3.1.10 \(\int (a+b \tan (c+d x))^2 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [10]

Optimal. Leaf size=112 \[ -\left (\left (2 a b B+a^2 C-b^2 C\right ) x\right )-\frac {\left (a^2 B-b^2 B-2 a b C\right ) \log (\cos (c+d x))}{d}+\frac {b (a B-b C) \tan (c+d x)}{d}+\frac {B (a+b \tan (c+d x))^2}{2 d}+\frac {C (a+b \tan (c+d x))^3}{3 b d} \]

[Out]

-(2*B*a*b+C*a^2-C*b^2)*x-(B*a^2-B*b^2-2*C*a*b)*ln(cos(d*x+c))/d+b*(B*a-C*b)*tan(d*x+c)/d+1/2*B*(a+b*tan(d*x+c)
)^2/d+1/3*C*(a+b*tan(d*x+c))^3/b/d

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Rubi [A]
time = 0.08, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3711, 3609, 3606, 3556} \begin {gather*} -\frac {\left (a^2 B-2 a b C-b^2 B\right ) \log (\cos (c+d x))}{d}-x \left (a^2 C+2 a b B-b^2 C\right )+\frac {b (a B-b C) \tan (c+d x)}{d}+\frac {B (a+b \tan (c+d x))^2}{2 d}+\frac {C (a+b \tan (c+d x))^3}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3556

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

Rule 3606

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 3609

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 3711

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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 1.25, size = 172, normalized size = 1.54 \begin {gather*} \frac {2 C (a+b \tan (c+d x))^3+3 (a B+b C) \left (i \left ((a+i b)^2 \log (i-\tan (c+d x))-(a-i b)^2 \log (i+\tan (c+d x))\right )-2 b^2 \tan (c+d x)\right )+3 B \left ((i a-b)^3 \log (i-\tan (c+d x))-(i a+b)^3 \log (i+\tan (c+d x))+6 a b^2 \tan (c+d x)+b^3 \tan ^2(c+d x)\right )}{6 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*(a + b*Tan[c + d*x])^3 + 3*(a*B + b*C)*(I*((a + I*b)^2*Log[I - Tan[c + d*x]] - (a - I*b)^2*Log[I + Tan[c
+ d*x]]) - 2*b^2*Tan[c + d*x]) + 3*B*((I*a - b)^3*Log[I - Tan[c + d*x]] - (I*a + b)^3*Log[I + Tan[c + d*x]] +
6*a*b^2*Tan[c + d*x] + b^3*Tan[c + d*x]^2))/(6*b*d)

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Maple [A]
time = 0.07, size = 135, normalized size = 1.21

method	result	size
norman	\(\left (-2 B a b -C \,a^{2}+b^{2} C \right ) x +\frac {\left (2 B a b +C \,a^{2}-b^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {b \left (B b +2 C a \right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b^{2} C \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {\left (a^{2} B -b^{2} B -2 C a b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(120\)
derivativedivides	\(\frac {\frac {b^{2} C \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {B \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+C a b \left (\tan ^{2}\left (d x +c \right )\right )+2 B a b \tan \left (d x +c \right )+C \,a^{2} \tan \left (d x +c \right )-b^{2} C \tan \left (d x +c \right )+\frac {\left (a^{2} B -b^{2} B -2 C a b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-2 B a b -C \,a^{2}+b^{2} C \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(135\)
default	\(\frac {\frac {b^{2} C \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {B \,b^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+C a b \left (\tan ^{2}\left (d x +c \right )\right )+2 B a b \tan \left (d x +c \right )+C \,a^{2} \tan \left (d x +c \right )-b^{2} C \tan \left (d x +c \right )+\frac {\left (a^{2} B -b^{2} B -2 C a b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-2 B a b -C \,a^{2}+b^{2} C \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(135\)
risch	\(-\frac {4 i C a b c}{d}-i B \,b^{2} x +\frac {2 i B \,a^{2} c}{d}-2 B a b x -C \,a^{2} x +C \,b^{2} x +i B \,a^{2} x +\frac {2 i \left (-3 i B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 i C a b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 i B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 i C a b \,{\mathrm e}^{2 i \left (d x +c \right )}+12 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 B a b +3 C \,a^{2}-4 b^{2} C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {2 i B \,b^{2} c}{d}-2 i C a b x -\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2} B}{d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C a b}{d}\)	\(324\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 120, normalized size = 1.07 \begin {gather*} \frac {2 \, C b^{2} \tan \left (d x + c\right )^{3} + 3 \, {\left (2 \, C a b + B b^{2}\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \tan \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.37, size = 119, normalized size = 1.06 \begin {gather*} \frac {2 \, C b^{2} \tan \left (d x + c\right )^{3} - 6 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} d x + 3 \, {\left (2 \, C a b + B b^{2}\right )} \tan \left (d x + c\right )^{2} - 3 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \tan \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.13, size = 194, normalized size = 1.73 \begin {gather*} \begin {cases} \frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 2 B a b x + \frac {2 B a b \tan {\left (c + d x \right )}}{d} - \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} - C a^{2} x + \frac {C a^{2} \tan {\left (c + d x \right )}}{d} - \frac {C a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {C a b \tan ^{2}{\left (c + d x \right )}}{d} + C b^{2} x + \frac {C b^{2} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {C b^{2} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**2*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1509 vs. \(2 (108) = 216\).
time = 1.25, size = 1509, normalized size = 13.47 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*C*a^2*d*x*tan(d*x)^3*tan(c)^3 + 12*B*a*b*d*x*tan(d*x)^3*tan(c)^3 - 6*C*b^2*d*x*tan(d*x)^3*tan(c)^3 + 3
*B*a^2*log(4*(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 + 1))*tan(d*x)^3*tan(c)^3 - 6*C*a*b*log(4*(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 + 1))*tan(d*x)^3*tan(c)^3 - 3*B*b^2*log(4*(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 + 1)
)*tan(d*x)^3*tan(c)^3 - 18*C*a^2*d*x*tan(d*x)^2*tan(c)^2 - 36*B*a*b*d*x*tan(d*x)^2*tan(c)^2 + 18*C*b^2*d*x*tan
(d*x)^2*tan(c)^2 - 6*C*a*b*tan(d*x)^3*tan(c)^3 - 3*B*b^2*tan(d*x)^3*tan(c)^3 - 9*B*a^2*log(4*(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 + 1))*tan(d*x)
^2*tan(c)^2 + 18*C*a*b*log(4*(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 + 1))*tan(d*x)^2*tan(c)^2 + 9*B*b^2*log(4*(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 + 1))*tan(d*x)^2*tan(c)^2 + 6*C*a
^2*tan(d*x)^3*tan(c)^2 + 12*B*a*b*tan(d*x)^3*tan(c)^2 - 6*C*b^2*tan(d*x)^3*tan(c)^2 + 6*C*a^2*tan(d*x)^2*tan(c
)^3 + 12*B*a*b*tan(d*x)^2*tan(c)^3 - 6*C*b^2*tan(d*x)^2*tan(c)^3 + 18*C*a^2*d*x*tan(d*x)*tan(c) + 36*B*a*b*d*x
*tan(d*x)*tan(c) - 18*C*b^2*d*x*tan(d*x)*tan(c) - 6*C*a*b*tan(d*x)^3*tan(c) - 3*B*b^2*tan(d*x)^3*tan(c) + 6*C*
a*b*tan(d*x)^2*tan(c)^2 + 3*B*b^2*tan(d*x)^2*tan(c)^2 - 6*C*a*b*tan(d*x)*tan(c)^3 - 3*B*b^2*tan(d*x)*tan(c)^3
+ 2*C*b^2*tan(d*x)^3 + 9*B*a^2*log(4*(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 + 1))*tan(d*x)*tan(c) - 18*C*a*b*log(4*(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 + 1))*tan(d*x)*tan(c) - 9*B*
b^2*log(4*(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 + 1))*tan(d*x)*tan(c) - 12*C*a^2*tan(d*x)^2*tan(c) - 24*B*a*b*tan(d*x)^2*tan(c) + 18*C*b^2*tan(d*
x)^2*tan(c) - 12*C*a^2*tan(d*x)*tan(c)^2 - 24*B*a*b*tan(d*x)*tan(c)^2 + 18*C*b^2*tan(d*x)*tan(c)^2 + 2*C*b^2*t
an(c)^3 - 6*C*a^2*d*x - 12*B*a*b*d*x + 6*C*b^2*d*x + 6*C*a*b*tan(d*x)^2 + 3*B*b^2*tan(d*x)^2 - 6*C*a*b*tan(d*x
)*tan(c) - 3*B*b^2*tan(d*x)*tan(c) + 6*C*a*b*tan(c)^2 + 3*B*b^2*tan(c)^2 - 3*B*a^2*log(4*(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 + 1)) + 6*C*a*b*lo
g(4*(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)/(ta
n(c)^2 + 1)) + 3*B*b^2*log(4*(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 + 1)) + 6*C*a^2*tan(d*x) + 12*B*a*b*tan(d*x) - 6*C*b^2*tan(d*x) + 6*C*a^2*tan(
c) + 12*B*a*b*tan(c) - 6*C*b^2*tan(c) + 6*C*a*b + 3*B*b^2)/(d*tan(d*x)^3*tan(c)^3 - 3*d*tan(d*x)^2*tan(c)^2 +
3*d*tan(d*x)*tan(c) - d)

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Mupad [B]
time = 8.80, size = 121, normalized size = 1.08 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B\,b^2}{2}+C\,a\,b\right )}{d}-x\,\left (C\,a^2+2\,B\,a\,b-C\,b^2\right )+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (C\,a^2+2\,B\,a\,b-C\,b^2\right )}{d}-\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (-\frac {B\,a^2}{2}+C\,a\,b+\frac {B\,b^2}{2}\right )}{d}+\frac {C\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x))^2,x)

[Out]

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

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