3.435 \(\int \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=147 \[ -\frac{b \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac{a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}+b x \left (3 a^2-b^2\right )-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\frac{(a+b \tan (c+d x))^3}{3 d}-\frac{a (a+b \tan (c+d x))^2}{2 d} \]

[Out]

b*(3*a^2 - b^2)*x + (a*(a^2 - 3*b^2)*Log[Cos[c + d*x]])/d - (b*(a^2 - b^2)*Tan[c + d*x])/d - (a*(a + b*Tan[c +
d*x])^2)/(2*d) - (a + b*Tan[c + d*x])^3/(3*d) - (a*(a + b*Tan[c + d*x])^4)/(20*b^2*d) + (Tan[c + d*x]*(a + b*
Tan[c + d*x])^4)/(5*b*d)

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Rubi [A]  time = 0.181611, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3566, 3630, 12, 3528, 3525, 3475} \[ -\frac{b \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac{a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}+b x \left (3 a^2-b^2\right )-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\frac{(a+b \tan (c+d x))^3}{3 d}-\frac{a (a+b \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^3,x]

[Out]

b*(3*a^2 - b^2)*x + (a*(a^2 - 3*b^2)*Log[Cos[c + d*x]])/d - (b*(a^2 - b^2)*Tan[c + d*x])/d - (a*(a + b*Tan[c +
d*x])^2)/(2*d) - (a + b*Tan[c + d*x])^3/(3*d) - (a*(a + b*Tan[c + d*x])^4)/(20*b^2*d) + (Tan[c + d*x]*(a + b*
Tan[c + d*x])^4)/(5*b*d)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 12

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

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 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 \tan ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac{\int (a+b \tan (c+d x))^3 \left (-a-5 b \tan (c+d x)-a \tan ^2(c+d x)\right ) \, dx}{5 b}\\ &=-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}+\frac{\int -5 b \tan (c+d x) (a+b \tan (c+d x))^3 \, dx}{5 b}\\ &=-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx\\ &=-\frac{(a+b \tan (c+d x))^3}{3 d}-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\int (-b+a \tan (c+d x)) (a+b \tan (c+d x))^2 \, dx\\ &=-\frac{a (a+b \tan (c+d x))^2}{2 d}-\frac{(a+b \tan (c+d x))^3}{3 d}-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\int (a+b \tan (c+d x)) \left (-2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=b \left (3 a^2-b^2\right ) x-\frac{b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac{a (a+b \tan (c+d x))^2}{2 d}-\frac{(a+b \tan (c+d x))^3}{3 d}-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}-\left (a \left (a^2-3 b^2\right )\right ) \int \tan (c+d x) \, dx\\ &=b \left (3 a^2-b^2\right ) x+\frac{a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}-\frac{b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac{a (a+b \tan (c+d x))^2}{2 d}-\frac{(a+b \tan (c+d x))^3}{3 d}-\frac{a (a+b \tan (c+d x))^4}{20 b^2 d}+\frac{\tan (c+d x) (a+b \tan (c+d x))^4}{5 b d}\\ \end{align*}

Mathematica [C]  time = 0.759874, size = 161, normalized size = 1.1 \[ \frac{-20 b^3 \left (b^2-3 a^2\right ) \tan ^3(c+d x)+30 a b^2 \left (a^2-3 b^2\right ) \tan ^2(c+d x)+60 b^3 \left (b^2-3 a^2\right ) \tan (c+d x)-3 \left (a^5+10 b^2 (a+i b)^3 \log (-\tan (c+d x)+i)+10 b^2 (a-i b)^3 \log (\tan (c+d x)+i)\right )+45 a b^4 \tan ^4(c+d x)+12 b^5 \tan ^5(c+d x)}{60 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^3,x]

[Out]

(-3*(a^5 + 10*(a + I*b)^3*b^2*Log[I - Tan[c + d*x]] + 10*(a - I*b)^3*b^2*Log[I + Tan[c + d*x]]) + 60*b^3*(-3*a
^2 + b^2)*Tan[c + d*x] + 30*a*b^2*(a^2 - 3*b^2)*Tan[c + d*x]^2 - 20*b^3*(-3*a^2 + b^2)*Tan[c + d*x]^3 + 45*a*b
^4*Tan[c + d*x]^4 + 12*b^5*Tan[c + d*x]^5)/(60*b^2*d)

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Maple [A]  time = 0.005, size = 198, normalized size = 1.4 \begin{align*}{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{b{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{3\,d}}+{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{\tan \left ( dx+c \right ){a}^{2}b}{d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{3}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d}}+{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) a{b}^{2}}{2\,d}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}b}{d}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{3}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.84267, size = 185, normalized size = 1.26 \begin{align*} \frac{12 \, b^{3} \tan \left (d x + c\right )^{5} + 45 \, a b^{2} \tan \left (d x + c\right )^{4} + 20 \,{\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{3} + 30 \,{\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2} + 60 \,{\left (3 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )} - 30 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \,{\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(12*b^3*tan(d*x + c)^5 + 45*a*b^2*tan(d*x + c)^4 + 20*(3*a^2*b - b^3)*tan(d*x + c)^3 + 30*(a^3 - 3*a*b^2)
*tan(d*x + c)^2 + 60*(3*a^2*b - b^3)*(d*x + c) - 30*(a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1) - 60*(3*a^2*b - b^
3)*tan(d*x + c))/d

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Fricas [A]  time = 1.82059, size = 321, normalized size = 2.18 \begin{align*} \frac{12 \, b^{3} \tan \left (d x + c\right )^{5} + 45 \, a b^{2} \tan \left (d x + c\right )^{4} + 20 \,{\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{3} + 60 \,{\left (3 \, a^{2} b - b^{3}\right )} d x + 30 \,{\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2} + 30 \,{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 60 \,{\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(12*b^3*tan(d*x + c)^5 + 45*a*b^2*tan(d*x + c)^4 + 20*(3*a^2*b - b^3)*tan(d*x + c)^3 + 60*(3*a^2*b - b^3)
*d*x + 30*(a^3 - 3*a*b^2)*tan(d*x + c)^2 + 30*(a^3 - 3*a*b^2)*log(1/(tan(d*x + c)^2 + 1)) - 60*(3*a^2*b - b^3)
*tan(d*x + c))/d

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Sympy [A]  time = 0.919811, size = 194, normalized size = 1.32 \begin{align*} \begin{cases} - \frac{a^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{a^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} + 3 a^{2} b x + \frac{a^{2} b \tan ^{3}{\left (c + d x \right )}}{d} - \frac{3 a^{2} b \tan{\left (c + d x \right )}}{d} + \frac{3 a b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{3 a b^{2} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac{3 a b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} - b^{3} x + \frac{b^{3} \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac{b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac{b^{3} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right )^{3} \tan ^{3}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(a+b*tan(d*x+c))**3,x)

[Out]

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

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Giac [B]  time = 7.13218, size = 2696, normalized size = 18.34 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(180*a^2*b*d*x*tan(d*x)^5*tan(c)^5 - 60*b^3*d*x*tan(d*x)^5*tan(c)^5 + 30*a^3*log(4*(tan(c)^2 + 1)/(tan(d*
x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^5*ta
n(c)^5 - 90*a*b^2*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(
d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^5*tan(c)^5 - 900*a^2*b*d*x*tan(d*x)^4*tan(c)^4 + 300*b^3*d*x*tan(d*x
)^4*tan(c)^4 + 30*a^3*tan(d*x)^5*tan(c)^5 - 135*a*b^2*tan(d*x)^5*tan(c)^5 - 150*a^3*log(4*(tan(c)^2 + 1)/(tan(
d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^4*
tan(c)^4 + 450*a*b^2*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + t
an(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^4*tan(c)^4 + 180*a^2*b*tan(d*x)^5*tan(c)^4 - 60*b^3*tan(d*x)^5*ta
n(c)^4 + 180*a^2*b*tan(d*x)^4*tan(c)^5 - 60*b^3*tan(d*x)^4*tan(c)^5 + 1800*a^2*b*d*x*tan(d*x)^3*tan(c)^3 - 600
*b^3*d*x*tan(d*x)^3*tan(c)^3 + 30*a^3*tan(d*x)^5*tan(c)^3 - 90*a*b^2*tan(d*x)^5*tan(c)^3 - 90*a^3*tan(d*x)^4*t
an(c)^4 + 495*a*b^2*tan(d*x)^4*tan(c)^4 + 30*a^3*tan(d*x)^3*tan(c)^5 - 90*a*b^2*tan(d*x)^3*tan(c)^5 - 60*a^2*b
*tan(d*x)^5*tan(c)^2 + 20*b^3*tan(d*x)^5*tan(c)^2 + 300*a^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(
d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^3*tan(c)^3 - 900*a*b^2*log
(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*t
an(c) + 1))*tan(d*x)^3*tan(c)^3 - 900*a^2*b*tan(d*x)^4*tan(c)^3 + 300*b^3*tan(d*x)^4*tan(c)^3 - 900*a^2*b*tan(
d*x)^3*tan(c)^4 + 300*b^3*tan(d*x)^3*tan(c)^4 - 60*a^2*b*tan(d*x)^2*tan(c)^5 + 20*b^3*tan(d*x)^2*tan(c)^5 + 45
*a*b^2*tan(d*x)^5*tan(c) - 1800*a^2*b*d*x*tan(d*x)^2*tan(c)^2 + 600*b^3*d*x*tan(d*x)^2*tan(c)^2 - 90*a^3*tan(d
*x)^4*tan(c)^2 + 450*a*b^2*tan(d*x)^4*tan(c)^2 + 120*a^3*tan(d*x)^3*tan(c)^3 - 540*a*b^2*tan(d*x)^3*tan(c)^3 -
90*a^3*tan(d*x)^2*tan(c)^4 + 450*a*b^2*tan(d*x)^2*tan(c)^4 + 45*a*b^2*tan(d*x)*tan(c)^5 - 12*b^3*tan(d*x)^5 +
120*a^2*b*tan(d*x)^4*tan(c) - 100*b^3*tan(d*x)^4*tan(c) - 300*a^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 -
2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 + 900*a*
b^2*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan
(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 + 1440*a^2*b*tan(d*x)^3*tan(c)^2 - 600*b^3*tan(d*x)^3*tan(c)^2 + 1440*a
^2*b*tan(d*x)^2*tan(c)^3 - 600*b^3*tan(d*x)^2*tan(c)^3 + 120*a^2*b*tan(d*x)*tan(c)^4 - 100*b^3*tan(d*x)*tan(c)
^4 - 12*b^3*tan(c)^5 - 45*a*b^2*tan(d*x)^4 + 900*a^2*b*d*x*tan(d*x)*tan(c) - 300*b^3*d*x*tan(d*x)*tan(c) + 90*
a^3*tan(d*x)^3*tan(c) - 450*a*b^2*tan(d*x)^3*tan(c) - 120*a^3*tan(d*x)^2*tan(c)^2 + 540*a*b^2*tan(d*x)^2*tan(c
)^2 + 90*a^3*tan(d*x)*tan(c)^3 - 450*a*b^2*tan(d*x)*tan(c)^3 - 45*a*b^2*tan(c)^4 - 60*a^2*b*tan(d*x)^3 + 20*b^
3*tan(d*x)^3 + 150*a^3*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 +
tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c) - 450*a*b^2*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 -
2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c) - 900*a^2*b*t
an(d*x)^2*tan(c) + 300*b^3*tan(d*x)^2*tan(c) - 900*a^2*b*tan(d*x)*tan(c)^2 + 300*b^3*tan(d*x)*tan(c)^2 - 60*a^
2*b*tan(c)^3 + 20*b^3*tan(c)^3 - 180*a^2*b*d*x + 60*b^3*d*x - 30*a^3*tan(d*x)^2 + 90*a*b^2*tan(d*x)^2 + 90*a^3
*tan(d*x)*tan(c) - 495*a*b^2*tan(d*x)*tan(c) - 30*a^3*tan(c)^2 + 90*a*b^2*tan(c)^2 - 30*a^3*log(4*(tan(c)^2 +
1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)) + 9
0*a*b^2*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2
*tan(d*x)*tan(c) + 1)) + 180*a^2*b*tan(d*x) - 60*b^3*tan(d*x) + 180*a^2*b*tan(c) - 60*b^3*tan(c) - 30*a^3 + 13
5*a*b^2)/(d*tan(d*x)^5*tan(c)^5 - 5*d*tan(d*x)^4*tan(c)^4 + 10*d*tan(d*x)^3*tan(c)^3 - 10*d*tan(d*x)^2*tan(c)^
2 + 5*d*tan(d*x)*tan(c) - d)