}{b}^{2}}{e}}-3\,{\frac{\arctan \left ( \tan \left ( ex+d \right ) \right ) a{b}^{4}}{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/e*a^4*b*tan(e*x+d)^4+1/3/e*tan(e*x+d)^3*a^5+4/3/e*tan(e*x+d)^3*a^3*b^2+3/2/e*tan(e*x+d)^2*a^4*b+3/e*tan(e*
x+d)^2*a^2*b^3-1/e*a^5*tan(e*x+d)+2/e*a^3*b^2*tan(e*x+d)+4/e*a*b^4*tan(e*x+d)-3/2/e*ln(1+tan(e*x+d)^2)*a^4*b-1
/e*ln(1+tan(e*x+d)^2)*a^2*b^3+1/2/e*ln(1+tan(e*x+d)^2)*b^5+1/e*arctan(tan(e*x+d))*a^5-2/e*arctan(tan(e*x+d))*a
^3*b^2-3/e*arctan(tan(e*x+d))*a*b^4

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Maxima [A]  time = 1.51236, size = 203, normalized size = 1.41 \begin{align*} \frac{3 \, a^{4} b \tan \left (e x + d\right )^{4} + 4 \,{\left (a^{5} + 4 \, a^{3} b^{2}\right )} \tan \left (e x + d\right )^{3} + 18 \,{\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \tan \left (e x + d\right )^{2} + 12 \,{\left (a^{5} - 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )}{\left (e x + d\right )} - 6 \,{\left (3 \, a^{4} b + 2 \, a^{2} b^{3} - b^{5}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right ) - 12 \,{\left (a^{5} - 2 \, a^{3} b^{2} - 4 \, a b^{4}\right )} \tan \left (e x + d\right )}{12 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^4*b*tan(e*x + d)^4 + 4*(a^5 + 4*a^3*b^2)*tan(e*x + d)^3 + 18*(a^4*b + 2*a^2*b^3)*tan(e*x + d)^2 + 12
*(a^5 - 2*a^3*b^2 - 3*a*b^4)*(e*x + d) - 6*(3*a^4*b + 2*a^2*b^3 - b^5)*log(tan(e*x + d)^2 + 1) - 12*(a^5 - 2*a
^3*b^2 - 4*a*b^4)*tan(e*x + d))/e

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Fricas [A]  time = 1.85719, size = 342, normalized size = 2.38 \begin{align*} \frac{3 \, a^{4} b \tan \left (e x + d\right )^{4} + 4 \,{\left (a^{5} + 4 \, a^{3} b^{2}\right )} \tan \left (e x + d\right )^{3} + 12 \,{\left (a^{5} - 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} e x + 18 \,{\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \tan \left (e x + d\right )^{2} + 6 \,{\left (3 \, a^{4} b + 2 \, a^{2} b^{3} - b^{5}\right )} \log \left (\frac{1}{\tan \left (e x + d\right )^{2} + 1}\right ) - 12 \,{\left (a^{5} - 2 \, a^{3} b^{2} - 4 \, a b^{4}\right )} \tan \left (e x + d\right )}{12 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^4*b*tan(e*x + d)^4 + 4*(a^5 + 4*a^3*b^2)*tan(e*x + d)^3 + 12*(a^5 - 2*a^3*b^2 - 3*a*b^4)*e*x + 18*(a
^4*b + 2*a^2*b^3)*tan(e*x + d)^2 + 6*(3*a^4*b + 2*a^2*b^3 - b^5)*log(1/(tan(e*x + d)^2 + 1)) - 12*(a^5 - 2*a^3
*b^2 - 4*a*b^4)*tan(e*x + d))/e

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Sympy [A]  time = 0.741305, size = 248, normalized size = 1.72 \begin{align*} \begin{cases} a^{5} x + \frac{a^{5} \tan ^{3}{\left (d + e x \right )}}{3 e} - \frac{a^{5} \tan{\left (d + e x \right )}}{e} - \frac{3 a^{4} b \log{\left (\tan ^{2}{\left (d + e x \right )} + 1 \right )}}{2 e} + \frac{a^{4} b \tan ^{4}{\left (d + e x \right )}}{4 e} + \frac{3 a^{4} b \tan ^{2}{\left (d + e x \right )}}{2 e} - 2 a^{3} b^{2} x + \frac{4 a^{3} b^{2} \tan ^{3}{\left (d + e x \right )}}{3 e} + \frac{2 a^{3} b^{2} \tan{\left (d + e x \right )}}{e} - \frac{a^{2} b^{3} \log{\left (\tan ^{2}{\left (d + e x \right )} + 1 \right )}}{e} + \frac{3 a^{2} b^{3} \tan ^{2}{\left (d + e x \right )}}{e} - 3 a b^{4} x + \frac{4 a b^{4} \tan{\left (d + e x \right )}}{e} + \frac{b^{5} \log{\left (\tan ^{2}{\left (d + e x \right )} + 1 \right )}}{2 e} & \text{for}\: e \neq 0 \\x \left (a + b \tan{\left (d \right )}\right ) \left (a^{2} \tan ^{2}{\left (d \right )} + 2 a b \tan{\left (d \right )} + b^{2}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**5*x + a**5*tan(d + e*x)**3/(3*e) - a**5*tan(d + e*x)/e - 3*a**4*b*log(tan(d + e*x)**2 + 1)/(2*e)
+ a**4*b*tan(d + e*x)**4/(4*e) + 3*a**4*b*tan(d + e*x)**2/(2*e) - 2*a**3*b**2*x + 4*a**3*b**2*tan(d + e*x)**3
/(3*e) + 2*a**3*b**2*tan(d + e*x)/e - a**2*b**3*log(tan(d + e*x)**2 + 1)/e + 3*a**2*b**3*tan(d + e*x)**2/e - 3
*a*b**4*x + 4*a*b**4*tan(d + e*x)/e + b**5*log(tan(d + e*x)**2 + 1)/(2*e), Ne(e, 0)), (x*(a + b*tan(d))*(a**2*
tan(d)**2 + 2*a*b*tan(d) + b**2)**2, True))

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Giac [B]  time = 7.67141, size = 3140, normalized size = 21.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*a^5*x*e*tan(x*e)^4*tan(d)^4 - 24*a^3*b^2*x*e*tan(x*e)^4*tan(d)^4 - 36*a*b^4*x*e*tan(x*e)^4*tan(d)^4 +
18*a^4*b*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 -
2*tan(x*e)*tan(d) + 1))*tan(x*e)^4*tan(d)^4 + 12*a^2*b^3*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*
e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1))*tan(x*e)^4*tan(d)^4 - 6*b^5*log(4*(ta
n(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d)
+ 1))*tan(x*e)^4*tan(d)^4 - 48*a^5*x*e*tan(x*e)^3*tan(d)^3 + 96*a^3*b^2*x*e*tan(x*e)^3*tan(d)^3 + 144*a*b^4*x*
e*tan(x*e)^3*tan(d)^3 + 15*a^4*b*tan(x*e)^4*tan(d)^4 + 36*a^2*b^3*tan(x*e)^4*tan(d)^4 - 72*a^4*b*log(4*(tan(d)
^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1)
)*tan(x*e)^3*tan(d)^3 - 48*a^2*b^3*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^
2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1))*tan(x*e)^3*tan(d)^3 + 24*b^5*log(4*(tan(d)^2 + 1)/(tan(x*e)^
4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1))*tan(x*e)^3*tan(d
)^3 + 12*a^5*tan(x*e)^4*tan(d)^3 - 24*a^3*b^2*tan(x*e)^4*tan(d)^3 - 48*a*b^4*tan(x*e)^4*tan(d)^3 + 12*a^5*tan(
x*e)^3*tan(d)^4 - 24*a^3*b^2*tan(x*e)^3*tan(d)^4 - 48*a*b^4*tan(x*e)^3*tan(d)^4 + 72*a^5*x*e*tan(x*e)^2*tan(d)
^2 - 144*a^3*b^2*x*e*tan(x*e)^2*tan(d)^2 - 216*a*b^4*x*e*tan(x*e)^2*tan(d)^2 + 18*a^4*b*tan(x*e)^4*tan(d)^2 +
36*a^2*b^3*tan(x*e)^4*tan(d)^2 - 24*a^4*b*tan(x*e)^3*tan(d)^3 - 72*a^2*b^3*tan(x*e)^3*tan(d)^3 + 18*a^4*b*tan(
x*e)^2*tan(d)^4 + 36*a^2*b^3*tan(x*e)^2*tan(d)^4 - 4*a^5*tan(x*e)^4*tan(d) - 16*a^3*b^2*tan(x*e)^4*tan(d) + 10
8*a^4*b*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2
*tan(x*e)*tan(d) + 1))*tan(x*e)^2*tan(d)^2 + 72*a^2*b^3*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)
^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1))*tan(x*e)^2*tan(d)^2 - 36*b^5*log(4*(tan
(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) +
1))*tan(x*e)^2*tan(d)^2 - 48*a^5*tan(x*e)^3*tan(d)^2 + 24*a^3*b^2*tan(x*e)^3*tan(d)^2 + 144*a*b^4*tan(x*e)^3*
tan(d)^2 - 48*a^5*tan(x*e)^2*tan(d)^3 + 24*a^3*b^2*tan(x*e)^2*tan(d)^3 + 144*a*b^4*tan(x*e)^2*tan(d)^3 - 4*a^5
*tan(x*e)*tan(d)^4 - 16*a^3*b^2*tan(x*e)*tan(d)^4 + 3*a^4*b*tan(x*e)^4 - 48*a^5*x*e*tan(x*e)*tan(d) + 96*a^3*b
^2*x*e*tan(x*e)*tan(d) + 144*a*b^4*x*e*tan(x*e)*tan(d) - 24*a^4*b*tan(x*e)^3*tan(d) - 72*a^2*b^3*tan(x*e)^3*ta
n(d) + 36*a^4*b*tan(x*e)^2*tan(d)^2 + 72*a^2*b^3*tan(x*e)^2*tan(d)^2 - 24*a^4*b*tan(x*e)*tan(d)^3 - 72*a^2*b^3
*tan(x*e)*tan(d)^3 + 3*a^4*b*tan(d)^4 + 4*a^5*tan(x*e)^3 + 16*a^3*b^2*tan(x*e)^3 - 72*a^4*b*log(4*(tan(d)^2 +
1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1))*tan
(x*e)*tan(d) - 48*a^2*b^3*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^
2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1))*tan(x*e)*tan(d) + 24*b^5*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 -
2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1))*tan(x*e)*tan(d) + 48*a^5*tan(
x*e)^2*tan(d) - 24*a^3*b^2*tan(x*e)^2*tan(d) - 144*a*b^4*tan(x*e)^2*tan(d) + 48*a^5*tan(x*e)*tan(d)^2 - 24*a^3
*b^2*tan(x*e)*tan(d)^2 - 144*a*b^4*tan(x*e)*tan(d)^2 + 4*a^5*tan(d)^3 + 16*a^3*b^2*tan(d)^3 + 12*a^5*x*e - 24*
a^3*b^2*x*e - 36*a*b^4*x*e + 18*a^4*b*tan(x*e)^2 + 36*a^2*b^3*tan(x*e)^2 - 24*a^4*b*tan(x*e)*tan(d) - 72*a^2*b
^3*tan(x*e)*tan(d) + 18*a^4*b*tan(d)^2 + 36*a^2*b^3*tan(d)^2 + 18*a^4*b*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d
)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1)) + 12*a^2*b^3*log(4*(tan
(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) +
1)) - 6*b^5*log(4*(tan(d)^2 + 1)/(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^
2 - 2*tan(x*e)*tan(d) + 1)) - 12*a^5*tan(x*e) + 24*a^3*b^2*tan(x*e) + 48*a*b^4*tan(x*e) - 12*a^5*tan(d) + 24*a
^3*b^2*tan(d) + 48*a*b^4*tan(d) + 15*a^4*b + 36*a^2*b^3)/(e*tan(x*e)^4*tan(d)^4 - 4*e*tan(x*e)^3*tan(d)^3 + 6*
e*tan(x*e)^2*tan(d)^2 - 4*e*tan(x*e)*tan(d) + e)