\(\int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx\) [436]

Optimal result

Integrand size = 21, antiderivative size = 94 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=-a \left (a^2-3 b^2\right ) x+\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}-\frac {2 a b^2 \tan (c+d x)}{d}-\frac {b (a+b \tan (c+d x))^2}{2 d}+\frac {(a+b \tan (c+d x))^4}{4 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3624, 3563, 3606, 3556} \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b \left (3 a^2-b^2\right ) \log (\cos (c+d x))}{d}-a x \left (a^2-3 b^2\right )-\frac {2 a b^2 \tan (c+d x)}{d}+\frac {(a+b \tan (c+d x))^4}{4 b d}-\frac {b (a+b \tan (c+d x))^2}{2 d} \]

[In]

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

[Out]

-(a*(a^2 - 3*b^2)*x) + (b*(3*a^2 - b^2)*Log[Cos[c + d*x]])/d - (2*a*b^2*Tan[c + d*x])/d - (b*(a + b*Tan[c + d*
x])^2)/(2*d) + (a + b*Tan[c + d*x])^4/(4*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 3563

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

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 3624

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {2 i (a+i b)^3 \log (i-\tan (c+d x))+2 (i a+b)^3 \log (i+\tan (c+d x))-12 a b^2 \tan (c+d x)-2 b^3 \tan ^2(c+d x)+\frac {(a+b \tan (c+d x))^4}{b}}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.29

method	result	size
norman	\(\left (-a^{3}+3 a \,b^{2}\right ) x +\frac {a \,b^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{d}+\frac {a \left (a^{2}-3 b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {b \left (3 a^{2}-b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(121\)
derivativedivides	\(\frac {\frac {\left (\tan ^{4}\left (d x +c \right )\right ) b^{3}}{4}+a \left (\tan ^{3}\left (d x +c \right )\right ) b^{2}+\frac {3 a^{2} b \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a^{3} \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a^{3}+3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(124\)
default	\(\frac {\frac {\left (\tan ^{4}\left (d x +c \right )\right ) b^{3}}{4}+a \left (\tan ^{3}\left (d x +c \right )\right ) b^{2}+\frac {3 a^{2} b \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+a^{3} \tan \left (d x +c \right )-3 a \,b^{2} \tan \left (d x +c \right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-a^{3}+3 a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(124\)
parallelrisch	\(-\frac {-\left (\tan ^{4}\left (d x +c \right )\right ) b^{3}-4 a \left (\tan ^{3}\left (d x +c \right )\right ) b^{2}+4 a^{3} d x -12 a \,b^{2} d x -6 a^{2} b \left (\tan ^{2}\left (d x +c \right )\right )+2 b^{3} \left (\tan ^{2}\left (d x +c \right )\right )+6 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b -2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3}-4 a^{3} \tan \left (d x +c \right )+12 a \,b^{2} \tan \left (d x +c \right )}{4 d}\)	\(132\)
parts	\(\frac {a^{3} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {b^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{2} b \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\)	\(134\)
risch	\(-3 i a^{2} b x +i b^{3} x -a^{3} x +3 a \,b^{2} x -\frac {6 i b \,a^{2} c}{d}+\frac {2 i b^{3} c}{d}-\frac {2 \left (-i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+6 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+12 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+10 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-i a^{3}+4 i a \,b^{2}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(304\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.20 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^{3} \tan \left (d x + c\right )^{4} + 4 \, a b^{2} \tan \left (d x + c\right )^{3} - 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} d x + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.70 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\begin {cases} - a^{3} x + \frac {a^{3} \tan {\left (c + d x \right )}}{d} - \frac {3 a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 a^{2} b \tan ^{2}{\left (c + d x \right )}}{2 d} + 3 a b^{2} x + \frac {a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 a b^{2} \tan {\left (c + d x \right )}}{d} + \frac {b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{3} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \tan ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^{3} \tan \left (d x + c\right )^{4} + 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{2} - 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1385 vs. \(2 (90) = 180\).

Time = 1.96 (sec) , antiderivative size = 1385, normalized size of antiderivative = 14.73 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/4*(4*a^3*d*x*tan(d*x)^4*tan(c)^4 - 12*a*b^2*d*x*tan(d*x)^4*tan(c)^4 - 6*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 -
2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 2*b^3*log(4*(t
an(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(
c)^4 - 16*a^3*d*x*tan(d*x)^3*tan(c)^3 + 48*a*b^2*d*x*tan(d*x)^3*tan(c)^3 - 6*a^2*b*tan(d*x)^4*tan(c)^4 + 3*b^3
*tan(d*x)^4*tan(c)^4 + 24*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan
(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 - 8*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(
d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 4*a^3*tan(d*x)^4*tan(c)^3 - 12*a*b^2*tan(d
*x)^4*tan(c)^3 + 4*a^3*tan(d*x)^3*tan(c)^4 - 12*a*b^2*tan(d*x)^3*tan(c)^4 + 24*a^3*d*x*tan(d*x)^2*tan(c)^2 - 7
2*a*b^2*d*x*tan(d*x)^2*tan(c)^2 - 6*a^2*b*tan(d*x)^4*tan(c)^2 + 2*b^3*tan(d*x)^4*tan(c)^2 + 12*a^2*b*tan(d*x)^
3*tan(c)^3 - 8*b^3*tan(d*x)^3*tan(c)^3 - 6*a^2*b*tan(d*x)^2*tan(c)^4 + 2*b^3*tan(d*x)^2*tan(c)^4 + 4*a*b^2*tan
(d*x)^4*tan(c) - 36*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^
2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 12*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^
2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 12*a^3*tan(d*x)^3*tan(c)^2 + 48*a*b^2*tan(d*x)^
3*tan(c)^2 - 12*a^3*tan(d*x)^2*tan(c)^3 + 48*a*b^2*tan(d*x)^2*tan(c)^3 + 4*a*b^2*tan(d*x)*tan(c)^4 - b^3*tan(d
*x)^4 - 16*a^3*d*x*tan(d*x)*tan(c) + 48*a*b^2*d*x*tan(d*x)*tan(c) + 12*a^2*b*tan(d*x)^3*tan(c) - 8*b^3*tan(d*x
)^3*tan(c) - 12*a^2*b*tan(d*x)^2*tan(c)^2 + 4*b^3*tan(d*x)^2*tan(c)^2 + 12*a^2*b*tan(d*x)*tan(c)^3 - 8*b^3*tan
(d*x)*tan(c)^3 - b^3*tan(c)^4 - 4*a*b^2*tan(d*x)^3 + 24*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) +
1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) - 8*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*
tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)*tan(c) + 12*a^3*tan(d*x)^2*ta
n(c) - 48*a*b^2*tan(d*x)^2*tan(c) + 12*a^3*tan(d*x)*tan(c)^2 - 48*a*b^2*tan(d*x)*tan(c)^2 - 4*a*b^2*tan(c)^3 +
4*a^3*d*x - 12*a*b^2*d*x - 6*a^2*b*tan(d*x)^2 + 2*b^3*tan(d*x)^2 + 12*a^2*b*tan(d*x)*tan(c) - 8*b^3*tan(d*x)*
tan(c) - 6*a^2*b*tan(c)^2 + 2*b^3*tan(c)^2 - 6*a^2*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(
d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) + 2*b^3*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan
(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1)) - 4*a^3*tan(d*x) + 12*a*b^2*tan(d*x) - 4*a^3*tan(c) + 12*a*b^2*
tan(c) - 6*a^2*b + 3*b^3)/(d*tan(d*x)^4*tan(c)^4 - 4*d*tan(d*x)^3*tan(c)^3 + 6*d*tan(d*x)^2*tan(c)^2 - 4*d*tan
(d*x)*tan(c) + d)

Mupad [B] (verification not implemented)

Time = 5.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.64 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a\,b^2-a^3\right )}{d}-\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{2}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{2}\right )}{d}+\frac {a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{d}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^2-3\,b^2\right )}{3\,a\,b^2-a^3}\right )\,\left (a^2-3\,b^2\right )}{d} \]

[In]

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

[Out]

(b^3*tan(c + d*x)^4)/(4*d) - (tan(c + d*x)*(3*a*b^2 - a^3))/d - (log(tan(c + d*x)^2 + 1)*((3*a^2*b)/2 - b^3/2)
)/d + (tan(c + d*x)^2*((3*a^2*b)/2 - b^3/2))/d + (a*b^2*tan(c + d*x)^3)/d + (a*atan((a*tan(c + d*x)*(a^2 - 3*b
^2))/(3*a*b^2 - a^3))*(a^2 - 3*b^2))/d