3.33 \(\int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=103 \[ -\frac{b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac{1}{2} a x \left (a^2-9 b^2\right )+\frac{9 a b^2 \tan (c+d x)}{2 d}-\frac{\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))^3}{2 d}+\frac{b^3 \tan ^2(c+d x)}{d} \]

[Out]

(a*(a^2 - 9*b^2)*x)/2 - (b*(3*a^2 - 2*b^2)*Log[Cos[c + d*x]])/d + (9*a*b^2*Tan[c + d*x])/(2*d) + (b^3*Tan[c +
d*x]^2)/d - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(2*d)

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Rubi [A]  time = 0.144593, antiderivative size = 103, 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 = {3516, 1645, 801, 635, 203, 260} \[ -\frac{b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac{1}{2} a x \left (a^2-9 b^2\right )+\frac{9 a b^2 \tan (c+d x)}{2 d}-\frac{\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))^3}{2 d}+\frac{b^3 \tan ^2(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^3,x]

[Out]

(a*(a^2 - 9*b^2)*x)/2 - (b*(3*a^2 - 2*b^2)*Log[Cos[c + d*x]])/d + (9*a*b^2*Tan[c + d*x])/(2*d) + (b^3*Tan[c +
d*x]^2)/d - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(2*d)

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rule 1645

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \sin ^2(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{x^2 (a+x)^3}{\left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{(a+x)^2 \left (-a b^2-4 b^2 x\right )}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=-\frac{\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}-\frac{\operatorname{Subst}\left (\int \left (-9 a b^2-4 b^2 x-\frac{a b^2 \left (a^2-9 b^2\right )+2 b^2 \left (3 a^2-2 b^2\right ) x}{b^2+x^2}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=\frac{9 a b^2 \tan (c+d x)}{2 d}+\frac{b^3 \tan ^2(c+d x)}{d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{a b^2 \left (a^2-9 b^2\right )+2 b^2 \left (3 a^2-2 b^2\right ) x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=\frac{9 a b^2 \tan (c+d x)}{2 d}+\frac{b^3 \tan ^2(c+d x)}{d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}+\frac{\left (a b \left (a^2-9 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 d}+\frac{\left (b \left (3 a^2-2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{1}{2} a \left (a^2-9 b^2\right ) x-\frac{b \left (3 a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac{9 a b^2 \tan (c+d x)}{2 d}+\frac{b^3 \tan ^2(c+d x)}{d}-\frac{\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{2 d}\\ \end{align*}

Mathematica [A]  time = 4.22673, size = 203, normalized size = 1.97 \[ \frac{b \left (-\frac{a \left (a^2-3 b^2\right ) \sin (2 (c+d x))}{2 b}+\left (3 a^2-b^2\right ) \cos ^2(c+d x)-\frac{a \left (a^2-3 b^2\right ) \tan ^{-1}(\tan (c+d x))}{b}+\left (\frac{a^3-6 a b^2}{\sqrt{-b^2}}+3 a^2-2 b^2\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+\left (\frac{6 a b^2-a^3}{\sqrt{-b^2}}+3 a^2-2 b^2\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )+6 a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^3,x]

[Out]

(b*(-((a*(a^2 - 3*b^2)*ArcTan[Tan[c + d*x]])/b) + (3*a^2 - b^2)*Cos[c + d*x]^2 + (3*a^2 - 2*b^2 + (a^3 - 6*a*b
^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + (3*a^2 - 2*b^2 + (-a^3 + 6*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2
] + b*Tan[c + d*x]] - (a*(a^2 - 3*b^2)*Sin[2*(c + d*x)])/(2*b) + 6*a*b*Tan[c + d*x] + b^2*Tan[c + d*x]^2))/(2*
d)

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Maple [B]  time = 0.052, size = 226, normalized size = 2.2 \begin{align*}{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{3}}{d}}+2\,{\frac{{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{a{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{9\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) a{b}^{2}}{2\,d}}-{\frac{9\,a{b}^{2}x}{2}}-{\frac{9\,a{b}^{2}c}{2\,d}}-{\frac{3\,b{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{b{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}x}{2}}+{\frac{{a}^{3}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x)

[Out]

1/2/d*b^3*sin(d*x+c)^6/cos(d*x+c)^2+1/2/d*b^3*sin(d*x+c)^4+1/d*sin(d*x+c)^2*b^3+2*b^3*ln(cos(d*x+c))/d+3/d*a*b
^2*sin(d*x+c)^5/cos(d*x+c)+3/d*a*b^2*cos(d*x+c)*sin(d*x+c)^3+9/2/d*sin(d*x+c)*cos(d*x+c)*a*b^2-9/2*a*b^2*x-9/2
/d*a*b^2*c-3/2/d*b*a^2*sin(d*x+c)^2-3/d*b*a^2*ln(cos(d*x+c))-1/2/d*a^3*sin(d*x+c)*cos(d*x+c)+1/2*a^3*x+1/2/d*a
^3*c

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Maxima [A]  time = 1.51541, size = 153, normalized size = 1.49 \begin{align*} \frac{b^{3} \tan \left (d x + c\right )^{2} + 6 \, a b^{2} \tan \left (d x + c\right ) +{\left (a^{3} - 9 \, a b^{2}\right )}{\left (d x + c\right )} +{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + \frac{3 \, a^{2} b - b^{3} -{\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(b^3*tan(d*x + c)^2 + 6*a*b^2*tan(d*x + c) + (a^3 - 9*a*b^2)*(d*x + c) + (3*a^2*b - 2*b^3)*log(tan(d*x + c
)^2 + 1) + (3*a^2*b - b^3 - (a^3 - 3*a*b^2)*tan(d*x + c))/(tan(d*x + c)^2 + 1))/d

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Fricas [A]  time = 2.16828, size = 342, normalized size = 3.32 \begin{align*} \frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} - 4 \,{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + 2 \, b^{3} -{\left (3 \, a^{2} b - b^{3} - 2 \,{\left (a^{3} - 9 \, a b^{2}\right )} d x\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (6 \, a b^{2} \cos \left (d x + c\right ) -{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(2*(3*a^2*b - b^3)*cos(d*x + c)^4 - 4*(3*a^2*b - 2*b^3)*cos(d*x + c)^2*log(-cos(d*x + c)) + 2*b^3 - (3*a^2
*b - b^3 - 2*(a^3 - 9*a*b^2)*d*x)*cos(d*x + c)^2 + 2*(6*a*b^2*cos(d*x + c) - (a^3 - 3*a*b^2)*cos(d*x + c)^3)*s
in(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{3} \sin ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**2*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sin(c + d*x)**2, x)

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Giac [B]  time = 3.22714, size = 3497, normalized size = 33.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/4*(2*a^3*d*x*tan(d*x)^4*tan(c)^4 - 18*a*b^2*d*x*tan(d*x)^4*tan(c)^4 - 6*a^2*b*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 + 4*b^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 + 2*a^3*d*x*tan(d*x)^4*tan(c)^2 - 18*a*b^2*d*x*tan(d*x)^4*tan(
c)^2 - 4*a^3*d*x*tan(d*x)^3*tan(c)^3 + 36*a*b^2*d*x*tan(d*x)^3*tan(c)^3 + 2*a^3*d*x*tan(d*x)^2*tan(c)^4 - 18*a
*b^2*d*x*tan(d*x)^2*tan(c)^4 + 3*a^2*b*tan(d*x)^4*tan(c)^4 + b^3*tan(d*x)^4*tan(c)^4 - 6*a^2*b*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)^2 + 4*b^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)^2 + 12*a^2*b*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 -
8*b^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 + 2*a^3*tan(d*x)^4*tan(c)^3 - 18*a*b^2*tan(d*x)^4*tan(c)^3 - 6*a^2*b
*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)^4 + 4*b^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)^4 + 2*a^3*tan(d*x)^3*tan(c)^4 - 1
8*a*b^2*tan(d*x)^3*tan(c)^4 - 4*a^3*d*x*tan(d*x)^3*tan(c) + 36*a*b^2*d*x*tan(d*x)^3*tan(c) + 4*a^3*d*x*tan(d*x
)^2*tan(c)^2 - 36*a*b^2*d*x*tan(d*x)^2*tan(c)^2 - 3*a^2*b*tan(d*x)^4*tan(c)^2 + 5*b^3*tan(d*x)^4*tan(c)^2 - 4*
a^3*d*x*tan(d*x)*tan(c)^3 + 36*a*b^2*d*x*tan(d*x)*tan(c)^3 - 18*a^2*b*tan(d*x)^3*tan(c)^3 + 6*b^3*tan(d*x)^3*t
an(c)^3 - 3*a^2*b*tan(d*x)^2*tan(c)^4 + 5*b^3*tan(d*x)^2*tan(c)^4 + 12*a^2*b*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)
- 8*b^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) - 12*a*b^2*tan(d*x)^4*tan(c) - 12*a^2*b*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 + 8*b^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 - 6*a^3*tan(d*x)^3*tan(c)^2 + 18*a*b^2*tan(d*x)^3*tan(c)^2 +
12*a^2*b*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)^3 - 8*b^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)^3 - 6*a^3*tan(d*x)^2*tan(c)^3
+ 18*a*b^2*tan(d*x)^2*tan(c)^3 - 12*a*b^2*tan(d*x)*tan(c)^4 + 2*a^3*d*x*tan(d*x)^2 - 18*a*b^2*d*x*tan(d*x)^2
+ 2*b^3*tan(d*x)^4 - 4*a^3*d*x*tan(d*x)*tan(c) + 36*a*b^2*d*x*tan(d*x)*tan(c) + 6*a^2*b*tan(d*x)^3*tan(c) - 2*
b^3*tan(d*x)^3*tan(c) + 2*a^3*d*x*tan(c)^2 - 18*a*b^2*d*x*tan(c)^2 + 30*a^2*b*tan(d*x)^2*tan(c)^2 - 2*b^3*tan(
d*x)^2*tan(c)^2 + 6*a^2*b*tan(d*x)*tan(c)^3 - 2*b^3*tan(d*x)*tan(c)^3 + 2*b^3*tan(c)^4 - 6*a^2*b*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 + 4*b^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 + t
an(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2 + 12*a*b^2*tan(d*x)^3 + 12*a^2*b*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)
- 8*b^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) + 6*a^3*tan(d*x)^2*tan(c) - 18*a*b^2*tan(d*x)^2*tan(c) - 6*a^2*b*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(c)^2 + 4*b^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(c)^2 + 6*a^3*tan(d*x)*tan(c)^2 - 18*a*b^2*tan(d*x)*tan(c)^2 + 12
*a*b^2*tan(c)^3 + 2*a^3*d*x - 18*a*b^2*d*x - 3*a^2*b*tan(d*x)^2 + 5*b^3*tan(d*x)^2 - 18*a^2*b*tan(d*x)*tan(c)
+ 6*b^3*tan(d*x)*tan(c) - 3*a^2*b*tan(c)^2 + 5*b^3*tan(c)^2 - 6*a^2*b*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)) + 4*b^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)) -
2*a^3*tan(d*x) + 18*a*b^2*tan(d*x) - 2*a^3*tan(c) + 18*a*b^2*tan(c) + 3*a^2*b + b^3)/(d*tan(d*x)^4*tan(c)^4 +
d*tan(d*x)^4*tan(c)^2 - 2*d*tan(d*x)^3*tan(c)^3 + d*tan(d*x)^2*tan(c)^4 - 2*d*tan(d*x)^3*tan(c) + 2*d*tan(d*x)
^2*tan(c)^2 - 2*d*tan(d*x)*tan(c)^3 + d*tan(d*x)^2 - 2*d*tan(d*x)*tan(c) + d*tan(c)^2 + d)