3.166 \(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^2 (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=363 \[ \frac {(a+b \tan (e+f x))^{m+1} \left (2 a^2 C d^2-a b d (m+3) (B d+2 c C)+b^2 (m+2) \left (d^2 (m+3) (A-C)+2 B c d (m+3)+2 c^2 C\right )\right )}{b^3 f (m+1) (m+2) (m+3)}+\frac {(c-i d)^2 (A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac {(c+i d)^2 (i A-B-i C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}-\frac {d \tan (e+f x) (2 a C d-b (B d (m+3)+2 c C)) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+2) (m+3)}+\frac {C (c+d \tan (e+f x))^2 (a+b \tan (e+f x))^{m+1}}{b f (m+3)} \]

[Out]

(2*a^2*C*d^2-a*b*d*(B*d+2*C*c)*(3+m)+b^2*(2+m)*(2*c^2*C+2*B*c*d*(3+m)+(A-C)*d^2*(3+m)))*(a+b*tan(f*x+e))^(1+m)
/b^3/f/(1+m)/(2+m)/(3+m)+1/2*(A-I*B-C)*(c-I*d)^2*hypergeom([1, 1+m],[2+m],(a+b*tan(f*x+e))/(a-I*b))*(a+b*tan(f
*x+e))^(1+m)/(I*a+b)/f/(1+m)+1/2*(I*A-B-I*C)*(c+I*d)^2*hypergeom([1, 1+m],[2+m],(a+b*tan(f*x+e))/(a+I*b))*(a+b
*tan(f*x+e))^(1+m)/(a+I*b)/f/(1+m)-d*(2*a*C*d-b*(2*c*C+B*d*(3+m)))*tan(f*x+e)*(a+b*tan(f*x+e))^(1+m)/b^2/f/(2+
m)/(3+m)+C*(a+b*tan(f*x+e))^(1+m)*(c+d*tan(f*x+e))^2/b/f/(3+m)

________________________________________________________________________________________

Rubi [A]  time = 1.15, antiderivative size = 360, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3647, 3637, 3630, 3539, 3537, 68} \[ \frac {(a+b \tan (e+f x))^{m+1} \left (2 a^2 C d^2-a b d (m+3) (B d+2 c C)+b^2 (m+2) \left (d^2 (m+3) (A-C)+2 B c d (m+3)+2 c^2 C\right )\right )}{b^3 f (m+1) (m+2) (m+3)}+\frac {(c-i d)^2 (A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac {(c+i d)^2 (i A-B-i C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}+\frac {d \tan (e+f x) (-2 a C d+b B d (m+3)+2 b c C) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+2) (m+3)}+\frac {C (c+d \tan (e+f x))^2 (a+b \tan (e+f x))^{m+1}}{b f (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*C*d^2 - a*b*d*(2*c*C + B*d)*(3 + m) + b^2*(2 + m)*(2*c^2*C + 2*B*c*d*(3 + m) + (A - C)*d^2*(3 + m)))*(
a + b*Tan[e + f*x])^(1 + m))/(b^3*f*(1 + m)*(2 + m)*(3 + m)) + ((A - I*B - C)*(c - I*d)^2*Hypergeometric2F1[1,
1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 + m))/(2*(I*a + b)*f*(1 + m)) + ((I*A -
B - I*C)*(c + I*d)^2*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)]*(a + b*Tan[e + f*x])^
(1 + m))/(2*(a + I*b)*f*(1 + m)) + (d*(2*b*c*C - 2*a*C*d + b*B*d*(3 + m))*Tan[e + f*x]*(a + b*Tan[e + f*x])^(1
+ m))/(b^2*f*(2 + m)*(3 + m)) + (C*(a + b*Tan[e + f*x])^(1 + m)*(c + d*Tan[e + f*x])^2)/(b*f*(3 + m))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3537

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

Rule 3539

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

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 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(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 - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}+\frac {\int (a+b \tan (e+f x))^m (c+d \tan (e+f x)) \left (A b c (3+m)-C (2 a d+b c (1+m))+b (B c+(A-C) d) (3+m) \tan (e+f x)+(2 b c C-2 a C d+b B d (3+m)) \tan ^2(e+f x)\right ) \, dx}{b (3+m)}\\ &=\frac {d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}-\frac {\int (a+b \tan (e+f x))^m \left (a d (2 b c C-2 a C d+b B d (3+m))-b c (2+m) (A b c (3+m)-C (2 a d+b c (1+m)))-b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) \tan (e+f x)-\left (b c (2+m) (2 b c C-2 a C d+b B d (3+m))+d \left (b^2 (B c+(A-C) d) (2+m) (3+m)-a (2 b c C-2 a C d+b B d (3+m))\right )\right ) \tan ^2(e+f x)\right ) \, dx}{b^2 (2+m) (3+m)}\\ &=\frac {\left (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}+\frac {d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}-\frac {\int (a+b \tan (e+f x))^m \left (-b^2 \left (A c^2-c^2 C-2 B c d-A d^2+C d^2\right ) (2+m) (3+m)-b^2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) (2+m) (3+m) \tan (e+f x)\right ) \, dx}{b^2 (2+m) (3+m)}\\ &=\frac {\left (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}+\frac {d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}+\frac {1}{2} \left ((A-i B-C) (c-i d)^2\right ) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac {1}{2} \left ((A+i B-C) (c+i d)^2\right ) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=\frac {\left (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}+\frac {d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}+\frac {\left ((i A+B-i C) (c-i d)^2\right ) \operatorname {Subst}\left (\int \frac {(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}-\frac {\left (i (A+i B-C) (c+i d)^2\right ) \operatorname {Subst}\left (\int \frac {(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac {\left (2 a^2 C d^2-a b d (2 c C+B d) (3+m)+b^2 (2+m) \left (2 c^2 C+2 B c d (3+m)+(A-C) d^2 (3+m)\right )\right ) (a+b \tan (e+f x))^{1+m}}{b^3 f (1+m) (2+m) (3+m)}-\frac {(i A+B-i C) (c-i d)^2 \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a-i b) f (1+m)}-\frac {(A+i B-C) (c+i d)^2 \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a-b) f (1+m)}+\frac {d (2 b c C-2 a C d+b B d (3+m)) \tan (e+f x) (a+b \tan (e+f x))^{1+m}}{b^2 f (2+m) (3+m)}+\frac {C (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^2}{b f (3+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 6.35, size = 505, normalized size = 1.39 \[ \frac {C (c+d \tan (e+f x))^2 (a+b \tan (e+f x))^{m+1}}{b f (m+3)}+\frac {\frac {d \tan (e+f x) (-2 a C d+b B d (m+3)+2 b c C) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}-\frac {\frac {i (a+b \tan (e+f x))^{m+1} \left (-\left (b^2 (m+2) (m+3) \left (A c^2-A d^2-2 B c d-c^2 C+C d^2\right )\right )-i b^2 (m+2) (m+3) \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {-i a-i b \tan (e+f x)}{b-i a}\right )}{2 f (m+1) (a+i b)}-\frac {i (a+b \tan (e+f x))^{m+1} \left (-b^2 (m+2) (m+3) \left (A c^2-A d^2-2 B c d-c^2 C+C d^2\right )+i b^2 (m+2) (m+3) \left (2 c d (A-C)+B \left (c^2-d^2\right )\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac {i a+i b \tan (e+f x)}{-i a-b}\right )}{2 f (m+1) (a-i b)}+\frac {(a+b \tan (e+f x))^{m+1} \left (-d \left (b^2 (m+2) (m+3) (d (A-C)+B c)-a (-2 a C d+b B d (m+3)+2 b c C)\right )-b c (m+2) (-2 a C d+b B d (m+3)+2 b c C)\right )}{b f (m+1)}}{b (m+2)}}{b (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(a + b*Tan[e + f*x])^(1 + m)*(c + d*Tan[e + f*x])^2)/(b*f*(3 + m)) + ((d*(2*b*c*C - 2*a*C*d + b*B*d*(3 + m)
)*Tan[e + f*x]*(a + b*Tan[e + f*x])^(1 + m))/(b*f*(2 + m)) - (((-(b*c*(2 + m)*(2*b*c*C - 2*a*C*d + b*B*d*(3 +
m))) - d*(b^2*(B*c + (A - C)*d)*(2 + m)*(3 + m) - a*(2*b*c*C - 2*a*C*d + b*B*d*(3 + m))))*(a + b*Tan[e + f*x])
^(1 + m))/(b*f*(1 + m)) + ((I/2)*(-(b^2*(A*c^2 - c^2*C - 2*B*c*d - A*d^2 + C*d^2)*(2 + m)*(3 + m)) - I*b^2*(2*
c*(A - C)*d + B*(c^2 - d^2))*(2 + m)*(3 + m))*Hypergeometric2F1[1, 1 + m, 2 + m, ((-I)*a - I*b*Tan[e + f*x])/(
(-I)*a + b)]*(a + b*Tan[e + f*x])^(1 + m))/((a + I*b)*f*(1 + m)) - ((I/2)*(-(b^2*(A*c^2 - c^2*C - 2*B*c*d - A*
d^2 + C*d^2)*(2 + m)*(3 + m)) + I*b^2*(2*c*(A - C)*d + B*(c^2 - d^2))*(2 + m)*(3 + m))*Hypergeometric2F1[1, 1
+ m, 2 + m, -((I*a + I*b*Tan[e + f*x])/((-I)*a - b))]*(a + b*Tan[e + f*x])^(1 + m))/((a - I*b)*f*(1 + m)))/(b*
(2 + m)))/(b*(3 + m))

________________________________________________________________________________________

fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C d^{2} \tan \left (f x + e\right )^{4} + {\left (2 \, C c d + B d^{2}\right )} \tan \left (f x + e\right )^{3} + A c^{2} + {\left (C c^{2} + 2 \, B c d + A d^{2}\right )} \tan \left (f x + e\right )^{2} + {\left (B c^{2} + 2 \, A c d\right )} \tan \left (f x + e\right )\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*d^2*tan(f*x + e)^4 + (2*C*c*d + B*d^2)*tan(f*x + e)^3 + A*c^2 + (C*c^2 + 2*B*c*d + A*d^2)*tan(f*x
+ e)^2 + (B*c^2 + 2*A*c*d)*tan(f*x + e))*(b*tan(f*x + e) + a)^m, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{2} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(d*tan(f*x + e) + c)^2*(b*tan(f*x + e) + a)^m, x)

________________________________________________________________________________________

maple [F]  time = 2.19, size = 0, normalized size = 0.00 \[ \int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{2} \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^2\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (e + f x \right )}\right )^{m} \left (c + d \tan {\left (e + f x \right )}\right )^{2} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**m*(c+d*tan(f*x+e))**2*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

Integral((a + b*tan(e + f*x))**m*(c + d*tan(e + f*x))**2*(A + B*tan(e + f*x) + C*tan(e + f*x)**2), x)

________________________________________________________________________________________