3.3.51 \(\int (a+b \tan ^2(c+d x))^3 \, dx\) [251]

Optimal. Leaf size=77 \[ (a-b)^3 x+\frac {b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {(3 a-b) b^2 \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3742, 398, 209} \begin {gather*} \frac {b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {b^2 (3 a-b) \tan ^3(c+d x)}{3 d}+x (a-b)^3+\frac {b^3 \tan ^5(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^2)^3,x]

[Out]

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

Rule 209

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

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3742

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[c*(ff/f), Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \left (a+b \tan ^2(c+d x)\right )^3 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^3}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (b \left (3 a^2-3 a b+b^2\right )+(3 a-b) b^2 x^2+b^3 x^4+\frac {(a-b)^3}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {(3 a-b) b^2 \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d}+\frac {(a-b)^3 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=(a-b)^3 x+\frac {b \left (3 a^2-3 a b+b^2\right ) \tan (c+d x)}{d}+\frac {(3 a-b) b^2 \tan ^3(c+d x)}{3 d}+\frac {b^3 \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.62, size = 102, normalized size = 1.32 \begin {gather*} \frac {\tan (c+d x) \left (\frac {15 (a-b)^3 \tanh ^{-1}\left (\sqrt {-\tan ^2(c+d x)}\right )}{\sqrt {-\tan ^2(c+d x)}}+b \left (45 a^2-15 a b \left (3-\tan ^2(c+d x)\right )+b^2 \left (15-5 \tan ^2(c+d x)+3 \tan ^4(c+d x)\right )\right )\right )}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^2)^3,x]

[Out]

(Tan[c + d*x]*((15*(a - b)^3*ArcTanh[Sqrt[-Tan[c + d*x]^2]])/Sqrt[-Tan[c + d*x]^2] + b*(45*a^2 - 15*a*b*(3 - T
an[c + d*x]^2) + b^2*(15 - 5*Tan[c + d*x]^2 + 3*Tan[c + d*x]^4))))/(15*d)

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Maple [A]
time = 0.05, size = 108, normalized size = 1.40

method	result	size
norman	\(\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) x +\frac {b \left (3 a^{2}-3 a b +b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}+\frac {\left (3 a -b \right ) b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}\)	\(88\)
derivativedivides	\(\frac {\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+a \,b^{2} \left (\tan ^{3}\left (d x +c \right )\right )-\frac {b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3}+3 a^{2} b \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )+b^{3} \tan \left (d x +c \right )+\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(108\)
default	\(\frac {\frac {b^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5}+a \,b^{2} \left (\tan ^{3}\left (d x +c \right )\right )-\frac {b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3}+3 a^{2} b \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )+b^{3} \tan \left (d x +c \right )+\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(108\)
risch	\(a^{3} x -3 a^{2} b x +3 a \,b^{2} x -b^{3} x +\frac {2 i b \left (45 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-90 a b \,{\mathrm e}^{8 i \left (d x +c \right )}+45 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+180 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-270 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+90 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+270 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-330 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+140 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+180 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-210 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+70 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+45 a^{2}-60 a b +23 b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 104, normalized size = 1.35 \begin {gather*} a^{3} x - \frac {3 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b}{d} + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a b^{2}}{d} + \frac {{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} b^{3}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 3*(d*x + c - tan(d*x + c))*a^2*b/d + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a*b^2/d + 1/15*(3
*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*b^3/d

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Fricas [A]
time = 3.60, size = 90, normalized size = 1.17 \begin {gather*} \frac {3 \, b^{3} \tan \left (d x + c\right )^{5} + 5 \, {\left (3 \, a b^{2} - b^{3}\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x + 15 \, {\left (3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^3*tan(d*x + c)^5 + 5*(3*a*b^2 - b^3)*tan(d*x + c)^3 + 15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*d*x + 15*(3
*a^2*b - 3*a*b^2 + b^3)*tan(d*x + c))/d

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Sympy [A]
time = 0.15, size = 126, normalized size = 1.64 \begin {gather*} \begin {cases} a^{3} x - 3 a^{2} b x + \frac {3 a^{2} b \tan {\left (c + d x \right )}}{d} + 3 a b^{2} x + \frac {a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a b^{2} \tan {\left (c + d x \right )}}{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 ^{2}{\left (c \right )}\right )^{3} & \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)**3,x)

[Out]

Piecewise((a**3*x - 3*a**2*b*x + 3*a**2*b*tan(c + d*x)/d + 3*a*b**2*x + a*b**2*tan(c + d*x)**3/d - 3*a*b**2*ta
n(c + d*x)/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)**2)**3, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (73) = 146\).
time = 0.92, size = 1027, normalized size = 13.34 \begin {gather*} \frac {15 \, a^{3} d x \tan \left (d x\right )^{5} \tan \left (c\right )^{5} - 45 \, a^{2} b d x \tan \left (d x\right )^{5} \tan \left (c\right )^{5} + 45 \, a b^{2} d x \tan \left (d x\right )^{5} \tan \left (c\right )^{5} - 15 \, b^{3} d x \tan \left (d x\right )^{5} \tan \left (c\right )^{5} - 75 \, a^{3} d x \tan \left (d x\right )^{4} \tan \left (c\right )^{4} + 225 \, a^{2} b d x \tan \left (d x\right )^{4} \tan \left (c\right )^{4} - 225 \, a b^{2} d x \tan \left (d x\right )^{4} \tan \left (c\right )^{4} + 75 \, b^{3} d x \tan \left (d x\right )^{4} \tan \left (c\right )^{4} - 45 \, a^{2} b \tan \left (d x\right )^{5} \tan \left (c\right )^{4} + 45 \, a b^{2} \tan \left (d x\right )^{5} \tan \left (c\right )^{4} - 15 \, b^{3} \tan \left (d x\right )^{5} \tan \left (c\right )^{4} - 45 \, a^{2} b \tan \left (d x\right )^{4} \tan \left (c\right )^{5} + 45 \, a b^{2} \tan \left (d x\right )^{4} \tan \left (c\right )^{5} - 15 \, b^{3} \tan \left (d x\right )^{4} \tan \left (c\right )^{5} + 150 \, a^{3} d x \tan \left (d x\right )^{3} \tan \left (c\right )^{3} - 450 \, a^{2} b d x \tan \left (d x\right )^{3} \tan \left (c\right )^{3} + 450 \, a b^{2} d x \tan \left (d x\right )^{3} \tan \left (c\right )^{3} - 150 \, b^{3} d x \tan \left (d x\right )^{3} \tan \left (c\right )^{3} - 15 \, a b^{2} \tan \left (d x\right )^{5} \tan \left (c\right )^{2} + 5 \, b^{3} \tan \left (d x\right )^{5} \tan \left (c\right )^{2} + 180 \, a^{2} b \tan \left (d x\right )^{4} \tan \left (c\right )^{3} - 225 \, a b^{2} \tan \left (d x\right )^{4} \tan \left (c\right )^{3} + 75 \, b^{3} \tan \left (d x\right )^{4} \tan \left (c\right )^{3} + 180 \, a^{2} b \tan \left (d x\right )^{3} \tan \left (c\right )^{4} - 225 \, a b^{2} \tan \left (d x\right )^{3} \tan \left (c\right )^{4} + 75 \, b^{3} \tan \left (d x\right )^{3} \tan \left (c\right )^{4} - 15 \, a b^{2} \tan \left (d x\right )^{2} \tan \left (c\right )^{5} + 5 \, b^{3} \tan \left (d x\right )^{2} \tan \left (c\right )^{5} - 150 \, a^{3} d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 450 \, a^{2} b d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 450 \, a b^{2} d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 150 \, b^{3} d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 3 \, b^{3} \tan \left (d x\right )^{5} + 30 \, a b^{2} \tan \left (d x\right )^{4} \tan \left (c\right ) - 25 \, b^{3} \tan \left (d x\right )^{4} \tan \left (c\right ) - 270 \, a^{2} b \tan \left (d x\right )^{3} \tan \left (c\right )^{2} + 360 \, a b^{2} \tan \left (d x\right )^{3} \tan \left (c\right )^{2} - 150 \, b^{3} \tan \left (d x\right )^{3} \tan \left (c\right )^{2} - 270 \, a^{2} b \tan \left (d x\right )^{2} \tan \left (c\right )^{3} + 360 \, a b^{2} \tan \left (d x\right )^{2} \tan \left (c\right )^{3} - 150 \, b^{3} \tan \left (d x\right )^{2} \tan \left (c\right )^{3} + 30 \, a b^{2} \tan \left (d x\right ) \tan \left (c\right )^{4} - 25 \, b^{3} \tan \left (d x\right ) \tan \left (c\right )^{4} - 3 \, b^{3} \tan \left (c\right )^{5} + 75 \, a^{3} d x \tan \left (d x\right ) \tan \left (c\right ) - 225 \, a^{2} b d x \tan \left (d x\right ) \tan \left (c\right ) + 225 \, a b^{2} d x \tan \left (d x\right ) \tan \left (c\right ) - 75 \, b^{3} d x \tan \left (d x\right ) \tan \left (c\right ) - 15 \, a b^{2} \tan \left (d x\right )^{3} + 5 \, b^{3} \tan \left (d x\right )^{3} + 180 \, a^{2} b \tan \left (d x\right )^{2} \tan \left (c\right ) - 225 \, a b^{2} \tan \left (d x\right )^{2} \tan \left (c\right ) + 75 \, b^{3} \tan \left (d x\right )^{2} \tan \left (c\right ) + 180 \, a^{2} b \tan \left (d x\right ) \tan \left (c\right )^{2} - 225 \, a b^{2} \tan \left (d x\right ) \tan \left (c\right )^{2} + 75 \, b^{3} \tan \left (d x\right ) \tan \left (c\right )^{2} - 15 \, a b^{2} \tan \left (c\right )^{3} + 5 \, b^{3} \tan \left (c\right )^{3} - 15 \, a^{3} d x + 45 \, a^{2} b d x - 45 \, a b^{2} d x + 15 \, b^{3} d x - 45 \, a^{2} b \tan \left (d x\right ) + 45 \, a b^{2} \tan \left (d x\right ) - 15 \, b^{3} \tan \left (d x\right ) - 45 \, a^{2} b \tan \left (c\right ) + 45 \, a b^{2} \tan \left (c\right ) - 15 \, b^{3} \tan \left (c\right )}{15 \, {\left (d \tan \left (d x\right )^{5} \tan \left (c\right )^{5} - 5 \, d \tan \left (d x\right )^{4} \tan \left (c\right )^{4} + 10 \, d \tan \left (d x\right )^{3} \tan \left (c\right )^{3} - 10 \, d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 5 \, d \tan \left (d x\right ) \tan \left (c\right ) - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a^3*d*x*tan(d*x)^5*tan(c)^5 - 45*a^2*b*d*x*tan(d*x)^5*tan(c)^5 + 45*a*b^2*d*x*tan(d*x)^5*tan(c)^5 - 1
5*b^3*d*x*tan(d*x)^5*tan(c)^5 - 75*a^3*d*x*tan(d*x)^4*tan(c)^4 + 225*a^2*b*d*x*tan(d*x)^4*tan(c)^4 - 225*a*b^2
*d*x*tan(d*x)^4*tan(c)^4 + 75*b^3*d*x*tan(d*x)^4*tan(c)^4 - 45*a^2*b*tan(d*x)^5*tan(c)^4 + 45*a*b^2*tan(d*x)^5
*tan(c)^4 - 15*b^3*tan(d*x)^5*tan(c)^4 - 45*a^2*b*tan(d*x)^4*tan(c)^5 + 45*a*b^2*tan(d*x)^4*tan(c)^5 - 15*b^3*
tan(d*x)^4*tan(c)^5 + 150*a^3*d*x*tan(d*x)^3*tan(c)^3 - 450*a^2*b*d*x*tan(d*x)^3*tan(c)^3 + 450*a*b^2*d*x*tan(
d*x)^3*tan(c)^3 - 150*b^3*d*x*tan(d*x)^3*tan(c)^3 - 15*a*b^2*tan(d*x)^5*tan(c)^2 + 5*b^3*tan(d*x)^5*tan(c)^2 +
180*a^2*b*tan(d*x)^4*tan(c)^3 - 225*a*b^2*tan(d*x)^4*tan(c)^3 + 75*b^3*tan(d*x)^4*tan(c)^3 + 180*a^2*b*tan(d*
x)^3*tan(c)^4 - 225*a*b^2*tan(d*x)^3*tan(c)^4 + 75*b^3*tan(d*x)^3*tan(c)^4 - 15*a*b^2*tan(d*x)^2*tan(c)^5 + 5*
b^3*tan(d*x)^2*tan(c)^5 - 150*a^3*d*x*tan(d*x)^2*tan(c)^2 + 450*a^2*b*d*x*tan(d*x)^2*tan(c)^2 - 450*a*b^2*d*x*
tan(d*x)^2*tan(c)^2 + 150*b^3*d*x*tan(d*x)^2*tan(c)^2 - 3*b^3*tan(d*x)^5 + 30*a*b^2*tan(d*x)^4*tan(c) - 25*b^3
*tan(d*x)^4*tan(c) - 270*a^2*b*tan(d*x)^3*tan(c)^2 + 360*a*b^2*tan(d*x)^3*tan(c)^2 - 150*b^3*tan(d*x)^3*tan(c)
^2 - 270*a^2*b*tan(d*x)^2*tan(c)^3 + 360*a*b^2*tan(d*x)^2*tan(c)^3 - 150*b^3*tan(d*x)^2*tan(c)^3 + 30*a*b^2*ta
n(d*x)*tan(c)^4 - 25*b^3*tan(d*x)*tan(c)^4 - 3*b^3*tan(c)^5 + 75*a^3*d*x*tan(d*x)*tan(c) - 225*a^2*b*d*x*tan(d
*x)*tan(c) + 225*a*b^2*d*x*tan(d*x)*tan(c) - 75*b^3*d*x*tan(d*x)*tan(c) - 15*a*b^2*tan(d*x)^3 + 5*b^3*tan(d*x)
^3 + 180*a^2*b*tan(d*x)^2*tan(c) - 225*a*b^2*tan(d*x)^2*tan(c) + 75*b^3*tan(d*x)^2*tan(c) + 180*a^2*b*tan(d*x)
*tan(c)^2 - 225*a*b^2*tan(d*x)*tan(c)^2 + 75*b^3*tan(d*x)*tan(c)^2 - 15*a*b^2*tan(c)^3 + 5*b^3*tan(c)^3 - 15*a
^3*d*x + 45*a^2*b*d*x - 45*a*b^2*d*x + 15*b^3*d*x - 45*a^2*b*tan(d*x) + 45*a*b^2*tan(d*x) - 15*b^3*tan(d*x) -
45*a^2*b*tan(c) + 45*a*b^2*tan(c) - 15*b^3*tan(c))/(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)

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Mupad [B]
time = 11.47, size = 115, normalized size = 1.49 \begin {gather*} \frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2\,b-3\,a\,b^2+b^3\right )}{d}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^3}{a^3-3\,a^2\,b+3\,a\,b^2-b^3}\right )\,{\left (a-b\right )}^3}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a\,b^2-\frac {b^3}{3}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x)^2)^3,x)

[Out]

(b^3*tan(c + d*x)^5)/(5*d) + (tan(c + d*x)*(3*a^2*b - 3*a*b^2 + b^3))/d + (atan((tan(c + d*x)*(a - b)^3)/(3*a*
b^2 - 3*a^2*b + a^3 - b^3))*(a - b)^3)/d + (tan(c + d*x)^3*(a*b^2 - b^3/3))/d

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